Physics - 1 Iit Material

PHYSICS ( 1 ST YEAR ) IIT MATERIAL
1. PHYSICAL WORLD
2. UNITS AND MEASUREMENTS
3. MOTION IN A STRAIGHT LINE
4. MOTION IN A PLANE
5. LAWS OF MOTION
6. WORK , ENERGY AND POWER
7. SYSTEM OF PARTICLES AND ROTATIONAL MOTION
8. OSCILLATIONS
9. GRAVITATION
10. MECHANICAL PROPERTIES OF SOLIDS
11. MECHANICAL PROPERTIES OF FLUIDS
12. THERMAL PROPERTIES OF MATTER
13. THERMODYNAMICS
14. KINETIC THEORY

1. PHYSICAL WORLD
2. UNITS & MEASUREMENTS
Physical Quantity:
 Any quantity which can be measured directly (or) indirectly (or) interms of which the laws of physics can
be expressed is called physical quantity.
 There are two types of physical quantities
1) Fundamental quantities2) Derived quantities
Fundamental Quantities: Physical quantities which cannot be expressed interms of any other physical
quantities are called fundamental physical quantities.
Ex. length, mass, time, temperature etc..
Derived Quantities: Physical Quantities which are derived from fundamental quantities are called
derived quantities.
Ex. Area, density, force etc...
Unit of physical qantity:
 A unit of measurement of a physical quantity is the standard reference of the same physical quantity
which is used for comparison of the given physical quantity.
Fundamental unit :The unit used to measure the fundamental quantity is called fundamental unit.
Ex: metre for length, kilogram for mass etc..
Derived unit : The unit used to measure the derived quantity is called derived unit.
Ex: m2 for area, gm cm-3 for density etc...
 The numerical value obtained on measuring a physical quantity is inversely proportional to the magnitude
of the unit chosen.
1
n  nU = constant
U
 n1U  n2U
1 2
Where n1 and n2 are the numerical values and U1 and U 2 are the units of same physical quantity in different
systems.
System of units
 There are four systems of units
1) F.P.S 2) C.G.S
3) M.K.S 4) SI
 Based on SI system there are three categories of
physical quantities.
1)fundamental quantities
2)supplementary quantities and
3)derived quantities
Fundamental Quantities and their SI Units
 There are seven fundamental quantities and two supplementary quantities in S. I. system. These
quantities along with their unit and symbols are given below:
S.No Physical Quantity SI unit Symbol
1. Length metre m
2. Mass kilogram kg
3. Time second s
4. Thermo dynamic
temperature kelvin K (or) 
5. Luminous
intensity candela Cd
6. Electric current ampere A
7. Amount of
substance
(or) quantity of
matter mole mol
Supplementary quantities
1. Plane angle radian rad
2. Solid angle steradian sr
Measurement of length
 The length of an object can be measured by using different units. Some particle units of length
are
angstrom( Ao )=10 10 m=108 cm
nanometre(nm)  109 m  10 A0
fermi  10 15 m
micron  10 6 m
X-ray unit  10 13 m
1 A.U. = distance between sun & earth = 1.496×1011 m
 One light year is the distance travelled by light in one year in vacuum . This unit is used in astronomy.
Light year  9.46  1015 m
parsec  3.26 light years  30.84  1015 m
Bohr radius  0.5 10 10 m
Mile=1.6 km
Measurement of mass:
The mass of an object can be measured by using different units.Some practical units of mass are
Quintal = 100 kg Metric ton = 1000 kg
Atomic mass unit (a.m.u) =  1.67  10 27 kg
Measurement of time:
One day = 86400 second
Shake 108 second
Abbreviations for multiples and sub multiples:
 MACRO Prefixes
MultiplierSymbol Prefix
101 da Deca
2
10 h Hecto
3
10 k Kilo
106 M Mega
9
10 G Giga
12
10 T Tera
1015 P Peta
1018 E Exa
10 21 Z Zetta
10 24 Y Yotta
 MICRO Prefixes
MultiplierSymbol Prefix
10-1 d deci
-2
10 c centi
10-3 m milli
10 -6  micro
-9
10 n nano
10-12 p pico
-15
10 f femto
-18
10 a atto
-21
10 z zepto
10-24 y yocto
Some important conversions:

5
 1kmph  ms 1
18
1 newton=105 dyne
1 joule=107 erg
1 calorie=4.18 J
1eV= 1.6 1019 J
1gcm 3  1000kgm 3
1 lit=1000cm3  103 m3
1KWH  36 105 J
1 HP=746 W
1 degree=0.017 rad
1cal g 1  4180JKg 1
1kgwt= 9.8 N
1 telsa=10 4 gauss
1Am 1  4 103 oersted
1 weber=108 maxwell
Some physical constants and their values:
 1 amu =1.67 1027 kg  931.5MeV
1 atm pressure = pressure exerted by 76cm of Hg column  1.013  105 Pa
Avagadro number (N)= 6.023  1023
Permittivity of free space= 8.854  1012 Fm 1 or C 2 / Nm2
Permeability of free space  0   4 107 Hm1
Joule’s constant (J)= 4.186Jcal 1
Planck’s constant(h)= 6.62  10 34 Js
Rydberg’s constant(R)= 1.0974  107 m 1
Boltzmann’s constant(KB)=1.38  1023 JK 1
Stefan’s constant    5.67 108Wm2 K 4
Universal gas constant(R)= 8.314Jmol 1 K 1
= 1.98cal mol 1 K 1
Wien’s constant(b)= 2.93 103 metre kelvin
Accuracy and precision of instruments :

 The numerical values obtained on measuring physical quantities depend upon the measuring instruments,
methods of measurement.
 Accuracy refers to how closely a measured value agrees with the true value.
 Precision refers to what limit or resolution the given physical quantity can be measured.
 Precision refers to closeness between the different observed values of the same quantity .
 High precision does not mean high accuracy.
 The difference between accuracy and precision can be understood and by the following example: Suppose
three students are asked to find the length of a rod whose length is known to be 2.250cm.The observations
are given in the table .
Measurement- Measurement- Measurement- Average

Student
1 2 3 length
A 2.25cm 2.27cm 2.26cm 2.26cm
B 2.252cm 2.250cm 2.251cm 2.251cm
C 2.250cm 2.250cm 2.251cm 2.250cm
It is clear from the above table , that the observations taken by a student A are neither precise nor
accurate. The observations of student B are more precise . The observations of student C are precise as
well as accurate.
Error:
 The result of every measurement by any measuring instrument contains some uncertainty. This uncertainty
in measurement is called error.
Mathematically
 Error = True value - Measured value
Correction =-error
 True value means, standard value free of errors.
 Errors are broadly classified into 3 types :
i) Systematic errors
ii) Random errors
iii) Gross errors
Systematic Errors
 The errors due to a definite cause and which follow a particular rule are called systematic errors. They
always occur in one direction (either +ve or -ve )
 Systematic errors with a constant magnitude are called constant errors.
The constant arised due to imperfect design, zero error in the instrument or any other such defects.
These are also called instrumental errors.
 Example for the error due to improper designing and construction.
If a screw gauge has a zero error of -4 head scale divisions, then every reading will be 0.004cm less than
the true value.
 The error arised due to external conditions like changes in environment, changes in temperature, pressure,
humidity etc.
Ex: Due to rise in temperature, a scale gets expanded and this results in error in measurement of length.
Imperfection in Experimental technique or Procedure:

 The error due to experimental arrangement, procedure followed and experimental technique is called
imperfection error.
Ex: In calorimetric experiments, the loss of heat due to radiation, the effect on weighing due to buoyancy
of air cannot be avoided.
Personal errors or observational errors:
 These are entirely due to the personal peculiarities of the experimenter. Individual bias, lack of proper setting of
the apparatus, carelessness in taking observations (without taking the required necessary precautions.) etc. are
the causes for these type of errors. A person may be habituated to hold his eyes (head) always a bit too far to the
right (or left) while taking the reading with a scale. This will give rise to parallax error.
 If a person keeps his eye-level below the level ofmercury in a barometer all the time, his readings will
have systematic error.
These errors can be minimised by obtaining several readings carefully and then taking their
arithmetic mean..
1
probable error 
no. of readings
Ex: Parallax error
Random Errors:
 They are due to uncontrolled disturbances which influence the physical quantity and the instrument.
these errors are estimated by statistical methods.
1
Random error 
no. of observations
Ex-:The errors due to line voltage changes and backlash error.
Backlash errors are due to screw and nut.
Gross Errors
 The cause for gross errors are improper recording, neglecting the sources of the error, reading the
instrument incorrectly, sheer carelessness
Ex: In a tangent galvanometer experiment, the coil is to be placed exactly in the magnetic meridian and
care should be taken to see that no any other magnetic material is present in the vicinity.
 No correction can be applied to these gross errors.
 When the errors are minimised, the accuracy increases.
The systematic errors can be estimated and observations can be corrected.
 Random errors are compensating type. A physical quantity is measured number of times and these
values lie on either side of mean value. These errors are estimated by statistical methods and accuracy is
achieved.
 Personal errors like parallax error can be avoided by taking proper care.
 The instrumental errors are avoided by calibrating the instrument with a standard reference and by
applying proper corrections.
Errors in measurement.
True Value :
 In the measurement of a physical quantity the arithmetic mean of all readings which is found to be very
close to the most accurate reading is to be taken as True value of the quantities.
1 n
If a1, a2 , a3 ..................an are readings then true value amean   ai
n i 1
Absolute Error :
 The magnitude of the difference between the true value of the measured physical quantity and the value of
individual measurement is called absolute error.
Absolute error =|True value - measured values|
ai  amean  ai
The absolute error is always positive.
Mean absolute error:
 The arithmetic mean of all the absolute errors is considered as the mean absolute error of the physical
quantity concerned.
a1  a2      an 1 n
amean 
n
  ai
n i 1
The mean absolute error is always positive.
Relative error:
 The relative error of a measured physical quantity is the ratio of the mean absolute error to the mean
value of the quantity measured.
a mean
Relative error= a mean
It is a pure number having no units.

Percentage error:
 amean 
a   100  %
 amean 
Relative error and percentage error give a measure of accuracy i.e. if percentage error increases
accuracy decreases.
EX. 1:Repetition in the measurements of a certain quantity in an experiment gave the following values: 1.29,
1.33, 1.34, 1.35, 1.32, 1.36, 1.30, and 1.33. Calculate the mean value, mean absolute error, relative error
and percentage error.
Sol. Here, mean value
1.29  1.33  1.34  1.35  1.32  1.36  1.30  1.33
xm 
8
= 1.3275=1.33 (rounded off to two places of decimal)
Absolute errors in measurement are
x1  1.33  1.29  0.04; x2  1.33  1.33  0.00;
x3  1.33  1.34  0.01; x4  1.33  1.35  0.02;
x5  1.33  1.32  0.01; x6  1.33  1.36  0.03;
x7  1.33  1.30  0.03; x8  1.33  1.33  0.00;

mean absolute error
0.04  0.00  0.01 0.02  0.01 0.03  0.03  0.00
xm 
8
= 0.0175
= 0.02 (rounded off to two places of decimal)
xm 0.02
Relative error    0.01503  0.02
xm 1.33
(rounded off to two places of decimal)
Percentage error = 0.01503100  1.503  1.5%
EX.2 : The length and breadth of a rectangle are (5.7  0.1) cm and (3.4  0.2) cm. Calculate the area
of the rectangle with error limits.
Sol. Here l   5.7  0.1 cm, b   3.4  0.2  cm
Area : A  l  b  5.7  3.4  19.38 cm 2  19 cm2
(rounding off to two significant figures)
A  l  b   0.1 0.2 
        
A  l b   5.7 3.4 
 0.34  1.14  1.48
  
 5.7  3.4  19.38
1.48 1.48
 A   A  19.38  1.48  1.5
19.38 19.38
(rounding off to two significant figures)
So, Area  19.0  1.5  cm 2
EX.3: The distance covered by a body in time  5.0  0.6  s is  40.0  0.4  m. Calculate the speed of
the body. Also determine the percentage error in the speed.
Sol. Here, s   40.0  0.4  m and t   5.0  0.6  s
s 40.0 s
 Speed v    8.0 ms 1 As v 
t 5.0 t
v s t
  
v s t
Here s  0.4 m, s=40.0 m, t  0.6 s, t=5.0 s
v 0.4 0.6
    0.13
v 40.0 5.0
 v  0.13  8.0  1.04
Hence, v   8.0  1.04  ms 1
  v
 Percentage error   v 100   0.13  100  13%
EX. 4: A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions of the circular scale.
[AIEEE 2011]
Sol. Main scale reading = 0 mm
Circular scale reading = 52 divisions
value of 1 main scale division 1
Least count =  mm
Total divisions on circular scale 100
Diameter of wire = M.S.R +( C.S.R x L.C)

1
 0  52  mm  0.52mm  0.052cm
100
EX.5: The current voltage relation of diode is given by I   e1000V /T  1 mA,where the applied voltage V
is in volt and the temperature T is in kelvin.If a student makes an error measuring  0.01V while
measuring the current of 5mA at 300K,what will be the error in the value of current in mA? (JEE
MAIN-2014)
Sol. I   e1000V /T  1 mA
dV=  0.01V, T=300K,I=5mA
I  1  e1000V /T
1000V
log  I  1 
T
dI 1000
 dV  dI=0.2mA
I 1 T
EX.6 : In an experiment the angles are required to be measured using an instrument. 29 divisions of
the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest
division of the main scale is half-a-degree(= 0.50 ), then the least count of the instrument is
(AIEEE-2009)
Valuesof mainscaledivision
Sol. Least count = No.of divisionsof vernier scale
1 1 10 10
= MSD     1 min
30 30 2 60
Combination of Errors:
 Error due to addition
If Z  A  B ;
Z  A  B (Max. possible error)
Z  Z   A  B    A  B 
A  B A  B
Relative error= Percentage error= 100
A B A B
 Error due to subtraction
If Z=A-B
Z  A  B (Max. possible error )
Z  Z   A  B    A  B 
A  B A  B
Relative error = Percentage error = 100
A B A B
 Whether it is addition or subtraction, absolute error is same.
 In subtraction the percentage error increases.
 Error due to Multiplication:
Z A B
If Z = AB then  
Z A B
Z
is called fractional error or relative error..
Z
Z  A   B 
Percentage error  100   100    100 
Z  A   B 
 Here percentage error is the sum of individual percentage errors.
A
 Error due to division: if Z 
B
Z A B
Maximum possible relative error  
Z A B
A B
Max. percentage error in division  100   100
A B
 Error due to Power:
Z A
If Z= An ; n
Z A
Ap Bq
 In more general form : If Z 
Cr
Z A B C
then maximum fractional error in Z is p q r
Z A B C
C
As we check for maximum error a +ve sign is to be taken for the term r
C
Maximum Percentage error in Z is
Z A B C
 100  p 100  q  100  r  100
Z A B C
EX.7: A physical quantity is represented by x =MaLbT-c. The percentage of errors in the measurements
of mass,length and time are  %,  %,  % respectively then the maximum percentage error is
x M L T
Sol. 100  a. 100 b. 100  c. 100
x M L T
 a  b  c
EX.8:Resistance of a given wire is obtained by measuring the current flowing in it and the voltage
difference applied across it. If the percentage errors in the measurement of the current and
the voltage difference are 3% each, then error in the value of resistance of the wire is
[AIEEE 2012]
V
Sol. R   log R  log V  log I 
I
R  V I 
 R 100    V  I  100 
 
= 3% + 3% = 6%
EX.9: Two resistors of resistances R1  100  3 ohm and R2   200  4 ohm are connected (a) in series,
(b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel
combination. Use for (a) the relation R  R1  R2 and for (b)
1 1 1 R ' R1 R2

   2  2
R ' R1 R2 and R '2 R1 R2
Sol. (a) The equivalent resistance of series combination
R  R1  R2  100  3 ohm   200  4  ohm
=  300  7  ohm .
(b) The equivalent resistance of parallel combination
R1 R2 200
R'    66.7 ohm
R1  R2 3
1 1 1
Then, from R '  R  R
1 2
R ' R1 R2

we get, R '2  R 2  R 2
1 2
'2 R1 '2 R2

R   R  2   R  2
'
R1 R2
2 2
 66.7   66.7 
  3   4  1.8
 100   200 
Then, R '   66.7  1.8  ohm
Significant Figures :
 A significant figure is defined as the figure, which is considered reasonably, trust worthy in number.
Ex:  = 3.141592654
(upto 10 digits)
=3.14 (with 3 figures )
=3.1416 (upto 5 digits )
 The significant figures indicate the extent to which the readings are reliable.
Rules for determining the number of
significant figures:
 All the non-zero digits in a given number are significant without any regard to the location of the decimal
point if any.
Ex: 18452 or 1845.2 or 184.52 all have the same number of significant digits,i.e. 5.
 All zeros occurring between two non zero digits are significant without any regard to the location of
decimal point if any.
Ex: 106008 has six significant digits.
106.008 or 1.06008 has also got six significant digits.
 If the number is less than one, all the zeros to the right of the decimal point but to the left of first non-zero
digit are not significant.
Ex: 0.000308
In this example all zeros before 3 are insignificant.
 a)All zeros to the right of a decimal point are significant if they are not followed by a non-zero digit.
Ex: 30.00 has 4 significant digits
 b) All zeros to the right of the last non-zero digit after the decimal point are significant.
Ex: 0.05600 has 4 significant digits
 c) All zeros to the right of the last non-zero digit in a number having no decimal point are not significant.
Ex: 2030 has 3 significant digits
Rounding off numbers:
 The result of computation with approximate numbers, which contain more than one uncertain digit,should
be rounded off.
Rules for rounding off numbers:
 The preceding digit is raised by 1 if the immediate insignificant digit to be dropped is more than 5.
Ex: 4728 is rounded off to three significant figures as 4730.
 The preceding digit is to be left unchanged if the immediate insignificant digit to be dropped is less than
5.
Ex: 4723 is rounded off to three significant figures as 4720
 If the immediate insignificant digit to be dropped is 5 then there will be two different cases
a) If the preceding digit is even then it is to be unchanged and 5 is dropped.
Ex: 4.7253 is to be rounded off to two decimal places. The digit to be dropped here is 5 (along with
3) and the preceding digit 2 is even and hence to be retained as two only
4.7253=4.72
b)If the preceding digit is odd, it is to be raised by 1
Ex: 4.7153 is to be rounded off to two decimal places. As the preceding digit ‘1’ is odd, it is to be
raised by 1.
4.7153=4.72
Rules for Arithmetic Operations with
significant Figures:
 In multiplication or division, the final result should retain only that many significant figures as are there in
the original number with the least number of significant figures.
Ex:1.2  2.54  3.26  9.93648 .But the result should be limited to the least number of significant digits-
that is two digits only. So final answer is 9.9.
 In addition or subtraction the final result should retain only that many decimal places as are there in the
number with the least decimal places.
Ex:2.2+4.08+3.12+6.38=15.78.Finally we should have only one decimal place and hence 15.78 is to
be rounded off as 15.8.
EX.10:The respective number of significant figures for the numbers 23.023,0.0003and
21 103 are (AIEEE-2010)
Sol.(i)All non -zero numbers are significant figures
(ii) If the number is less than one,zero between the decimal and first non zero digit are not significant.
(iii) Powers of 10 is not a significant figure.
 5,1,2
S.No. Physical Quantity Formula Dimensional Formula SI Unit
1. Displacement,
Wave length,
Radius of gyration,  M 0 L1 T 0  m
Circumference,
Perimeter,Light year,
2. Mass  M 1 L0 T 0  kg
total time
3. Period of oscillation, no.of oscillations
Time,  M 0 L0 T 1  s
Time constant T = Capacity  Resistance
1
4. Frequency Reciprocal of time period n   M 0 L0T 1  hertz ( Hz)
T
5. Area A = length  breadth  M 0 L2T 0  m2
6. Volume V=length  breadth  height  M 0 L3T 0  m3
mass
7. Density d=  M 1 L3T 0  kgm-3
volume
mass
8. Linear mass density λ=  M 1 L1T 0  kgm-1
length
displacement
v=
9. Speed, Velocity time  M 0 L1T 1  ms-1
change in velocity
10. Acceleration a=  M 0 L1T 2  ms-2
time
11. Linear momentum P= mass  velocity  M 1 L1T 1  kgms-1
12. Force F = Mass  acceleration  M 1 L1T 2  N

1 1 1
13. Impulse J= Force  time  M L T  Ns
14. Work,Energy,PE, KE, W = Force  displacement
Strain energy, P.E= mgh
1
Heat energy KE = (Mass) (velocity)2  M 1 L2T 2  J(or) N.m
2
1
SE= ×Stress×Strain×volume
2
Work
15. PoEX.r P=  M 1 L2T 3  watt
time
Force
16. Pressure , Stress,
Area
Stress
Modulus of elasticity (Y, , k) Y=  M 1 L1T 2  pascal or Nm 2
Strain
change in dimension
17. Strain = original dimension  M 0 L0T 0  no units
work
18. Strain energy density E=  M 1 L1T 2  Jm-3
volume
length of arc
19. Angular displacement θ=  M 0 L0T 0  rad
radius
angular dispacement
20. Angular velocity ω=  M 0 L0T 1  rads-1
time
change in angular velocity
21. Angular acceleration α=  M 0 L0T 2  rads-2
time
22. Angular momentum L=linear momentum
 perpendicular distance  M 1 L2T 1  Js
energy
23. Planck's constant h=  M 1 L2T 1  Js
frequency
24. Angular impulse Torque  time  M 1 L2T 1  Js
25. Torque τ=force×  distance  M 1 L2T 2  Nm
26. Acceleration due to
weight
gravity(g) g=  M 0 LT 2  ms-2 or Nkg-1
mass
2
Force   distence 
27. Universal gravitational G=  M 1 L3T 2  Nm2 kg-2
Mass1  Mass 2
Constant
28. Moment of inertia I=Mass  (radius of gyration)2  M 1 L2T 0  kgm2

dv
29. Velocity gradient =  M 0 L0T 1  S 1
dx
surface energy force
30. Surface tension, S= =  M 1 L0T 2  Nm-1 or Jm-2
change in area length
Surface energy
Spring constant
force
Force constant K=
elongation
tangential stress
31. Coefficient of viscosity η= velocity gradient  M 1 L1T 1  Pa s (or) Nm 2 s
32. Gravitational potential Gravitational field  distance  M 0 L2T 2  J/Kg
1 2 2
33. Heat energy msθ  M L T  joule
34. Temperature θ  M 0 L0T 0 1  kelvin( K)
heat energy
35. Specific heat capacity S (or) C= mass×temp.  M 0 L2T 2 1  Jkg-1 K-1
dQ
36. Thermal capacity =mass×sp.ht  M 1 L2T 2 1  JK-1
dθ
37. Latent heat (or)
heat energy
Calorific value L=  M 0 L2T 2  Jkg-1
mass
38. Water equivalent W=Mass  specific heat  M 1 L0T 0  kg
l A   V
39. Coefficient of thermal  ;  ; V   M 0 L0T 0 1  K-1
l  A
expansion
PV
40. Universal gas constant R=  M 1 L2T 2 1mol 1  Jmol-1K-1
nT
R
41. Gas constant (for 1 gm) r=  M 0 L2T 2 1mol 1  Jkg-1K-1
Mol.wt
42. Boltzmann’s constant
R
(for 1 Molecule) k=  M 1 L2T 2 1  JK-1molecule-1
Avagadro number
W
43. Mechanical equivalent J  M 0 L0T 0  no SI units
H
of heat
Qd
44. Coefficient of thermal K=  M 1 L1T 3 1  Js-1 m-1 K-1 (or) Wm-1 K-
A Δθt
1
conductivity
dQ heat energy
45. Entropy =  M 1 L2T 2 1  JK-1
T temperature
ΔE
46. Stefan's constant σ=  M 1 L0T 3 4  Js-1m-2K-4 (or) Wm-2K-4
ΔAΔTθ 4
dθ temp×time
R= =
47. Thermal resistance  dQ  Heat  M 1 L2T 3 1  KsJ-1
 
 dt 
d
( or) R=
KA
Change in temp dθ
48. Temperature gradient =  M 0 L1T 0 1  Km-1
length dl
Change in pressure dp
49. Pressure gradient =  M 1 L2T 2  pascal m-1
length dl
Energy ΔE
50. Solar constant =  M 1 L0T 3  Js-1m-2 (or) Wm-2
area × time AT
51. Enthalpy heat ( Q )  M 1 L2T 2  joule
0 0
52. Pole strength m=IL ( or)  M LT A  Am
Magnetic Momement
Mag.Length
53. Magnetic moment M= 2 l ×m  M 0 L2T 0 A Am2
m
54. Magnetic intensity (or) H=  M 0 L1T 0 A Am-1
4πd 2
Magnetising field
Magnetic moment
55. Intensity of magnetisation I=  M 0 L1T 0 A Am-1
Volume
ur ur
56. Magnetic flux  = B×A  M 1 L2T 2 A1  Wb
=(Magnetic induction  Area)
ur  Magnetic flux F
57. Magnetic induction B  =  M 1 L0T 2 A1  Tesla (or) Wbm-2 (or) NA-
A Area il
1 -1
m
4πFd 2
58. Magnetic permeability μ=  M 1 L1T 2 A2  Hm-1
m1m 2
I
59. Magnetic susceptibility χ=  M 0 L0T 0  no units
H
60. Electric current I  M 0 L0T 0 A A
61. Charge Q =Current  Time  M 0 L0TA  C
62. Electric dipole moment P=Charge  Distance  M 0 L1 AT  Cm
63. Electric field strength (or)
Force
Electric field intensity E=  M 1 LT 3 A1  NC -1
Charge
64. Electrical flux ( E ) Electrical intensity  area  M 1 L3T 3 A1  Nm2 C-1
Work
65. Electric potential (or) V=  M 1 L2T 3 A1  V
Charge
Potential difference
Pot.diff
66. Electrical resistance R=  M 1 L2T 3 A2  
Current
1 1
67. Electrical conductance C= =  M 1 L2T 3 A2  mho (or) Siemen (S)
R Resistance
68. Specific resistance (or

RA
Resistivity  (or) s ρ=  M 1 L3T 3 A2  Ohm-m
l
1
69. Electrical conductivity  = R esistivity  M 1 L3T 3 A2  Ohm-1 m-1 (or) Siemen m-
1
70. Current density

( current per unit area J = Electrical intensity
of cross section)  Conductivity  M 0 L2T 0 A  Am-2
 Current 
or  Area 
 
Q Charge
71. Capacitance C= =  M 1 L2T 4 A2  F
V Potential
dε Voltage×Time
L= =
72. Self (or) Mutual  dI  Current  M 1 L2T 2 A2  H(or) Wb/amp
 
 dt 
inductance
q1q 2
73. Electrical permittivity ε=  M 1 L3T 4 A2  farad/m
4πFd 2
74. Surface charge density  M 0 L2T 1 A1  Cm-2
Charge
Area
Light energy
75. Luminous flux  M 1 L2T 3  lumen
Time
ΔE  Luminous flux 
76. Intensity of illumination I= =   M 1 L0T 3  lumen m-2 (or) lux.
ΔtΔA  Area 
(or) Iluminance
1
77. Focal poEX.r P=  M 0 L1T 0  dioptre
Focal length
1
78. Wave number =  M 0 L1T 0  m-1
λ
(Propagation constant)
Z2 e4 m
79. Rydberg’s constant R=  M 0 L1T 0  m-1
8ε 02 ch 3
 Dimensions of a physical quantity are the powers to which the fundamental quantities are to be raised to
represent that quantity.
Dimensional Formula :
 An expression showing the powers to which the fundamental quantities are to be raised to represent the
derived quantity is called dimensional formula of that quantity.
In general the dimensional formula of a quantity can be written as  M x L yT z  . Here x,y,z are dimensions.
Dimensional Constants:
 The physical quantities which have dimensions and have a fixed value are called dimensional constants.
Ex:Gravitational constant (G), Planck's constant (h), Universal gas constant (R), Velocity of light in
vacuum (c) etc.,
Dimensionless Quantities:
 Dimensionless quantities are those which do not have dimensions but have a fixed value.
(a):Dimensionless quantities without units.
Ex:Pure numbers,angle trigonometric functions , logarthemic functions etc.,
(b)Dimensionless quantities with units.
Ex:Angular displacement - radian, Joule's constant etc.,
Dimensional variables:
 Dimensional variables are those physical
quantities which have dimensions and do not have fixed value.
Ex:velocity, acceleration, force, work, power.etc.
Dimensionless variables:
 Dimensionless variables are those physical quantities which do not have dimensions and do not have
fixed value.,
Ex: Specific gravity, refractive index, Coefficient of friction, Poisson's Ratio etc.,
Limitationas of Dimensional analysis method
 Dimensionless quantities cannot be determined by this method. Constant of proportionality cannot be
determined by this method. They can be found either by experiment (or) by theory.
 This method is not applicable to trigonometric, logarithmic and exponential functions.
 In the case of physical quantities which are
dependent upon more than three physical quantities, this method will be difficult.
 In some cases, the constant of proportionality also possesses dimensions. In such cases we cannot use
this system.
 If one side of equation contains addition or subtraction of physical quantities, we cannot use
this method.
EX.11: Let  0  denote the dimensional formula of permittivity of vacuum .If M is mass ,L is
length,T is time and A is electric current,then(JEE-MAIN 2013)
1 q1q2
Sol. From coulomb’s law F  4 R 2
0
q1q2
0 
4 FR 2
Substituting the units
2
0 
c2

 AT 
  M 1 L3T 4 A2 
N  m2 MLT 2   L2  
EX.12:The dimensional formula of magnetic field strength in M, L, T and C (coulomb) is given as
(AIEEE 2008)
Sol. From F = Bqv
2
F  MLT 
B  1
  M 1 L0T 1C 1 
qv C  LT 
Physical Quantities Having Same
Dimensional Formulae:
 Distance, Displacement, radius,wavelength, radius of gyration [L]
 Speed, Velocity, Velocity of light  LT 1 
 acceleration ,acceleration due to gravity, intensity of gravitational field, centripetal acceleration  LT 2 
 Impulse, Change in momentum  M LT 1 
 Force, weight, Tension,energy gradient, Thrust  M LT 2 
 Work, Energy, Moment of force or Torque, Moment of couple  M L2 T 2 
 
 Force constant, Surface Tension, Spring constant, surface energy i.e. Energy per unit area  M T 2 
 Angular momentum, Angular impulse, Planck's constant  M L2 T 1 
 Angular velocity, Frequency, angular frequency,Velocity gradient,
Decay constant, rate of disintegration [T–1]
 Stress, Pressure, Modulus of Elasticity, Energy density M L1 T 2 
 Latent heat, Gravitational potential  L2 T 2 
 Specific heat, Specific gas constant L2 T 2  1 
 Thermal capacity, Entropy, Boltzmann constant, Molar thermal capacity, M L2 T 2  1 
 Wave number, Power of a lens, Rydberg’s constant  L1 

L
 Time, RC, , LC  T 
R
 Power, Rate of dissipation of energy,  ML2T 3 
 Intensity of sound, Intensity of radiation [ MT 3 ]
 Electric potential, potential difference, electromotive force [ ML2T 3 I 1 ]
 Intensity of magnetic field, Intensity of magnetization I L1 
 Electric field and potential gradient  MLT 3 A1 
 Rydberg’s constant and propagation constant  M 0 L1T 0 

 Strain , Poisson’s ratio, refractive index, dielectric constant, coefficient of friction, relative permeability,
magnetic susceptibility, electric susceptibility, angle, solid angle, trigonometric ratios,logarithm function,
exponential constant are all dimensionless.
L
 If L,C and R stands for inductance, capacitance and resistance respectively then , LC , RC and
R
time  M 0 L0T 
 Coefficient of linear expansion, coefficient of superficial expansion and coefficient of cubical
expansion,temperature coefficient of resistance  M 0 L0T 0 K 1 
 Solar constant and poynting vector  ML0T 3 

Principle of homogeneity:
 It states that only quantities of same dimensions can be added, subtracted and equated.
a a  ct 2
EX.13: The dimensional formula of in the equation P  where P = pressure, x=
b bx
displacement and t = time
2
 a   ct 
Sol.  P   
 bx   bx 
 
a
By principle of Homogeneity,  bx  should represent pressure
a 1 a 2
  b  L   ML T    b    MT 
1 2
 
Uses of dimensional analysis method:
 To check the correctness of the given equation. (This is based on the principle of homogeneity)
 To convert one system of units into another system.
 To derive the equations showing the relation between different physical quantities.
1 2
EX.14:Check whether the relation S  ut  at is dimensionally correct or not, where symbols
2
have their usual meaning.
1 2
Sol. we have S  ut  at . checking the dimensions on both sides, LHS=  S    M 0 LT
1 0
 ,
2
1 
RHS= ut    2 at    LT  T    LT  T 
2 1 2 2
  M 0 LT
1 0
   M 0 LT
1 0
   M 0 L1T 0 
we find LHS=RHS.
Hence, the formula is dimensionally correct.
EX.15:Young’s modulus of steel is 19  1010 N / m 2 . Express it in dyne / cm 2 . Here dyne is the CGS
unit of force.
Sol. The SI unit of Young’s modulus is N / m 2 . .
 5 
N  19 1010  10 dyne   dyne 
Given Y  19  10
10
 102 cm 2   19  1011  2 
m2   
  cm 
EX.16 : For a particle to move in a circular orbit uniformly, centripetal force is required, which
depends upon the mass (m), velocity (v) of the particle and the radius (r) of the circle. Express
centripetal force in termsof these quantities
Sol. According to the provided information,
let F  m a vb r c .  F  km a vb r c
b
  M 1 LT
1 2
   M a  LT 1  Lc 
 
  M 1 LT
1 2
   M a Lb  cT  b 
using principle of homogeneity we have
a = 1 ,b + c = 1 ,b = 2
on solving we have a = 1, b = 2, c = -1
using these values we get F = km1v 2 r 1
mv 2
F k
r
Note: The value of the dimensionless constant k is to be found experimentally.
EX.17: Derive an expression for the time period of a simple pendulum of mass(m), length (l) at a
place where acceleration due to gravity is (g).
Sol. Let the time period of a simple pendulum depend
upon the mass of bob m, length of pendulum l ,
and acceleration due to gravity g, then
t  m a l b g c  t  km al b g c
c
M 0 L0T 1  M a Lb  LT 2   M 0 L0T 1  M a Lb  cT 2 c
comparing the powers of M, L, and T on
both sides, we get a = 0, b + c = 0, -2c=1
1
2
0 l l
 a = 0, b = 1/2 and c = -1/2. Putting these values, we get T  km 1
2
T  k
g
,
g
which is the required relation.
EX.18: If C is the velocity of light, h is Planck’s constant and G is Gravitational constant are taken
as fundamental quantities, then the dimensional formula of mass is.(Eamcet - 2014)
Sol. C   LT 1   (1) ; h   ML2T 1   (2)
1 3
G   M L T  2   (3)
Solving (2) and (3)
h  ML2T 1 
   M 2 L1T 1 
G  M 1 L3T 2 
Substituting (1) in above
1 1 1
h M 2  M  h 2 G 2 C 2 
    
G C  
EX.19: If E, M, J and G respectively denote energy, mass, angular momentum and universal gravi-
EJ 2
tational constant, the quantity, which has the same dimensions as the dimensions of
M 5G 2
(Eamcet - 2013)
EJ 2
Sol. D.F. of
M 5G 2
Substituting D.F. of E, J, M, and G in above formula
2
ML2T 2  ML2T 1 

5 1 3 2 2   M 0 L0T 0 
M  M L T 
 1  y
EX.20: In the equation  p    k T where p is the pressure, y is the distance, k B is Boltzmann con-
  B
stant and T is the temperature. Dimensions of  are (Med- 2013)
1 y
Sol. 
p  k BT
 Dimensional formulae of k B  Dimensional formulae of T 
Dimension of    =  Dimensional formulae of p Dimensional formulae of y 
 ML2T 3  T 
   M 0 L2T 0 
 ML T   L 
1 2
 Dimensions of M,L,T in  are 0,2,0

  a
EX.21: The vander Waal’s equation for n moles of a real gas is  p  V 2 V b  nRT where p is pres-
 
sure, V is volume, T is absolute temperature, R is molar gas constant a, b and c are vander
Waal’s constants. The dimensional formula for ab is (Med- 2012)
a
Sol. By principle of homogenity of dimensions P can added to P only. It means 2 also gives pressure.
V
1 1 2 0 3 0
Dimension formulae for pressure  P    M L T  and Volume V    M L T 
a
Since = pressure
V2
a a
   M 1 L1T 2   0 6 0   M 1 L1T 2 
M LT0 
0 3
M LT
 a   M 1 L5T 2 
similarly, b will have same dimensions as volume V  b  volume
 b   M 0 L3T 0 
  ab    M 1 L5T 2   M 0 L3T 0    M 1 L8T 2 

  a
EX.21: The vander Waal’s equation for n moles of a real gas is  p  V 2 V b  nRT where p is pres-
 
sure, V is volume, T is absolute temperature, R is molar gas constant a, b and c are vander
Waal’s constants. The dimensional formula for ab is (Med- 2012)
a
Sol. By principle of homogenity of dimensions P can added to P only. It means 2 also gives pressure.
V
1 1 2 0 3 0
Dimension formulae for pressure  P    M L T  and Volume V    M L T 
a
Since = pressure
V2
a a
   M 1 L1T 2   0 6 0   M 1 L1T 2 
 M 0 3 0
L T  M LT
 a   M 1 L5T 2 
similarly, b will have same dimensions as volume V  b  volume
 b   M 0 L3T 0 
  ab    M 1 L5T 2   M 0 L3T 0    M 1 L8T 2 

EX.22:A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the
diameter of a wire of length 5.6 cm. The main scale reading is 1 mm and 47th circular division
coincides with the main scale. Find the curved surface area of the wire in cm 2 to appropriate
significant figures.(Use  = 22/7)
1 mm
Sol. Least Count =  0.01 mm
100
Diameter = MSR + CSR(LC) = 1 mm+47 (0.01) mm = 1.47 mm
22
Surface area =  Dl   1.47  56mm 2
7
= 2.58724 cm 2 = 26cm 2
EX.23: In Searle’s experiment, the diameter of the wire as measured by a screw gauge of least
count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm.
When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by
a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s
modulus of the material of the wire from these data.
Sol.Maximum percentage error in Y is given by
W L
Y    Y   2  D   x  L
 D2 x  Y   D  x L
4
 0.001   0.001   0.1 
 2     0.0489
 0.05   0.125   110 
EX.24:The side of a cube is measured by vernier calipers (10 divisions of the vernier scale coincide
with 9 divisions of the main scale, where 1 division of main scale is 1 mm). The main scale reads
10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736
g. Find the density of the cube in appropriate significant figures.
Sol.Least count of vernier calipers
1 division of main scale 1
   0.1 mm
Number of divisions in vernier scale 10
The side of cube = 10 mm + 1 0.1 mm  1.01 cm

Mass 2.736 g 3
Now, density = Volume  1.013 cm3  2.66 g cm
Accuracy, precision, types of errors and combination of errors

EX 25. T he accur acy in the measur ement of the diameter of hydr ogen atom as 1.06 x 10-10 m is
1
1) 0.01 2)106 x 10-10 3) 4)0.01 x 10-10
106
d 0.011010 1
Sol :  10
 key-3
d 1.06 10 106
EX 26. The length of a rod is measured as 31.52 cm. Graduations on the scale are up to
1) 1 mm 2) 0.01 mm 3) 0.1 mm 4) 0.02 cm
Sol : 0.01cm is the least count of varnier caliperse.
key-3
EX 27. If L   20  0.01 m and B  10  0.02  m then L/B is
1)  2  0.03 m 2)  2  0.015 m
3)  2  0.01 m 4)  2  0.005 m
x L B  L B 
Sol :    x  x   
x L B L B
20  0.01 0.02 
 
10  20 10 
x  x   2  0.005  m
key-4
EX 28. The radius of a sphere is measured as 10  0.02%  cm . The error in the measurement of its
volume is
1) 25.1cc 2)25.12cc 3)2.51cc 4)251.2cc
4 3 v r
Sol : V   r   100  3  100
3 v r
r
v  3   v , key-3
r
EX 29. If length and breadth of a plate are  40  0.2  cm and  30  0.1 cm , the absolute error in
measurement of area is
1) 10cm2 2) 8cm2 3) 9 cm 2 4) 7 cm 2
A l  b  l  b 
Sol : A  lb  A

l

b
 A  A  
b 
 l
A  bl  l b  10cm 2 , key-1

EX 30. If the length of a cylinder is measured to be 4.28 cm with an error of 0.01 cm, the percentage
error in the measured length is nearly
1) 0.4 % 2) 0.5 % 3) 0.2 % 4) 0.1 %
l 0.01
Sol :  100   100  0.2%
l 4.28
key-3
EX 31. When 10 observations are taken, the random error is x. When 100 observations are taken,
the random error becomes
1) x/10 2) x 2 3) 10 x 4) x
1 X1 N2 10
Sol : X   X  N  100
N 2 1
key-1
EX 32. If L1   2.02  0.01 m and L2  1.02  0.01 m then L1  2 L2 is (in m)
1) 4.06  0.02 2) 4.06  0.03
3) 4.06  0.005 4) 4.06  0.01
Sol : L1  2 L2  2.02  2 1.02  4.06
L1  2L2  0.01  2  0.01  0.03
key-2
EX 33. A body travels uniformly a distance of  20.0  0.2  m in time  4.0  0.04  s . The velocity of
the body is
1)  5.0  0.4  ms 1 2)  5.0  0.2  ms 1
3)  5.0  0.6  ms 1 4)  5.0  0.1 ms 1
S V S T
Sol : V    
T V S T
key-4
Significant figures & Rounding off

EX 34. If the value of 103.5 kg is rounded off to three significant figures, then the value is
1) 103 2) 103.0 3) 104 4) 10.3
Sol :If last digit is 5, if the preceding digit is odd then it should be increased by adding 1 and last digit 5
has to be ignored.
key-3
EX 35. The number of significant figures in 6.023 1023 mole 1 is
1) 4 2) 3 3) 2 4) 23
Sol :Use limitation of significant figures
key-1
EX 36. The side of a cube is 2.5 metre. The volume of the cube to the significant figures is
1) 15 2) 16 3) 1.5 4) 1.6
Sol : V  l 3 and rounded off to minimum significant
key-2
Units and dimensional formulae
EX 37. If the unit of length is doubled and that of mass and time is halved, the unit of energy will be
1) doubled 2)4 times 3)8 times 4) same
2 2
E2 M 2  L2   T2 
Sol : E  M  L   T 
1 1  1   1 
key-3
EX 38. Given M is the mass suspended from a spring of force constant. k. The dimensional formula
1/ 2
for  M / k  is same as that for
1) frequency 2) time period 3) velocity 4) wavelength
Sol :Here [k] = force/ length = ML0T 2
1/ 2
M 
Hence    M 0 L0T
k 
key-2
EX 39. The dimensional formula for the product of two physical quantities P and Q is [ ML2T 2 ]. The
P
dimensional formula of Q is [ MT 2 ]. Then P and Q respectively are(2001 M)
1) Force and velocity 2) Momentum and displacement
3) Force and displacement 4) Work and velocity
P
Sol : PQ  ML2T 2 ----(1);  MT 2 -----(2)
Q
(1) X (2) = P2  M 2 L2T 4
2 2
 P  MLT 2  FORCE (1) / (2) = Q = L
key-3
EX 40. If the unit of length is doubled and that of mass and time is halved, the unit of energy will be
1) doubled 2)4 times 3)8 times 4) same
2 2
E2 M 2  L2   T2 
Sol :     
E1 M 1  L1   T1 
key-3
EX 41. Given M is the mass suspended from a spring of force constant. k. The dimensional formula
1/ 2
for  M / k  is same as that for
1) frequency 2) time period
3) velocity 4) wavelength
Sol :Here [k] = force/ length = ML0T 2
1/ 2
M 
Hence    M 0 L0T , key-2
k 
EX 42. The dimensional formula for the product of two physical quantities P and Q is [ ML2T 2 ]. The
P
dimensional formula of Q is [ MT 2 ]. Then P and Q respectively are(2001 M)
1) Force and velocity
2) Momentum and displacement
3) Force and displacement
4) Work and velocity
Sol :
P
PQ  ML2T 2 ----(1);  MT 2 -----(2)
Q
(1) X (2) = P2  M 2 L2T 4
2 2
 P  MLT 2  FORCE (1) / (2) = Q = L
key-3
EX 43. If minute is the unit of time, 10 ms-2 is the unit of acceleration and 100 kg is the unit of mass,
the new unit of work in joule is
1) 105 2) 106 3) 6 x 106 4) 36x 106
W2 M 2 a2 2T2 2
Sol : W  Ma T ;
2 2 
W1 M1a12T12
key-4
EX 44. The magnitude of force is 100 N. What will be its value if the units of mass and time are
doubled and that of length is halved?
1) 25 2)100 3) 200 4) 400
Sol : n1[M1L1T12 ]  n2 [ M 2 L2T22 ]
key-1
EX 45. If force (F), work (W) and velocity (V) are taken as fundamental quantities then the dimensional
formula of Time (T) is (2007 M)
1) [ W 1 F 1V 1 ] 2) [ W 1 F 1V 1 ]
3) [ W 1 F 1V 1 ] 4) [ W 1 F 1V 1 ]
Sol : T  F xW yV z ; M 0 L0T 1 [ MLT 2 ]x [ ML2T 2 ] y [ LT 1 ]z

key-4
EX 46. The error in the measurement of the length of the simple pendulum is 0.2 % and the error in
L
time period 4%. The maximum possible error in measurement of 2 is
T
1) 4.2% 2) 3.8% 3) 7.8% 4) 8.2%
L x L T
Sol :Let x  ; = 2
T2 x L T
key-4
EX 47. The least count of a stop watch is (1/5) s. The time of 20 oscillations of a pendulum is measured
to be 25 s. The maximum percentage error in this measurement is
1) 8 % 2) 1 % 3) 0.8 % 4) 16 %
1/ 5 25 T
Sol :. T  and T  ; % error  100
20 20 T
key-3
EX 48. The diameter of a wire as measured by a screw gauge was found to be 1.002 cm, 1.004 cm
and 1.006 cm. The absolute error in the third reading is
1) 0.002 cm 2) 0.004 cm
3) 1.002 cm 4) zero
Sol :. x3  x3  xmean
key-1
EX 49. Force and area are measured as 20 N and 5m2 with errors 0.05 N and 0.0125m2. The
maximum error in pressure is (SI unit)
1) 4  0.0625 2) 4  0.05
3) 4  0.125 4) 4  0.02
F p F A  F A 
Sol :. p   p  F  A  p  p  F  A 
A
key-4
EX 50. The length and breadth of a rectangular object are 25.2cm and 16.8cm respectively and have
been measured to an accuracy of 0.1cm. Relative error and percentage error in the area of the
object are
1) 0.01 & 1% 2) 0.02 & 2%
3) 0.03 & 3% 4) 0.04 & 4%
a l b
Sol :. a  l  b ;  
a l b
a  l b 
 100      100
a  l b 
key-1
3 -3
x
EX 51. Dimensional analysis of the equation  Velocity  =  Pressure difference  2 .  density  2
gives the value of x as: (1986 E)
1) 1 2) 2 3) 3 4)-3
Sol :17. Substitute dimension formulae
key-3
EX 52. For the equation F =Aavbdc where F is force, A is area, v is velocity and d is density, with the
dimensional analysis gives the following values for the exponents. (1985 E)
1) a=1, b = 2, c =1 2) a =2, b =1, c= 1
3)a =1, b =1, c= 2 4) a = 0, b =1 , c = 1
a b c
Sol :18. F = Aa vb d c ; MLT 2   L2   LT 1   ML3  comparing the powers on both sides
key-1
EX 53. The length of pendulum is measured as 1.01m and time for 30 oscillations is measured as
one minute 3 seconds. Error in length is 0.01 m and error in time is 3 secs. The percentage error
in the measurement of acceleration due to gravity is. (Engg. - 2012)
1) 1 2) 5 3) 10 4) 15
l g l T
Sol :19. T  2 ; 100  100  2 100
g g l T
key-3
EX 54. The Energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as
fundamental quantities. The dimensions of universal gravitational constant in the dimensional
formula of Planks constant (h) is (Eng - 2008)
1) 0 2) -1 3) 5/3 4) 1
Sol : h  E , L, G
a b c
ML2T 1   ML2T 2   ML2T 1   M 1L3T 2 
key-1
EX 55. A body weighs 22.42 g and has a measured volume of 4.7 cc the possible errors in the
measurement of mass and volume are0.01g and 0.1 cc. Then the maximum percentage error
in the density will be(Med- 2010)
1) 22% 2) 2.2% 3) 0.22% 4) 0.022%
M
Sol :. The density of d  ; % Error of density
V
Δd ΔM ΔV
×100= ×100+ ×100 , key-2
d M V
EX 56. If energy E, velocity v and time T are taken as fundamental quantities, the dimensional
formula for surface tension is (Med-2009)
1)  Ev 2T 2  2)  E 2 vT 2 
3)  Ev 2T 1  4)  E 2 v 2T 1 
a b c
Sol :  S    E    v   T 
a b c
 MT 2    ML2T 2    LT 1   T 
Comparing the powers on both sides we get a,b,c
key-1
EX 57. The measured mass and volume of a body are 53.63 g and 5.8 cm3 respectively, with possible
errors of 0.01 g and 0.1 cm3. The maximum percentage error in density is about
1) 0.2% 2) 2% 3) 5% 4) 10%
M   m V 
Sol :Density   ;  100   m  V  100
V  
key-2
EX 58. A vernier calipers has 1 mm marks on the main scale . It has 20 equal divisions on the vernier
scale,which match with 16main scale divisions. For this vernier calipers the least count is
1) 0.02mm 2) 0.05 mm 3) 0.1mm 4) 0.2mm
Sol :. 16 M.S.D = 20 V.S.D  1V .S .D  4 / 5 M .S .D
L.C = 1M.S.D - 1 V.S.D
key-4
EX 59. The resistance of metal is given by V=IR. The voltage in the resistance is V   8  0.5  V
and current in the resistance is I   2  0.2  A, the value of resistance with its percentage error
is
1)  4  16.25%   2)  4  2.5%  
3)  4  0.04%   4)  4  1%  
V R  V I 
Sol :. R ; 100     100
I R  V I 
 R 
Resistance =  R  R  100 
 
key-1
EX 60. In an experiment, the values of refractive indices of glass were found to be 1.54, 1.53, 1.44,
1.54, 1.56 and 1.45 in successive measurements
i) mean value of refractive index of glass ii) mean absolute error
iii) relative error and iv) percentage error are respectively,
1)1.51,0.04,0.03,3% 2)1.51,0.4,0.03,3 %
3)15.1,0.04,0.03,3% 4)15.1,0.04,0.3,3 %
   i 
mean   ; mean   mean

Sol : ;
6 6
mean 
relative % error in  =   100 key-1
mean
 4 2 L 
g
EX 60. A student performs an experiment for determination of   2  ,L  1m, and he commits
 T 
an error of L for T he tajes the time of n oscillations with the stop watch of least count T .For
which of the following data the measurement of g will be most accurate?
1) L  0.5, T  0.1, n  20
2) L  0.5, T  0.1, n  50
3) L  0.5, T  0.01, n  20
4) L  0.5, T  0.05, n  50
g l T
Sol :  2 ( l and T are least, and the number of readings are maximum)
g l T
key- 4
EX 61. A rectangular metal slab of mass 33.333 has its length 8.0 cm, breadth 5.0 cm and thickness
1mm. The mass is measured with accuracy up to 1 mg with a sensitive balance. The length and
breadth are measured with vernier calipers having a least count of 0.01 cm. The thickness is
measured with a screw gauge of least count 0.01 mm. The percentage accuracy in density
calculated from the above measurements is
1) 13 % 2)130 % 3)1.6 % 4)16 % m
Sol : Percentage error gives percentage accuracy d 
lbh
d m l b h
relative error,    
d m l b h
 d 
and calculate   100
 d 
key- 3
  z / K
EX 62. In the relation p  e ; P is pressure, K is Boltzmann’s constant, Z is distance and

 is temperature. The dimensional formula of  will be
0 2 0 1 2 1
1)  M L T  2)  M L T 
0 1 0 2 1
3)  ML T  4)  M L T 
 z 
Sol :    1 ; 20)Here  A  IT 2 and  B   KT
 k 
key-1
EX 63. The heat generated in a circuit is given by Q = i2 Rt joule , where ‘i’ is current, R is
resistance and t is time. If the percentage errors in measuring i, R and t are 2%, 1% and 1%
respectively, the maximum error in measuring heat will be
1) 2 % 2) 3 % 3) 4 % 4) 6 %
Q 2 i R t
Sol : Q  i 2 Rt ; Q 100  i 100  R 100  t 100
key-4
EX 64. You measure two quantities as A=1.0m  0.2m, B=2.0m  0.2m. We should report correct
value for AB as
1)1.4m  0.4m 2) 1.41m  0.15m
3) 1.4m  0.3m 4) 1.4m  0.2m
Sol : Y  AB  1.0 2.0  1.414m
y 1  A B  1  0.2 0.2  0.6
        
y 2 A B  2  10 2.0  2  2.0
Rounding off to one significant digit y =0.2 cm
key-4
EX 65. Which of the following measurement is most precise ?
1) 5.00 mm 2) 5.00 cm 3) 5.00 m 4) 5.00 km
Sol : All given measurement are correct upto two decimal places. As here 5.00 mm has the smallest unit and
the error in 5.00 mm is least (commonly taken as 0.01 mm if not specified), hence, 5.00 mm is mpst
precise.
key-1
EX 66. The mean length of an object is 5 cm. Which of the following measurement is most accurate
?
1) 4.9 cm 2) 4.805 cm3) 5.25 cm 4) 5.4 cm
Sol : Given length
Now, checking the errors with each options one by one, we get
l1 = 5 - 4.9 = 0.1 cm
l 2 = 5 - 4.805 = 0.195 cm
l3 =5.25 - 5 = 0.25 cm
l 4 = 5.4 - 5 = 0.4 cm
Error l1 is least
Hence, 4.9 cm is most precise.
key-1
JEE MAIN PREVIOUS YEAR QUESTIONS
TOPIC-1…..Unit of Physical Quantities
1. The density of a material in SI unit is 128 kg m−3 In certain units in which the unit of
length is 25 cm and the unit of mass is 50 g, the numerical value of density of the
material is:
[10 Jan. 2019 ]
(a) 40 (b) 16 (c) 640 (d) 410
128kg
sol. (a) Density of material in SI unit,= m3
Density of material in new system

128(50g)(20) 128
= = (20) = 40units
(25cm)3 (4)3 64
2. A metal sample carrying a currentalong X-axis with density Jx is subjected to a

magnetic field Bz (along z‐axis). The electric field Ey developed along Y‐axis is directly
proportional to Jx as well as Bz . The constant of proportionality has SI unit
[Online April 25, 2013]
𝑚2 𝑚3 𝑚2 𝐴𝑠
(a) (b) (c) (d) 𝑚3
𝐴 𝐴𝑠 𝐴𝑠
sol. (b) According to question Ey ∝ Jx BZ

Ey C m3
Constant of proportionality K = B =J =
Z Jx x As
E I
[As B = C (speed of light) and J = Area]
TOPIC-2…..Dimensions of Physical Quantities
1 𝐸 1
3. The quantities = , 𝑦 = 𝐵 and 𝑧 = 𝐶𝑅 are defined where 𝐶‐capacitance, 𝑅‐
√𝜇0 𝜀0
Resistance, 𝑙‐length, 𝐸‐Electric field, 𝐵‐magnetic field and 𝜀0 , 𝜇0 , ‐ free space

permittivity and permeability respectively. Then : [Sep. 05, 2020 (II)]
(a) 𝑥, 𝑦 and 𝑧 have the same dimension.
(b) Only 𝑥 and 𝑧 have the same dimension.
(c) Only 𝑥 and 𝑦 have the same dimension.
(d) Only 𝑦 and 𝑧 have the same dimension.
1
sol. (a) We know that Speed of light, 𝑐 = =𝑥
√𝜇0 𝜀0
𝐸
Also, 𝑐 = 𝐵 = 𝑦
Time constant, 𝛤 = 𝑅𝑐 = 𝑡
𝑙 𝑙
𝑧 = 𝑅𝑐 = 𝑡 = Speed
Thus, 𝑥, 𝑦, 𝑧 will have the same dimension of speed.
4. Dimensional formula for thermal conductivity is (here 𝐾 denotes the temperature):

[Sep. 04, 2020 (I)]
(a) MLT −2 K (b) MLT −2 K −2 (c) MLT −3 K (d) MLT −3 K −1
 dQ 
 
k  
dQ dT dt 
sol. (d) From formula,  kA
dt dx  dT 
A 
 dx 
[ML2 T −3 ]
[𝑘] = = [MLT −3 K −1 ]
[L2 ][KL−1 ]
5. A quantity x is given by (𝐼𝐹𝑣 2 /𝑊𝐿4 ) in terms of moment of inertia 𝐼, force 𝐹, velocity

𝑣, work 𝑊 and Length 𝐿. The dimensional formula for 𝑥 is same as that of:
[Sep. 04, 2020 (II)]
(a) planck’s constant (b) force constant
(c) energy density (d) coefficient of viscosity
sol. (c) Dimension of𝛤orce F= M1 L1 T −2
Dimension of velocity 𝑉 = L1 T −1
Dimension of work = M1 L2 T −2
Dimension of length = L
Moment of inertia = ML2
2
𝐼𝐹𝑣 2 (M1 L2 )(M1 L1 T−2 )(L1 T−2 )
𝑥= =
𝑊𝐿4 (M1 L2 T−2 )(L4 )
M1 L−2 T−2
= = M1 L−1 T −2 = Energy density
L3
6. Amount of solar energy received on the earth’s surface per unit area per unit time is
defined a solar constant. Dimension of solar constant is: [Sep. 03, 2020 (II)]
(a) ML2 𝑇 −2 (b) ML0 𝑇 −3 (c) M 2 L0 𝑇 −1 (d) ML𝑇 −2
Energy
sol. (b) Solar constant = TimeArea
Dimension of Energy, 𝐸 = ML2 T −2

Dimension of Time = T
Dimension of Area = L2
M1 L2 T−2
Dimension of Solar constant = = M1 L0 T −3
TL2
7. If speed V, area A and force 𝐹 are chosen as fundamental units, then the dimension of
Young’s modulus will be: [Sep. 02, 2020 (I)]
(a) 𝐹A2 V −1 (b) 𝐹A2 V −3 (c) 𝐹A2 V −2 (d) 𝐹A−1 V 0
stress
sol. (d) Young’s modulus, 𝑌 = strain
𝐹 𝛥ℓ
⇒𝑌= / = 𝐹A−1 V 0
A ℓ0
8. If momentum(P), area (A) and time (T) are taken to be the fundamental quantities then
the dimensional formula for energy is : [Sep. 02, 2020 (II)]
(a) [P 2 AT −2 ] (b) [PA−1 T −2 ] (c) [PA1/2 T −1 ] (d) [P1/2 AT −1 ]
sol. (c) Energy 𝐸 ∝ 𝐴𝑎 𝑇 𝑏 𝑃𝑐
or, 𝐸 = 𝑘𝐴𝑎 𝑇 𝑏 𝑃𝑐 ---(i)
where 𝑘 is a dimensionless constant and 𝑎, 𝑏 and 𝑐 are the exponents.
Dimension of momentum, 𝑃 = 𝑀1 𝐿1 𝑇 −1
Dimension of area, 𝐴 = 𝐿2
Dimension of time, 𝑇 = 𝑇 1
Putting these values in equation (i), we get 𝑀1 𝐿2 𝑇 −2 = 𝑀𝑐 𝐿2𝑎+𝑐 𝑇 𝑏−𝑐
by comparison
𝑐=1
2𝑎 + 𝑐 = 2
𝑏 − 𝑐 = −2
𝑐 = 1 , 𝑎 = 1/2, 𝑏 = −1
𝐸 = 𝐴1/2 𝑇 −1 𝑃1
9. Which of the following combinations has the dimension of electrical resistance (∈0 is the
permittivity of vacuum and 𝜇0 is the permeability of vacuum)? [12 April 2019 I]
𝜇 𝜇0 𝜀 𝜀
(a) √ 𝜀 0 (b) (c) √𝜇0 (d) 𝜇0
0 𝜀0 0 0
𝜇 𝜇02
sol. (a) √ 𝜀 0 = √𝜀 = 𝜇0𝐶
0 0 𝜇0
𝜇0 𝑐 → M𝐿T −2 A−2 × LT1 =ML2 T 3 A2

Dimensions of resistance
10. In the formula X = 5YZ 2 , X and Z have dimensions of capacitance and magnetic field,
respectively. What are the dimensions of Y in SI units? [10 Apri12019 II]
(a) [M 3 L2 T 8 A4 ] (b) [M1 L2 T 4 A2 ] (c) [M 2 L0 T ⊲ A2 ] (d) [M 2 L2 T 6 A3 ]
sol. (a) X = 5YZ 2
𝑋
⇒ 𝑌 ∝ 𝑍 2 --(i)
𝑄 𝑄2 [𝐴2 𝑇 2 ]
𝑋 = Capacitance = V = = [𝑀𝐿2 𝑇 −2 ]
𝑊
X= [M1 2 4 2 ]
LT A
𝐹
𝑍 = 𝐵 = 𝐼𝐿 [⋅.⋅ 𝐹 = 𝐼𝐿𝐵]
Z = [MT 2 A1 ]
[𝑀−1 𝐿−2 𝑇 4 𝐴2 ]
𝑌=
[𝑀𝑇 −2 𝐴−1 ]2
Y = [M 3 L2 T 8 A4 ] (Using (i))
∈0
11. In SI units, the dimensions of √𝜇 is: [8 April 2019 I]
0
(a) A−1 T1 M1 L3 (b) AT 2 M −1 L−1 (c) AT −3 ML3/2 (d) A2 T 3 M −1 L−2

𝜀 𝜀02 𝜀0
sol. (d) [√𝜇0 ] = √𝜇 =[ ] = 𝜀0 𝐶[𝐿𝛤 1 ] × [𝜀0 ]
0 0 𝜀0 √ 𝜇0 𝜀 0
1
[⋅.⋅ = 𝐶]
√𝜇0 𝜀0
𝑞2
𝐹=
4𝜋𝜀0 𝑟 2
[𝐴𝑇]2
⇒ [𝜀0 ] = = [𝐴2 𝑀−1 𝐿−3 𝑇 4 ]
[𝑀𝐿𝑇 −2 ] × [𝐿2 ]
𝜀0
[√ ] = [𝐿𝑇 −1 ] × [𝐴2 𝑀−1 𝐿−3 𝑇 4 ]
𝜇0
= [𝑀−1 𝐿−2 𝑇 3 𝐴2 ]
12. Let 𝑙, 𝑟, 𝑐 and 𝑣 represent inductance, resistance, capacitance and voltage, respectively.
ℓ
The dimension of in SI units will be: [12 Jan. 2019 II]
𝑟𝑐𝑣
(a) [𝐿A−2 ] (b) [A−1 ] (c) [LTA] (d) [LT 2 ]

sol. (b) As we know,
ℓ
[r] = [T] and [cv] = [AT]
ℓ T
[ ] = [ ] = [A−1 ]
rcv AT
𝑥2
13. The force of interaction between two atoms is given by = 𝛼𝛽 exp (− 𝛼𝑘𝑇) ; where 𝑥 is
the distance, k is the Boltzmann constant and T is temperature and 𝛼 and 𝛽 are two
constants. The dimensions of 𝛽 is: [11 Jan. 2019 I]
(a) M 0 L2 T −4 (b) M 2 L𝑇 −4 (c) MLT −2 (d) M 2 L2 T −2
sol. (b) Force of interaction between two atoms,
−x 2
𝛤 = 𝛼𝛽e ( )
𝛼kT
Since exponential terms are dimensionless
x2
[ ] = M 0 L0 T 0
𝛼kT
L2
⇒ = M 0 L0 T 0
[𝛼]ML2 T −2
⇒ [𝛼] = M −1 T 2
[𝛤] = [𝛼][𝛽]
MLT −2 = M −1 T 2 [𝛽]
⇒ [𝛽] = M 2 LT −4
14. If speed (V), acceleration (A) and force (𝐹) are considered as fundamental units, the
dimension of Young’s modulus will be: [11 Jan. 2019 II]
(a) V −2 A2 𝐹 −2 (b) V −2 A2 𝐹 2 (c) V −4 A−2 𝐹 (d) V −4 A2 𝐹
sol. (d) Let [Y] = [V]a [𝛤]b [A]c
[ML−1 T −2 ] = [LT −1 ]a [MLT −2 ]b [LT −2 ]c
[ML−1 T −2 ] = [M b La+b+c T −a−2b−2c ]
Comparing power both side of similar terms we get,
b = 1 , a + b + c = −1, −a − 2b − 2c = −2
solving above equations we get:
a = −4, b = 1, c = 2
so [Y] = [V −4 𝛤A2 ] = [V −4 A2 𝛤]
ℎ𝑐 5
15. A quantity 𝑓 is given by 𝑓 = √ where c is speed of light, G universal gravitational
𝐺
constant and ℎ is the Planck’s constant. Dimension of 𝑓 is that of: [9 Jan. 2019 I]
(a) area (b) energy (c) momentum (d) volume
sol. (b) Dimension of [ℎ] = [𝑀𝐿2 𝑇 1 ]
[C] = [𝐿𝑇 1 ]
[𝐺] = [𝑀1 𝐿3 𝑇 2 ]
Hence dimension of
ℎ𝐶 5 [𝑀𝐿2 𝑇 −1 ] ⋅ [𝐿5 𝑇 −5 ]
[√ ]=
𝐺 [𝑀 −1 𝐿3 𝑇 −2 ]
= [𝑀𝐿2 𝑇 2 ] = energy
16. Expression for time in terms of G (universal gravitational constant), h (Planck’s constant)
and c (speed of light) is proportional to: [9 Jan. 2019 II]
hc5 c3 Gh Gh
(a) √ (b) √Gh (c) √ c5 (d) √ c3
G
sol. (c) Let t ∝ Gx hy Cz

Dimensions ofG = [M −1 L3 T −2 ],
h = [ML2 T −1 ] and C = [LT −1 ]
[T] = [M −1 L3 T −2 ]x [ML2 T −1 ]y [LT −1 ]z
[M 0 L0 T1 ] = [M −x+y L3x+2y+z T −2x−y−z ]
By comparing the powers ofM, L, T both the sides
−x + y = 0 ⇒ x = y
3x + 2y + z = 0 ⇒ 5x + z = 0 (i)
−2x − y − z = 1 ⇒ 3x + z = −1 (ii) Solving eqns. (i) and (ii),
1 5 Gh
x = y = 2, z = − 2 t ∝ √ C5
17. The dimensions of stopping potential 𝑉0 in photoelectric effect in units of Planck’s

constant ′ℎ’, speed oflight‘c’ and Gravitational constant ‘ 𝐺’ and ampere 𝐴 is:
[8 Jan. 2019 I]
(a) ℎ𝑈3 𝐺 2/3 𝑐 𝑈3 𝐴−1 (b) ℎ2/3 𝑐 5/3 𝐺 1/3 𝐴−1 (c) ℎ2/3 𝐶 1/3 𝐺 4/3 𝐴−1 (d) ℎ2 𝐺 3/2 𝐶 1/3 𝐴−1
sol. (None)
Stopping potential (𝑉0 ) ∝ ℎ 𝑥 𝐼 𝑦 𝐺 𝑍 𝐶 𝑟
Here, ℎ = Planck’s constant = [𝑀𝐿2 𝑇 −1 ]
𝐼 = current = [𝐴]
𝐺 = Gravitational constant = [𝑀1 𝐿3 𝑇 2 ]
and 𝑐 = speed oflight = [𝐿𝑇 1 ]
𝑉0 = potentia1 = [𝑀𝐿2 𝑇 3 𝐴1 ]
[𝑀𝐿2 𝑇 3 𝐴1 ] = [𝑀𝐿2 𝑇 1 ]x [𝐴]y [𝑀1 𝐿3 𝑇 2 ]z [𝐿𝑇 1 ]𝑟
𝑀 𝑥−𝑧 ; 𝐿2𝑥+3𝑧+𝑟 ; 𝐼 ←𝑥−2𝑧−𝑟 ; 𝐴𝑦
Comparing dimension ofM, L, T, A, we get
y = −1, x = 0, z = −1 , r = 5
𝑉0 ∝ ℎ0 𝐼 −1 𝐺 −1 𝐶 5
𝐵2
18. The dimensions of 2𝜇 , where 𝐵 is magnetic field and 𝜇0 is the magnetic permeability of
0
vacuum, is: [8 Jan. 2019 II]

(a) MLT −2 (b) ML2 T −1 (c) ML2 T −2 (d) ML−1 T −2
B2
sol. (d) The quantity 2𝜇 is the energy density ofmagnetic field.
0
𝐵2 Energy 𝛤orce×disp1acement
⇒ [2𝜇 ] = 𝑉𝑜𝑙𝑢𝑚𝑒 = (disp1acement)3
0
𝑀𝐿2 𝑇 −2
=[ ] = 𝑀𝐿−1 𝑇 −2
𝐿3
19. The characteristic distance at which quantum gravitational effects are significant, the
Planck length, can be determined from a suitable combination of the fundamental
physical constants G, h and c. Which of the following correctly gives the Planck length?
1
1
2 Gh 2
(a) G hc (b) ( c3 ) (c) G 2 h2C (d) Gh2 c 3
sol. (b) Plank length is a unit of length, 𝑙𝑝 = 1.616229 × 10−35 m
ℎ𝐺
𝑙𝑝 = √
𝑐3
20. Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities
instead of mass, length and time. In terms of these, the dimensions of mass would be:
(a) [M] = [T −1 C−2 h] (b) [M] = [T −1 C2 h]
(c) [M] = [T −1 C−2 h−1 ] (d) [M] = [TC−2 h]
sol. (a) Let mass, related as M ∝ T x Cy hz
M1 L0 T 0 = (T ′ )x (L1 T1 )y (M1 L2 T1 )z
M1 L0 T 0 = M z Ly+2z + T xyz
z=1
y + 2z = 0 x − y − z = 0
y = −2 x + 2 − 1 = 0
x = −1
M = [T1 C2 h1 ]
21. A, B, C and D are four different physical quantities having different dimensions. None of
them is dimensionless. But we know that the equation AD = C ln (BD) holds true. Then
which of the combination is not a meaningful quantity? [Online Apri110, 2016]
C AD2 A (A−C)
(a) BD − (b) A2 − B2 C2 (c) B − C (d)
C D
sol. (d) Dimension ofA ≠ dimension of(C)

Hence A‐C is not possible.
22. In the following I refers to current and other s𝛾nbols have their usual meaning, Choose
the option that corresponds to the dimensions of electrical conductivity:
(a) M −1 L−3T 3 I (b) M −1 L−3 T 3 I2 (c) M −1 L3 T 3 I (d) ML−3 T −3 I2
sol. (b) We know that resistivity
RA
p=
ℓ
1 ℓ
Conductivity = resistivity = RA
ℓI
= (⋅.⋅ V = RI)
VA
[L][I] W W
= [ML2 T−2
..V= =
| [I][T] |×[L2 ] q it
= [M −1 L−3 T 3 ][I2 ] = [M −1 L−3 T 3 I2 ]
23. If electronic charge e, electron mass m, speed of light in vacuum c and Planck’s constant
h are taken as fundamental quantities, the permeability of vacuum 𝜇0 can be expressed in
units of: [Online April 11, 2015]
h hc h mc2
(a) (me2 ) (b) (me2 ) (c) (ce2 ) (d) ( he2 )
sol. (c) Let 𝜇0 related with 𝑒, 𝑚, 𝑐 and ℎ as follows.

𝜇0 = 𝑘𝑒 𝑎 𝑚𝑏 𝑐 𝑐 ℎ𝑑
[𝑀𝐿𝛤 2 𝐴−2 ] = [𝐴𝑇⌋𝑎 [𝑀]𝑏 [𝐿𝛤 1 ]𝑐 [𝑀𝐿2 𝛤 1 ]𝑑
= [𝑀𝑏+𝑑 𝐿𝑐+2𝑑 𝑇 𝑙−𝑐−𝑑 𝐴𝑎 ]
On comparing both sides we get
𝑎 = −2 (i)
𝑏 + 𝑑 = 1 (ii)
𝑐 + 2𝑑 = 1 (iii)
𝑎 − 𝑐 − 𝑑 = −2 (iv)
By equation (i), (ii), (iii) & (iv) we get,
𝑎 = −2, 𝑏 = 0, 𝑐 = −1 , 𝑑 = 1
ℎ
[𝜇0 ] = [ ]
𝑐𝑒 2
24. If the capacitance of a nanocapacitor is measured in terms of a unit ‘u’ made by
combining the electric charge ‘e’, Bohr radius ‘a0’, Planck’s constant‘h’ and speed of
light‘ c’ then: [Online Apri110, 2015]
e2 h hc e 2C e2 a0
(a) u = (b) u = e2 a (c) u = ha (d) u =
a0 0 0 hc
sol. (d) Let unit ‘u’ related with 𝑒, 𝑎0 , ℎ and 𝑐 as follows.

[𝑢] = [𝑒]𝑎 [𝑎0 ]𝑏 [ℎ]𝑐 [𝐶]𝑑
Using dimensional method,
[𝑀 −1 𝐿−2 𝑇 ⊢4 𝐴+2 ] = [𝐴1 𝑇 1 ]𝑎 [𝐿]𝑏 [𝑀𝐿2𝛤 1 ]𝑐 [𝐿𝛤 1 ]𝑑
[𝑀−1 𝐿−2 𝑇 ⊢4 𝐴+2 ] = [𝑀𝑐 𝐿𝑏+2𝑐+𝑑 𝛤 𝑟−𝑐−𝑑 𝐴𝑎 ]
𝑎 = 2, 𝑏 = 1 , 𝑐 = −1 , 𝑑 = −1
𝑒 2 𝑎0
𝑢=
ℎ𝑐
25. From the following combinations of physical constants (expressed through their usual
symbols) the only combination, that would have the same value in different systems of
units, is: [Online Apri112, 2014]
ch e2
(a) 2𝜋𝜀2 (b) 2𝜋𝜀 2 (me = mass of electron)
o o Gme
𝜇0 𝜀 o G 2𝜋√𝜇0 𝜀o h
(c) (d)
c2 he2 ce2 G
sol. (b) The dimensional formulae of

e = [M 0 L0 T1 A1 ]
𝜀0 = [M −1 L3 T 4 A2 ]
G = [M −1 L3 T −2 ] and me = [M1 L0 T 0 ]
e2
Now, 2𝜋𝜀 2
0 Gme
[M 0 L0 T1 A1 ]2
=
2𝜋[M −1 L−3 T 4 A2 ][M −1 L3 T −2 ][M1 L0 T 0 ]2
[T 2 A2 ]
=
2𝜋[M −1−1+2 L−3+3 T 4−2 A2 ]
[T 2 A2 ] 1
= =
2𝜋[M L T A ] 2𝜋
0 0 2 2
1
is dimensionless.
2𝜋
e2
Thus the combination 2𝜋𝜀 2 would have the same value in different systems of units.
0 Gme
𝜇
26. In terms of resistance R and time T, the dimensions ofratio 𝜀 of the permeability 𝜇 and
permittivity 𝜀 is: [Online April 11, 2014]

(a) [RT −2 ] (b)[R2 T −1 ] (c) [R2 ] (d) [R2 T 2 ]
sol. (c) Dimensions of 𝜇 = [MLT −2 A−2 ]
Dimensions of∈ = [M −1 L−3 T 4 A2 ]
Dimensions ofR = [ML2 T −3A−2 ]
Dimensionsof𝜇 [MLT −2 A−2 ]
=
Dimensionsof ∈ [M −1 L−3 T 4 A2 ]
= [M 2 L4 𝛤 −6 A−4 ] = [R2 ]
27. Let [∈0 ] denote the dimensional formula ofthe permittivity of vacuum. If M =mass,
L =length, T = time and A = electric current, then: [2013]
(a) ∈0 = [M −1 L−3 T 2 A] (b) ∈0 = [M −1 L−3 T 4 A2 ]
(c) ∈0 = [M1 L2 T1 A2 ] (d) ∈0 = [M1 L2 T1 A]
1q q
sol. (b) As we know, 𝛤 = 4𝜋𝜀1 R22
0
q1 q 2
⇒ 𝜀0 =
4𝜋𝛤R2
C2 [AT]2
Hence, 𝜀0 = N.m2 = [MLT−2 ][L2 ]
= [M −1 L−3 T 4 A2 ]
28. If the time period 𝑡 ofthe oscillation of a drop ofliquid of density 𝑑, radius 𝑟, vibrating
under surface tension 𝑠 is given by the formula 𝑡 = √𝑟 2𝑏 𝑠 𝑐 𝑑 𝑎/2 . It is observed that the
𝑑
time period is directly proportional to √ 𝑠 . The value of 𝑏 should therefore be:

3 3 2
(a) 4 (b) √3 (c) 2 (d) 3
sol. (c)
29. The dimensions of angular momentum, latent heat and capacitance are, respectively.
(a) ML2 T1 A2 , L2 T −2 , M −1 L−2 T 2 (b) ML2 T −2 , L2 T 2 , M −1 L−2 T 4 A2
(c) ML2 T −1 , L2 T −2 , ML2 TA2 (d) ML2 T −1 , L2 T −2 , M −1 L−2 T 4 A2
Q ML2 T−2
sol. (d) Angular momentum = m × v × r = ML2 T −1 Latent heat L = m = = L2 T −2
M
Charge
Capacitance C = = M −1 L−2 T 4 A2
P.d
30. Given that 𝐾 =energy, 𝑉 =velocity, 𝑇 =time. Ifthey are chosen as the fundamental units,
then what is dimensional formula for surface tension? [Online May 7, 2012]
(a) [𝐾𝑉 −2 T −2 ] (b) [𝐾 2 𝑉 2 𝛤 −2 ] (c) [𝐾 2 𝑉 −2 𝛤 −2 ] (d) [𝐾𝑉 2 𝑇 2 ]
𝐹 𝐹 ℓ 𝑇2
sol. (a) Surface tension, 𝑇 = = ℓ . ℓ . 𝑇2
ℓ
𝑇2
(As, 𝐹. ℓ = 𝐾(energy); = 𝑉 −2)
ℓ2
Therefore, surface tension = [𝐾𝑟 2 𝛤 −2 ]
31. The dimensions of magnetic field in M, L, T and C (coulomb) is given as [2008]

(a) [MLT‐l C −1] (b) [MT2 C−2 ] (c) [MT −1 C−1 ] (d) [MT −2 C−1 ]
sol. (c) Magnitude of Lorentz formula 𝐹 = 𝑞𝑣𝐵 sin 𝜃
𝐹 𝑀𝐿𝑇 −2
𝐵= = = [𝑀𝑇 −1 𝐶 −1 ]
𝑞𝑣 𝐶 × 𝐿𝑇 −1
32. Which of the following units denotes the dimension ML2 Q2 , where 𝑄 denotes the electric
charge? [2006]
(a) Wb/m2 (b) Henry (H) (c) H/m2 (d) Weber (Wb)
𝜑 𝐵𝐴 [𝑀𝑇 −1 𝑄 −1 𝐿2 ]
sol. (b) Mutual inductance = = [Henry] = [𝑄𝑇 −1 ]
= 𝑀𝐿2 𝑄 −2
𝐼 𝐼
33. Out of the following pair, which one does NOT have identical dimensions? [2005]
(a) Impulse and momentum
(b) Angular momentum and planck’s constant
(c) Work and torque
(d) Moment of inertia and moment of a force
sol. (d) Moment of Inertia, 𝐼 = 𝑀𝑅 2
[𝐼] = [𝑀𝐿2 ]
Moment of force, 𝛤⃗ = 𝑟⃗ × 𝐹⃗
→ ′r = [𝐿][𝑀𝐿𝑇 −2 ] = [𝑀𝐿2 𝑇 −2 ]
34. Which one of the following represents the correct dimensions of the coefficient of
viscosity? [2004]
(a) [ML−1 T −1 ] (b) [MLT −1 ] (c) [ML−1 T −2 ] (d) [ML−2 T −2 ]
sol. (a) According to, Stokes law,
𝐹
𝐹 = 6𝜋𝜂𝑟𝑣 ⇒ 𝜂 =
6𝜋𝑟𝑣
[𝑀𝐿𝑇 −2 ]
𝜂= ⇒ 𝜂 = [𝑀𝐿−1 𝑇 −1 ]
[𝐿][𝐿𝑇 −1 ]
1
35. Dimensions of 𝜇 , where symbols have their usual meaning, are [2003]
0 𝜀o
(a) [L−1 T] (b) [L−2 T 2 ] (c) [L2 T −2 ] (d) [LT −1 ]

sol. (c) As we know, the velocity of light in free space is given by
1 1
𝑐= = 𝑒 2 = 𝑍12 𝑇 2
√ 𝜇0 𝜀 𝑜 𝜇0 8 0
1
= 𝐶 2 [m/s]2
𝜇0 𝜀𝑜
= [𝐿𝛤 1 ]2
= [𝑀0 𝐿2 𝛤 2 ]
36. The physical quantities not having same dimensions are [2003]
(a) torque and work (b) momentum and planck’s constant
(c) stress and young’s modulus (d) speed and (𝜇0 𝜀o )−1/2
sol. (b) Momentum, = 𝑚𝑣 = [𝑀𝐿𝛤 1 ]
Planck’s constant,
𝐸 [𝑀𝐿2 𝑇 −2 ]
ℎ= = = [𝑀𝐿2 𝛤 1 ]
𝑣 [𝑇 −1 ]
37. Identify the pair whose dimensions are equal [2002]

(a) torque and work (b) stress and energy
(c) force and stress (d) force and work
sol. (a) Work 𝑊 = 𝐹⃗ ⋅ 𝑠⃗ = 𝐹𝑠 cos 𝜃
𝐴⃗ ⋅ 𝐵
⃗⃗ = 𝐴𝐵 cos 𝜃
= [𝑀𝐿𝑇 −2 ][𝐿] = [𝑀𝐿2 𝑇 −2 ];
Torque, 𝛤⃗ = 𝑟⃗ × 𝐹⃗ ⇒/𝜏 = 𝑟𝐹 sin 𝜃
𝐴⃗ × 𝐵
⃗⃗ = 𝐴𝐵 sin 𝜃
= [𝐿][𝑀𝐿𝑇 −2 ] = [𝑀𝐿2 𝑇 −2 ]
TOPIC-3…..Errors in Measurements
38. A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of
the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a
displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved,
and the least count of the screw gauge, are respectively:
[Sep. 06, 2020 (I)]
(a) Negative, 2 𝜇m (b) Positive, 10 𝜇m (c) Positive, 0.1 mm (d) Positive, 0.1 𝜇m
sol. (b) Given: No. of division on circular scale of screw gauge = 50 Pitch = 0.5 mm
Least count of screw gauge
Pitch
=
No. ofdivisiononcircularscale
0.5
= mm = 1 × 105 m = 10𝜇m
50
And nature of zero error is positive.
39. The density of a solid metal sphere is determined by measuring its mass and its diameter.
𝑥
The maximum error in the density of the sphere is (100) %. Ifthe relative errors in
measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is .
[NA Sep. 06, 2020 (I)]
sol. (1050)
𝑀 𝑀 6
Density, p = =4 𝐷 3
⇒ p = 𝜋 𝑀𝐷−3
𝑉 𝜋( )
3 2
𝛥p 𝛥𝑚 𝛥𝐷
%(p)= + 3 ( 𝐷 ) = 6 + 3 × 1.5 = 10.5%
𝑚
𝛥p 1050 𝑥
%(p)= % = (100) %
100
𝑥 = 1050.00
40. A student measuring the diameter of a pencil of circular cross‐section with the help of a
Vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65
mm, The average ofthese four reading is 5.5375 mm and the standard deviation ofthe
data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:
[Sep. 06, 2020 (II)]
(a) (5.5375 ± 0.0739) mm (b) (5.5375 ± 0.0740) mm
(c) (5.538 ± 0.074) mm (d) (5.54 ± 0.07) mm
sol. (d) Average diameter, 𝑑av = 5.5375 mm
Deviation of data, 𝛥𝑑 = 0.07395 mm
As the measured data are up to two digits after decimal, therefore answer should be in
two digits after decimal. 𝑑 = (5.54 ± 0.07) mm
𝑎2 𝑏 2
41. A physical quantity 𝑧 depends on four observables 𝑎, 𝑏, 𝑐 and 𝑑, as 𝑧 = . The
√𝑐𝑑3 3
percentages of error in the measurement ofa, 𝑏, 𝑐 and 𝑑 are 2%, 1.5%, 4% and 2.5%
respectively. The percentage oferror in 𝑧 is:
[Sep. 05, 2020 (I)]
(a) 12.25% (b) 16.5% (c) 13.5% (d) 14.5%
𝑎2 𝑏 2/3
sol. (d) Given: 𝑍 =
√𝑐𝑑3
Percentage error in 𝑍,
𝛥𝑍 2𝛥𝑎 2 𝛥𝑏 1 𝛥𝑐 3𝛥𝑑
= = + + +
𝑍 𝑎 3 𝑏 2 𝑐 𝑑
2 1
= 2 × 2 + 3 × 1.5 + 2 ×4+3× 2.5 = 14.5%.
42. Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of
an object is measured. It should correctly be recorded as : [Sep. 03, 2020 (I)]
(a) 2.12lcm (b) 2.124cm (c) 2.125cm (d) 2.123cm
sol. (a) Thickness = M. S.Reading + Circular Scale Reading (L.C.)
Pitch 0.1
Here LC = Circularsca1edivision = = 0.002 cm per division
50
So, correct measurement is measurement of integral multiple of L.C.
43. The least count of the main scale of a Vernier calipers is 1 mm. Its vernier scale is divided
into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching
each other, the 7th division of vernier scale coincides with a division of main scale and
the zero of vernier scale is lying right side ofthe zero ofmain scale. When this vernier is
used to measure length ofa cylinder the zero of the Vernier scale between 3.1 cm and 3.2
cm and 4th VSD coincides with a main scale division. The length ofthe cylinder is: (VSD
is vernier scale division)
[Sep. 02, 2020 (I)]
(a) 3.2 cm (b) 3.21 cm (c) 3.07 cm (d) 2.99 cm
sol. (c) L.C. of Vernier calipers = 1 MSD‐I VSD
9
= (1 − 10) × 1 = 0.1mm = 0.01 cm
Here 7th division ofvernier scale coincides with a division of main scale and the zero of
Vernier scale is lying right side of the zero of main scale.
Zero error = 7 × 0.1 = 0.7mm = 0.07 cm.
Length of the cylinder = measured value‐ zero error
= (3.1 + 4 × 0.01) − 0.07 = 3.07 cm.
44. If the screw on a screw‐gauge is given six rotations, it moves by 3 mm on the main scale.
If there are 50 divisions on the circular scale the least count of the screw gauge is:
[9 Jan. 2020 I]
(a) 0.00lcm (b) 0.02mm (c) 0.01 cm (d) 0.00l mm
sol. (d) When screw on a screw‐gauge is given six rotations, it moves by 3mm on the main
scale
3
Pitch = 6 = 0.5 mm
Pitch 0.5mm
Least count L.C. = =
𝐶𝑆𝐷 50
1
= mm = 0.01𝑚𝑚 = 0.00lcm
100
45. For the four sets of three measured physical quantities as given below. Which of the
following options is correct?
[9 Jan. 2020 II]
(A) A1 = 24.36, B1 = 0.0724, C1 = 256.2
(B) A2 = 24.44, B2 = 16.082, C2 = 240.2
(C) A3 = 25.2, B3 = 19.2812, C3 = 236.183
(D) A4 = 25, B4 = 236.191 , C4 = 19.5
(a) A4 + B4 + C4 < A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2
(b) A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4
(c) A4 + B4 + C4 < A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3
(d) A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 < A4 + B4 + C4
sol. (None)
𝐷1 = 𝐴1 + 𝐵1 + 𝐶1 = 24.36 + 0.0724 + 256.2 = 280.6𝐷2 = 𝐴2 + 𝐵2 + 𝐶2
= 24.44 + 16.082 + 240.2 = 280.7𝐷3 = 𝐴3 + 𝐵3 + 𝐶3
= 25.2 + 19.2812 + 236.183 = 280.7
𝐷4 = 𝐴4 + 𝐵4 + 𝐶4 = 25 + 236.191 + 19.5 = 281
None of the option matches.
46. A simple pendulum is being used to determine the value ofgravitational acceleration 𝑔 at
a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s
resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in 𝑔 is:
[8 Jan. 2020 II]
(a) 5.40% (b) 3.40% (c) 4.40% (d) 2.40%
sol. (c) Given, Length of simple pendulum, 𝑙 = 25.0 cm
Time of 40 oscillations, 𝑇 = 50𝑠
Time period of pendulum
ℓ
𝑇 = 2𝜋√
𝑔
4𝜋 2ℓ 4𝜋 2ℓ
⇒ 𝑇2 = ⇒𝑔= 2
𝑔 𝑇
𝛥𝑔 𝛥𝑙 2𝛥𝑇
⇒Fractional error in g = = +
𝑔 𝑙 𝑇
𝛥𝑔 0.1 1
⇒ =( ) + 2 ( ) = 0.044
𝑔 25.0 50
𝛥𝑔
Percentage error in 𝑔 = × 100 = 4.4%
𝑔
47. In the density measurement of a cube, the mass and edge length are measured as
(10.00 ± 0.10) kg and (0.10 ± 0.01)m, respectively. The error in the measurement of
density is: [9 April 2019 I]
(a) 0.01k𝑔/m3 (b) 0.10k𝑔/m3 (c) 0.013k𝑔/m3 (d) 0.07k𝑔/m3
𝑀 𝑀
sol. (Bonus) 6 = = 𝑙3 = 𝑀𝑙 −3
𝑉
𝛥𝛿 𝛥𝑀 𝛥𝑙 0.10 0.01
= +3 = + 3( ) = 0.31k𝑦m3
6 𝑀 𝑙 10.00 0.10
48. The area of a square is 5.29 cm2 . The area of 7 such squares taking into account the
significant figures is: [9 April 2019 II]
(a) 37cm2 (b) 37.030cm2 (c) 37.03cm2 (d) 37.0cm2
sol. (d) 𝐴 = 7 × 5.29 = 37.03cm2
The result should have three significant figures, so
𝐴 = 37.0cm2
49. In a simple pendulum experiment for determination of acceleration due to gravity (g) ,
time taken for 20 oscillations is measured by using a watch of 1 second least count. The
mean value of time taken comes out to be 30 s. The length of pendulum is measured by
using a meter scale ofleast count 1 mm and the value obtained is 55.0 cm. The percentage
error in the determination of g is close to: [8 April 2019 II]
(a) 0.7% (b) 0.2% (c) 3.5% (d) 6.8%
sol. (d) We have
ℓ ℓ
𝑇 = 2𝜋√𝑔 or 𝑔 = 4𝜋 2 𝑇 2
𝛥𝑔 𝛥𝑅 𝛥𝑇
× 100 = × 100 + 2 × 100
𝑔 𝑄 𝑇
0.1 1
= × 100 + 2 ( ) × 100
55 30
= 0.18 + 6.67 = 6.8%
50. The least count of the main scale of a screw gauge is 1 mm. The minimum number of
divisions on its circular scale required to measure 5 𝜇m diameter ofa wire is:
[12 Jan. 2019 I]
(a) 50 (b) 200 (c) 100 (d) 500
sol. (b) Least count of main scale of screw gauge = lmm
Least count of screw gauge
Pitch
=
Numberofdivisiononcircularscale
10−3
5 × 10−6 =
N
⇒ N = 200
51. The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm
and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate
significant figures? [10 Jan. 2019 II]
(a) 4264 ± 81cm3 (b) 4264.4 ± 81.0cm3
(c) 4260 ± 80cm3 (d) 4300 ± 80cm3
sol. (c)
52. The pitch and the number of divisions, on the circular scale for a given screw gauge are
0.5 mm and 100 respectively. When the screw gauge is fully tightened without any
object, the zero of its circular scale lies 3 division below the mean line. The readings of
the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the
thickness of the sheet is: [9 Jan. 2019 II]
(a) 5.755mm (b) 5.950mm (c) 5.725mm (d) 5.740mm
sol. (c) Least count of screw gauge,
Pitch
LC =
No. ofdivision
= 0.5 × 10−3 = 0.5 × 10−2 mm + ve error = 3 × 0.5 × 10−2 mm
= 1.5 × 10−2 mm = 0.015mm
Reading = MSR + CSR −(+ve error)
= 5.5mm + (48 × 0.5 × 10−2 ) − 0.015
= 5.5 + 0.24 − 0.015 = 5.725 mm
53. The density of a material in the shape of a cube is determined by measuring three sides of
the cube and its mass. If the relative errors in measuring the mass and length are
respectively 1.5% and 1%, the maximum error in determining the density is: [2018]
(a) 2.5% (b) 3.5% (c) 4.5% (d) 6%
sol. (c) = 1.5 %+3 (1%) = 4.5%
54. The percentage errors in quantities P, Q, Rand S are 0.5%, 1%, 3% and 1.5% respectively
P3 Q2
in the measurement of a physical quantity A = . The maximum percentage error in
√RS
the value ofA will be

[Online Apri116, 2018]
(a) 8.5% (b) 6.0% (c) 7.5% (d) 6.5%
sol. (d) Maximum percentage error in A
= 3(% error in P)+2(% error in Q)
1
+ 2(% error in R)+l(% error in S)
1
= 3 × 0.5 + 2 × 1 + × 3 + 1 × 1.5
2
= 1.5 +2+ 1.5 + 1.5 = 6.5%
55. The relative uncertainty in the period of a satellite orbiting around the earth is 10−2. If
the relative uncertainty in the radius ofthe orbit is negligible, the relative uncertainty in
the mass of the earth is [Online Apri116, 2018]
(a) 3 × 1𝜎 − 2 (b) 10−2 (c) 2 × 10−2 (d) 6 × 10−2
sol. (c) From Kepler’s law, time period of a satellite,
r3 4𝜋 2 3
T = 2𝜋√Gm T 2 = r
GM
Relative uncertainty in the mass of the earth

𝛥M 𝛥T 𝛥r
| |=2 = 2 × 10−2 (4𝜋 & G constant and relative uncertainty in radius
M T r
negligible)
56. The relative error in the determination of the surface area of a sphere is 𝛼. Then the
relative error in the determination Of its volume is [Online April 15, 2018]
2 2 3
(a) 3 𝛼 (b) 3 𝛼 (c) 2 𝛼 (d) 𝛼
𝛥s 𝛥r
sol. (c) Relative error in Surface area, = 2× = 𝛼 and
s r
𝛥v 𝛥r
relative error in volume, = 3×
v r
Relative error in volume w.r. t. relative error in area,

𝛥v 3
= 𝛼
v 2
57. In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance
of 0.25 cm. There are 100 circular scale divisions. The thickness ofa wire measured by
this screw gauge gives a reading of4 main scale divisions and 30 circular scale divisions.
Assuming negligible zero error, the thickness of the wire is: [Online Apri115, 2018]
(a) 0.0430 cm (b) 0.3150 cm (c) 0.4300 cm (d) 0.2150 cm
Va1ueof1partonmainsca1e
sol. (d) Least count = Numberofpartsonverniersca1e
0.25
= 5×100 cm = 5 × 10−4 cm
Reading = 4 × 0.05cm + 30 × 5 × 10−4 cm

= (0.2 + 0.0150)cm = 0.2150 cm(Thickness of wire)
58. The following observations were taken for determining surface tension T of water by
capillary method: Diameter of capillary, D = 1.25 × 1𝑇 2 m , rise of water, h = 1.45 ×
𝑟ℎ𝑔
102 m. Using g = 9.80m/s 2 and the simplified relation, 𝑇 = × 103 N/m, the possible
2
error in surface tension is closest to : [2017]

(a) 2. 4% (b) 10% (c) 0.15% (d) 1.5%
𝑟ℎ𝑔
sol. (d) Surface tension, 𝑇 = × 103
2
Relative error in surface tension,

𝛥𝑇 𝛥𝑟 𝛥ℎ
= + + 0 (g, 2 & 103 are constant) Percentage error
𝑇 𝑟 ℎ
𝛥𝑇
100 × = ( )100 (. )
𝑇
= (0.8 + 0.689)
=(1.489)= 1.489%≅ 1.5%
59. A physical quantity P is described by the relation P = 𝑎1/2 𝑏 2 c 3 𝑑 −4 . If the relative errors
in the measurement of 𝑎, 𝑏, 𝑐 and 𝑑 respectively, are 2%, 1%, 3% and 5%, then the
relative error in P will be: [Online April 9, 2017]
(a) 8% (b) 12% (c) 32% (d) 25%
sol. (c) Given, P = a1/2 b2 c 2 d→↓ ,
Maximum relative error,
𝛥P 1 𝛥a 𝛥b 𝛥c 𝛥d
= +2 +3 +4
P 2 a b c d
1
= × 2 + 2 × 1 + 3 × 3 + 4 × 5 = 32%
2
60. A screw gauge with a pitch of0.5 mm and a circular scale with 50 divisions is used to
measure the thickness of a thin sheet ofAluminium. Before starting the measurement, it is
found that wen the two jaws of the screw gauge are brought in contact, the 45th division
coincides with the main scale line and the zero of the main scale is barely visible. What is
the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division
coincides with the main scale line? [2016]
(a) 0.70 mm (b) 0.50 mm (c) 0.75mm (d) 0.80mm
0.5
sol. (d) L.C. = = 0.01mm
50
Zero error = 5 × 0.01 = 0.05 mm

(Negative) Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm
61. A student measures the time period of100 oscillations of a simple pendulum four times.
The data set is90 s, 91s, 95s, and 92 s. Ifthe minimum division in the measuring clock is
1 s, then the reported mean time should be: [2016]
(a) 92 ± 1.8s (b) 92 ± 3s (c) 92 ± 1.5s (d) 92 ± 5.0s
|𝛥T1 |+|𝛥T2 |+|𝛥T3 |+|𝛥T4 |
sol. (c) 𝛥T = 4
2+1+3+0
= = 1.5
4
As the resolution of measuring clock is 1.5 therefore the mean time should be 92 ± 1.5
L
62. The period of oscillation of a simple pendulum is T = 2𝜋√g. Measured value ofL is 20.0
cm known to 1 mm accuracy and time for 100 oscillations ofthe pendulum is found to be
90 s using a wrist watch ofls resolution. The accuracy in the determination of g is:
[2015]
(a) 1% (b) 5% (c) 2% (d) 3%
𝐿
sol. (d) As, g = 4𝜋 2 𝑇 2
𝛥𝑔 𝛥𝐿 𝛥𝑇
So, × 100 = × 100 + 2 × 100
𝑔 𝐿 𝑇
0.1 1
= × 100 + 2 × × 100 = 2.72 = 3%
20 90
63. Diameter of a steel ball is measured using a Vernier calipers which has divisions of 0.1
cm on its main scale (MS) and 10 divisions ofits vernier scale (VS) match 9 divisions on
the main scale. Three such measurements for a ball are given as: [Online Apri110, 2015]
S.No. MS(cm) VS divisions
1. 0.5 8
2. 0.5 4
3. 0.5 6
If the zero error is‐0.03 cm, then mean corrected diameter is:
(a) 0.52 cm (b) 0.59 cm (c) 0.56 cm (d) 0.53 cm
0.1
sol. (b) Least count = = 0.01 cm
10
𝑑1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm

𝑑2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
𝑑3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
0.61+0.57+0.59
Mean diameter = 3
= 0.59 cm
64. The current voltage relation of a diode is given by I = (e1000V/T − 1)mA, where the
applied voltage V is in volts and the temperature T is in degree kelvin. Ifa student makes
an error measuring ±0.01V while measuring the current of 5 mA at 300 K, what will be
the error in the value of current in mA? [2014]
(a) 0.2mA (b)0.02mA (c) 0.5mA (d) 0.05mA
sol. (a) The current voltage relation of diode is 𝐼 = (𝑒 1000𝑉/𝑇 − 1)mA (given)
When, 𝐼 = 5𝑚𝐴, e1000 𝑉𝑙𝜏=6𝑚𝐴
Also, 𝑑𝐼 = (𝑒 1000 𝑉/𝜏)×1000
𝑇
Error = ±0.01 (By exponential function)

1000
= (6𝑚𝐴) × × (0.01) = 0.2mA
300
65. A student measured the length ofa rod and wrote it as 3.50 cm. Which instrument did he
use to measure it? [2014]
(a) A meter scale.
(b) A vernier caliper where the 10 divisions in vernier scale matches with 9 division in
main scale and main scale has 10 divisions in 1 cm.
(c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
(d) A screwgauge having 50 divisions in the circular scale and pitch as 1 mm.
sol. (b) Measured length of rod = 3.50 cm
For Vernier Scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD‐I VSD = 0.1 mm
66. Match List‐I(Event) with List‐II(Order of the time interval for happening of the event)
and select the correct option from the options given below the lists:
List‐I List‐ II
(1) Rotation period of earth (i) 105 s
(2) Revolution period of earth (ii) 107 s
(3) Period of light wave (iii) 10−15 s
(4) Period of sound wave (iv) 10−3 s
(a) (1) ‐(i), (2) − (ii), (3) ‐(iii), (4) − (iv)
(b) (1) ‐(ii), (2) − (i), (3) ‐(iv), (4) ‐(iii)
(c) (1) ‐(i), (2) ‐(ii), (3) ‐(iv), (4) ‐(iii)
(d) (1) ‐(ii), (2) − (i), (3) ‐(iii), (4) − (iv)
sol. (a) Rotation period of earth is about 24 hrs = 105 s
Revolution period of earth is about 365 days = 107 s
Speed of light wave C = 3 × 108 m/s
Wavelength of visible light of spectrum
𝜆 = 4000 − 7800A
1
C = f𝜆 (and T  )
f
Therefore period of light wave is 10−15 s (approx)
67. In the experiment of calibration of voltmeter, a standard cell of e. m. f. 1.1 volt is

balanced against 440 cm ofpotential wire. The potential difference across the ends of
resistance is found to balance against 220 cm of the wire. The corresponding reading of
voltmeter is 0.5 volt. The error in the reading of voltmeter will be:
(a) −0.15 volt (b) 0.15 volt (c) 0.5 volt (d) −0.05 volt
sol. (d) In a voltmeter
V∝𝑙
V = k𝑙
Now, it is given E = 1.1 volt for 𝑙1 = 440 cm
and V = 0.5 volt for 𝑙2 = 220 cm
Let the error in reading of voltmeter be 𝛥V then,
1.1 = 400K and (0.5 − 𝛥V) = 220 K.
1.1 0.5−𝛥V
⇒ 440 = 220
𝛥V = −0.05 volt
68. An experiment is performed to obtain the value of acceleration due to gravity g by using
a simple pendulum of length L. In this experiment time for 100 oscillations is measured
byusing a watch of 1 second least count and the value is 90.0 seconds. The length L is
measured byusing a meter scale ofleast count 1 mm and the value is 20.0 cm. The error
in the determination ofg would be:
(a) 1.7% (b) 2.7% (c) 4.4% (d) 2.27%
sol. (b) According to the question.
𝛥𝑡 1
𝑡 = (90 ± 1) or, = 90
𝑡
𝛥𝑙 0.1
𝑙 = (20 ± 0.1) or, =
𝑙 20
𝛥𝑔
%=?
𝑔
As we know,
𝑙 4𝜋 2 𝑙
𝑡 = 2𝜋√ ⇒ 𝑔 = 2
𝑔 𝑡
𝛥𝑔 𝛥𝑙 𝛥𝑡 0.1 1
or, = ± ( 𝑙 + 2 𝑡 ) = ( 20 + 2 × 90) = 0.027
𝑔
g
%  2.7%
g
69. Resistance of a given wire is obtained by measuring the current flowing in it and the
voltage difference applied across it. If the percentage errors in the measurement of the
current and the voltage difference are 3% each, then error in the value of resistance of
the wire is [2012]
(a) 6% (b) zero (c) 1% (d) 3%
sol. (a) According to ohm’s law, V = IR
𝑉
R=
𝐼
Abso1uteerror
Percentage error = × 102
Measurement
𝛥𝑉 𝛥𝐼
where, × 100 = × 100 = 3%
𝑉 𝐼
𝛥𝑅 𝛥𝑉 𝛥𝐼
then, × 100 = × 102 + × 102
𝑅 𝑉 𝐼
=3%+3% = 6%
70. A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading: 58.5 degree
Vernier scale reading: 09 divisions
Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the
Vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism
from the above data is [2012]
(a) 58.59 degree (b) 58.77 degree (c) 58.65 degree (d) 59 degree
sol. (c) Reading of Vernier = Main scale reading+ Vernier scale reading × least count.
Main scale reading = 58.5
Vernier scale reading = 09 division
least count of Vernier = 0. 5∘ /30
0.5∘
Thus, R = 58. 5∘ + 9 × 30
R = 58.65∘
71. N divisions on the main scale of a Vernier calipers coincide with (𝑁 + 1) divisions of the
Vernier scale. Ifeach division of main scale is ‘a’ units, then the least count of the
instrument is [Online May 19, 2012]
a 𝑁 a
(a) a (b) 𝑁 (c) 𝑁+1 × a (d) 𝑁+1
sol. (d) No. of divisions on main scale = 𝑁

No. of divisions on Vernier scale = 𝑁 + 1
size of main scale division = 𝑎
Let size of Vernier scale division be 𝑏
then we have
𝑎𝑁
𝑎𝑁 = 𝑏(𝑁 + 1) ⇒ b =
𝑁+1
𝑎𝑁
Least count is a—b = a − 𝑁+1
𝑁+1−𝑁 𝑎
= 𝑎[ ]=
𝑁+1 𝑁+1
72. A student measured the diameter of a wire using a screw gauge with the least count 0.001
cm and listed the measurements. The measured value should be recorded as
[Online May 12, 2012]
(a) 5.3200 cm (b) 5.3 cm (c) 5.32 cm (d) 5.320cm
sol. (d) The least count(L.C.) of a screw guage is the smallest length which can be measured
accurately with it.
1
As least count is 0.001 cm = 1000 cm
Hence measured value should be recorded upto 3 decimal places i.e., 5.320 cm
73. A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading: 0 mm
Circular scale reading : 52 divisions
Given that l mm on main scale corresponds to 100 divisions of the circular scale. The
diameter of wire from the above data is [2011]
(a) 0.052 cm (b) 0.026 cm (c) 0.005 cm (d) 0.52cm
1 1
sol. (a) Least count, L.C. = 100 mm = 30 MSD
1
Diameter ofwire = 𝑀𝑆𝑅 + 𝐶𝑆𝑅 × 𝐿. C. = 30 × 0. 5∘ = 1 minute.
1 mm = 0.1 cm
= 0 +× 52 = 0.52mm = 0.052 cm1
74. The respective number of significant figures for the numbers 23.023, 0.0003 and
2.1 × 10−3 are [2010]
(a) 5, 1, 2 (b) 5, 1, 5 (c) 5, 5, 2 (d) 4, 4, 2
sol. (a) Number of significant figures in 23.023 = 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10−3 = 2
75. In an experiment the angles are required to be measured using an instrument, 29 divisions
of the main scale exactly coincide with the 30 divisions of the Vernier scale. If the
smallest division of the main scale is half‐ a degree (= 0. 5∘ ) , then the least count of the
instrument is: [2009]
(a) half minute (b) one degree (c) half degree (d) one minute
75. (d) 30 Divisions of V.S. coincide with 29 divisions of M.S.
29
I V.S. D = 30 MSD
L.C. = 1 MSD‐ IVSD

29
= 1 MSD — 30 MSD
Reading = [M. S. R. +C. S. R. × L. C. ] −(zero error)

= [3 + 35 × 0.01] − (−0.03) = 3.38 mm
1
 MSD
30
1
  0.50  1min ute
30
76. A body of mass m = 3.513 kg is moving along the x‐axis with a speed of 5.00ms−1 . The
magnitude ofits momentum is recorded as [2008]
(a) 17.6kgms −1 (b) 17.565 kg ms−1 (c) 17.56kgms−1 (d) 17.57kgms−1
sol. (a) Momentum, 𝑝 = 𝑚 × 𝑣
Given, mass of a body = 3.513kg
speed of body = (3.513) × (5.00) = 17.565kgm/s
= 17.6 (Rounding offto get three significant figures)
77. Two full turns of the circular scale of a screw gauge cover a distance of l mm on its main
scale. The total number of divisions on the circular scale is 50. Further, it is found that the
screw gauge has a zero error of‐0.03 mm. While measuring the diameter ofa thin wire, a
student notes the main scale reading of3 mm and the number ofcircular scale divisions in
line with the main scale as 35. The diameter of the wire is [2008]
(a) 3.32mm (b) 3.73 mm (c) 3.67mm (d) 3.38mm
sol. (d) Least count of screw gauge = 0.01 mm
0.5
mm
50
Reading = [M.S.R. + C.S.R.  L.C.]  ( zero error)
= [3 + 35  0.01]  (  0.03 )=3.38 mm
Chapter 3 - MOTION IN A STRAIGHT LINE
 Motion in a straight line deals with the motion of an object which changes its position with time
along a straight line.
 The study of the motion of objects without considering the cause of motion is called kinematics.
Rest and Motion
If the position of a body does not change with time with respect to the surroundings then it is said
to be at rest, if not it is said to be in motion.
Distance and Displacement
 Distance is the actual path covered by a moving particle in a given interval of time while
displacement is the change in position vector,i.e., a vector joining initial to final position. If a
particle moves from A to B as shown in Fig. the distance travelled is s while displacement is
r uur ur
 r  rf  ri
ur ur ur
R  A B
ur
B
O ur A
A
 Distance is a scalar while displacement is a vector, both having same dimensions[L] and SI unit
is metre.
x x
h
A r O B s h
h
(a)
AB
Distance = r ( c)
(b) (d)
Displacement = 2r
Distance = 2h Distance = s distance = h + 2x
Displacement = 0 Displacement = s displacement = h
 The magnitude of displacement is equal to minimum possible distance between two positions; so
Distance  Displacement
 For motion between two points displacement is single valued while distance depends on actual
path and so can have many values.
 For a moving particle distance can never decrease with time while displacement can.
 Decrease in displacement with time means body is moving towards the initial position.
 For a moving particle distance can never be negative or zero while displacement can be negative.
 In general, magnitude of displacement is not equal to distance. However, it can be so if the
motion is along a straight line without change in direction.
 Magnitude of displacement is less than the distance travelled in case of curvilinear motion.
Ex : If an object turns through an angle  along a circular path of radius r from point A to point
B then
i) distance d  r
  x 
ii)displacement 2 x  2r sin  / 2  Q sin 2  r 
 
d
B x x A
r r
– –
22
EX.1:An athlete completes one round of a circular track of radius R in 40 s. What will be his
displacement at the end of 2 min 20 s? (2010E)
Sol. The time = 2 min 20s = 140s
A R R B
In 40 seconds athlete completes = 1 round

In 140 seconds athlete will completes
140
= round =3.5 rounds
40
The displacement in 3 rounds = 0
So net displacement =2R
EX.2 : If the position of a particle along Y axis is represented as a function of time t by the
equation y(t)=t3 then find displacement of the particle during the period t to t  t
Sol. Position at time t is y  t   t 3
3
Position at time t  t is y  t  t    t  t 
3 3
 displacement of the particle from t to t  t is y  t  t   y  t    t  t   t
2 3
 t 3  3t 2 t  3t  t    t   t 3
2 3
 3t 2 t  3t  t    t 
Average Speed and Average Velocity:
 Average speed or velocity is a measure of overall 'fastness' of motion during a specified interval of
time.
 The average speed of a particle for a given 'interval of time' is defined as the ratio of distance
travelled to the time taken while average velocity is defined as the ratio of displacement to time
taken.
r
 Thus, if a particle in time interval t after travelling distance r is displaced by  r
Distance travelled r
Average speed = Time taken , i.e., Vavg  ........(i)
t
r
Displacement uuur  r
Average velocity = Time , i.e, Vavg  ............(ii)
t
 Average speed is a scalar while average velocity is a vector both having same units (m/s) and
dimensions  LT -1  .
 For a given time interval average velocity is single valued while average speed can have many
values depending on path followed.
 During the motion if the body comes back to its initial position.
ur r
V avg = 0 (Q  r  0 )
but Vavg  0 and finite (Q r  0 )
 For a moving body average speed can never be negative or zero while average velocity can be
negative.
 If a graph is plotted between distance (or displacement) and time, the slope of chord during a
given interval of time gives average speed (or ) average velocity
Δr
Vavg = = tan = slope of chord
Δt
Instantaneous speed and Instantaneous velocity
 Instantaneous velocity is defined as rate of change of position of the particle with time. If the
r
position r of a particle at an instant t changes by rr in a small time interval t
r r r
r dr
V  lim 
t o t dt
 The magnitude of velocity is called speed,i.e
r
Speed= velocity i.e, V  V
 Velocity is a vector while speed is a scalar , both having same units (m/s) and dimensions  LT -1  .
 If during motion velocity remains constant throughout a given interval of time, then the motion
r r
is said to be uniform and for uniform motion, V=constant=Vavg
 If velocity is constant, speed will also be constant. However , the converse may or may not be
true,i.e, if speed=constant,velocity may or may not be constant as velocity has a direction in
addition to magnitude which may or may not change,e.g, in case of uniform rectilinear motion
r
V  constant and V  constant
While in case of uniform circular motion
r
V=constant but V is not constant , due to change in direction.
 velocity can be positive or negative as it is a vector but speed can never be negative as it is
r
magnitude of velocity, i.e, V  V
dr
 As by definition V  , the slope of displacement-versus time graph gives velocity , i.e,
dt
dr
V  tan   slope of r  t curve
dt
velocity
dA
v
t Time dt Time
(a) (b)
dr
 As by definition V  ; i.e, dr  Vdt
dt
and from fig. Vdt  dA
So, dA  dr i.e, r   dA   Vdt

i.e, area under velocity-time graph gives displacement while without sign gives distance.
 Average speed is the total distance divided by total time
Total distance travelled
v avg =
Total time taken
 If a body travels a distance s 1 in time t1 , s 2 in time t 2 and s 3 in time t3 then the average speed
s1 + s 2 + s3
is v avg = t + t + t
1 2 3
 If an object travels distances s1, s2, s3 etc. with speeds v1, v2, v3 respectively in the same direction.
Then
s1  s 2  s 3
Average speed = s 1  s 2  s 3
v1 v2 v3
 If an object travels first half of the total journey with a speed v1 and next half with a speed v2
s+s 2s 2 2v1v 2
v avg = = = =
s s s s 1 1 v1 + v 2
then its average speed is + + +
v1 v 2 v1 v 2 v1 v 2
 If a body travels first 1/3 rd of the distance with a speed v1 and second 1/3rd of the distance with
a speed v2 and last 1/3rd of the distance with a speed v3, then the average speed
s s s
+ +
vavg = 3 3 3
s s s
+ +
3v1 3v 2 3v3
3v1v2 v3
vavg 
v1v2  v2 v3  v3v1
 If an object travels with speeds v1, v2, v3 etc., during time intervals t1, t2, t3 etc.,
v1t1  v2 t 2  v3t 3  ....
then its average speed  t1  t 2  t3  ....
If t1 = t2 = t3 = .... = t, then
If t1 = t2 = t3 = .... = t, then
v1t  v2 t  v3 t  ..... v1  v2  ...
vavg  
nt n
i.e.The average speed is equal to the arithmetic mean of individual speeds.
 The actual path length traversed by a body is called distance.
r
Note : If v = v x $i + v y $j + v z k$varies with time t then
uur
uur r  r dr 
dr = vdt Q v = 
 dt 
ur
rf uur t2 r
Integrating on both sides then  ur dr =  vdt
ri t1
 displacement of the particle from time t1 to t 2 is given by

r ur r t2 t2 t2
s = rf - ri =  v x îdt +  v y ˆjdt +  v z kdt
ˆ
t1 t1 t1
EX.3: A particle starting from point A, travelling upto B with a speed S, then upto point ‘C’ with
a speed 2S and finally upto ‘A’ with a speed 3S, then find its average speed.
C
O
120
B
A
Total dis tan ce travelled
Sol. Average speed = Total time taken
2S t2
C
2/3
5/6 /2 B
t3
S
3S t1
A
Total distance travelled = AB + BC + CA = 2r

Total time taken is
AB BC CA r 2r 5r
T  t1  t 2  t 3       (arc length = radius  angle)
V1 V2 V3 2S 6S 18S
2 r
Vavg   1.8 S
r 2 r 5r
 
2S 6S 18S
EX.4: For a man who walks 720 m at a uniform speed of 2 m/s, then runs at a uniform speed of 4
m/s for 5 minute and then again walks at a speed of 1 m/s for 3 minutes. His average speed
is
s1
Sol. Where s1 = 720 m and t1  v  360s  6 min .
1
s2 = (4)(5)(60) = 1200m, t2 = 300 s s3 = (1)(3)(60) = 180 m, t3 = 180 s

s1  s2  s3 720  1200  180
vavg 
t1  t 2  t 3 = 360  300  180
 v avg  2.5m / s
EX.5: A particle is at x = +5 m at t = 0, x = -7m at t = 6s and x = +2m at t = 10s. Find the average
velocity of the particle during the interval (a) t = 0 to t = 6s; (b) t = 6s to t = 10s, (c) t = 0 to
t = 10s.
Sol. x1 = +5m, t1 = 0, x2 = -7m; t2 = 6s, x3 =+2m, t3= 10s
a) The average velocity between the times t = 0 to t = 6s is
r x  x1 7  5
v1  2   2m / s
t 2  t1 60
b) The average velocity between the times t2 = 6s to t3 = 10s is
r x  x 2 2   7  9
v2  3    2.25m / s
t3  t 2 10  6 4
c) The average velocity between times t1 = 0 to t3 = 10s is
r x x 25
v3  3 1   0.3m / s
t 3  t1 10  0
EX.6: A particle traversed one third of the distance with a velocity v0. The remaining part of the
distance was covered with velocity v1 for half the time and with a velocity v2 for the remaining
half of time. Assuming motion to be rectilinear, find the mean velocity of the particle averaged
over the whole time of motion.
A
t1 t /2
2 t /2
2
B
Sol. v0 C v1 D v2
S
S S
For AC;  v0 t1  t1  ---(1)
3 3v0
2S
For CB;  CD  DB
3
2S t  t  4S
  v1  2   v2  2   t 2 
3 2 2 3  v1  v2 
Since, average velocity is defined as
S 2S

v avg 
Total Displacement
 3 3
S
Total Time t t =
t1  2  2 t1 + t 2
2 2
3v0  v1  v2 
 vavg 
4v0  v1  v2
Acceleration
 The rate of change of velocity is equal to acceleration.
Average and Instantaneous acceleration
 If the velocity of a particle changes ( either in magnitude or direction or both) with time the
motion is said to be accelerated or non-uniform. In case of non-uniform motion if change in
ur
velocity is V in time interval t , then
r uur uur
uuur Δv v - v
a avg = = 2 1 .......(1)
Δt t 2 - t1
Instantaneous acceleration or simply acceleration is defined as rate of change of velocity with time at
r ur
a given instant. So if the velocity of a particle v at time t changes by V in a small time interval t
then
ur ur
r ΔV dV
a = lim = .......... (2)
Δt 0 Δt dt
Regarding acceleration it is worth noting that:
 It is a vector with dimensions  LT -2  and SI units  m / s 2  .
 If acceleration is zero, velocity will be constant and the motion will be uniform. However, if
acceleration is constant(uniform), motion is non-uniform and if acceleration is not constant
then both motion and acceleration are non-uniform.
r r
 
As by definition V  ds / dt 
r r r
r dv d  ds  d 2 s
So, a = dt = dt  dt  = dt 2 .........(3)
 
r
i.e., if s is given as a function of time, second derivative of displacement w.r.t. time gives acceleration.
 If velocity is given as a function of position, then by chain rule
dv dv dx dv  dx 
a= = . or a = v  as = v
dt dx dt dx  dt 
r uur
  
As acceleration a = dv / dt , the slope of velocity-time graph gives acceleration.
Velocity - Time Graph
B
V2
V
V1 A 
t
t
O t1 t2
r
r V
a avg   tan 
t
r
aavg = slope of the line joining two points in v-t graph.
r
 The slope of a versus t curve, i.e, d a / dt is a measure of rate of non-uniformity of
acceleration(usually it is known as JERK).
 Acceleration can be positive or negative. Positive acceleration means velocity is increasing with
time while negative acceleration called retardation means velocity is decreasing with time.
Equations of motion :
 If a particle starts with an initial velocity u, acceleration a and it gains velocity v in time t then
dv
a= or dv = a dt
dt
v t V t
or u dv = a 0 dt or  Vu = a  t0
or v = u + at .......(1A)
r r r
In vector form v = u + a t ........(1B)
ds
 Now again by definition of velocity, Eqn. (1A) reduces to = u + at
dt
s t
or  ds =   u + at  dt
0 0
or
1
s = ut + at 2 ........(2A)
2
r r 1r 2
In vector form s = ut + at .............(2B)
2
From eqns. (1A) and (2A), we get
2
s=u
v - u + 1 a v - u
 
a 2  a 
or 2as = 2uv - 2u 2 + v 2 + u 2 - 2uv
i.e., v 2 = u 2 + 2as .......(3)
 In scalar form
rr rr rr rr
v.v = u.u + 2a.s or v2 = u 2 + 2a.s
 Distance travelled = average speed x time
uv
s  t .............(4)
 2 
 1
 Distance travelled in nth second s n = u + a  n -  ..... (5)
 2
 If acceleration and velocity are not collinear, v can be calculated using
2 1/ 2
v =  u 2 +  at  + 2uat cosθ 
 
at sinθ
with tan = .......(6)
u + atcosθ
 
v at


u
Graphs
Characteristics of s-t and v-t graphs
 Slope of displacement time graph gives velocity.
 Slope of velocity-time graph gives acceleration.
 Area under velocity-time graph gives displacement
Let us plot v-t and s-t graphs of some standard results. To draw the following graphs assume that the
particle has got either a one-dimensional motion with uniform velocity or with constant acceleration.
 If a particle starts from rest and moves with uniform acceleration ‘a’ such that it travels distances
sm and sn in the mth and nth seconds
a
then Sn = 0 +  2n -1  1 (u=0)
2
a
Sm = 0 +  2m -1   2 
2
sn  sm
subtracting eq (2) from eq (1) a  n  m .
 
 A particle starts from rest and moves along a straight line with uniform acceleration. If ‘s’ is the
distance travelled by it in n seconds and sn is the distance travelled in the nth second, then
s n  2n  1
 (fraction of distance fallen in nth second during free fall )
s n2
 Moving with uniform acceleration, a body crosses a point 'x' with a velocity 'u' and another point
‘y’ with a velocity ‘v’. Then it will cross the mid point of ‘x’ and ‘y’ with velocity
 v1 
u v y
x
S S
v12 - u 2 v 2 - v12
In this case acceleration a = =
2S 2S
v2  u2
 on solving, v1  .
2
S.No Situation v-t graph s-t graph Interpretation
1 Uniform motion V-t graph s-t graph i)Slope of s-t graph =
V S v = constant.
S=Vt ii) In s-t- graph s = 0 at
V= constant t=0
t
I) u=0, i.e.,v=0 and t=0

2 Uniformly V S ii) u=0, i.e., slope of s-t
accelerated graph at t=0, should
motion with V= at be zero
u=0 and iii) a or slope of v-t
s=0 at t = 0 t
graph is constant
3 Uniformly I) u  0 , i.e., v or
accelerated V slope of s-t graph at
S 2
motion with v=u+at s=ut+1/2 at t=0 is not zero
u  0 and u ii) v or slope of s-t
s  s0 at t  0 graph gradually
goes on increasing
4 Uniformly V I) s = s0 at t=0
S
accelerated v=u+at
motion with u
s0 _
u  0 and
s  s0 at t  0
V I)Slope of s-t graph at
5 Uniformly S
retarded motion t=0 gives u.
till velocity be u ii) Slope of s-t graph at
v=u-at
t=t0 becomes zero
comes zero
t0 iii) In this case u can’t
t0 be zero.
6 Uniformly I)At time t=t0, v=0 or
V
retarded then S slope of s-t graph is
accelerated in zero.
u
opposite ii) In s-t graph slope or
direction t0 velocity first decreases
O then increases with
t0
opposite sign.
 If a bullet looses (1/n)th of its velocity while passing through a plank, then the minimum no. of
such planks required to just stop the bullet is .
u u
u
n v=0
x x x
Let m be the no of planks required to stop the bullet

2
 u 2
  u -  - u = 2ax  1 0 2 - u 2 = 2amx   2
 n
Dividing eq (2) with eq (1)
02 - u 2 2amx
2
=
 u 2
2ax
u -  -u
 n 
u2 1 1 n2
m 2
 2
m 2

 
2 u  1   n 1  n 2  n 2  1  2n
u  u   1  1   1  
 n  n  n 
n2
m
2n  1
1
 The velocity of a bullet becomes of the initial velocity while penetrating a plank. The number
n
of such planks required to stop the bullet.
u u
u
n v=0
x x x
2
u 2
   - u = 2ax  1 0 2 - u 2 = 2amx   2
n
n2
From eq (1) and eq (2) ; m
n2  1
1
 A bullet loose of its velocity while penetrating a distance x into the target. The further distance
n
travelled before coming to rest.
u
u u
n v=0
x y
1
Let x is the distance covered by the bullet to loose the th of initial velocity
n
 v 2 - u 2 = 2as
2
 u 2
 u -  - u = 2ax  1
 n 
Let y is the further distance covered by the bullet to come to rest
 0 2 - u 2 = 2a  x + y    2
 From eq (1) and (2)
  n  12 
y  x 
 2n 
 
th
1
 If the velocity of a body becomes   of its initial velocity after a displacement of ‘x’ then it
n
will come to rest after a further displacement of
u
u v=0
n
x y
2 2
u 2 2 u
   - u = 2ax  1  0 -   = 2ay   2
n n
x
From eq. (1) and eq (2) ; y 2
n 1
EX.7. A body covers 100cm in first 2 seconds and 128cm in the next four seconds
moving with constant acceleration. Find the velocity of the body at the end of 8sec?
Sol. distance in first two seconds is
1 S1 S2
s1 = ut1 + at12
2 t1 t2
1
100 = 2u + a  4 ......(1)
2
distance in (2+4)sec from starting point is
1 2 1
s1 + s 2 = u  t1 + t 2  + a  t1 + t 2  228 = 6u + a  36 ......(2)
2 2
from eq (1) and (2)
We get a = -6 cm/s2 , sub a=-6 in eq - (1)
1
 100  2u   6  4
2
2u = 112 u = 56 cm/s
v = u + at = 56 - 6 x 8  v  8cm / s
EX.8. A car starts from rest and moves with uniform acceleration ‘a’. At the same instant from
the same point a bike crosses with a uniform velocity ‘u’. When and where will they meet ?
What is the velocity of car with respect to the bike at the time of meeting ?
1 2
Sol. s = at  1 , s = ut   2 
car 2 bike
if they meet at the same point

s = s
car bike
1 2 2u
at = ut  t 
2 a
2u 2u 2
s bike  ut  u 
a a
v car  at  2u
r r r
vcar w.r.t. bike at the time of meeting, vcb = vcar - vbike = 2u - u = u
r r
EX.9: What does d v /dt and d v / dt represent? can these be equal ? can:
r r
(a) d v /dt=0 while d v / dt  0
r r
(b) d v / dt  0 while d v / dt  0 ?
r r r
Sol. d v / dt represents time rate of change of speed as v  u , while d v / dt represents magnitude
of acceleration.
If the motion of a particle has translational acceleration (without change in direction)
r
r r dv d r
as v  v n$,   v n$
dt dt  
r
d v $d r
or  n v [as n$ is constant]
dt dt
r
dv d r
or  v   0
dt dt
In this case both these will be equal and not equal to zero.
(a) The given condition implies that:
r r
d v / dt  0, i.e., acc.  0 while d v / dt  0 , i.e., speed =constant.
This actually is the case of uniform circular motion. In case of uniform circular motion
r
dv r u 2
=a= = constant ¹ 0
dt r
r d r
while v =constant. i.e., v 0
dt
r r r r
(b) d v / dt  0 means a = 0 , i.e., a = 0
r r r
 
or d v / dt  0 or v = constant
r
And when velocity v is constant speed will be constant,
r d r
i.e., speed  v =constant or v 0
dt
r
dv d r
So, it is not possible to have dt  0 while v  0.
dt
EX.10:In a car race, car A takes time t less than car B and passes the finishing point with a
velocity v more than the velocity with which car B passes the point . Assuming that the
cars start from rest and travel with constant accelerations a 1 and a 2 respectively , the
V
value of t
is
Sol. The distance covered by both cars is same
Thus, s1=s2=s
If the cars take time t1 and t2 for the race and their velocities at the end of race be v1 and v2 , then
it is given that
v1  v2  v and t 2  t1  t
v v1  v2 2a1s  2a 2s a1  a 2 v
     a 1a 2
Now, t t 2  t1 2s

2s 1

1 t \
a2 a1 a2 a1
EX.11 : A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed
again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires
1sec. How long the drunkard takes to fall in a pit 13m away from the start?
Sol. Distance of the pit from the start =13-5=8m
Time taken to move first 5m=5sec
5 steps (i.e., 5m) forward and 3steps(i.e.,3m) backward means that net distance moved =5-3=2m
and time taken during this process =5+3=8sec

8×8
 Time taken in moving 8m = = 32sec
2
 Total time taken to fall in the pit =32+5=37sec
EX.12 : An  particle travels inside a straight hollow tube 2m long of a particle accelerator
under uniform acceleration. How long is the particle in the tube if it enters at a speed of
1000 m/s and leaves at 9000 m/s. What is its acceleration during this interval ?
Sol. Let ‘a’ be the uniform acceleration of  -particle. According to given problem s = 2m,
v = 9000 m/s and u = 1000 m/s. Since v 2 - u2 = 2aS,
2 2
 9000   1000   2a  2   a = 2 x 107 m/s2
Let the particle remains in the tube for time ‘t’, then v  u  at
v  u 9000  1000
t  7
 4 104 s
a 2  10
EX.13:A car starts from rest and moves with uniform acceleration of 5 m/s2 for 8 sec. If the
acceleration ceases after 8 seconds then find the distance covered in 12s starting from rest.
Sol. The velocity after 8 sec v = 0 + 5 x 8 = 40 m/s
Distance covered in 8 sec
1
s 0  0   5  64  160m
2
After 8s the body moves with uniform velocity and distance covered in next 4s with uniform
velocity.
s = vt = 40 x 4 = 160 m
The distance covered in12 s =160 +160 =320m.
EX.14 : A car is moving with a velocity of 40 m/s. The driver sees a stationary truck ahead at a
distance of 200 m. After some reaction time t the breaks are applied producing a (reaction)
retardation of 8m/s2. What is the maximum reaction time to avoid collision ?
Sol. The car before coming to rest (v = 0)
2
v 2  u 2  2as  0   40   2  8  s
 s  100m
The distance travelled by the car is 100m
To avoid the clash, the remaining distance 200 - 100 = 100m must be covered by the car with
uniform velocity 40 m/s during the reaction time t .
100
 40  t  2.5s
t
The maximum reaction time t  2.5s
EX.15 : Two trains one travelling at 54 kmph and the other at 72 kmph are headed towards one
another along a straight track. When they are 1/2 km apart, both drivers simultaneously
see the other train and apply their brakes. If each train is decelerated at the rate of 1 m/s2,
will there be a collision ?
Sol. Velocity of the first train is 54 kmph = 15 m/s
Distance travelled by the first train before coming to rest
u 2 225
s1    112.5m
2a 2
Velocity of the second train is 72 kmph= 20 m/s
Distance travelled by the second train before coming to rest
u 2 400
s2    200m
2a 2
Total distance travelled by the two trains before coming to rest = s1 + s2 = 112.5 + 200 = 312.5m
Because the initial distance of separation is 500m which is greater than 312.5m, there will be no
collision between the two trains.
EX.16. A bus accelerates from rest at a constant rate ‘  ’ for some time, after which it decelerates at
a constant rate ‘  ’ to come to rest. If the total time elapsed is t seconds. Then evaluate follow-
ing parameters from the given graph
a) the maximum velocity achieved
b) the total distance travelled graphically and
c) Average velocity
V
Vmax A
 
O t1 t2 B t
vmax vmax
Sol. a)  = Slope of line OA = t  t1  
1
vmax vmax
 = Slope of line AB = t  t 2  
2
vmax vmax  

t  t1  t 2    v max  
    
  
v max   t
 
b) Displacement = area under the v-t graph
= area of OAB
1 1 1
=  base  height   t1  t 2  v max  tv max
2 2 2
1    2
=  t
2     
total displacement 1    v
vavg     t  max
total time 2     2
EX.17: A body starts from rest and travels a distance S with uniform acceleration, then moves
uniformly a distance 2S and finally comes to rest after moving further 5S under uniform
retardation. Find the ratio of average velocity to maximum velocity
v
Vmax
Sol.
O t1 t2 t3 t
area of ΔOAC S  1 Vmax t1 (or)t1  2S ;
2 Vmax
2S
area of ABCD 2S  Vmax t 2 (or) t 2 
Vmax
area of ΔBDE 5S  1 Vmax t 3 (or) t 3  10S

2 Vmax
S  2S  5S 8S
Total displacement Vavg  2S 2S 10S

1 4S
Vavg  ;  
Total time Vmax Vmax Vmax V m ax
V avg 8S 4
 
V m ax 14S 7
EX.18: The acceleration-displacement (a - x) graph of a particle moving in a straight line is as
shown. If the particle starts from rest,then find the velocity of the particle when displace-
ment of the particle is 12m. 2
a m/s
2 8 10 12 x(m)
Sol. v 2 - u 2 = 2ax
v2 - u 2 v2
 ax =  (Q u=0 )
2 2
 v  2  area under a  x graph 
Area under a-x graph =
Area of OAE + Area of rectangle ABEF
+ Area of trapezium BFGC + Area of CGD
1 1 1
Area   2 2  6  2   2  4 2   2  4  24
2 2 2
 v  2  24  4 3m / s
EX.19: Velocity-time graph for the motion of a certain body is shown in Fig. Explain the nature
of this motion. Find the initial velocity and acceleration and write the equation for the
variation of displacement with time. What happens to the moving body at point B ? How
will the body move after this moment ?
v
5A
10 15 t
B
C
Sol. The velocity -time graph is a straight line with -ve slope, the motion is uniformly retarding one
upto point B and uniformly accelerated after with-ve side of velocity axis .
At point B the body stops and then its direction of velocity reversed.
The initial velocity at point A is v 0 = 7ms -1
The acceleration
Δv vt - v0 0-7
a= = = = -0.6364ms-2  -0.64ms-2
Δt Δt 11
The equation of motion for the variation of displacement with time is
1
s = 7t - 0.64t 2  7t  0.32t 2
2
EX.20: A particle starts from rest and accelerates as shown in the graph. Determine
a) the particle’s speed at t = 10s and at t = 20s
b) the distance travelled in the first220s.
a m/s
2
1
15 20
t sec
-1 5 10
-2
-3
Sol. a) Upto 10 sec the particle moves with uniform acceleration, hence the velocity at t = 10s,
v1 = u + at = 0 + 2 x 10 = 20m/s
From t = 10s to t = 15s the acceleration is zero,so The velocity of the particle at t=15s is 20m/s
Velocity at t = 20sec, v2 = v1 + at
= 20 + (-3)5 = 5 m/s
b) Distance travelled in first 10 sec
1 1 2
s1  ut  a1t 2  0 10   2  10   100m Distance travelled when t = 10 sec to t = 15 sec
2 2
s 2  vt  20  5  100m Distance travelled when t = 15 sec to t = 20 sec
1
s3  20  5    3  52  62.5m
2
Total distance travelled in 20 sec = s1 + s2 + s3 = 100 + 100 + 62.5 = 262.5 m
EX.21: Velocity (v) versus displacement (x) plot of a body moving along a straight line is as shown
in the graph. The corresponding plot of acceleration (a) as a function of displacement (x) is
(EAM-2014) v
Velocity
0 100 200 x
Displacement
a a
1) 2)
0 100 200 O 100 200 X
Displacement x Displacement
3) O 100 200
4)
X
O 100 200 X
Displacement Displacement
Sol. From the given graph equation for velocity v = kx
on differentiation
dv
 kv ------ (i)
dt
dv
= k(kx) = k2x; a = k2x and v = -kx + v 0
dt
dv
on differentiation  kv  k  kx  v 0  a = k2x - kv0 -------(ii)
dt
So, according to the eq. (ii) the shape of a-x graph is as below
0 x
Equations of Motion for Variable Acceleration :
 When acceleration ‘a’ of the particle is a function of time i.e., a = f(t)
dv
  f  t   dv  f  t  dt
dt
Integrating both sides within suitable limits, we have
v t t
 dv   f  t  dt  v  u   f  t  dt
u 0 0
 When acceleration ‘a’ of the particle is a function of distance a = f(x)

dv dv dx
  f x   f x
dt dx dt
v x x
2 2
 vdv   f  x  dx  v  u  2  f  x  dx
u x0 x0
EX.22: A particle located at x = 0, at time t = 0, starts moving along the positive x-direction with
a velocity v that varies as v = α x . The displacement of particle varies with time as
dx dx
Sol.  x   dt
dt x
x t
dx
On integrating, we get    dt
0 x 0
[ at t = 0, x = 0 and let at any time t, particle be at x]
x
 x1/2  
   t or x1/2  t  x  t 2
1/ 2  0 2
EX.23 : The acceleration (a) of a particle moving in a straight line varies with its displacement (S)
as a = 2S. The velocity of the particle is zero at zero displacement. Find the corresponding
velocity-displacement equation.
dv
Sol. a = 2S  v  2S  vdv  2SdS
ds
v s v S
 v2   S2 
  vdv   2SdS     2 
0 0  2 0  2 0
v2
  S2  v   2S
2
dv
EX.24: An object moving with a speed of 6.25 m/s, is decelerated at a rate given by = -2.5 v ,
dt
where v is instantaneous speed. The time taken by the object, to come to rest, would be
(AIEEE-2011)
dv 1
Sol.  2.5 v (or) dv  2.5dt
dt v
On integrating, within limits
(v1 = 6.25 m/s to v2 = 0)
v2  0 t
1/ 2
 v dv  2.5 dt
v1  6.25m/s 0
1/2
0 2   6.25
2   v1/ 2     2.5  t (or) t  2s
6.25
2.5
EX.25:The velocity of a particle moving in the positive direction of the X-axis varies as V  K S
where K is a positive constant. Draw V-t graph.
Sol. V  K s
dS S dS t
K S    K dt
dt 0
S 0
1
 2 S  Kt and S= K 2t 2
4
dS 1 2 1
V   K 2t  K 2t V  t
dt 4 2
V  t
 The V-t graph is a straight line passing through the origin.
Acceleration due to gravity
 The uniform acceleration of a freely falling body towards the centre of earth due to earth’s gravitational
force is called acceleration due to gravity
 It is denoted by ‘g’
 Its value is constant for all bodies at a given place. It is independent of size, shape, material,
nature of the body.
 Its value changes from place to place on the surface of the earth.
 It has maximum value at the poles of the earth. The value is nearly 9.83 m/s2.
 It has minimum value at equator of the earth. The value is nearly 9.78 m/s2.
g earth
 On the surface of the moon, g moon 
6
 The acceleration due to gravity of a body always directed downwards towards the centre of the
earth, whether a body is projected upwards or downwards.
 When a body is falling towards the earth, its velocity increases, g is positive.
 The acceleration due to gravity at the centre of earth is zero.
Motion under gravity

Equation of motion for a body projected vertically downwards :
 When a body is projected vertically downwards with an initial velocity u from a height h
then a = g, s = h
1 2 g
a) v = u + gt b) h = ut +gt c) v2 - u2 = 2gh d) Sn = u + (2n-1)
2 2
 In case of freely falling body u = 0, a = +g
1 2  1
a) v = gt b) S  gt c) v 2  2gS d) Sn  g  n  
2  2
 For a freely falling body, the ratio of distances travelled in 1 second, 2seconds, 3 seconds, .... = 1
: 4 : 9 : 16....
 For a freely falling body, the ratio of distances travelled in successive seconds = 1 : 3 : 5 : 9 ....
 A freely falling body passes through two points A and B in time intervals of t1 and t2 from the
g 2 2
start, then the distance between the two points A and B is =
2

t 2  t1 
 A freely falling body passes through two points A and B at distances h1 and h2 from the start, then
the time taken by it to move from A to B is
2h 2 2h1 2
T=
g

g

g
 h 2  h1 
 Two bodies are dropped from heights h1 and h2 simultaneously. Then after any time the distance
between them is equal to (h2~ h1).
 A stone is dropped into a well of depth ‘h’, then the sound of splash is heard after a time of ‘t’.
2h
Time taken by the body to reach water, t1 
g
h
Time taken by sound to travel a distance ‘h’, t2  V
sound
2h h
 Time to hear splash of sound is t  t1  t2 = +
g vsound
 A stone is dropped into a river from the bridge and after ‘x’ seconds another stone is projected
down into the river from the same point with a velocity of ‘u’. If both the stones reach the water
simultaneously, then S1 t   S 2 t  x 
1 2 1 2
gt  u  t  x   g  t  x 
2 2
 A body dropped freely from a multistoried building can reach the ground in t1 sec. It is stopped in
its path after t2 sec and again dropped freely from the point. The further time taken by it to reach
the ground is t 3  t12  t 2 2 .
H1 t2 H2
t1 H3 t3=?
We know that H1 = H2 + H3
1 2 1 2 1 2
 gt1  gt 2  gt 3
2 2 2
 t12  t 2 2  t 32  t3  t12  t 22
Equations of motion of a body Projected Vertically up :
 Acceleration (a) = -g
1 2 g
a) v = u - gt gt
b) s = ut - c) v2 - u2 = -2gh d) sn = u - (2n-1)
2 2
0
 Angle between velocity vector and acceleration vector is 180 until the body reaches the highest point.
 At maximum height, v = 0 and a = g
u2
 H max   H max  u 2 (independent of mass of the body)
2g
 A body is projected vertically up with a velocity ‘u’ from ground in the absence of air resistance
‘R’. then.  t a  t d 
u
i) t a  t d 
g
2u
ii) Time of flight T  t a  t d 
g
 A body is projected vertically up with a velocity ‘u’ from ground in the presence of constant air
resistance ‘R’. If it reaches the ground with a velocity ‘v’, then
a) Height of ascent = Height of descent
mu
b) Time of ascent t a 
mg  R
mv
c) Time of descent t d 
mg  R
d) t a  t d
v mg  R
e)  v  u
u mg  R
f) For a body projected vertically up under air resistance, retardation during its motion is > g
 At any point of the journey, a body possess the same speed while moving up and while moving
down.
g
 Irrespective of velocity of projection, all the bodies pass through a height in the last second of
2
g
ascent. Distance traveled in the last second of its journey u  .
2
 The change in velocity over the complete journey is ‘2u’ (downwards)
 If a vertically projected body rises through a height ‘h’ in nth second, then in (n-1)th second it will
rise through a height (h+g) and in (n+1)thsecond it will rise through height (h-g).
th th
 If velocity of body in nth second is ‘v’ then in  n  1 second it is (v + g) and that in  n  1 second
is (v - g) while ascending
 A body is dropped from the top edge of a tower of height ‘h’ and at the same time another body is
projected vertically up from the foot of the tower with a velocity ‘u’.
u=0
a) The separation between them after ‘t’ seconds is =  h  ut 

h
b) The time after which they meet t 
u
 gh 2 
c) The height at which they meet above the ground =  h  
 2u 2 
u
d) The time after which their velocities are equal in magnitudes is t 
2g
e) If they meet at mid point, then the velocity of thrown body is u  gh and its velocity of
meeting is zero
f) Ratio of the distances covered when the magnitudes of their velocities are equal is 1 : 3.
 A body projected vertically up crosses a point P at a height ‘h’ above the ground at time ‘ t1 ’
seconds and at time t2 seconds ( t1 and t2 are measured from the instant of projection) to same
point while coming down.
1
h  ut  gt 2 ; gt 2  2ut  2h  0
2
t2
h t1
This is quadratic equation in t

2u
Sum of the roots, t1  t2  (time of flight)
g
1
Velocity of projection , u  g  t1  t 2 
2
2h
Product of the roots, t1t2 
g
1
Height of P is h  gt1t 2
2
1 2
Maximum height reached above the ground H  g  t1  t 2 
8
g  t 2  t1 
Magnitude of velocity while crossing P is
2
 A body is projected vertically up with velocity u1 and after ‘t’ seconds another body is projected
vertically up with a velocity u2 .
a) If u2  u1 , the time after which both the bodies will meet with each other is h1  h2
1 2 1 2
u1 x  gx  u2  x  t   g  x  t 
2 2
1
u 2 t  gt 2
x 2 for the first body..
 u 2  u1   gt
u t  u t 
b) If u1  u 2  u , the time after which they meet is    for the first body and g  2
g 2  
for the second body.
u 2 gt 2
Height at which they meet = 
2g 8
 A rocket moves up with a resultant acceleration a . If its fuel exhausts completely after time ' t '
seconds, the maximum height reached by the rocket above the ground is
h  h1  h2
1 2 1 2 2  v2 a 2 t 2 
= at + a t  2 2g = 2g  ,  v  at 
h =
2 2g  
1  a
h  at 2  1  
2  g
 An elevator is accelerating upwards with an acceleration a. If a person inside the elevator throws
2u
a particle vertically up with a velocity u relative to the elevator, time of flight is t 
ga
2u
 In the above case if elevator accelerates down, time of flight is t 
ga
 The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that
instant. A particle may be momentarily at rest and yet have non-zero acceleration.
For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at
that instant continues to be the acceleration due to gravity.
EX.26: Drops of water fall at regular intervals from the roof of a building of height h = 16m. The
first drop striking the ground at the same moment as the fifth drop is ready to leave from
the roof. Find the distance between the successive drops.
2h
Sol. Step-I : Time taken by the first drop to touch the ground = t 
g
16  2  2
For h = 16m, t  4
g g
 1  1 2
Time interval between two successive drops is t   t   t 
 n 1   4  g
Where n = number of drops
Step-II :
1 2 1 2
Distance travelled by 1st drop S1  g  4t   g 16     16m
2 2 g
1 2 1 2
Distance travelled by 2nd drop S2  g  3 t   g  9     9 m
2 2 g
1 2 1 2
Distance travelled by 3rd drop S3  g  2 t   g  4     4 m
2 2 g
1 2 1  2 
Distance travelled by 4th drop S4  g   t   g    1m
2 2  g 
Distance between 1st and 2nd drops= S1  S2  16  9  7 m
Distance between 2nd and 3rd drops= S 2  S 3  9  4  5m
Distance between 3rd and 4th drops= S 3  S 4  4  1  3m
Distance between 4th and 5th drops= S 4  S 5  1  0  1m
EX.27: A body falls freely from a height of 125m (g = 10 m/s2). After 2 sec gravity ceases to act
Find time taken by it to reach the ground ?
1 1 2
Sol. 1) Distance covered in 2s under gravity S1  gt 2  10  2   20m
2 2
Velocity at the end of 2s V = gt = 10  2 = 20 m/s
Now at this instant gravity ceases to act, there after velocity becomes constant.The remaining
distance which is 125 - 20 = 105m is covered by body with constant velocity20 m/s.Time taken
to cover 105 m with constant velocity is given by
S 105
t1   t1   5.25s
V 20
Total time taken = 2 + 5.25 = 7.25 s
EX.28: A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens
out. T hen he descends with a net r etar dation of 2 m/s2. His velocity when he reaches the
ground is 8 m/s. Find the height at which he gets out of the aeroplane ?
Sol. Distance he falls before the parachute opens is
1
S1  g  100  490m
2
Then his velocity , u=gt = 98.0 m/s
Velocity on reaching ground = v=8 m / s
retardation = 2 m / s 2
v 2  u 2  2aS 2  S 2  2385m
Total distance S  S1  S 2 = 2385 + 490
=2875 m( height of aeroplane)
EX.29: If a freely falling body covers half of its total distance in the last second of its journey.
Find its time of fall
Sol. Suppose t is the time of free fall
g g 2
Sn 
2
 2n  1 and S
2
n 
g
 2n  1 
Sn 2 1
 2 g 2
S n  
2
 

n 2  4n  2  0 and n  2  2 sec 
EX.30: A body is projected vertically up with a velocity u. Its velocity at half of its maximum
height and at 3/4th of its maximum height are
Sol. From v2 - u2 = 2aS, here a = –g; when S = H/2, then
u2 u
v 2  u 2  2  g  v
4g 2
2 2 3u 2 u
When S = 3H/4, then v  u  2  g  4 2g v
  2
EX.31: A stone is allowed to fall from the top of a tower 300m height and at the same time
another stone is projected vertically up from the ground with a velocity 100 m/s. Find when
and where the two stones meet ?
300-x
300 m
Sol.
x
height of the tower, h= 300m

Suppose the two stones meet at a height x from ground after t seconds
Sr
t
u r , ur  u  0  u , S r  h
h 300
t   3sec
u 100
height of the stone from the point of projection is
1 2 1
h1  ut  gt  100  3   9.8  9  255.9m
2 2
EX.32: A stone is dropped from certain height above the ground. After 5s a ball passes through a
pane of glass held horizontally and instantaneously loses 20% of its velocity. If the ball takes
2 more seconds to reach the ground, the height of the glass above the ground is
Sol. In 5s velocity gained v = gt = 50 m/s.
80
Velocity after passing through the glass pane  50  40m / s
100
1 2 1 2
Height of the glass pane above the ground is h  ut  gt = 40  2   10   2  = 100m
2 2
Body Projected Vertically up from a Tower
 A body projected vertically up from a tower of height ‘h’ with a velocity ‘u’ (or) a body dropped
from a rising balloon (or) a body dropped from an helicopter rising up vertically with constant
velocity ‘u’ reaches the ground exactly below the point of projection after a time ‘t’. Then
+X -X
T
O -h
h W
E
R
1
(a) Height of the tower is h   ut  gt 2
2
(b) Time taken by the body to reach the ground
u  u 2  2gh
t=
g
(c) The velocity of the body at the foot of the tower v = u 2  2 gh

(d) Velocity of the body after ‘t’ sec. is
v  u  gt
1 2
(e) Distance between the body and balloon after this time = gt
2
EX.33: A ball is thrown vertically upwards from the top of a tower. Velocity at a point ‘h’ m
vertically below the point of projection is twice the downward velocity at a point ‘h’ m
vertically above the point of projection. The maximum height reached by the ball above the
top of the tower is (MED-2012)
Sol. If AB is the tower then according to the problem, velocity at ‘P’ is given as twice the velocity at
‘Q’
C
H Q
h
A
T h
O
VP  u 2
 2 gh  ; VQ  u 2
 2 gh  ; VP  2  VQ  W P
E
R
B
10gh
u 2
 2gh   2   2
u 2  2gh  u 
3
 10gh 
u   2
3  5h
From the top of the tower maximum height reached H  H 
2g 2g 3
EX.34: From a tower of height H, a particle is thrown vertically upwards with a speed u. The
time taken by the particle to hit the ground is n times that taken by it to reach the highest
point of its path. The relation between H, u and n is ( jee main- 2014)
u
Sol. Time taken to reach the maximum height t1 
g
If t2 is the time taken to hit ground
1 2
i.e.,  H  ut 2  gt
2
 nu  1  n 2 u 2 
But t2 = nt1 ; So,  H  u  g   2 g  g 2 
   
nu 2 1  n 2 u 2 
H     2 gH  nu 2  n  2 
g 2 g  
EX.35 : A balloon starts from rest, moves vertically upwards with an acceleration g/8 ms-2. A
stone falls from the balloon after 8 s from the start. Further time taken by the stone to
reach the ground (g = 9.8 ms-2) is
Sol. The distance of the stone above the ground about which it begins to fall from the balloon is
1g
h    82  4g
28
g
The velocity of the balloon at this height can be obtained from v = u + at; v  0    8  g
8
This becomes the initial velocity (u) of the stone as the stone falls from the balloon at the height
h.
u '  g
1 1 2
For the total motion of the stone h = -u t + gt
2
1
 4g  gt  gt 2 and t 2  2 t  8  0
2
Solving for ‘t’ we get t = 4 and -2s. Ignoring negative value of time, t = 4s.
 Three bodies are projected from towers of same height as shown. 1st one is projected vertically up
with a velocity ‘u’. The second one is thrown down vertically with the same velocity and the third
one is dropped as a freely falling body. If t , t , t are the times taken by them to reach ground, then,
1 2 3
u=u u=0
h h h
1 12 2 3
For 1st body, h  ut1  gt1  1
2
1 2
For 2nd body, h  ut2  gt2   2
2
1 2
For 3rd body, h  gt3   3
2
from 1  t2 +  2   t1
1
 h  t1  t2   gt1t2  t1  t2 
2
1
i) Height of the tower h  gt1t2   4
2
ii) From eq (3) & (4), t3  t1t2   5
iii) Equating R.H.S of (1) & (2),
1
velocity of projection u  g  t1  t 2    6 
2
2u
iv) Time difference between first two bodies to reach the ground t    7
g
Relative Motion in one dimension
 Velocity of one moving body with respect to other moving body is called Relative velocity.
 A and B are two objects moving uniformly with average velocities vA and vB in one dimension,
say along x-axis having the positions xA(0) and xB(0) at t = 0.
 If xA(t) and xB(t) are positions of objects A and B at time t then
x A  t   x A  0   vA t ; x B  t   x B  0   vB t
 The displacement from object A to B is given by
x BA  t   x B  t   x A  t 
  x B  0   x A  0     v B  v A  t
 x BA  0    v B  v A  t
r r r
Velocity of A w.r.t B is vAB = v A - vB
r
v AB = v 2A + v 2B - 2v A v Bcosθ
r r r
 Two bodies are moving in a straight line in same direction then, v AB = v A - v B (Q 00 )
 Two bodies are moving in a straight line in opposite
r r r
direction then, v AB = v A + v B , (Q 1800 )
 If two bodies move with same velocity and in same direction, then position between them does
not vary with time.
 If two bodies move with unequal velocity and in same direction, then position between them first
decreases to minimum and then increases.
 If the particles are located at the sides of n sided symmetrical polygon with each side a and each
particle moves towards the other, then time after which they meet is
Initial separation
T
Re lative velocity of approach
a a
T T
       
v  v cos   v 1  cos   
 n    n 
a
T
and 
2v sin 2  
n
Shortcut to solve the problems
2a a
For Triangle n = 3  T  ; For Square n = 4  T 
3v v
2a
For hexagon, n = 6  T 
v
EX.36: A passenger is at a distance ‘d’ from a bus, when the bus begins to move with a constant
acceleration a. Then find the minimum constant velocity with which the passenger should
run towards the bus so as to catch is
1 2
Sol. S passenger  d  Sbus ; vt  d  at
2
v  v 2  2ad
at 2  2vt  2d  0  t 
a
for minimum velocity v 2  2ad  0  v  2ad
EX.37. Two trains, each travelling with a speed of 37.5kmh-1, are approaching each other on the
same straight track. A bird that can fly at 60kmph flies off from one train when they are 90
km apart and heads directly for the other train. On reaching the other train, it flies back to
the first and so on. Total distance covered by the bird before trains collide is
Sol. Relative speed of trains  37.5  37.5  75kmh 1
S r 90 6
Time taken by them to meet t  u   h
r 75 5
6
Distance travelled by the bird, x  Vbird  t 60   72km (Q Vbird  60kmh 1 ) ,
5
EX.38: On a two-lane road, car A is travelling with a speed of 36 kmph. Two cars B and C
approach car A in opposite directions with speed of 54 kmph each. At a certain instant,
when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C
does. What minimum acceleration of car B is required to avoid an accident?
Sol. Velocity of a car A, VA  36km / h  10m / s
Velocity of car B, VB  54km / h  15m / s
Velocity of car C, VC  54km / h  15m / s
Relative velocity of car B with respect to car A, VBA  VB  VA  15  10  5m / s
Relative velocity of car C with respect to car A, VCA  VC  VA  15  10  25m / s
At a certain instance, both cars B and C are at the same distance from car A i.e., s  1km  1000m
1000
Time taken (t) by car C to cover 1000m is t  40s
25
The acceleration produced by car B is
1 1600
1000  5  40  at 2  a   1m / s 2
2 1600
EX.39: A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant
acceleration. At the same instant another body passes through x = 0 moving in the positive x-
direction with a constant speed. The position of the first body is given by x1  t  after time ‘t’
and that of the second body by x2  t  after the same time interval. Which of the following
graphs correctly describes  x1  x2  as a function of time ‘t’ ? [AIEEE - 2008]
x1  x2  x1  x2 
1) 2)
O t O t
x1  x2  x1  x2 
3) 4)
O t t
O
1 2 1 2
Sol. As, x1  t  = at and x2  t   vt  x 1 - x 2 = at - vt (parabola)
2 2
Clearly, graph (2) represents it correctly.
EX.40.A particle has an initial velocity 3iˆ  4ˆj and an acceleration of 0.4iˆ  0.3jˆ . It speed after 10s
is (AIEEE-2009)
r r r
Sol. v = u + at
   
 3iˆ  4 ˆj  0.4iˆ  0.3 ˆj 10  3iˆ  4 ˆj  4iˆ  3 ˆj  7iˆ  7 ˆj
r
v = 49 + 49  7 2
EX.41. A body is moving along the circumference of a circle of radius ‘R’ and completes half of
the revolution. Then, the ratio of its displacement to distance is
1)  : 2 2) 2:1 3) 2 :  4) 1:2
C
Sol. .
A B
Displacement : Distance   R : 2 R key-3

EX.42. A body completes one round of a circle of radius ‘R’ in 20 second. The displacement of the
body after 45 second is
R
1) 2) 2 R 3) 2 R 4) 2R
2
Sol. In 40sec body completes two revolutions.
 
In 5 sec it covers 1/4 th of the circle and angle traced is . So displacement s  2 R sin
2 2
key-2
EX.43. If a body covers first half of its journey with uniform speed v1 and the second half of the
journey with uniform speed v2, then the average speed is
2 v 1v 2 v 1v 2
1) v1  v2 2) v  v 3) v  v 4) v1v2
1 2 1 2
s1  s 2 2v1v 2
Sol. . Average speed = t  t  v = v + v key-2
1 2 1 2
EX.44. A car is moving along a straight line, say OP in figure. It moves from O to P in 18 s and
return from P to Q in 6 s. What are the average velocity and average speed of the car in
going from O to P and back to Q?
1) 10 m s  1 , 20 m s  1 2) 20 m s  1 ,10 m s  1 3) 10 m s  1 ,10 m s  1 4) 20 m s  1 , 20 m s  1
total displacement s1  s 2
Sol. . v avg = = t  t key-1
total time 1 2
EX.45. For a body moving with uniform acceleration ‘a’, initial and final velocities in a time
interval ‘t’ are ‘u’ and ‘v’ respectively. Then, its average velocity in the time interval ‘t’ is
 at   at 
1)  v  at  2)  v   3)  v  at  4)  u  
 2  2
v  u v  v  at
Sol. vavg   key-2
2 2
EX.46. Two cars 1 & 2 starting from rest are moving with speeds v1 and v 2 m/s (v1 > v 2 ) . Car 2 is
ahead of car ‘1’ by s meter when the driver of the car ‘1’ sees car ‘2’. What minimum
retardation should be given to car ‘1’ to avoid collision. (2002 A)
2 2
1)
v1  v 2
2)
v1  v 2
3)
 v1  v 2  4)
 v1  v 2 
s s 2s 2s
2 2
Sol. urel  v1  v2 ; vrel  0 ; v
rel u rel  2as
Key-4
EX.47. If a car covers 2 / 5th of the total distance with v1 speed and 3/ 5th distance with v2 then
average speed is
1 v1  v 2 2v1v 2 5v1v 2
1) v1v 2 2) 3) v  v 4) 3v  2v \
2 2 1 2 1 2
s1  s2
total distance  s s
Sol. avg speed= total time 1
 2
v1 v2
key-4
EX.48. The coordinates of a moving particle at any time ‘t’ are given by x  t 3 and y   t 3 . The
speed of the particle at time ‘t’ is given by
1)  2  2 2) 3t  2   2 3) 3t 2  2   2 4) t 2  2   2
dx dy 2 2
Sol. vx  , vy  ; v  vx  vy
dt dt
key-3
EX.49. A car, starting from rest, accelerates at the rate f through a distance S, then continues at
constant speed for time t and then decelerate at the rate (f/2) to come to rest. If the total
distance travelled is 15S, then
1 2 1 2 1 2
1) S  ft 2) S  ft 3) S  ft 4) S  ft
6 72 4
Sol.
s
v 2 - o 2 = 2fs

2 2 f  ® s 3 = 2s
o - v = -2 .s3 
2 
 s 2 = 12s ; 12s = vt also v = 2fs
1 2
 12s = 2fs.t ; 144s2 = 2fs.t ; s = ft
72
key-3
EX.50. An armored car 2m long and 3 m wide is moving at 10ms-1 when a bullet hits it in a
1   3
direction making an angle tan  4  with the length of the car as seen by a stationary observer..
 
The bullet enters one edge of the car at the corner and passes out at the diagonally opposite
corner. Neglecting any interaction between the car and the bullet and effect of gravity, the
time for the bullet to cross the car is
1) 0.20 s 2) 0.15 s 3) 0.10 s 4) 0.50 s
Sol.
2m
0
3m vb sin 37 10m/s
vb -1 3
 
=tan  
4
=370
0
vb cos 37 2m
Relative velocity along x-axis is  v b cos37 0 -10 
Distance travelled by bullet along x- axis is
 v cos37
b
0
-10  t = 2 .........(1)
Distance travelled by bullet along y-axis is
 v sin37  t = 3 ..........(2)
b
0
Solving equation (1) and (2) we get t=0.20s key-1

EX.51. Two particles start simultaneously from the same point and move along two straight lines. One
with uniform velocity v and other with a uniform acceleration a. If  is the angle between the lines
of motion of two particles then the least value of relative velocity will be at time given by
v v v v
1) sin  2) cos  3) tan  4) cot 
a a a a
Sol. At any time velocity of first car is V and that of second car is v = v + at = 0 + at
2
v rel = v 2 +  at  - 2vat cosα
d 2 vcosα
v rel is minimum if
dt
 vr  = 0 ; t =
a
key-2
EX.52. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the
speed of 1500 km/h relative to the jet plane. What is the speed of the later with respect to an
observer on ground?
1) 100kmph 2) 1000kmph 3) 10kmph 4) 11kmph
Sol. Velocity of jet aeroplane = 500 $
j
velocity of fuel w.r.to plane  1500 $j
uur uur uur uur
 v f - v p = -1500 $j ; v f = v p -1500 $j
 500 $ j  1500 $
j  1000 $
j
 speed of fuel w.r.to ground is -1000km/hr..
key-2
EX.53. A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin
and positive direction upwards for all quantities, which one of the following is correct?
1) x < 0, v < 0, a > 0 2) x > 0, v < 0, a < 0 3) x > 0, v < 0, a > 0 4) x > 0, v > 0, a < 0
Sol. As the lift is coming in downward direction dis-placement will be negative. We have to see
whether the moton is accelerating or retarding.
We know that due to downward motion displace-ment will be negative. When the lift reaches 4th
floor is about to stop hence, motion is retarding in nature hence, x < 0,a > 0. As displacement is in
negative direction, velocity will also be negative i.e., V < 0.
key-1
EX.54. A vehicle travels half the distance ‘l’ with speed v1 and the other half with speed v2, then its
average speed is
v1  v 2 2v1  v 2 2v1v 2 L  v1  v 2 
1) 2) v  v 3) v  v 4)
2 1 2 1 2 v1v 2
Sol. Total time = t1 + t2
l l 1 1 1 
=     
2v1 2v 2 2  v1 v 2 
key-3
EX.55. The displacement of a particle is given by x = (t -2)2 if where x is in metre and t
in second. The distance coverred by the particle in first 4 seconds is
1) 4 m 2) 8 m 3) 12 m 4) 16m
2
Sol. Given x = (t - 2)
dx d 2
velocity, v=   t  2  2  t  2 m / s
dt dt
dv d
Acceleration, a=   2  t  2  
dt dt 
= 2 [l - 0] = 2 m/s2
key-2
EX.56. At a metro station, a girl walks up a stationary escalator in time t1. If she remains stationary
on the escalator, then the escalator take her up in time t2. The time taken by her to walk up
on the moving escalator will be
l) (t1+t2)/2 2 ) t1t2/(t2-t1) 3) t1t2 /(t2+t1) 4) t1 - t2
L
Sol. Velocity of girl vg = t
1
L
Velocity of escalator v e = t
2
L L
Net velocity of the girl v g +v e = t  t
1 2
L L L tt
  t 1 2
t t1 t 2 t1  t 2
key-3
EX.57. A body moving along a straight line traversed one third of the total distance with a
velocity 4 m/sec in the first stretch. In the second stretch the remaining distance is covered
with a velocity 2 m/sec for some time t0 and with 4m/s for the remaining time. if the average
velocity is 3 m/sec, find the time for which body moves with velocity 4 m/sec in second
stretch:
3 t0
A) t0 B) t0 C) 2t0 D)
2 2
s/3 s
Sol. t1  
4 12
2s 2s
 2  t 0   4  kt 0   t 0  2  4k  or t 0 
3 3  2  4k 
s
Average velocity = t  t  t k
1 0 0
s 6  2  4k 
 
s 2s 1  k   5  6k 

12 3  2  4k 
1
vav  5  6k   12  24k gives k 
2
t0
Required time = kt 0  . key-D
2
EX.58. For motion of an object along the x-axis, the velocity v depends on the displacement x
as v  3 x 2  2 x , then what is the acceleration at x = 2 m?
A) 48 ms 2 B) 80 ms 2 C) 18 ms 2 D) 10 ms 2
Sol. Given v  3x 2  2 x; differentiating v , we get
dv dx
  6x  2  6 x  2 v
dt dt
 a   6 x  2   3 x 2  2 x  Now put x= 2 m
2
 a   6  2  2  (3  2   2  2)  80 ms 2 key-B
EX.59. A police party is chasing a dacoit in a jeep which is moving at a constant speed v . The
dacoit is on a motorcycle. When he is at a distance x from the jeep, he accelerates from rest
at a constant rate  ? Which of the following relations is true if the police is able to catch
the dacoit ?
A) v 2   x B) v 2  2 x C) v 2  2 x D) v 2   x
Sol. If police is able to catch the dacoit after time t, then
1 
vt  x   t 2. This gives t 2  vt  x  0
2 2
v  v 2  2ax
or t  For t be real , v 2  2ax key-C

EX.60. A point moves in a straight line so that its displacement x metre time t second is given
by x 2  1  t 2 . Its acceleration in ms 2 at time t second is
1 t 1 t2 1 1
A) B) C)  D) 
x3/ 2 x3 x x3 x x2
1/ 2
Sol. x 2  1  t 2 or x  1  t 2 
dx 1 1/ 2 1/ 2
 1  t 2  2t  t 1  t 2 
dt 2
d2x 1/ 2  1 3/ 2
a  2  1  t 2   t    1  t 2  2t
dt  2
1 t2
  key-C
x x3
EX.61. A 2m wide truck is moving with a uniform speed v0  8 ms 1 along a straight horizontal
road. A pedestrain strarts to cross the road with a uniform speed v when the truck is 4 m
away from him. The minimum value of v so that he can cross the road safely is
A) 2.62 ms 1 B) 4.6 ms 1 C) 3.57 ms 1 D) 1.414 ms 1
v sin
2m 8
Sol.
4m 
v cos
2
Time of crosing  v sin 
4
Time in which truck just able to catch the man = 8  v cos 
2 4
For safe crosing v sin  = 8  v cos 
or 16  2 v cos   4v sin 
16
or v 
cos   2 sin 
For v minimum cos   2sin  is maximum
d
so ,  cos   2 sin    0
d
  sin   2 cos   0
 tan   2
1 2
 cos   ,sin  
5 5
16 5 16
v min  
Now 5 5 key-C
 3.57 m / s
EX.62. The velocity of a particle along a straight line increases according to the linear law v =
v0+kx, where k is a constant. Then
A) the acceleration of the particle is k  v0  kx 
1  v1 
B) the particle takes a time k log e  v  to attain a velocity v1
 0
C) velocity varies linearly with displacement with slope of velocity displacement curve equal
to k.
D) data is insufficient to arrive at a conclusion.
dv g
Sol. Acceleration =  v  0  kx
dt
  dx  g
Q x   v   v  a  kv  k  v  kx 
 dt  0
dv dv
Further, a   kv   kv
dt dt
dv
  kdt
v
v1 t
dv 1 v 
 v v 0 dt  t  k loge  v10 
 k
0
dv
Since, v=v0+kx. Hence slope of velocity displacement curve is k key-ABC
dx
EX.63. Two particles P and Q move in a straight line AB towards each other. P starts from A
with velocity u1 and an acceleration a1. Q starts from B with velocity u2 and acceleration
a2. They pass each other at the midpoint of AB and arrive at the other ends of AB with
equal velocities.
2 12u  u 
A) They meet at midpoint at time t   a  a 
1 2
4  u 2  u1  a1u 2  a 2 u1 
B) The length of path specified i.e., AB is l  a 1  a2 
2
C) They reach the other ends of AB with equal velocities if  u 2  u1  a1  a 2   8  a 1u 2  a 2 u1 

D) They reach the other ends of AB with equal velocities if
 u 2  u1  a1  a 2   8  a 2 u1  a1u 2 
u1a1 u2a2
Sol. .
l/2 l/2
l 1 l 1 2
 u1 t  a1t 2 ...(1) and    u1t   a 2  t
2 2 2 2
l 1
  u 2 t  a 2 t 2 ....(2)
2 2
 u 2  u1 
subtracting (1) and (2) , we get t  2   ....(3)
 a1  a 2 
4  u 2  u1 
2  1 2
Substituting (3) in (1) or(2) and rearranging, we get. l  a u  a 2 u1  ...(4)
 a1  a 2 
Since the particle P & Q reach the other ends of A and B with equal velocities say v
For particle P v 2  u12  2a 2 l ... (5)
For particle Q v 2  u 22  2a 2 l ...(6)
Substracting and then substituting value of l and rearranging, we get
 u 2  u1  a1  a 2   8  a1u 2  a 2 u1 
key-ABC
EX.64. A particle moves along a straight line so that its velocity depends on time as v  4t  t 2 .
Then for first 5s.
A) Average velocity is 25 / 3 ms 1
B) Average speed is 10 ms 1
C) Average velocity is 5 / 3 ms 1
D) Acceleration is 4 ms 2 at t  0
5 5
  4t  t  dt
2
r  vdt  2 t3  125
v 0
 0
 2t  3  50 
Sol. Average velocity 5 5

 0
 3  25  5
 dt
0
 dt
0
5 5 3 5 3
For average speed, let us put v= 0, which gives t  0 amd t  4s
 average speed =
4 5 4 5
2
 vdt   vdt
0 4
 (4t  t
0
)dt   vdt
4  2 t3 
4
 2 t3 
5
  2 t   2t  
5
5  3  0  3 4

 dt
0
5
4 5
 2 t3   2 t3 
 2t  3    2t  3 
 0  4 13 1
  ms
5 5
For acceleration :
dv d
a   4t  t 2   4  2t At t  0, a  4 ms 2 key-C,D
dt dt
EX.65. A particle moves with an initial velocity v0 and retardation  v , where v is velocity at any time t.
v0
A) The particle will cover a total distance

1
B) The particle will come to rest after time

C) The particle will continue to move for a long time.
v0 1
D) The velocity of particle will become after time
e 
0 x0
dv
Sol. v. dx   v  or  dv    dx
v0 0
v0
v0   x0  x0  ;

v t
dv dv
  v (or )     dt
dt v0
v 0
v  v0 e  t  or  v  0 for t  
v0 1 .key-A,C,D
v when t 
e 
t3
EX.66. A particle is moving along X–axis whose position is given by x  4  9t  . Mark the
3
correct statement(s) in relation to its motion.
A) direction of motion is not changing at any of the instants
B) direction of motion is changing at t = 3 s
C) for 0 < t < 3 s, the particle is slowing down
D) for 0 < t < 3 s, the particle is speeding up.
Sol. The particle’s velocity is getting zero at t = 3 s, where it changes its direction of motion.
For 0 < t < 3 s, V is negative, a is positive, so particle is slowing down.
For t < 3, both V and a are positive, so the particle is speeding up.
key-B,C
PASSAGE TYPE QUESTIONS
Passage-1
A train starts from rest with constant acceleration, a  1m / s 2 . A passenger at a distance ‘S’
from the train runs at his maximum velocity of 10 m/s to catch the train at the same moment
at which the train starts.
67. If S=25.5 m and passenger keeps running, find the time in which he will catch the train:
A) 5 sec B) 4 sec C) 3 sec D) 2 2 sec.
68. Find the critical distance ‘Sc’ for whcih passenger will take the ten seconds time to catch the
train:
A) 50m B) 35m C) 30m D) 25m
69. Find the speed of the train when the passenger catches it for the critical distacne:
A) 8 m/s B) 10 m/s C) 12 m/s D) 15m/s
Sol. 67. At time t, Xt and Xp are coordinates of train and passenger respectively.
1
X t  a1t 2 and X p  v P t  S
2
If passenger catches the train,
Xt = Xp
1 2 v P  v P2  2a1S
or a1t  v P t  S or t 
2 a1
2
10  10   2 1 25.5 
  3seconds
1
68. The critical distacne ‘Sc’ for which passenger will take the ten seconds time to catch the train is
v 2P
given by Sc 
2a1
The time is 10 seconds, if v 2P  2a1S  0
2
v 2 10 
Sc  P   50m
2a t 2 1
vP
69. For critical distance, passenger catches the train in time, t  a So, required velocity of train =
t
a t .t
v 
 a t  P   VP / 2  10m / sec
 at 
key-67-.C 68- A 69- B
Passage-2
A body is moving with uniform velocity of 8 ms 1 . When the body just crossed another
body, the second one starts and moves with uniform acceleration of 4 ms 2 .
70. The time after which two bodies meet will be
A) 2 s B) 4 s C) 6 s D) 8 s
71. The distance covered by the second body when they meet is
A) 8 m B) 16 m C) 24 m D) 32 m
Sol. 70. Let they meet after time t , then the distance travelled by both in time t should be
same
1
s  8t  4t 2  t  4 s
2
71. s  8t  8  4  32 m
key- 70-. B 71-. D
EX.72. An elevator in which a man is standing is moving upwards with a speed of 10 ms 1 . If
the man drops a coin from a height of 2.45 m from the floor of elavator, it reaches the floor
of the elavator after time ( g  9.8 ms 2 )
A) 2 s B) 1/ 2 s C) 2 s D) 1/ 2 s
Sol. Let the initial relative velocity, rerlative acceleration and relative displacement of the with
respect to the floor of the lift be ur , ar and sr , then sr  ur t  1/ 2  ar t 2
and ur  uc  ul  10  10  0
ar  ac  al   9.8  0  9.8 ms 2
sr  sc  sl  2.45m
2.45  0  t   1/ 2  9.8 t 2
or t 2  1/ 2 or t  1 2 s
key-B
EX.73. A body is thrown vertically upwards from A, the top of a tower. It reaches the ground in time
t1 . If it is thrown vertically downward from A with the same speed, it reaches the ground in time
t 2 . If it is allowed to fall freely from A, then the time it takes to reach the ground is given by
t1  t2 t1  t2 t1
A) t  B) t  C) t  t1t2 D) t 
2 2 t2
Sol. Suppose the body be projected vertically upwards from A with a speed u0 .
1 2
Using equation s  ut    at , we get
2
1 2
For first case:  h  u0t1    gt1 (i)
 2
1 2
For second case:  h  u0t2    gt2 ( ii )
2
1
( i )- ( ii )  0  u0  t2  t1     g  t2  t1 
2 2
2
 
1
 u0    g  t1  t2  ( iii )
2
Putting the value of u0 in ( ii ), we get
1 1
 h     g  t1  t2  t2    gt22
2 2
1
h gt1t2 ( iv)
2
For third case: u  0, t  ?
1 1
 h  0  t    gt 2 or h    gt 2 ( v)
2 2
Combining Eq. ( iv ) and ( v), we get
1 2 1
gt  gt1t2 or t  t1t2
2 2
key-C
EX.74. The decelaration experienced by a moving motor boat, after its engine is cut-off is
dv
given by  kv 3 , where k is constant.If v0 is the magnitude of the velocity at cut-off,
f,
dt
the magnitude of the velocity at a time t after the cut-off is
v0
A) v0 / 2 B) v C) v0 e  kt D)
2v02 kt  1
dv
Sol. Here  kv 3
dt
dv dv
v t
or 3   kdt or  3
   kdt
v v0 v 0
v
 1  1 2 2 v02 v0

or  2v 2    kt or  2
 2
  kt , or v  2
or v 
 V0 2v 2v0 1  2v0 kt 2v02 kt  1
EX.75. A jet plane starts from rest at S = 0 and is subjected to the acceleration shown. Determine
the speed of the plane when it has travelled 60 m.
A) 46.47 m / s B) 36.47 m / s C) 26.47 m / s D) 16.47 m / s
a(m/s)2
22.5
150 S(m)
dv s v
Sol. a  v   a.ds   v.dv  v 2  u 2
ds 0 u
 Area under a : s curve  v 2  u 2

1  1 2
  2  150  22.5   2  90 13.5   v  0
 
2
 v  75  22.5  45  13.5
 v  75  22.5  45 13.5  46.47
EX.76. The relation between time t and distance x is t = a x 2 + b x . Wheree a and b are constants.
The retardation is
A) 2a v 3 B) 2b v 3 C) 2a b v3 D) 2b 3 v3
1 
Sol. t   x 2   n 1  2 xv   v   2 x   
v 
0  2  x.a  v.v    a 0   2 x    a  2 v 2
2 v 2 2 v 2
a   2 v3
2 x   1
v
EX.77. The motion of a body falling from rest in a resisting medium is described by the equation
dv
= a -b v where a and b are constants. The velocity at any time t is given by
dt
a b -bt
A) v = (1 -e ) B) v= (e )
-b t
b a
a b bt
C) v = (1 + e ) D) v= e
-b t
dv b v dv t a
Sol.  a  bv    dt
dt 0 a  bv 0
 2  v a  bv
  ln  a  bv   0  t  ln   bt
 b  a
a
 a  bv  ae  bt  v  1  e  bt 
b
EX.78 A train stops at two stations s distance apart and takes time t on the journey from one station
to the other. Its motion is first of uniform acceleration a and then immediately of uniform
retardation b, then
1 1 t2 1 1 t2 1 1 t2 1 1 t2
A) - = B) + = C) + = D) - =
a b s a b s a b 2s a b 2s
v2 v2 v2  1 1 
Sol. s      
2a 2b 2  a b 
   ab 
Again t    v  t  
a b ab
t2 1 1
s   
2 t2 1 1
1 1 a b   
2   2s a b
a b
EX.79. A ball is thrown from the top of a tower in vertically upward direction. Velocity at
apoint h metre below the point of projection is twice of the velocity at a point h metree
above the point of projection. find the maximum height reached by the ball above the top
of tower.
A) 2 h 2 B) 3 h C) (5/3) h D) (4/3) h
u
Sol. H  ; given v2  2v1
2g
(i) A to B: v12  u 2  2 gh
(ii) A to C v22  u 2  2 g  h 
(iii) solving (i), (ii),(iii) we get the value of u 2 as 10g/h/3 and then we get the value of H by
u2
using H 
2g
key-C
EX.80. A parachutist drops first freely from an aeroplane for 10 s and then his parachute
opens out. Now he descends with a net retardation of 2.5 ms 2 . If he bails out of the plane
at a height of 2495 m and g  10 ms 2 , his velocity on reaching the ground will be
A) 5 ms 1 B) 10 ms 1 C) 15 ms 1 D) 20 ms 1
Sol. The velocity v acquired by the parachutist after 10 s.
v  u  gt  0  10 10  100ms 1
1 2 1 2
Then, s1  ut  gt  0   10 10  500m
2 2
The distance travelled by the parachutist under retadation is
s2  2495  500  1995 m
Let vg be the velocity on reaching the ground. Then vg2  v 2  2as2
2
or vg2  100   2   2.5   1995 or vg  5 ms 1
key-A
EX.81. The velocity-time plot a particle moving on a straight line is shown in figure.
–1
v(ms )
10
0 t(s)
10 20 30
–10
–20
A) The particle has a constant acceletration
B) The particle has never tuned around
C) The particle has zero displacement
D) The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20
s
Sol. Since the graph is astraight line, its slope is constant, it means acceleration of the particle is
constant.
Velocity of the particle changes from positive to negative at t  10 s , so particle changes
direction at this time.
The particle has zero displacement up to 20 s, but not for the entire motion.
The average speed in the interval of 0 to 10 s is the same as the average speed in the internal of
10 s to 20 s because distance covered in both time interval is same. key-AD
EX.82. The displacement of a particle as a function of time is shown in figure. It indicates
S
O t
1 2 4 5
A) The particle starts with a certain velocity, but the motion is retarded and finally the particle
stops
B) The velocity of the particle decreases
C) The acceleration of the particle is in opposite direction to the velocity
D) The particle starts with a constant velocity, the motion is accelerated and finally the particle
moves with another constant velocity.
Sol. Initially at orogin, slope is not zero, so the particle has some initial velocity but with time we
see that slope is decreasing and finally the slope necomes zero, so the particle stops finally.
key-ABC
EX.83. The displacement-time graph of a moving particle with constant acceleration is shown in
the figure. The velocity time graph is given by
x(m)
t(s)
0 1 2
v v
v v
A) t B) t C) t D) t
1 2 0 1 2 0 1 2
0 0 2
1
Sol. At t  0 , slope of the x-t graph is zero; hence, velocity is zero at t  0 . As time increases, slope increases
in negative direction; hence, velocity increases in negative deirection. At point’I’, slope changes suddenly
from negative to positive value: hence, velocity changes suddenly from negative to positive and then
velocity starts decreasing and becomes zero at’2’, option ( A) represents all these clearly.
EX.84 The velocity-time graph of abody is given in figure. The maximum accceleration in ms 2
is
v(ms–1)
A) 4
60
B) 3
C) 2 20
D) 1 0 t(s)
20 30 40 70
Sol. Maximum acceleration will befrom 30 to 40 s, because slope in this interval maximum
v2  v1 60  20
a   4 ms 2
t2  t1 40  30
EX.85. The velocity-time graph of a body is shown in figure. The ratio of magnitude of average
acceleration during the intervals OA and AB is
–1
v(ms )
A) 1
B) 1/2
D C
40
C) 1/3
D) 3 30° E 60°
O A B t(s)
0 1
Sol. During OA , acceleration = tan 30  ms 2
3
During AB , acceleration =  tan 600   3 ms 2 .
1/ 3 1
required ratio = 
3 3
EX.86. The following graph shows the variation of velocity of a rocket with time. Then the
maximum height attained by the rocket is
v(ms–1)
1000
120
t(s)
0 10
100
A) 1.1 km B) 5 km C) 55 km D) None
Sol. Maximum height will be attained at 110 s. Because after 110 s, velocity becomes negative and
1
rocket will start coming down. Area from 0 to 110 s is  110  1000  55, 000 m  55 km
2
key-C
EX.87. The velocity-time graph of a particle moving in a straight line is shown in figure. The
acceleration of the particle at t  9 is
–1
v(ms )
15
10
t(s)
2 4 6 8 10 12
A) Zero B) 5 ms 2 C) 5 ms 2 D) 2 ms 2
v2  v1 5  15 2
Sol. Acceleration between 8 to 10 s( or at t= 9 s) . a  t  t  10  8  5m / s
2 1
key-C
EX.88. A ball is dropped vertically from a height d above the ground. It hits the ground and
d
bounces up vertically to a height . Neglecting subsequent motion and air resistance, its
2
velocity v varies with height h above the ground as: [2004]
V
V
d
h
A) B)
d h
V
V
d
h d
h
C) D)
Sol. (i) For uniformly accelerated/decelerated motion

v2= u2  2gh
i.e., v – h graph will be a parabola Because equation is quadratiC).
(ii) Initially velocity is downwards (–ve) and then after collision it reverses its direction with lesser
magnitude. I.e., velocity is upwards (+ve). Graph A) satisfies both these conditions.
Therefore, correct answer is A).
Note that time t = 0 corresponds to the point on the graph where h = d
v
at t = 0, h = d
2
d 1  2: increases
h downwards
3 1 At 2 velocity changes
Collision takes its direction
place here 2 2 3V decreases upwards
MOTION IN A STRAIGHT LINE
TOPIC-1 ….Distance, Displacement & Uniform Motion

1. A particle is moving with speed 𝑣 = 𝑏√𝑥 along positive 𝑥‐axis. Calculate the speed of the
particle at time 𝑡 = 𝜏(assume that the particle is at origin at t = 0).
[12 Apr. 2019 II]
𝑏 2𝛤 𝑏 2𝛤 𝑏 2𝛤
(a) (b) (c) 𝑏 2 𝜏 (d)
4 2 √2
sol.  b  Given, v  b x
or dx / dt  bx (1/2)
x t
or x 1/2 dx  bdt
0 0
or x (1/2)
/ (1/ 2)  6t
b 2t 2
or x 
4
Differentiating w. r. t. time, we get
𝑑𝑥 𝑏 2 ×2𝑡
= (𝑡 = 𝜏)
𝑑𝑡 4
𝑏 2𝛤
or 𝑣 = 2
2. All the graphs below are intended to represent the same motion. One of them does it incorrectly.
Pick it up.
[2018]
(a) (b) (c) (d)

sol. (b) Graphs in option (c) position‐time and option (a) velocity‐position are corresponding to
velocity‐time graph option (d) and its distance‐time graph is as given below. Hence distance‐time
graph option (b) is incorrect.
3. A car covers the first half of the distance between two places at 40 km/h and other half at 60
km/h. The average speed of the car is [Online May 7, 2012]
(a) 40 km/h (b) 45 km/h (c) 48km/h (d) 60km/h
Tota1distancetrave11ed x
sol. (c) Average speed = =T
Tota1timetaken
x
= x x = 48km/h
+
2 × 40 2 × 60
4. The velocity of a particle is 𝑣 = 𝑣0 + 𝑔𝑡 + 𝑓𝑡 2 . If its position is 𝑥 = 0 at 𝑡 = 0, then its

displacement after unit time (𝑡 = 1) is [2007]
(a) 𝑣0 + 𝑔𝑙2 + 𝑓 (b) 𝑣0 + 2𝑔 + 3𝑓 (c) 𝑣0 + 𝑔𝑙2 + 𝑓𝑙3 (d) 𝑣0 + 𝑔 + 𝑓
𝑑𝑥
sol. (c) We know that, 𝑣 = 𝑑𝑡
⇒ 𝑑𝑥 = 𝑣𝑑𝑖
𝑥 𝑡
Integrating, ∫0 𝑑 𝑥 = ∫0 𝑣 𝑑𝑡
𝑡
𝑡 𝑔𝑡 2 𝑓𝑡 3
or 𝑥 = ∫0 (𝑣0 + 𝑔𝑡 + 𝑓𝑡 2 ) 𝑑𝑡 = [𝑣0 𝑡 + + ]
2 3 0
𝑔𝑡 2 𝑓𝑡 3
or, 𝑥 = 𝑣0 𝑡 + +
2 3
𝑔 𝑓
At 𝑡 = 1, 𝑥 = 𝑣0 + 2 + 3
5. A particle located at 𝑥 = 0 at time 𝑡 = 0, starts moving along with the positive 𝑥‐direction with
a velocity’v’ that varies as 𝑣 = 𝛼 √𝑥. The displacement of the particle varies with time as [2006]
(a) 𝑡 2 (b) 𝑡 (c) 𝑡1/2 (d) 𝑡 3
sol. (a) 𝑣 = 𝛼 √𝑥,
𝑑𝑥 𝑑𝑥
⇒ = 𝛼 √𝑥 ⇒ = 𝛼𝑑𝑡
𝑑𝑡 √𝑥
Integrating both sides,

𝑥 𝑡 𝑥
𝑑𝑥 2√𝑥
∫ = 𝛼 ∫ 𝑑 𝑡; [ ] = 𝛼[𝑡]𝑡0
0 √𝑥 0 1 0
𝛼2 2
⇒ 2√𝑥 = 𝛼𝑡 ⇒ 𝑥 = 𝑡
4
TOPIC-2 ….Non-Uniform Motion
6. The velocity(v) and time (t) graph of a body in a straight line motion is shown in the figure. The
point 𝑆 is at 4.333 seconds. The total distance covered by the body in 6 s is:
[05 Sep. 2020 (II)]
37 49
(a) m (b) 12 m (c) 11 m (d) m
3 4
sol. (a)
1 13
𝑂𝑆 = 4 + =
3 3
1 5
𝑆𝐷 = 2 − =
3 3
Distance covered by the body = area of v‐t graph = ar(𝑂𝐴𝐵𝑆) + ar (SCD)
1 13 1 5 32 5 37
= ( + 1) × 4 + × × 2 = + = m
2 3 2 3 3 3 3
8. The distance x covered by a particle in one dimensional motion varies with time 𝑡 as 𝑥 2 =
𝑎𝑡 2 + 2𝑏𝑡 + 𝑐. If the acceleration of the particle depends on 𝑥 as 𝑥𝑛, where 𝑛 is an integer, the
value of n is .
[NA 9 Jan 2020 I]
sol. (3) Distance X varies with time t as 𝑥 2 = 𝑎𝑡 2 + 2𝑏𝑡 + 𝑐
𝑑𝑥
⇒ 2𝑥 = 2𝑎𝑡 + 2𝑏
𝑑𝑡
𝑑𝑥 𝑑𝑥 (𝑎𝑡 + 𝑏)
⇒𝑥 = 𝑎𝑡 + 𝑏 ⇒ =
𝑑𝑡 𝑑𝑡 𝑥
𝑑 2𝑋 𝑑𝑥 2
⇒𝑥 2+( ) =𝑎
𝑑𝑡 𝑑𝑡
𝑑𝑥 2 𝑎𝑡 + 𝑏 2
𝑑2𝑋 𝑎−( ) 𝑎−( 𝑥 )
⇒ = 𝑑𝑡 =
𝑑𝑡 2 𝑥 𝑥
𝑎𝑥 2 − (𝑎𝑡 + 𝑏)2 𝑎𝑐 − 𝑏 2
= =
𝑥3 𝑥3
⇒ a ∝ x 3 Hence, n = 3
9. A bullet of mass 20g has an initial speed of 1 m/s, just before it starts penetrating a mud wall
ofthickness 20 cm. Ifthe wall offers a mean resistance of 2.5 × 10−2 N, the speed ofthe bullet after
emerging fii om the other side of the wall is close to: [10Apr. 2019 II]
(a) 0.1 m/s (b) 0.7 m/s (c) 0.3 m/s (d) 0.4 m/s
sol. (b) From the third equation of motion
v 2 − u2 = 2aS
But, 𝑎 = F/m
𝐹
𝑣 2 = 𝑢2 − 2 ( ) 𝑆
𝑚
2.5 × 10−2 20
⇒ 𝑣 2 = (1)2 − (2) [ ]
20 × 10−3 100
1
⇒ v2 = 1 −
2
1
⇒𝑣=
m/s = 0.7m/s
√2
10. The position of a particle as a function of time 𝑡, is given by 𝑥(𝑡) = 𝑎𝑡 + 𝑏𝑡 2 − 𝑐𝑡 3
where, 𝑎, 𝑏 and 𝑐 are constants. When the particle attains zero acceleration, then its velocity
will be:
𝑏2 𝑏2 𝑏2 𝑏2
(a) 𝑎 + 4𝑐 (b) 𝑎 + 3𝑐 (c) 𝑎 + (d) 𝑎 + 2𝑐
𝑐
sol. (b) 𝑥 = 𝑎𝑡 + 𝑏𝑡 2 − 𝑐𝑡 3
𝑑𝑥 𝑑
Velocity, 𝑣 = = 𝑑𝑡 (𝑎𝑡 + 𝑏𝑡 2 + 𝑐𝑡 3 )
𝑑𝑡
= 𝑎 + 2𝑏𝑡 − 3𝑐𝑡 2
𝑑𝑣 𝑑
Acceleration, = 𝑑𝑡 (𝑎 + 2𝑏𝑡 − 3𝑐𝑡 2 )
𝑑𝑡
𝑏
or 0 = 2𝑏 − 3𝑐 × 2𝑡 ⋅. 𝑡 = (3𝑐)
𝑏 𝑏 2
and 𝑣 = 𝑎 + 2𝑏 (3𝑐) − 3𝑐 (3𝑐)
11. A particle starts fii om origin O from rest and moves with a uniform acceleration along the
positive 𝑥‐axis. Identify all figures that correctly represents the motion qualitatively
(𝑎 =acceleration, 𝑣 =velocity,𝑥 =displacement, 𝑡 =time)
[8 Apr. 2019 II]
(a) (B), (C) (b) (A) (c) (A), (B), (C) (d) (A), (B), (D)
sol. (d) For constant acceleration, there is straight line parallel to t‐axis on 𝑎 − 𝑡.
Inclined straight line on 𝑣 − 𝑡 , and parabola on 𝑥 − 𝑡.
12. A particle starts from the origin at time t = 0 and moves along the positive 𝑥‐axis. The graph of
velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s?
[10 Jan. 2019 II]
(a) 10 m (b) 6 m (c) 3m (d) 9m

sol. (d) Position of the particle,
S = area under graph (time t = 0 to 5s)
1
= × 2 × 2 + 2 × 2 + 3 × 1 = 9m
2
13. In a car race on straight road, car A takes a time t less than car B at the finish and passes
finishing point with a speed v more than of car B. Both the cars start from rest and travel with
constant acceleration a1 and a2 respectively. Then v is equal to:
[9 Jan. 2019 II]
2a1 a2 a1 +a2
(a) t (b) √2a1 a2 t (c) √a1 a2 t (d) t
a1 +a2 2
sol. (c) Let time taken by A to reach finishing point is t 0 Time taken by B to reach finishing point
= t0 + t
u=0
vA = a1 t 0
vB = a2 (t 0 + t)
vA − vB = v
⇒ v = a1 t 0 − a2 (t 0 + t) = (a1 − a2 )t 0 − a2 t … (i)
1 2 1
xB = xA = a t = a (t + t)2
2 1 0 2 2 0
⇒ √a1 t0 = √a2 (t 0 + t)
⇒ (√a1 − √a2 )t 0 = √a2 t
√a 2 t
⇒ to =
√a1 − √a 2
Putting this value of t 0 in equation (i)
√a 2 t
v = (a1 − a2 ) − a2 t
√a1 − √a 2
= (√a1 + √a2 )√a2 t − a2 t = √a1 a2 t + a2 t − a2 t
or, v = √a1 a2 t
14. An automobile, travelling at 40 km/h, can be stopped at a distance of 40 m by applying brakes.

If the same automobile is travelling at 80 kmph, the minimum stopping distance, in metres, is
(assume no skidding)
(a) 75m (b) 160m (c) 100m (d) 150m
sol. (b) According to question, 𝑢1 = 40 kmni, 𝑣1 = 0 and 𝑠1 = 40m
using 𝑣 2 − 𝑢2 = 2𝑎𝑠; 02 − 402 = 2𝑎 × 40 (i)
Again, 02 − 802 = 2𝑎𝑠 (ii)
From eqn. (i) and(ii)
Stopping distance, 𝑠 = 160m
15. The velocity‐time graphs of a car and a scooter are shown in the figure. (i) the difference between
the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will
catch up with the scooter are, respectively
0 5 10 15 20 25
Time in (s) →
(a) 337.5m and 25 s (b) 225.5m and 10 s

(b) 112.5m and 22.5 s (d) 112.5m and 15 s
𝑣−𝑢 1
sol. (c) Using equation, 𝑎 = and 𝑆 = 𝑢𝑡 + 2 𝑎𝑡 2
𝑡
1 (45)
Distance travelled by car in 15 sec = 2 (15)2
15
675
= m
2
Distance travelled by scooter in 15 seconds = 30 × 15 = 450 (distance =speed ×time)
Difference between distance travelled by car and scooter in 15 sec, 450 − 337.5 = 112.5m
Let car catches scooter in time t;
675
+ 45(𝑡 − 15) = 30𝑡
2
337.5 +45t − 675 = 30t ⇒ 15t = 337.5
⇒ t = 22.5 sec
16. A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a
point P in a field at a distance d from highway (point M) as shown in the figure. Speed of the
car in the field is half to that on the highway. What should be the distance RM, so that the time
taken to reach P is minimum?
d d d
(a) (b) (c) (d) d
√3 2 √2
sol. (a) Let the car turn of the highway at a distance x from the point M. So, RM = 𝑥
And if speed of car in field is 𝑣, then time taken bythe car to cover the distance 𝑄𝑅 = 𝑄𝑀 −
𝑥 on the highway,
𝑄𝑀−𝑥
𝑡1 = (i)
2𝑣
Time taken to travel the distance 𝑅𝑃′ in the field

√𝑑2 +𝑥 2
𝑡2 = ..... (ii)
𝑣
𝑄𝑀−𝑥 √𝑑2 +𝑥 2
Total time elapsed to move the car from 𝑄 to 𝑃𝑡 = 𝑡1 + 𝑡2 = +
2𝑣 𝑣
𝑑𝑡
For’t’ to be minimum =0
𝑑𝑥
1 1 𝑥
[− + ]=0
𝑣 2 √𝑑2 + 𝑥 2
𝑑 𝑑
or 𝑥 = √22 =
−1 √3
17. Which graph corresponds to an object moving with a constant negative acceleration and a positive
velocity?
sol. (c) According to question, object is moving with constant negative acceleration
i.e., a = − constant (C)
vdv
= −C
dx
vdv =‐Cdx
v2 v2 k
= −Cx + k x = − 2C + C
2
Hence, graph (3) represents correctly.
18. The distance travelled by a body moving along a line in time 𝑡 is proportional to 𝑡 3 . The
acceleration‐time (𝑎, 𝑡) graph for the motion of the body will be
sol. (b) Distance along a line i.e., displacement (s) = 𝑡 3 (𝑠 ∝ 𝑡 3 given)
By double differentiation of displacement, we get acceleration.
𝑑𝑠 𝑑𝑡 3 𝑑𝑣 𝑑3𝑡 2
𝑉 = 𝑑𝑡 = = 3𝑡 2 and 𝑎 = = = 6𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡
𝑎 = 6𝑡 or 𝑎 ∝ 𝑡
Hence graph (b) is correct.
19. The graph of an object’s motion (along the x‐axis) is shown in the figure. The instantaneous
velocity of the object at points 𝐴 and 𝐵 are 𝑣𝐴 and 𝑣𝐵 respectively. Then
(a) 𝑣𝐴 = 𝑣𝐵 = 0.5m/s (b) 𝑣𝐴 = 0.5m/s < 𝑣𝐵

(c) 𝑣𝐴 = 0.5m/s > 𝑣𝐵 (d) 𝑣𝐴 = 𝑣𝐵 = 2m/s
𝛥𝑥
sol. (a) Instantaneous velocity 𝑣 = 𝛥𝑡
𝛥𝑥𝐴 4𝑚
From graph, 𝑣𝐴 = 𝛥𝑡𝐴
= 8𝑠
= 0.5m/s
𝛥𝑥𝐵 8𝑚
and 𝑣𝐵 = = 16𝑠 = 0.5m/s
𝛥𝑡𝐵
i.e., 𝑣𝐴 = 𝑣𝐵 = 0.5m/s
𝑑𝑣
20. An object, moving with a speed of 6.25m/s, is decelerated at a rate given by = −2.5√𝑣
𝑑𝑡
where 𝑣 is the instantaneous speed. The time taken by the object, to come to rest, would be:
[2011]
(a) 2 s (b) 4 s (c) 8 s (d) 1 s
𝑑𝑣
sol. (a) Given, = −2.5√𝑣
𝑑𝑡
𝑑𝑣
⇒ = −2.5𝑑𝑡
√𝑣
Integrating,
1 0
𝑣 + /2
⇒[ ] = −2.5[𝑡]𝑡0
(/2 )
625
⇒ −2(6.25)1 /2 = −2.5𝑡
⇒ −2 × 2.5 = −2.5𝑡
⇒ 𝑡 = 25
21. A body is at rest at 𝑥 = 0. At 𝑡 = 0, it starts moving in the positive x‐direction with a constant
acceleration. At the same instant another body passes through 𝑥 = 0 moving in the positive
x‐direction with a constant speed. The position of the first body is given by 𝑥1 (𝑡) after time ‘t’;
and that of the second body by 𝑥2 (𝑡) after the same time interval. Which of the following graphs
correctly describes (𝑥1 − 𝑥2 ) as a function of time ‘t’?
[2008]
sol. (b) For the body starting from rest, distance travelled (𝑥1 ) is given by
1
𝑥1 = 0 + 𝑎𝑡 2
2
1
⇒ 𝑥1 = 2 𝑎𝑡 2
For the body moving with constant speed
𝑥2 = 𝑣𝑡
1 2
𝑥1 − 𝑥2 = 𝑎𝑡 − 𝑣𝑡
2
at 𝑡 = 0, 𝑥1 − 𝑥2 = 0
This equation is of parabola.
𝑣
For < 𝑎 ; the slope is negative
𝑣
For = 𝑎 ; the slope is zero
𝑣
For > 𝑎 ; the slope is positive
These characteristics are represented by graph (b).
22. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant
𝑓
speed for time t and then decelerates at the rate to come to rest. If the total distance traversed is
2
15 S , then
[2005]
1 1 1
(a) 𝑆 = 6 𝑓𝑡 2 (b) 𝑆 = 𝑓𝑡 (c) 𝑆 = 4 𝑓𝑡 2 (d) 𝑆 = 72 𝑓𝑡 2
22. (d) Let car starts from 𝐴 from rest and moves up to point 𝐵 with acceleration 𝑓.
1
Distance, 𝐴𝐵 = 𝑆 = 2 𝑓𝑡12
Distance, 𝐵𝐶 = (𝑓𝑡1 )𝑡
𝑢2 (𝑓𝑡 )2
1
Distance, 𝐶𝐷 = 2𝑎 = 2(𝑓/2) = 𝑓𝑡12 = 2𝑆
Total distance, 𝐴𝐷 = 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 = 15𝑆

𝐴𝐷 = 𝑆 + 𝐵𝐶 + 2𝑆
⇒ 𝑆 + 𝑓𝑡1 𝑡 + 2𝑆 = 15𝑆
⇒ 𝑓𝑡1 𝑡 = 12𝑆 (i)
1
𝑓𝑡12 = 𝑆 ............ (ii)
2
t
Dividing (i) by (ii), we get 𝑡1 = 6
1 𝑡 2 𝑓𝑡 2
⇒ 𝑆 = 𝑓( ) =
2 6 72
23. A particle is moving eastwards with a velocity of 5 ms −1 . In 10 seconds the velocity changes to 5
ms−1 northwards. The average acceleration in this time is
[2005]
1 1
(a) ms−2 towards north (b) ms−2 towards north‐ east
2 √2
1
(c) ms −2 towards north‐ west (d) zero
√2
sol. (c) 𝑣2
Initial velocity, 𝑣1 = 5𝑖̂,

Final velocity, 𝑣2 = 5𝑗̂,
Change in velocity 𝛥𝑣 = (𝑣2 − 𝑣1 )
= √𝑣12 + 𝑣22 + 2𝑣1 𝑣2 cos 90
= √52 + 52 + 0 = 5√2m/s
[As |𝑣1 | = |𝑣2 | = 5m/s]
𝛥𝑣
Avg. acceleration = 𝑡
5√2 1
= = m/s 2
10 √2
5
tan 𝜃 =
= −1
−5
which means 𝜃 is in the second quadrant. (towards north‐west)
24. The relation between time t and distance x is 𝑡 = 𝑎𝑥 2 + 𝑏𝑥 where 𝑎 and 𝑏 are constants. The
acceleration is
[2005]
(a) 2𝑏𝑣 3 (b) −2𝑎𝑏𝑣 2 (c) 2𝑎𝑣 2 (d) −2𝑎𝑣 3
sol. (d) Given, 𝑡 = 𝑎𝑥 2 + 𝑏𝑥;
Diff. with respect to time(t)
𝑑 𝑑 𝑑𝑥 𝑑𝑥
(𝑡) = 𝑎 (𝑥 2 ) + 𝑏 = 𝑎. 2𝑥 + b. v.
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
⇒ 1 = 2𝑎𝑥𝑣 + 𝑏𝑣 = 𝑣(2𝑎𝑥 + 𝑏)(𝑣 =velocity)
1
2𝑎𝑥 + 𝑏 = .
𝑣
Again differentiating, we get
𝑑𝑥 1𝑑𝑣
2𝑎 +0=− 2
𝑑𝑡 𝑣 𝑑𝑡
𝑑𝑣 𝑑𝑥
⇒𝑎= = −2𝑎𝑣 3 (⋅.⋅ 𝑑𝑡 = 𝑣)
𝑑𝑡
25. An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If
the car is going twice as fast i.e., 120 km/h, the stopping distance will be
[2004]
(a) 60m (b) 40m (c) 20m (d) 80m
sol. (d) In first case speed,
5 50
𝑢 = 60 × m/s = m/s
18 3
𝑑 = 20m,
Let retardation be a then
0−𝑢2 =−2𝑎𝑑
2
or 𝑢 = 2𝑎𝑑 … (i)
5
In second case speed, 𝑢′ = 120 × 18
100
= m/s
3
and 0 − 𝑢′2 = −2𝑎𝑑’
or 𝑢′2 = 2𝑎𝑑′ … (ii)
(ii) divided by (i) gives,
𝑑′
4= ⇒ 𝑑 ‘ = 4 × 20 = 80m
𝑑
26. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same
car is moving at a speed of 100 km/hr, the minimum stopping distance is
[2003]
(a) 12m (b) 18m (c) 24m (d) 6m
sol. (c) Fir first case : Initial velocity,
5
𝑢 = 50 × m/s,
18
𝑣 = 0, s = 6m, 𝑎 = 𝑎
Using, 𝑣 2 − 𝑢2 = 2𝑎𝑠
2
5 2
⇒ 0 − (50 × ) = 2 × 𝑎 × 6
18
5 2
⇒ − (50 × ) = 2 × 𝑎 × 6
18
250 × 250
𝑎=− ≈= −16ms−2 .
324 × 2 × 6
Case‐2 : Initial velocity, 𝑢 = 100km/hr
5
= 100 × m/ sec
18
𝑣 = 0, 𝑠 = 𝑠, 𝑎 = 𝑎
As 𝑣 2 − 𝑢2 = 2𝑎𝑠
5 2
⇒ 02 − (100 × ) = 2𝑎𝑠
18
5 2
⇒ − (100 × ) = 2 × (−16) × 5
18
500 × 500
𝑠= = 24m
324 × 32
27. If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it
penetrate more before coming to rest?
[2002]
(a) 1 cm (b) 2 cm (c) 3 cm (d) 4cm.
sol. (a) In first case
𝑢
𝑢1 = 𝑢; 𝑣1 = 2, 𝑠1 = 3 cm, 𝑎1 =?
Using, 𝑣12 − 𝑢12 = 2𝑎1 𝑠1 (i)

𝑢 2
( ) − 𝑢2=2×𝑎×3
2
−𝑢2
⇒ 𝑎=8
In second case: Assuming the same retardation
−𝑢2
𝑢2 = 𝑢𝑙2; 𝑣2 = 0; =? 𝑎2 =
8
𝑣22 − 𝑢22 = 2𝑎2 × 𝑠2 (ii)
𝑢2 −𝑢2
0− = 2( ) × 𝑠2
4 8
⇒ 𝑠2 = 1 cm
28. Speeds of two identical cars are 𝑢 and 4𝑢 at the specific instant. The ratio ofthe respective
distances in which the two cars are stopped from that instant is
[2002]
(a) 1 : 1 (b) 1 :4 (c) 1 :8 (d) 1 : 16
sol. (d) For first car
𝑢1 = 𝑢, 𝑣1 = 0, 𝑎1 = −𝑎, 𝑠1 = 𝑠1
As 𝑣12 − 𝑢12 = 2𝑎1 𝑠1
⇒ −𝑢2 = −2𝑎𝑠1
⇒ 𝑢2 = 2𝑎𝑠1
⇒ 𝑠1 = u/2a (i)
For second car
𝑢2 = 4𝑢, 𝑣1 = 0, 𝑎2 = −𝑎, 𝑠2 = 𝑠2
𝑣22 − 𝑢22 = 2𝑎2 𝑠2
⇒ −(4𝑢)2 = 2(−𝑎)𝑠2
⇒ 16𝑢2 = 2𝑎𝑠2
8𝑢2
⇒ 𝑠2 = (ii)
𝑎
Dividing(i) and(ii),
𝑠1 𝑢2 𝑎 1
= ⋅ 2=
𝑠2 2𝑎 8𝑢 16
TOPIC-3 …..Relative Velocity
29. Train 𝐴 and train 𝐵 are running on parallel tracks in the opposite directions with speeds of 36
km/hour and 72 km/hour, respectively. A person is walking in train 𝐴 in the direction opposite
to its motion with a speed of 1.8 km/hour. Speed (in ms−1 ) ofthis person as observed from train
𝐵 will be close to : (take the distance between the tracks as negligible)
[2 Sep. 2020 (I)]
(a) 29.5 ms −1 (b) 28.5 ms −1 (c) 31.5 ms −1q (d) 30.5 ms −1
sol. (a) According to question, train 𝐴 and 𝐵 are running on parallel tracks in the opposite direction.
𝑉𝐴 = 36km/h = 10m/s
72k
𝑉𝐵 = −72 kimh = −20m/s

𝑉𝑀𝐴 = −1.8knyh = −0.5m/s
𝑉man,𝐵 = 𝑉man,𝐴 + 𝑉𝐴,𝐵
= 𝑉 + 𝑉 − 𝑉 = −0.5 + 10 − (−20)
man, 𝐴 𝐴 𝐵
= −0.5 + 30 = 29.5m/s.
30. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train oflength
120 m travels at a speed of 30 km/h. The ratio of times taken by the passenger train to
completely cross the freight train when: (i) they are moving in same direction, and (ii) in the
opposite directions is:
[12 Jan. 2019 II]
11 5 3 25
(a) (b) (c) (d)
5 2 2 11
sol. (a)
31. A person standing on an open ground hears the sound of a jet aero plane, coming from north at an
angle 60o with ground level. But he finds the aero plane right vertically above his position. If 𝑣
is the speed of sound, speed of the plane is:
[12 Jan. 2019 II]
√3 2𝑣 𝑣
(a) 𝑣 (b) (c) 𝑣 (d)
2 √3 2
sol. (d)
R (Observer)
Distance, PQ = vp × t (Distance =speed ×time) Distance, QR = V. t
PQ
cos 60∘ =
QR
1 vp × t v
= ⇒ vp =
2 V. t 2
32. A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same
instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has
acceleration 4 m/s2 . The car will catch up with the bus after a time of:
(a) √110s (b) √120s (c) 10√2s (d) 15 s
sol. (c) → 4m/sec → 2m/sec 2
2
200 m
Given, uC = uB = 0, aC = 4m/s2 , aB = 2m/s 2
hence relative acceleration, aCB = 2m/sec 2
1
Now, we know, s = ut + 2 at 2
1
200 = 2 × 2t 2 u = 0
Hence, the car will catch up with the bus after time t = 10√2 second
33. A person climbs up a stalled escalator in 60 s. If standing on the same but escalator running with
constant velocity he takes 40 s. How much time is taken by the person to walk up the moving
escalator?
(a) 37 s (b) 27 s (c) 24 s (d) 45 s
1
sol. (c) Person’s speed walking only is
60
1
Standing the escalator without walking the speed is
40
Walking with the escalator going, the speed add.
1 1 15
So, the person’s speed is + 40 = 120
60
120
So, the time to go up the escalator t = = 24 second.
5
34. A goods train accelerating uniformly on a straight railway track, approaches an electric pole
standing on the side of track. Its engine passes the pole with velocity 𝑢 and the guard’s room
passes with velocity 𝑣. The middle wagon of the train passes the pole with a velocity.
𝑢+𝑣 1
(a) 2
(b) 2
√𝑢2 + 𝑣 2 (c) √𝑢𝑣 (d)
Sol. (d) Let 𝑆 be the distance between two ends a be the constant acceleration
As we know 𝑣 2 − 𝑢2 = 2𝑎𝑆
v2  u 2
Or aS 
2
Let 𝑣 be velocity at mid point.
𝑆
Therefore, 𝑣𝑐2 − 𝑢2 = 2𝑎 2
𝑣𝑐2 = 𝑢2 + 𝑎𝑆
𝑣 2 − 𝑢2
𝑣𝑐2 = 𝑢2 +
2
𝑢2 + 𝑣 2
𝑣𝑐 = √
2
TOPIC-4 ….Motion Under Gravity
35. A helicopter rises from rest on the ground vertically upwards with a constant acceleration 𝑔. A
food packet is dropped from the helicopter when it is at a height ℎ. The time taken by the packet
to reach the ground is close to [𝑔 is the accelertion due to gravity] :
[5 Sep. 2020 (I)]
2 ℎ ℎ ℎ 2ℎ
(a) 𝑡 = 3 √(𝑔) (b) 𝑡 = 1.8√𝑔 (c) 𝑡 = 3.4√(𝑔) (d) 𝑡 = √3𝑔
sol. (c) For upward motion of helicopter,

𝑣 2 = 𝑢2 + 2𝑔ℎ ⇒ 𝑣 2 = 0 + 2𝑔ℎ ⇒ 𝑣 = √2𝑔ℎ
Now, packet will start moving under gravity.
Let ’t’ be the time taken by the food packet to reach the ground.
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
1 1
⇒ −ℎ = √2𝑔ℎ𝑡 − 𝑔𝑡 2 ⇒ 𝑔𝑡 2 − √2𝑔ℎ𝑡 − ℎ = 0
2 2
𝑔
√2𝑔ℎ±√2𝑔ℎ+4× 2 ×ℎ
or, 𝑡 = 𝑔
2×
2
2𝑔ℎ 2ℎ
or, 𝑡 = √ (1 + √2) ⇒ 𝑡 = √ 𝑔 (1 + √2)
𝑔
ℎ
or, 𝑡 = 3.4√𝑔
36. A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds
ℎ
and reaches height 2. The velocity versus height of the ball during its motion may be represented
graphically by: (graph are drawn schematically and on not to scale)

[4 Sep. 2020 (I)]
sol. (c) For uniformly accelerated/ deaccelerated motion :

𝑣 2 = 𝑢2 ± 2𝑔ℎ
As equation is quadratic, so, v‐h graph will be a parabola
Initially velocity is downwards (‐ve) and then after collision it reverses its direction with lesser
magnitude, i. e. velocity is upwards (+ve) .
Note that time 𝑡 = 0 corresponds to the point on the graph where ℎ = 𝑑.
Next time collision takes place at 3.
1
37. A ball is dropped from the top of a 100 m high tower on a planet. In the last s before hitting
2
the ground, it covers a distance of 19 m. Acceleration due to gravity (in m/s 2) near the surface on
that planet is .
[NA 8 Jan. 2020 II]
sol. (08.00) Let the ball takes time t to reach the ground
1
Using, 𝑆 = 𝑢𝑡 + 2 𝑔𝑡 2
1
⇒ 𝑆 = 0 × 𝑡 + 𝑔𝑡 2
2
200
⇒ 200 = 𝑔𝑡 2 [2𝑆 = 100𝑚] ⇒ 𝑡 = √ ...(i)
𝑔
1 1
In last 𝑠, body travels a distance of19 𝑚, so in (𝑡 − 2) distance travelled = 81
2
1 1 2
Now, 𝑔 (𝑡 − 2) = 81
2
1 2
𝑔 (𝑡 − ) = 81 × 2
2
1 81 × 2
⇒ (𝑡 − ) = √
2 𝑔
1 1
= (√200 − √81 × 2) using (i)
2 √𝑔
⇒ √𝑔 = 2(10√2 − 9√2)
⇒ √𝑔 = 2√2
g = 8𝑚𝑙𝑠 2
38. A body is thrown vertically upwards. Which one of the following graphs correctly represent the
velocity vs time?
[2017]
sol. (a) For a body thrown vertically upwards acceleration remains constant (𝑎 = −𝑔)
and velocity at anytime 𝑡 is given by 𝑉 = 𝑢 − 𝑔𝑡
During rise velocity decreases linearly and during fall velocity increases linearly and direction is
opposite to each other.
Hence graph (a) correctly depicts velocity versus time.
39. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed
of10 m/s and 40 m/s respectively. Which of the following graph best represents the time
variation of relative position of the second stone with respect to the first?
(Assume stones do not rebound after hitting the ground and neglect air resistance, take g =
10m/s 2 )
[2015]
(The figures are schematic and not drawn to scale)
sol. (b) y1 = 10t − 5t 2 ; y2 = 40t − 5t 2 for y1 = −240m, t = 8s

y2 − y1 = 30t for t < 8s.
for t > 8s,
1
y2 − y1 = 240 − 40t − gt 2
2
40. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken
by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path.
The relation between H, u and n is:
[2014]
(a) 2gH = n2 u2 (b) gH = (n − 2)2 u2 d (c) 2gH = nu2 (n‐2) (d) gH = (n − 2)u2
sol. (c) Speed on reaching ground
𝑣 = √𝑢2 + 2𝑔ℎ
Now, 𝑣 = 𝑢 + 𝑎𝑡
⇒ √𝑢2 + 2𝑔ℎ = −𝑢 + 𝑔𝑡
41. Consider a rubber ball freely falling from a height ℎ = 4.9m onto a horizontal elastic plate.
Assume that the duration ofcollision is negligible and the collision with the plate is totally elastic.
Then the velocity as a function of time and the height as a function of time will be :
[2009]
sol. (b) For downward motion 𝑣 = −𝑔𝑡

The velocity of the rubber ball increases in downward direction and we get a straight line between
𝑣 and 𝑡 with a negative slope.
1
Also applying 𝑦 − 𝑦0 = 𝑢𝑡 + 2 𝑎𝑡 2
1 1
y  h   gt 2  y  h  gt 2
2 2
The graph between 𝑦 and 𝑡 is a parabola with 𝑦 = ℎ at 𝑡 = 0.
As time increases 𝑦 decreases.
For upward motion.
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity
is reversed and the magnitude remains the same.
Here 𝑣 = 𝑢 − 𝑔𝑡 where 𝑢 is the velocityjust after collision.
As 𝑡 increases, 𝑣 decreases. We get a straight line between 𝑣 and 𝑡 with negative slope.
1
Also y  ut  gt 2
2
All these characteristics are represented by graph (b).
42. A parachutist after bailing out falls 50m without friction. When parachute opens, it decelerates at 2
m/s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out?
[2005]
(a) 182 m (b) 91 m (c) 111 m (d) 293m
sol. (d) Initial velocity of parachute after bailing out,
𝑢 = √2𝑔ℎ
u = √2 × 98 × 50 = 14√5
The velocity atground v=3 m/s
𝑣 2 − 𝑢2 32 − 980
𝑆= = ≈ 243m
2×2 4
Initially he has fallen 50 m.
Total height from where he bailed out = 243 + 50 = 293m
43. A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground.
𝑇
What is the position of the ball at second
3
[2004]
8ℎ 7ℎ
(a) meters from the ground (b) meters from the ground
9 9
ℎ 17ℎ
(c) meters from the ground (d) meters from the ground
9 18
1
Sol. (a) We have 𝑠 = 𝑢𝑡 + 2 𝑔𝑡 2
1
⇒ ℎ = 0 × 𝑇 + 2 𝑔𝑇 2
1
⇒ ℎ = 2 𝑔𝑇 2
𝑇
Vertical distance moved in time is
3
1 𝑇 2 1 𝑔𝑇 2 ℎ
ℎ↑ = 𝑔 ( ) ⇒ ℎ ↑ = × =
2 3 2 9 9
ℎ 8ℎ
Position of ball from ground = ℎ − 9 = 9
44. From abuilding two balls A and B are thrown such that A is thrown upwards and 𝐵 downwards
(both vertically). If v𝐴 and v𝐵 are their respective velocities on reaching the ground, then
[2002]
(a) v𝐵 > v𝐴 (b) v𝐴 = v𝐵 (c) v𝐴 > v𝐵
(d) their velocities depend on their masses.
sol. (b)
Ball 𝐴 is thrown upwards with velocity 𝑢 from the building. During its downward journey when
it comes back to the point of throw, its speed is equal to the speed of throw (𝑢) . So, for the
journey of both the balls from point 𝐴 to 𝐵.
We can apply 𝑣 2 − 𝑢2 = 2𝑔ℎ.
As 𝑢, 𝑔, ℎ are same for both the balls, 𝑣𝐴 = 𝑣𝐵
Chapter 4 -- MOTION I N A PLANE
Relative Velocity
ur ur
 If body A is moving with a velocity VA w.r.t. ground and body B is moving with velocity VB w.r.t.
ground then
r r r
1) The relative velocity of body 'A' w.r.t. 'B' is given by VAB = VA -VB
r r r
2) The relative velocity of body 'B' w.r.t. 'A' is given by VBA = VB -VA
r r r r
3) Both VA -VB and VB -VA are equal in magnitude but opposite in direction.
r r r r
V AB = -V BA and V AB = VBA = VA 2 +VB 2 -2VA VB cosθ
4) For two bodies moving in same direction, magnitude of relative velocity is equal to the difference
r r
of magnitudes of their velocities. (θ = 00 , cos 0 = 1)  VAB =VA -VB , VBA =VB  VA
5) For two bodies moving in opposite directions, magnitude of relative velocity is equal to the sum
of the magnitudes of their velocities. (  1800 ; cos 1800  -1)
r r
 V AB  V BA = VB  VA
6) Relative displacement of A w.r.t. B is
r r r r
X A B  X A G  X BG Where X AG  displacement of ‘A’ w.r.t ground
r
and X BG  displacement of ‘B’ w.r.t ground
r r r
7) Relative velocity of A w.r.t. B is V AB  V AG  V BG
r r r
8) Relative acceleration of A w.r.t. B is a AB  a AG  a BG
9) Two trains of lengths l1 and l2 are moving on parallel tracks with speeds v1 and v2 (v1 > v2 )
w.r.t ground. The time taken to cross each other
Srel l1 + l2
when they move in same direction is t1 = V = v - v
rel 1 2
Srel l1 +l2
when they move in opposite direction is t 2 = V = v + v
rel 1 2
Application:
Relative Motion on a moving train
ur
If a boy in a train is running with velocity V BT relative to train and train is moving with velocity
ur ur
V TG relative to ground, then the velocity of the boy relative to ground V BG will be given by
ur ur ur
V BG  V BT  V TG
ur ur ur
So, if boy in a train is running along the direction of train. V BG  V BT  V TG
If the boy in train is running in a direction opposite to the motion of train, then
ur ur ur
V BG  V BT  V TG
EX.1: When two objects move uniformly towards each other, they get 4 metres closer each second
and when they move uniformly in the same direction with original speed, they get 4 metres
closer each 5s. Find their individual speeds.
Sol. Let their speeds be v1 and v2 and let v1 > v2.
In First case :
4
Relative velocity, v1 + v 2 = = 4 m / s ....(1)
1
In Second case:
4
Relative velocity = v1 - v 2 = = 0.8 m / s...(2)
5
solving eqns.(1) and (2), weget v1  2.4ms 1 , v2  1.6ms 1
EX.2: A person walks up a stationary escalator in time t1 . If he remains stationary on the escalator,
then it can take him up in time t2. How much time would it take for him to walk up the
moving escalator?
Sol. Let L be the length of escalator .
L
Speed of man w.r.t. escalator is v ME = t
1
L
Speed of escalator v E = t
2
1 1
Speed of man with respect to ground would be v M = v ME + v E = L  + 
 t1 t 2 
L t1 t 2
 The desired time is t = v = t + t .
M 1 2
EX.3: Two ships A and B are 10km apart on a line running south to north. Ship A farther north
is streaming west at 20km/h and ship B is streaming north at 20km/h. What is their distance
of closest approach and how long do they take to reach it?
Sol.
VB  20km / h
A C
10km V BA
450
V BA  20 2km / h B
450
VA  20 km / h
ur ur ur
Velocity of B relative to A is V BA  V B  V A
ur 2 2
V BA   20    20   20 2km / h
ur
i.e., V BA is 20 2 km/h at an angle of 450 from east towards north.
uuur
A is at rest and B is moving with VBA in the direction shown in Fig.
Therefore the minimum distance between ships
 1 
S min  AC  AB sin 450  10   km  5 2km
 2
BC 5 2 1
and time taken is t  Vur BA   h  15 min
20 2 4
Rain umbrella Concept

uv uv
 If rain is falling with a velocity V R and man moves with a velocity V M relative to ground, he will
uv uuv uuv
observe the rain falling with a velocity V RM  VR  VM .
uv
Case - I : If rain is falling vertically with a velocity V R and an observer is moving horizontally
uv
with velocity V M , then the velocity of rain relative to observer will be :
VR 
V RM
VR
 V M

VM VM
V RM  VR  VM
The magnitude of velocity of rain relative to man is VRM  VR2  VM2
If  is the angle made by the umbrella with horizontal, then, tan   VR

VM
If  is the angle made by the umbrella with vertical, then, tan   V M

VR
Case - II : When the man is moving with a velocity VM1 relative to ground towards east(positive

ur
x-axis), and the rain is falling with a velocity V R relative to ground by making an angle  with
r
vertical(negative z-axis). Then the velocity of rain relative to man V RM is as shown in figure.
1
r r
VR  VRx iˆ  VRy kˆ ; V M 1  V M 1 iˆ
V Rx  V M 1
and ta n   ...... ( 2 )
VRy
N

W E
VR  VR
M1 VR y S
VRM
1
VM 1 VM 1
VM 1
VR
x
ur
Case - III : If the man speeds up, at a particular velocity V M 2 , the rain will appear to fall vertically
ur ur ur uuur
with V RM 2 , then V RM 2  V R  VM 2 as shown in figure.
VRM
2
VRM VR
2
VM
2
VM VM
2 2
Case - IV : If the man increases his speed further, he will see the rain falling with a velocity as
shown in figure.
VRM
3
 
VRM
3 VR y
VM VR
3 VR x
VM 3
VM
3
ur ur ur VM  VR
tan   3 x
V RM 3  V R  V M 3 ; VR y
EX.4: Rain is falling vertically with a speed of 20ms 1 ., A person is running in the rain with a
velocity of 5 ms 1 and a wind is also blowing with a speed of 15 ms 1 (both from the west)
The angle with the vertical at which the person should hold his umbrella so that he may not
get drenched is :
r uur )
Sol. VRain  V R  20  k  
r r r r
V Man  V M  5 iˆ , VWind  VW  15 iˆ
r
Resultant velocity of rain and wind is VRW  20kˆ  15 iˆ
r r

Now, velocity of rain relative to man is VRW  VM  20kˆ  15iˆ  5iˆ  20kˆ  10iˆ   
vertical

20kˆ

10 iˆ
1 1
Tan     Tan 1
2 2
EX.5: To a man walking at the rate of 3km/h the rain appears to fall vertically.When he increases
his speed to 6km/h it appears to meet him at an angle of 450 with vertical. Find the angle
made by the velocity of rain with the vertical and its speed.
VRM 450  VR
y
V RM VR
Sol : 3 3
6
VM  3kmph
0 3 3
From the diagram Tan45  .........(1) and Tan  y ..............(2)
y
0 3 1 3
From (1) and (2)   450   sin 45  V , 2  V
R R
VR  3 2kmph
EX.6: Rain is falling, vertically with a speed of 1m/s .Wind starts blowing after sometime with
a speed of 1.732 m/s in east to west direction.In which direction should a boy waiting at a
bus stop hold his umbrella.?
vw N
 W E
vr
R
S
Sol. If R is the resultant of velocity of rain ( VR ) and velocity of wind ( VW ) then
2
R = v2r + v 2w = 12 + 1.732  ms-1 = 2ms-1
vw 3
The direction  that R makes with the vertical is given by tanθ = = = 600 Therefore,
vr 1
the boy should hold his umbrella in the vertical plane at an angle of about 600 with the vertical
towards the east.
EX.7: Rain is falling vertically with a speed of 1m/s . A woman rides a bicycle with a speed of
1.732 m/s in east to west direction. What is the direction in which she should hold her
umbrella ?
Sol. In Fig. vr represents the velocity of rain and vb , the velocity of the bicycle, the woman is riding.
Both these velocities are with respect to the ground.Since the woman is riding a bicycle, the
velocity of rain as experienced by
Vb N

W E
Vr
vrb
 vb S
here is the velocity of rain relative to the velocity of the bicycle she is riding. That is vrb = vr - vb
This relative velocity vector as shown in Fig. makes an angle  with the vertical.It is given by
v 3
Tan θ = b =  θ = 60o
v 1
r
Therefore,the woman should hold her umbrella at an angle of about 600 with the vertical towards
the west.
Motion of a Boat in the River
Boat motion is classified into three categories based on angle between VBR and VR they are
1) Down stream (   00 ):
VBR
VR
Resultant velocity of the boat = VBR  VR
The time taken for the boat to move a distance `d’ along the direction of flow of water is.
d
t1 
VBR  VR .............(1)
2) Up stream (   1800 ) :
Resultant velocity of the boat = VBR  VR
The time taken for the boat to move a distance `d’ opposite to the direction of flow of water is.
d
t2 
VBR  VR .......................(2)
t1 VBR  VR
From equation (1) and (2) t  V  V
2 BR R
time taken by person to go down stream a distance `d’ and come back is
d d
T  t1  t2  
VBR  VR VBR  VR
3) General approach :
Suppose the boat starts at point A on one bank with velocity VBR and reaches the other bank at
point D
C VR B D
VBRcos
d V BR

VBR sin  A
The component of velocity of boat anti parallel to the flow of water is VBR sin 
The component of velocity of boat perpendicular to the flow of water is VBR cos 
d
 The time taken by the boat to cross the river is, t  V cos 
BR
 Along the flow of water, distance travelled by the boat (or) drift is x  (VR  VBR Sin )t
 d 
x  (V R  V B R sin )  
 V B R cos 
 (a) The boat reaches the other end of the river to the right of B if VR >VBRsin 
(b) The boat reaches the other end of the river to the left of B if VR < VBRsin 
(c) The boat reaches the exactly opposite point on the bank if VR = VBRsin 
Motion of a Boat Crossing the River in Shortest Time
B VR C
VBR
d

A
ur ur
If V BR , V R are the velocities of a boat and river flow respectively then to cross the river in
shortest time, the boat is to be rowed across the river i.e., along normal to the banks of the river.
d
i) Time taken to cross the river, t= V where d = width of the river. This time is independent of
BR
velocity of the river flow
ii) Velocity of boat w.r.t. ground has a magnitude of VB = VBR2 +VR2
-1  VR 
iii) The direction of the resultant velocity is θ = tan   with the normal.
 VBR 
 d 
iv)The distance (BC) travelled downstream =VR   is called drift
 vBR 
Motion of a Boat Crossing the River in Shortest Distance
C VR B
VBR d
A
i) The boat is to be rowed upstream making some angle ' θ with normal to the bank of the river
V 
which is given by θ = sin-1  R 
 VBR 
ii) The angle made by boat with the river flow (or) bank is = 900 +θ
iii) Velocity of boat w.r.t. ground has a magnitude of VB = VBR2 -VR2
d
iv) The time taken to cross the river is t =
VBR2-VR2
Note : VBR = Relative velocity of the boat w.r.t river (or) velocity of boat in still water.
EX.8: A boat is moving with a velocity v bw = 5 km/hr relative to water. At time t = 0. the boat
passes through a piece of cork floating in water while moving down stream.If it turns back
at time t1  30 min.
a) when the boat meet the cork again ?
b) The distance travelled by the boat during this time.
t=0
Sol.
C
VW VW
Let AB =d is the distance travelled by boat along down stream in ` t1 ’ sec and it returns back and
it meets the cork at point C after ` t 2 ’ sec.
 Let AC=x is the distance travelled by the cork during  t1  t2  sec.
d  VB  VW  t1................(1)
d  x  VB  Vw  t2 .............  2 
and x  Vw  t1  t2  .............  3
Substitute (1) and (3) in (2) weget t1  t2
 The boat meets the cork again after T  2t1  60 min and the distance (AB+BC) travelled by
the boat before meets the cork is
D  2d  x
D  2(VB  Vw ) t1  Vw 2t1
D  2VB t1  2Vwt1  2Vwt1
30
D  2VB t1  2  5   5km
60
EX.9: A swimmer crosses a flowing stream of width `d’ to and fro normal to the flow of the river
in time t1 . The time taken to cover the same distance up and down the stream is t2. If t3. is
the time the swimmer would take to swim a distance 2d in still water, then relation between
t1 , t2. & t3.
Sol : Let v be the river velocity and u be the velocity of swimmer in still water. Then
d d 2ud
t2    2 2 ........(ii )
 d  u v u v u v
t1  2   .....(i) 2d
2 2
 u v  and t 3  ..............(iii )
u
from equation (i) ,(ii) and (iii) t12  t2t3  t1  t2t3

EX.10: Two persons P and Q crosses the river starting from point A on one side to exactly
opposite point B on the other bank of the river.The person P crosses the river in the shortest
path. The person Q crosses the river in shortest time and walks back to point B. Velocity of
river is 3 kmph and speed of each person is 5kmph w.r.t river.If the two persons reach the
point B in the same time, then the speed of walk of Q is.
For person (P) : For person(Q) :
C B B x C
VB VB
Sol :
A VW
A VW
d d d d d
tp    tQ   , t p  tQ  t
2
VB  Vw 2 2
5 3 2 4 VB 5
d d x d
  , But x  VW
4 5 Vman VB
d d VW d d d 3d
  ,  
4 5 VBVman 4 5  5  Vman
1 1 3 1 3
  , 
4 5 5Vman 20 5Vman
Vman 
 3 20   12kmph
5
 When a body is moving in a plane
a) A body can have any angle between velocity and acceleration
b) If the angle between velocity and acceleration is acute, velocity increases.
c) If the angle between velocity and acceleration is obtuse, velocity decreases.
d) If the angle between velocity and acceleration is a right angle, velocity remains constant.
e) A body can have constant speed and changing velocity
f) A body cannot have constant velocity and changing speed.
Projectiles :
Oblique Projectile :
 Any body projected into air with some velocity at an angle ‘  ’ [   (900 and 00)] with the
horizontal is called an oblique projectile.
 Horizontal component of velocity ux  u cos  , remains constant throughout the journey..
Vertical component of velocity u y  u sin  , gradually decreases to zero and then gradually
increases to u sin  . It varies at the rate ‘g’.
y vy
v
u sin  
u vx H
 x
R
u cos 
Vertical component of velocity u y  u sin  , gradually decreases to zero and then gradually
increases to u sin  . It varies at the rate ‘g’.
horizontal component of acceleration, a x  0
vertical component of acceleration , a y   g
At the Point of Projection
(a) Horizontal component of velocity u x = u cos
(b)Vertical component of velocity u y  u sin 
r
(c) velocity vector u   u cos   $i   u sin   $
j
(d) Angle between velocity and acceleration is  90   
 At any instant ‘t’
r r
Velocity vector ( v ) is v = v x $i + v y $j
Here vx = u cosθ and v y = u y + a y t = usinθ - gt
r
Hence v = ucosθ î + ( usinθ - gt) ˆj
2 2
magnitude of velocity is given by v = v2x +v2y =  ucosθ +  usinθ-gt 
v
-1  y  -1  u sin  - gt 
direction of velocity is given by   tan    tan  
 vx   u cos  
r
Displacement vector ( s )
r
displacement s  x $i  y $
j here
horizontal displacement during a time t x  uxt   u cos   t
1 2 1 2
vertical displacement during a time t y  u y t  gt   u sin   t  gt
2 2
Equation of projectile
 g  2 2
y   tan   x   2 2  x  Ax  Bx
 2u cos  
g
Where A and B are constants A  tan , B 
2u cos2 
2
Time of flight (T)

uy u sin 
Time of ascent  ta  = Time of descent  td   
g g
2u y 2u sin 
Time of flight T = ta + td = 
g g
 During time of flight
1) angle between velocity and acceleration vectors changes from  90 0    to  90 0    .
2) change in momentum is 2mu sin  .(In general, change in momentum P  mgT  )
3) vertical displacement is 0.
0 0
4) The angle between velocity and acceleration during the rise of projectile is 180 >  >90
5) The angle between velocity and acceleration during the fall of projectile is 00<  <900
Maximum height (H)
u 2y u 2 sin 2 
H  
2g 2g
 At maximum height
1) The vertical component of velocity becomes zero.
2) The velocity of the projectile is minimum at the highest point and is equal to u cos 
and is horizontal.
3) Acceleration is equal to acceleration due to gravity ‘g’, and it always acts vertically
downwards.
4) The angle between velocity and acceleration is 900.
Range (R):
2ux u y
 R  u xT  (or)
g
2u sin  u 2 sin 2
R   u cos   T  u cos   
g g
1) Range is maximum when   450
u2
2) Maximum range, Rmax 
g
R Max u2
3) When ‘R’ is maximum, HMax= 4
= 4g
4) For given velocity of projection range is same for complimentary angles of projection
i.e 1   2  90 
0
 Relation between H, T and R
H g H tan  R g
1) 2
 (b)  (c) 2 
T 8 R 4 T 2 tan 
gT 2 gT 2 2R
2) R  and if   450 then R  2  T  g
2 tan 
 If y  Ax  Bx 2 represents equation of a projectile then
-1
1) Angle of projection  = tan (A)
r
2) Initial velocity u 

g 1  A2 
2B
A
3) Range of the projectile R =
B
A2
4) Maximum height H =
4B
2A2
5) Time of flight T  
Bg
 If horizontal and vertical displacement of projectile are respectively x  at and y  bt  ct 2 then

1 b
1) angle of projection   tan  
a
2) velocity of projection u  a 2  b 2
3) acceleration of projectile = 2c
b2
4) maximum height reached =
4c
ab
5) horizontal range =
c
 In case of complimentary angles of projection
1) If T1 and T2 are the times of flight then
T1 2R
i) T  tan  ii) T1T2  T T  R
2 g  1 2
2)If H1 and H 2 are maximum heights then
H1 2 u2
i) H  tan  H
ii) 1  H 2  iii) R  4 H1H 2 iv) Rmax  2  H1  H 2 
2 2g
 If a man throws a body to a maximum distance ‘R’ then he can project the body to maximum
vertical height R/2.
 If a man throws a body to a maximum distance ‘R’ then maximum height attained by it in its path
is R/4.
At the point of striking the ground
1) Horizontal component of velocity = u cos 
2) Vertical component of velocity = -u sin 
3) Speed of projection is equal to striking speed of projectile.
4) Angle of projection is equal to the striking angle of projectile
5) If the angle of projection with the horizontal is  then angle of deviation is 2 
 The projectile crosses the points A, D in time interval t1 seconds and B,C in time interval t 2 seconds
2 2 8h
then t1  t2  (h is the distance between BC and AD)
g
B C
h
D
A
 A projectile is fired with a speed u at an angle  with the horizontal. Its speed when its
direction of motion makes an angle  with the horizontal. v = u cos  sec 
u sin  u
 v cos 
 v
u cos 
Q v cos   u cos  v  u cos  sec 
 If a body is projected with a velocity u making an angle  with the horizontal, the time after
u cosecθ u
which direction of velocity is perpendicular to the initial velocity is t  g

g sin 
and its velocity at that instant is v = u cot 
 The path of projectile as seen from another projectile
Suppose two bodies A and B are projected simultaneously from the same point with initial velocities
u and u at angles 1 and  2 with horizontal.
1 2
The instantaneous positions of the two bodies are given by
1 2
Body A : x1  u1 cos 1t , y1  u1 sin 1t  gt
2
1 2
Body B : x2  u2 cos  2t , y2  u2 sin  2t  gt
2
x   u1 cos 1  u2 cos  2  t
y   u1 sin 1  u2 sin  2  t
y u1 sin 1  u2 sin  2
slope = x  u cos   u cos 
1 1 2 2
y
i) If u1 sin 1  u 2 sin  2 ( initial vertical components) then slope 0
x
y
x
The path is a horizontal straight line
ii) If u1 cos 1  u 2 cos  2 ( initial horizontal components)
y
y
Then slope 
x
x
The path is a vertical straight line
1 u sin 
 For a projectile, ‘y’ component of velocity at th of the maximum height is
n n
1
 Resultant velocity at a height of of maximum height
nth
2
2 2  usinθ 
2 (n  1) cos 2   1
v = v + v =  ucosθ  + 
x y  u
 n  n
1  cos 2 
If n=2, velocity of a projectile at half of maximum height = u
2
 For a projectile, w.r.t stationary frame path (or) trajectory is a parabola.
 Path of projectile w.r.t frame of another projectile is a straight line
 Acceleration of a projectile relative to another projectile is zero
 A body is projected vertically up from a topless car relative to the car which is moving horizontally
relative to earth
a) If car velocity is constant, ball will be caught by the thrower.
b) If car velocity is constant, path of ball relative to the ground is a parabola and relative to this
car is straight up and then straight down
c) If the car accelerates, ball falls back relative to the car
d) If the car retards ball falls forward relative to the car
 If a gun is aimed towards a target and the bullet is fired, the moment when the target falls, the
bullet will always hit the target irrespective of the velocity of the bullet if it is with in the range.
Note : If air resistance is taken into consideration then
a) trajectory departs from parabola.
b) time of flight may increase or decrease.
c) the velocity with which the body strikes the ground decreases
d) maximum height may decrease.
e) striking angle increases
f) range decreases.
ur
 A particle is projected with a velocity u  ai$  b $j then the radius of curvature of the trajectory of
the particle at the
3/2
(i) point of projection is r=
a 2
+b 2 
(ii) Highest point is r=
a2
ga g
 velocity 2 u 2 cos 2 
 Expression for radius of curvature is r r
normalacceleration g cos3 
r
α is angle made by v with horizontal
EX.11: A bullet fired at an angle of 300 with the horizontal hits the ground 3.0 km away. By
adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the
muzzle speed to be fixed, and neglect air resistance.
Sol . weare given that angle of projection with the horizontal,   300 , horizontal range R = 3km.
u 02 s in 2 
R 
g
u02
2
u s in 6 0
0
0
u2
0 3 or g  2 3km
3  
g g 2
Since the muzzle speed (u0 ) is fixed
u02
Rmax   2 3  2 1.732  3.464km
g
so, it is not possible to hit the target 5km away.
EX.12: A cannon and a target are 5.10 km apart and located at the same level. How soon will the
shell launched with the initial velocity 240 m/s reach the target in the absence of air drag ?
Sol . Here, u0 =240 ms 1 , R =5.10 km =5100m,
g  9.8ms 2 ,   ?
u02 sin 2
R
g
Rg
sin 2  2    300 or 600
u0
2u0 sin
using, T 
g
2  240  0.5
When   300 , T1   24.5s
9.8
0 2  240  0.867
When   60 , T2   42.46s
9.8
EX.13: The ceiling of a long hall is 20 m high. What is the maximum horizontal distance that a
 2
ball thr own with a speed of 40ms-1 can go without hitting the ceiling of the hall g  10ms ?
Sol. : Here, H =20 m, u  40ms 1.
Suppose the ball is thrown at an angle  with the horizontal.
2
Now H 
u 2 sin 2 
 20 
 40 sin 2 
2g 2  10
or , sin   0.5    300
2 0
u 2 sin 2  40   sin 60
Now R  
g 10
2
 40   0.866
  138.56m
10
EX.14: A ball projected with a velocity of 10m/s at angle of 300 with horizontal just clears two
vertical poles each of height 1m. Find separation between the poles.
P R
0
30
O Q S X
1 2 1
Sol. h  uyt  gt  10sin30  t   10 t
0 2
2 2
1  5t  5t 2  t  0.72 s, 2.76s
are the instants at which projectile crosses the poles.
 separation between poles = OS - OQ
 u cos   t2  t1 
= 10cos 300  2.76  0.72   17.7m

EX.15: A body is projected with velocity u at an angle of projection  with the horizontal.The
body makes 300 with horizontal at t = 2 second and then after 1 second it reaches the
maximum height. Then find
a) angle of projection b) speed of projection.
usinθ-gt
Sol. During the projectile motion, angle at any instant t is such that tanα=
ucosθ
For t = 2 seconds,   300
1 usinθ-2g
= ---------1
3 ucosθ
For t = 3 seconds, at the highest point   0o
usinθ-3g
0=
ucosθ
usinθ=3g ------------(2)
using eq. (1) and eq. (2)
ucosθ= 3g......................(3)
Eq. (2)  eq.(3) give   600 squaring and adding equation (2) and (3)
u  20 3 m / s.
EX.16: A particle is thrown over a triangle from one end of horizontal base and grazing the
vertex falls on the other end of the base.If  and  are the base angles and  be the angle
of projection, prove that tan   tan   tan  .
Sol.: The situation is shown in figure.From figure,we have
A(x,y)
y
  
x R-x x
O
y y
tan   tan   
x Rx
yR
tan   tan   _______(1)
x R  x
 x
But equation of trajectory is y  x tan  1  
 R
 yR 
tan          (ii )
 x  R  x  
From Eqs. (i) and (ii), tan   tan   tan 
2
EX.17: The velocity of a projectile at its greatest height is times its velocity, at half of its
5
greatest height, find the angle of projection.
2 1  cos2 
Sol.: u cos  u
5 2
2 2 2 2  1  cos 2  
Squaring on both sides u cos   u  
5  2 
2 2 1
10 cos 2   2  2 cos 2   8cos   2  cos      600
4
EX.18: A foot ball is kicked off with an initial speed of 19.6 m/s to have maximum range. Goal
keeper standing on the goal line 67.4 m away in the direction of the kick starts running opposite
to the direction of kick to meet the ball at that instant. What must his speed be if he is to catch
the ball before it hits the ground?
2
u 2 sin 2 19.6   sin 90
Sol.: R 
g 9.8
or R= 39.2 metre.
Man must run 67.4 m -39.2m=28.2m
in the time taken by the ball to come to ground Time taken by the ball.
2u sin  2 19.6  sin 450 4
t  
g 9.8 2
t  2 2  2 1.41  2.82sec.
28.2m
Velocity of man   10m / sec.
2.82sec
EX.19: A body projected from a point `0’ at an angle  , just crosses a wall `y’ m high at a distance
`x’ m from the point of projection and strikes the ground at `Q’ beyond the wall as shown,
then find height of the wall Y
y

O x R-x Q X
R=range
gx 2
Sol . we know that the equation of the trajectory is y  x tan   can be written as
2u 2 cos 2 
 gx 2  sin 
y  x tan    2 2 
 2u cos   sin 
x 2 tan 
2
gx tan   y  x tan  
y  x tan   u 2 sin 2
u 2 (2sin  cos  ) g
 x u 2 sin 2
 y  x tan  1   [Q R  ]
 R g
EX.20: A particle is projected with a velocity of 10 2 m/s at an angle of 450 with the horizontal .
Find the interval between the moments when speed is 125 m/s  g  10m / s 
2
Sol. t
v  125 m / s
ux  10 2 cos 450  10m / s, u y  10 2 sin 450  10m / s

v 2  v x2  v 2y

125  100  v 2y  v y  5m / s Q vx  ux 
2v y 25
The required time interval is t    1s
g 10
EX.21: A projectile of 2 kg has velocities
3 m/s and 4 m/s at two points during its flight in the uniform gravitational field of the earth.
If these two velocities are  to each other then the minimum KE of the particle during its
flight is
Sol. V1 cos   V2 cos (90   )
-1
3 ms
3 cos   4 sin  
3
tan  
4
0
1 90 -
KE min  mv12 cos 2  -1
2 4 ms
2
1  4  9 16
  2  32      5.76 J
2 5 25
EX.22: In the absence of wind the range and maximum height of a projectile were R and H. If
wind imparts a horizontal acceleration a =g/4 to the projectile then find the maximum
range and maximum height.
Sol . H 1  H ( u sin  remains same )
T1  T
1 1g 2
R1  u xT  aT 2  R  T
2 24
1 2
= R  gT  R  H R1  R  H H 1  H
8
 If a body is projected with a velocity
r uur uur uur
u  ai  bj  ck
r r r
( i  east j  north k  vertical ) then
u x  a 2  b 2 ; u y  c
2c
T  ;H 
c2
,R 
2  
a2  b2 c
g 2g g
EX.23: A particle is projected from the ground with an initial speed v at an angle  with
horizontal.The average velocity of the particle between its point of projection and highest
point of trajectory is [EAM 2013]
Y
Sol. H
 X
R/2
r r
r v + u ucosθiˆ + (ucosθiˆ + usinθj)ˆ v
v avg = = , v av = 1 + 3cos 2θ
2 2 2
 Horizontal projectile
When a body is projected horizontally with a velocity from a point above the ground level, it is
called a Horizontal Projectile.
 Path of the Horizontal Projectile is parabola
u

R
2h
a) Time of descent t  g
(is independent of u)
b) The horizontal displacement (or) range
2h
R  u g
c) The velocity of projectile at any instant of time is v = u 2 + g 2 t 2
1  gt 
The direction of velocity   tan  
u
d) The velocity with which it hits the ground v = u 2 + 2gh
1
 2 gh 
e) The angle at which it strikes the ground   tan  u 
 
f) If  is angle of elevation of point of projection from the point where body hits the ground then
h gt 2 /2 gt tanθ
tanα = = =  tanα =
R ut 2u 2
 is the angle with which body reaches the ground
Case (i) : If the body is projected at an angle  in upward direction from the top of the tower,then
u

2u sin 
a) The time taken by projectile to reach same level as point of projection is T 
g
1 2
b) The time taken by projectile to reach ground is calculated from h   u sin   t  gt
2
c) The horizontal distance from foot of the tower where the projectile lands is given by
x  u cos  t
d) The velocity with which it strikes the ground v  u 2  2 gh
e) The angle at which it strikes the ground
 u 2 sin 2 θ + 2gh 
-1  -usinθ + gt  -1
α = tan  ucosθ  (or) α = tan  
   ucosθ 
Case (ii) : If the body is projected at angle  from top of the tower in the downward direction,
then

u
1 2
a) The time taken by projectile to reach ground is calculated from h   u sin   t  gt
2
b) The horizontal distance from foot of the tower where the projectile lands is given by
x  u cos  t
c) The velocity with which it strikes the ground v  u 2  2 gh
1
 u 2 sin 2   2 gh 
d) The angle at which it strikes the ground   tan  u cos 

 
 When an object is dropped from an aeroplane moving horizontally with constant velocity
a) Path of the object relative to the earth is parabola
b) Path of the object relative to pilot is a straight line vertically down.
 Two bodies are projected horizontally from top of the tower of height h in opposite directions
with velocities u1 and u2 then
a) The time after which their velocity vectors are making an angle  with each other
u1u2 
t  cot
g 2
b) The distance between them when their velocity vectors are making an angle  with
u1u 2 
each other x   u1  u 2  cot
g 2
c) The time after which their position vectors
2 u1u2 
are making an angle  with each other  cot
g 2
d) The distance between them when their
displacement vectors are making an angle  with each other is
2 u1u2 
x   u1  u2  cot
g 2
 Two tall towers having heights h1 and h2 are separated by a distance d. A person throws a ball
horizontally with velocity u from the top of the first tower to reach the top of the second tower
then
h1  h2 
h1
h2
2  h1  h2 
a) Time taken t 
g
b) Horizontal distance travelled d  ut
 A ball rolls off from the top of a stair case with a horizontal velocity u. If each step has a height ‘h’
and width “b” then the ball will just hit the nth step, directly if n equal to
1 2
nb = ut and nh = gt
2
h b
1
n
R
2hu 2
n =
gb 2
 From the top of the tower of height h , one stone is thrown towards east with velocity u 1 and
another is thrown towards north with velocity u 2 . The distance between them after striking the
ground,
2h
d  t u12  u22 , t 
g
EX.24: A ball is thrown from the top of a tower of 61 m high with a velocity 24.4 ms 1 at an
elevation of 300 above the horizontal . What is the distance from the foot of the tower to
the point where the ball hits the ground?
Sol. :
u sin 
u

u cos 
1 2
h  gt  ( u sin  ) t  t  5 sec onds
2
Also, d   u cos   t  105.65m
EX.25: A particle is projected from a tower as shown in figure, then find the distance from the
foot of the tower where it will strike the ground.  g  10m / s 
2
37 0
1500 m 500
m/s
3
Sol.:
500 1
u y  u sin   sin 37 0 s = ut + at 2
3 2
 500  1
1500   sin 37  t  10t 2
 3  2
500  3  2
1500    t  5t
3 5
300  20t  t 2  t  20 s  horizontal distance = ( u cos  ) t
500  4  4000
   10  m
3 5 3
EX.26: A golfer standing on the ground hits a ball with a velocity of 52 m/s at an angle  above
5
the horizontal if tan   find the time for which the ball is at least 15m above the ground?
12
 g  10m / s  2
Sol. v y = u y 2 - 2gy , u y = usinθ
5 5
vy  52  52   2 10 15
13 13
= 16 × 25 - 300 = 10
2v y 2 ×10
Δt = = = 2s
10 10
EX.27: Two paper screens A and B are separated by a distance of 100m. A bullet pierces A and B.
The hole in B is 10 cm below the hole in A. If the bullet is travelling horizontally at the time
of hitting the screen A, calculate the velocity of the bullet when it hits the screen A. Neglect
resistance of paper and air.
Sol. : The situation is shown in Fig.
u P x Q
0.1 m
100 m B
A
2  h1  h2  2  0.1
d  u  100  u  u  700m / s.
g 9.8
EX.28: A boy aims a gun at a bird from a point, at a horizontal distance of 100m. If the gun can
impart a velocity of 500m/sec to the bullet, at what height above the bird must he aim his
gun in order to hit it?
Sol : x = vt or 100 = 500× t ; t  0.2sec.
1 2
Now h  0   10   0.2  = 0.20m = 20cm.
2
EX.29: An enemy plane is flying horizontally at an altitude of 2 km with a speed of 300 ms-1. An
army man with an anti - aircraft gun on the ground sights enemy plane when it is directly
overhead and fires a shell with a muzzle speed of 600ms-1. At what angle with the vertical
should the gun be fired so as to hit the plane?
Sol. Let G be the position of the gun and E that of the enemy plane flying horizontally with speed.
E P
u
v0
vy
0
(90 -)
 Ground
G
vx
u = 300ms-1, when the shell is fired with a speed v0 , v x = v 0 cosθ

The shell will hit the plane, if the horizontal distance EP travelled by the plane in time t = the
distance travelled by the shell in the horizontal direction in the same time, i.e.
u × t = v x × t or u = v x  u = v0cosθ
u 300
or cosθ = v = 600 = 0.5 or
0
  600
Therefore, angle with the vertical  900    300.
EX.30: From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in
opposite directins. If their velocities are perpendicular to each other just before they strike
the ground, find the height of tower.
2h
Sol. Time taken to reach ground t 
g
uur uur
at time of reaching ground respective velocities are v1 = u1$i + gt $j, v 2 = -u 2 $i + gt $j
uur uur u1u2

Given v1.v 2 = 0 , t 
g
2h uu uu
  1 2 h 1 2
g g 2g
is the height of the tower.
EX.31: From points A and B, at the respective heights of 2m and 6m, two bodies are thrown
simultaneously towards each other, one is thrown horizontally with a velocity of 8m/s and
the other, downward at an angle 450 to the horizontal at an initial velocity v0 such that the
bodies collide in flight. The horizontal distance between points A and B equal to 8m . Then
find
a) The initial velocity V0 of the body thrown at an angle 450
b) The time of flight t of the bodies before colliding
c) The coordinate (x,y) of the point of collision (consider the bottom of the tower
A as origin) is
B
450

4m
V0
A
Sol : 2m 2m
8m
4 1
a) From diagram tan    tan   .................(1)
8 2
r ˆ v = -vr cos450 î - vr sin450 ˆj
v A = 8i, B 0 0
r  v  v
vBA =  - 0 -8  i - 0 j
 2  2
r
Direction of vBA
v0 2
tanθ =
2 v 0 + 8 2 .................(2)
 
From eq (1) and eq (2) 2v0 = v0 + 8 2 , v0 = 11.28m / s
r  v0  v
b) v BA =  - - 8  î - 0 ˆj
 2  2
r
Q v0 = 8 2 Þ v BA = -16iˆ - 8jˆ
r
| v BA | t = SBA   16 
2
  8 
2
t  82  4 2
80 1
t   t  0.5s
320 4
c)
x = v x t =  8  0.5 = 4
1 2 1 1
y' = gt = ×10× = 1.25
2 2 4
y = 2 - y' = 0.75
Motion of a Projected Body on an inclined plane :
 A body is projected up the inclined plane from the point O with an initial velocity v0 at an angle
 with horizontal.
v0 x
x
A


O B
x'
y'
x
y
g sin 
g cos
 
O g B
x1
y1
a) Acceleration along x  axis , a x = -gsinα

b) Acceleration along y  axis , a y = -g c o s α
c) Component of velocity along x  axis u x = v0cos  θ - α 
d) Component of velocity along y  axis u y = v0sin  θ -α
2v0sin  θ - α 
e) Time of flight T =
gcosα
f) Range of projectile (OA)
v20 2v02sin  θ - α  cosθ
R= sin  2θ - α  - sinα  . (or) R=
gcos2α  gcos 2α

For maximum range  2    
2
v 20 1- sinα 
 R max =
gcos 2 α
g) T 2 g  2 Rmax
horizontal range (OB) x  R cos 
Down the plane : Here, x and y-directions are down the plane and perpendicular to plane
respectively
y
u x
 g sin
 
g g cos
 
O
ux  u cos(   ), ax  g sin  u y  u sin(   ), a y   g cos 

Proceeding in the similar manner , weget the following results
2u sin(   )
T ,
g cos 
u2
R sin  2     sin  
g cos 2  
EX.32: A particle is projected horizontally with a speed “u” from the top of plane inclined at an
angle “  ” with the horizontal. How far from the point of projection will the particle strike
the plane ?
Sol:
u

y
R

x
x
y 
R  x2  y 2   tan  
x 
2
 x 2   x tan    x 1  tan 2   x sec 
1 y 1 gt 2
x  ut; y  gt 2 ; 
2 x 2 ut
gt 2u
tan   ; t tan 
2u g
2u 2
x  ut  tan  ;
g
2u 2
R  tan  sec 
g
EX.33: A projectile has the maximum range of 500m. If the projectile is now thrown up on an
inclined plane of 300 with the same speed , what is the distance covered by it along the
inclined plane?
Sol:
u2
R max =
g
u2
 500 = or u = 500g
g
v 2 - u 2 = 2gs
0 - 500g = 2×  -gsin300  × x
x = 500m.
EX. 34: The displacement of the point of a wheel initially in contact with the ground when the
wheel rolls forward quarter revolution where perimeter of the wheel is 4 m, is (Assume
the forward direction as x-axis)
1 2
1) (  2) 2  4 along Tan with x - axis

1 2
2) (  2) 2  4 along Tan with x - axis
 2
1 2
3) (  2) 2  4 along Tan with x - axis

1 2
4) (  2) 2  4 along Tan with x - axis
 2
P 2
r
Sol. . 
P
 cm
1  y 
S y 2  x 2 and   tan  x  key-2
 
EX.35: A particle starts from the origin at t  0s with a velocity of 10.0 ˆj m/s and moves in the
 
xy  plane with a constant acceleration of 8iˆ  2 ˆj ms 2 . Then y  coordinate of the par--
ticle in 2 sec is
1) 24 m 2) 16 m 3) 8 m 4) 12 m
r r 1r 2
Sol. . r  v0t  at , y  t   t 2  10t key-1
2
EX.36: A car moving at a constant speed of 36 kmph moves north wards for 20 minutes then due
1
to west with the same speed for 8 minutes. what is the average velocity of the car during
3
this run in kmph
1) 27.5 2) 40.5 3)20.8 4) 32.7
 
v t v t
Sol. . vavg  11 22 key-1
t1  t2
EX 37: Velocity of a particle at time t = 0 is 2ms 1 . A constant acceleration of 2 ms 2 acts on
the particle for 1 second at an angle of 600 with its initial velocity . Find the magnitude of
velocity at the end of 1 second.
1) 3 m / s 2) 2 3 m / s 3) 4 m / s 4) 8 m / s
Sol. . v  vx iˆ  v y ˆj : vx  ux  axt , v y  u y  a y t

ax  a cos  , a y  a sin  key-2
EX.38: An aeroplane moving in a circular path with a speed 250 km/h. The change in velocity
in half of the revolution is.
1) 500km/h 2) 250km/h 3) 120 km/h 4) zero

Sol. . V  2V sin 6. v  v12  v22 key-1
2
EX.39: A ship is moving due east with a velocity of 12 m/sec, a truck is moving across on the ship
with velocity 4m/sec. A monkey is climbing the vertical pole mounted on the truck with a
velocity of 3m/sec. Find the velocity of the monkey as observed by the man on the shore (m/
sec)
1) 10 2) 15 3) 13 4)20
Sol. . V  Vx 2  Vy 2  Vz 2 key-3
EX.40: A man is walking due east at the rate of 2Kmph. The rain appears to him to come down
vertically at the rate of 2kmph. The actual velocity and direction of rainfall with the vertical
respectively are (2008 M)
1
1) 2 2kmph, 450 2) kmph,300 3) 2 kmph, 00 4) 1kmph, 900
2
Vm
Sol. . VR  V 2  Vm2 ; Tan  key-1
V
EX.41: A boat takes 2 hours to travel 8km and back in still water lake.With water velocity of 4
kmph, the time taken for going upstream of 8km and coming back is
1) 160 minutes 2) 80 minutes 3) 320 minutes 4) 180 minutes
88
Sol. . VB   8kmph
2
d d
t  t1  t2   key-1
vB  vr vB  vr
EX.42: A ball is thrown with a velocity of u making an angle  with the horizontal. Its velocity
vector normal to initial vector (u) after a time interval of
u sin  u u u cos 
1) 2) 3) 4)
g g cos  g sin  g
Sol. u  (u cos  )iˆ  (u sin  ) ˆj
v  (u cos  )iˆ  (u sin   gt ) ˆj ; u .v  0 key-3
EX.43: A number of bullets are fired in all possible directions with the same initial velocity u. The
maximum area of ground covered by bullets is
2 2 2
 u2   u2  u  u 

1)  g  
2)  2 g  3)    4)   
    g  2g 
2
Sol. . Max area   Rmax  key-1
EX.44: An aeroplane flies along a straight line from A to B with a speed v0 and back again with
the same speed v0 . A steady wind v is blowing. If AB = l then
2v0l
a) total time for the trip is v 2  v 2 if wind blows along the line AB
0
2l
b) total time for the trip is , if wind blows perpendicular to the line AB
v02  v 2
c) total time for the trip decrease because of the presence of wind
d) total time for the trip increase because of the presence of wind
1) a , b , d are correct 2) a , b , c are correct
3) only a , d are correct 4) only b , d are correct
Sol. . a,b,d are correct
When wind blows along the line AB,
t  t A B  t B  A
l l 2lv0
t   t 
v  v0 v0  v v  v2
2
0
If wind blows perpendicular to AB
v0
v v v0
A 1 B A 1 B
v v
t  t A B  t B  A
vt  v02  v 2 vt  v02  v 2
l l
t A B  tB A 
2 2
v v
0 v  v2
2
0
2l
Hence t 
v  v2
2
0
If the wind were not present then total time
2l
taken for the trip would have been t  v
0
i.e. the total time for the trip increases because of the presence of wind. key-2
ur uur
EX.45: Two particles A and B move with constant velocity v1 and v2 along two mutually
perpendicular straight lines towards intersection point O as shown in figure. At moment t =
0 particles were located at distance l1 and l2 respectively from O. Then minimum distance
1
between the particles and time taken are respectively
A Vt
1
S l1
2 B
l2 O
V2t
l1v2  l2v1 l1v1  l2 v2 l1v1  l2v2 l1v2  l2 v1
1) , 2) ,
v12  v22 v12  v22 v12  v22 v12  v22
l1v2  l2v1 l1  l1v1  l2 v2  l1 l1v2  l2 v1 l2  l1v1  l2 v2  l2

3)
v12  v22 l2 , v12  v22 l2
  4)
v12  v22 
l1 , v12  v22 l1
Sol. . Let the separation between the particles be minimum at time t, Then
1
V1t
A
S
l1
B
2
l2 O
V2t
Since OB  l2  v 2t and OA  l1  v 1t and

2 2
AB 2  OB 2  OA2  s 2   l1  v1t    l2  v2t 
ds d 2
For s to be minimum
dt
 0 or
dt
s 0  
ds
 2 s  2(l1  1t )  1  2(l2  2t )  2  0
dt
l1v1  l2 v2
l1v1  v12t  l2 v2  v22t  0  t 
v12  v22
2 2
2
  l v l v     l v  l v 
S min  l   1 12 22 2  v1   l2  v2  1 12 22 2  
  v1  v2     v1  v2  
2
l v  l v  l1v2  l2 v1
S 2
min  1 2 2 1  S min  key-3
v12  v22 v12  v22
EX.46: The distance between two moving particles P and Q at any time is a. If vr be their relative
velocity and if u and v be the components of vr , along and perpendicular to PQ.The closest
distance between P and Q and time that elapses before they arrive at their nearest distance
is
2 2
a  v  vr   vr  av  u
1) , a 1   2)  v  v  , a 1  
v  u r  vr 
avr avr av au
3) , 4) v , v 2
v u2 r r
Sol. . Assuming P to be at rest, particle Q is moving with velocity vr , in the direction shown in

figure. components of vr along and perpendicular to PQ are u and v respectively, In the figure
u v
sin   , cos  
vr vr
a
u v1


Q
The closest distance between the particles is PR.
v av
Smin  PR  PQ cos    a     S min 
 vr  vr
Time after which they arrive at their nearest distance is
v
a 
QR  PQ  sin   vr   au key-2
t  
vr vr vr vr2
EX.47: Two stones are projected from the top of a tower in opposite direction, with the same
velocity V but at 300 & 600 with horizontal respectively.The relative velocity of first stone
relative to second stone is
2V V
1) 2v 2) 2v 3) 4)
3 2
Sol.
V V
0 0
30 60
ur
V 1  V cos 30iˆ  V sin 30 ˆj
ur
V 2  V cos 60iˆ  V sin 60 ˆj key-2
ur ur ur
V 12  V 1  V 2
EX.48: A motor boat going down stream comes over a floating body at a point A. 60 minutes later it
turned back and after some time passed the floating body at a distance of 12 km from the point A.
Find the velocity of the stream assuming constant velocity for the motor boat in still water.
1) 2 Km/hr 2) 3 Km/hr 3) 4 Km/hr 4) 6 Km/hr
Sol. . d   vB  vW  t1   1 ;
 
d  12  v  v t    2  ;
B w 2
12  v  t  t      3
w 1 2
solve above equtions key-2
EX.49: It is raining at a speed of 5 m / s 1 at an angle 370 to vertical, towards east.A man is
moving to west with a velocity of 5 m / s 1 . The angle with the vertical at which he has to
hold the umbrella to protect himself from rain is.
1) Tan 1  2  to west 2) Tan 1  2  to east
1  1  1  1 
3) Tan   to south 4) Tan   to east
2 2

37
0 VR-VM
VR=5
y
Sol. .
x -VM=5
x 3 y x  VM
sin 37   ; cos 37  ; Tan  key-1
VR 5 VR y
EX.50: Rain, pouring down at an angle  with the vertical has a speed of 10ms 1 . A girl runs
against the rain with a speed of 8ms 1 and sees that the rain makes an angle  with the
vertical, then relation between  and  is
8  10sin  8  10 sin 
1) tan   2) tan   3) tan   tan  4) tan   cot 
10 cos  10 cos 
Sol.

VR  VG

VR=10
y
x
VG  8
x y x  VG
sin   ; x  VR sin  cos   ; y  VR cos  ; Tan  key-2
VR VR y
EX.51: A particle when fired at an angle   600 along the direction of the breadth of a rectangular
building of dimension 9m  8m  4m so as to sweep the edges. Find the range of the projectile.
8 4
1) 8 3 2) 4 3 3) 4)
3 3
Sol.
(x,h) (x+b,h)
w

x b = 2h x
gx 2
y  h  x tan   ,
2u 2 cos 2 
R u 2 sin 2
R  x  x  2h  x   h and R 
2 g
2
R  1R 
using equations 1 and 2 h    h  tan     h  tan 
 2  R  2 
 
 R  2h cot  
2 key-1
putting   600 , h  4m then R  8 3m
EX.52: .The direction of projectile at certain instant is inclined at angle  to the horizontal after
t sec.If it is inclined at an angle  then the horizontal component of velocity is
g gt t gt
1) 2) 3) 4)
tan   tan  tan   tan  g(tan   tan  ) (tan   tan  )
gt
Tan  Tan 
Vy  gt u cos 
Vy
Sol. . Tan  ; Tan  ; gt key-2
u cos  u cos  u cos   u cos   u cos  
Tan  Tan
EX.53:Two bodies are projected from the same point with same speed in the directions making
an angle 1 and  2 with horizontal and strike at the same point in the horizontal plane
t12  t22
through a point of projection. If t1 and t2 are their time of flights. Then 2 2
t1  t2
tan 1   2  sin 1   2  sin 1   2  sin 2 1   2 

1) tan    2) sin    3) sin    4)
 1 2  1 2  1 2 sin 2 1   2 
Sol. . 1   2  900  sin 1   2   1

2u sin 1 2u sin  2
t1  , t2  key-2
g g
EX.54: An object in projected up the inclined at the angle shown in the figure with an initial
velocity of 30ms 1 . The distance x up the incline at which the object lands is
-1
30 ms
X
30
30
1) 600 m 2) 104m 3) 60 m 4) 208 m
Sol. . From figure   300 ,   600 key-2

2
2u cos  sin    
R
g cos 2 
EX.55: A projectile fired with velocity u at right angle to the slope which is inclined at an angle 
with horizontal. The expression for R is
2u 2 2u 2 u2 2 2u 2
1. tan  2. sec  3. tan  4. tan  sec 
g g g g
 1  900   , 1    900
Sol. .    key-4
2u 2 cos  1 sin  1   
R
g cos 2 
EX.56: In figure shown below, the time taken by the projectile to reach from A to B is t then, the
distance AB is equal to
B
0
60 0
30
A C
ut 3ut
1) 2) 3) 3ut 4) 2ut
3 2
Sol. . u x  u cos   u cos 600
AC
x  AC  u x .t ; from figure cos 300 
AB
AC key-1
AB 
cos 300
EX.57: Two particles are projected in air with speed 0 at angles 1 and 2 (both acute) to the
horizontal, respectively. If the height reached by the first particle is greater than that of the
second, then thick the right choices
1) angle of projection : 1 > 2
2) time of flight: T2 > T1
3) horizontal range: R1 > R2
4) total energy: U1> U2
Sol. . H1 > H2
sin 1  sin θ 2 or θ1  2
T1 > T2
R1 sin 21
 1
R 2 sin 22
1
U1  KE  PE  m1v 02
2
1
U 2  KE  PE  m 2 v 02
2
If m1 = m2 then U1 = U2
m1 > m2 then U1 > U2
m1 < m2 then U1 < U2
key-1,2,3
EX.58: A particle slides down friction less parabolic (y = x2) track (A - B - C) starting from rest at
point A. Point B is at the vertex of parabola and point C is at a height less than that of point
A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point
at P, then A .
A
P

C
-x2 -x1 B -x0 x

(x=0)
1) KE at P = KE at B
2) height at P = height at A
3) total energy at P = total energy at A
4) time of travel from A to B = time of travel from B to P
Sol. . In this type of question, nature of track is very important of consider, as friction is not in this
track, total energy of the particle will remain constant throughout the journey.
(KE)B > (KE)P
(PE)P < (PE)A
As, height of p < Height of A
Hence, path length AB > path length of BP
Hence, time of travel from A to B  Time of travel from B to P..
key-3
ur r r ur r r
EX.59: Two particles having position vectors r1  (3i  5 j )m and r2  (5i  3 j )m aree
ur uur
moving with velocities V1  (4$i  4 $
j )ms 1 and V2  (ai$ 3 $
j )ms 1 . If they collide after
2 seconds , the value of ‘ a’ is
1) 2 2) 4 3) 6 4) 8
Sol. . r1  v1t  r2  v2t ; 2. 4t  5t 2  0
key-3
EX.60: A particle starts from origin at t = 0 with a constant velocity 5 $i ms 1 and moves in xy

plane under action of a force which produces a constant acceleration of 3$i  2 $ 
j ms 2 . The
y - coordinate of the particle at the instant its x co-ordinate is 84 m in m is
1) 6 2) 36 3) 18 4) 9
 1
Sol. . r  ut  at 2
2
equate x coordinate to 84 to find time t
key-2
MOTION IN A PLANE
1. Two vectors ⃗A and ⃗B have equal magnitudes. The magnitude of (A + B) is ‘n’ times the
magnitude of (A − B) .The angle between ⃗A and ⃗B is: [10 Jan. 2019 II]
n2 −1 n−1 n2 −1 n−1
(a) cos−1 [n2+1] (b) cos−1 [n+1] (c) sin−1 [n2 +1] (d) sin−1 [n+1]
Sol. (a) Let magnitude of two vectors ⃗Aand ⃗B = a
⃗ +B
|A ⃗ −B
⃗ | = √a2 + a2 + 2a2 cos 𝜃 and |A ⃗ | = √a2 + a2 − 2a2 [ cos (180∘ − 𝜃)]
= √a2 + a2 − 2a2 cos 𝜃
and according to question,
⃗ + ⃗B| = n|A
|A ⃗ − ⃗B|
a2 +a2 +2a2 cos 𝜃

or, = n2
a2 +a2 −2a2 cos 𝜃
ℋ ′ (1 + 1 + 2 cos 𝜃) 2 (1 + cos 𝜃)
⇒ n ⇒ = n2
↙ (1 + 1 − 2 cos 𝜃) (1 − cos 𝜃)
using componendo and dividendo theorem, we get
n2 − 1
𝜃 = cos1 ( )
n2 + 1
2 ⃗ = (î + ĵ) and B

Let A ⃗ such that A
⃗ = (î − ĵ) . The magnitude of a coplanar vector C ⃗ .C
⃗ =B ⃗ =
⃗ .C
⃗A. ⃗B is given by [Online April 16, 2018]
5 10 20 9
(a) √9 (b) √ 9 (c) √ 9 (d) √12
Sol. (a) If ⃗C = aî+bj then ⃗A. ⃗C = ⃗A. ⃗B
a + b = 1 (i)
⃗ =A
⃗ .C
B ⃗ .B
⃗
2a − b = 1 (ii)
Solving equation (i) and(ii) we get

1 2
a = 3, b = 3
⃗ | = √1 + 4 = √5
Magnitude of coplanar vector, |C 9 9 9
3. A vector A is rotated by a small angle 𝛥𝜃 radian (𝛥𝜃 << 1) to get a new vector B. In that case
|B − A| is : [Online April 11, 2015]
𝛥𝜃2
(a) |A|𝛥𝜃 (b) |B|𝛥𝜃 − |A| (c) |A| (1 − ) (d) 0
2
Sol (a) Arc length = radius × angle
⃗ − 𝐴| = |𝐴|𝛥𝜃
So, |𝐵
4. ⃗ =𝐵
If 𝐴 × 𝐵 ⃗ × 𝐴, then the angle between A and B is [2004]
𝜋 𝜋 𝜋
(a) (b) (c) 𝜋 (d)
2 3 4
Sol. ⃗ −𝐵
(c) 𝐴 × 𝐵 ⃗ ×𝐴=0⇒𝐴×𝐵
⃗ +𝐴×𝐵
⃗ =0
⃗ =0
𝐴×𝐵
Angle between them is 0, 𝜋, or 2 𝜋
from the given options, 𝜃 = 𝜋
TOPIC-1 ….Vectors
5. Two forces P and Q, ofmagnitude 2𝛤 and 3𝛤, respectively, are at an angle 𝜃 with each other. If
the force Q is doubled, then their resultant also gets doubled. Then, the angle 𝜃 is:
[10 Jan. 2019 II]
(a) 120∘ (b) 60∘ (c) 90∘ (d) 30∘

sol. (a) Using, R2 = P 2 + Q2 + 2PQ cos 𝜃
4 𝛤 2 + 9𝛤 2 + 12𝛤 2 cos 𝜃 = R2
When forces Q is doubled,
4 𝛤 2 + 36𝛤 2 + 24𝛤 2 cos 𝜃 = 4R2
4 𝛤 2 + 36𝛤 2 + 24𝛤 2 cos 𝜃
= 4(13𝛤 2 + 12𝛤 2 cos 𝜃) = 52𝛤 2 + 48𝛤 2 cos 𝜃
12𝛤 2 1
cos 𝜃 = − 2
= − ⇒ 𝜃 = 120o
24𝛤 2
6. Two vectors ⃗A and ⃗B have equal magnitudes. The ma𝒳ii mde of (A + B) is ‘n’ times the
⃗ and B
ma𝑔𝑛𝑖𝑡𝑢de of (A − B) . The angle between A ⃗ is:
[10 Jan. 2019 II]
n2 −1 n−1 n2 −1 n−1
(a) cos−1 [n2+1] (b) cos−1 [n+1] (c) sin−1 [n2 +1] (d) sin−1 [n+1]
sol. (a) Let magnitude of two vectors ⃗Aand ⃗B = a
⃗ + ⃗B| = √a2 + a2 + 2a2 cos 𝜃 and

|A
⃗ −B
|A ⃗ | = √a2 + a2 − 2a2 [ cos (180∘ − 𝜃)]
= √a2 + a2 − 2a2 cos 𝜃
and according to question,
⃗ + ⃗B| = n|A
|A ⃗ − ⃗B|
a2 +a2 +2a2 cos 𝜃

or, = n2
a2 +a2 −2a2 cos 𝜃
(1 + 1 + 2 cos 𝜃) 2 (1 + cos 𝜃)
⇒ n ⇒ = n2
(1 + 1 − 2 cos 𝜃) (1 − cos 𝜃)
using componendo and dividendo theorem, we get
n2 − 1
𝜃 = cos1 ( )
n2 + 1
7. Let ⃗A = (î + ĵ) and ⃗B = (î − ĵ) . The magnitude of a coplanar vector ⃗C such that ⃗A. ⃗C = ⃗B. ⃗C =
⃗ .B
A ⃗ is given by
5 10 20 9
(a) √9 (b) √ 9 (c) √ 9 (d) √12
sol. (a) If ⃗C = aî+bj then ⃗A. ⃗C = ⃗A. ⃗B
a + b = 1 (i)
⃗ =A
⃗ .C
B ⃗ .B
⃗
2a − b = 1 (ii)
Solving equation (i) and(ii) we get
1 2
a = 3, b = 3
⃗ | = √1 + 4 = √5
Magnitude of coplanar vector, |C 9 9 9
8. A vector A is rotated by a small angle 𝛥𝜃 radian (𝛥𝜃 << 1) to get a new vector B. In that case
|B − A| is :
𝛥𝜃2
(a) |A|𝛥𝜃 (b) |B|𝛥𝜃 − |A| (c) |A| (1 − ) (d) 0
2
sol. (a) Arc length = radius × angle
⃗ − 𝐴| = |𝐴|𝛥𝜃
So, |𝐵
9. ⃗ =𝐵
If 𝐴 × 𝐵 ⃗ × 𝐴, then the angle between A and B is
[2004]
𝜋 𝜋 𝜋
(a) (b) (c) 𝜋 (d)
2 3 4
sol. ⃗ −𝐵
(c) 𝐴 × 𝐵 ⃗ ×𝐴=0⇒𝐴×𝐵
⃗ +𝐴×𝐵
⃗ =0
⃗ =0
𝐴×𝐵
Angle between them is 0, 𝜋, or 2 𝜋
from the given options, 𝜃 = 𝜋
TOPIC-2 ….Motion in a plane with constant acceleration
10. A balloon is moving up in air vertically above a point A on the ground. When it is at a height ℎ1 , a
girl standing at a distance 𝑑 (point B) from A (see figure) sees it at an angle 45o with respect to
the vertical. When the balloon climbs up a further height ℎ2 , it is seen at an angle 60o with
respect to the vertical ifthe girl moves further bya distance 2.464𝑑∞intC). Then the height ℎ2 is
(given tan 30o = 0.5774):
[Sep. 05, 2020 (I)]
A ← d → B + 2.464d ∗ C
(a) 1.464𝑑 (b) 0.732𝑑 (c) 0.464𝑑 (d) 𝑑
sol. (d) From figure/ trigonometry
ℎ1
= tan 45∘ ℎ1 = 𝑑
𝑑
ℎ1 +ℎ2
And, = tan 30∘
𝑑+2.464𝑑
⇒ (ℎ1 + ℎ2 ) × √3 = 3.46𝑑
3.46𝑑 3.46𝑑
⇒ (ℎ1 + ℎ2 ) = ⇒ 𝑑 + ℎ2 =
√3 √3
ℎ2 = 𝑑
11. Starting from the origin at time 𝑡 = 0, with initial velocity 5𝑗ms −1 , a particle moves in the 𝑥 − 𝑦
plane with a constant acceleration of (10î+4j) ms−2 . At time 𝑡, its coordiantes are (20m, 𝑦0 m) .
The values oft and 𝑦0 are, respectively:
[Sep. 04, 2020 (I)]
(a) 2 s and 18 m (b) 4 s and52 m (c) 2 sand24m (d) 5 s and25 m
sol. (a) Given: 𝑢

⃗ = 5𝑗n𝜈s
Acceleration, 𝑎 = 10𝑖̂ + 4𝑗 and
final coordinate (20, 𝑦0 ) in time 𝑡.
1
𝑆𝑥 = 𝑢𝑥 𝑡 + 2 𝑎𝑥 𝑡 2 [⋅.⋅ 𝑢𝑥 = 0]
1
⇒ 20 = 0 + × 10 × 𝑡 2 ⇒ 𝑡 = 2s
2
1
𝑆𝑦 = 𝑢𝑦 × 𝑡 + 𝑎𝑦 𝑡 2
2
1
𝑦0 = 5 × 2 + × 4 × 22 = 18m
2
12. The position vector of a particle changes with time according to the relation 𝑟(t) = 15t 2 𝑖̂ +
(4 − 20t 2 )𝑗̂. What is the magnitude ofthe acceleration at 𝑡 = 1?
[9 April 2019 II]
(a) 40 (b) 25 (c) 10 (d) 50
sol. (d) 𝑟 = 15𝑡 2 𝑖̂ + (4 − 20𝑡 2 )𝑗̂
𝑑𝑟→
𝑣= =30tî‐40tj
𝑑𝑡
𝑑𝑣→
Acceleration, → 𝑎 = =30î‐40j
𝑑𝑡
𝑎 = √302 + 402 = 50m/s2

13. A particle moves from the point (2.0î+ 4.0j) m , at t = 0, with an initial velocity (5.0î+ 4.0j)
ms−1 . It is acted upon by a constant force which produces a constant acceleration(4.0𝑖̂ +
4.0ĵ)ms−2. What is the distance of the particle from the origin at time 2s?
[11 Jan. 2019 II]
(a) 15m (b) 20√2m (c) 5m (d) 10√2m

1
sol. (b) As s  ut  at 2
2
⃗S = (5î + 4ĵ)2 + 1(4î+4j)4

2
=10î+8j+8î+8j
rf − ⃗⃗ri = 18î + 16ĵ
[as s = change in position = rf − ⃗⃗ri ]
rr = 20î + 20ĵ
|rr | = 20√2
14. A particle is moving with a velocity𝑣⃗ = K (𝑦 î+xj), where K is a constant. The general equation
for its path is:
[9 Jan. 2019 I]
(a) 𝑦 = 𝑥 2 + constant (b) 𝑦 2 = 𝑥 + constant
(c) 𝑦 2 = 𝑥 2 + constant (d) 𝑥𝑦 = constant
sol. (c) From given equation,
⃗V =K(yî+xj)
dx dy
= ky and = kx
dt dt
dy dx x dy
Now / dt = y = dx , ⇒ ydy = xdx
dt
Integrating both side
y2 = x2 + c
TOPIC-3 …Projectile Motion
15. A particle starts from the origin at 𝑡 = 0 with an initial velocity of 3.0î m/s and moves in the x‐y
plane with a constant acceleration (6.0î+ 4.0j) m/s 2 . The x coordinate of the particle at the instant
when its y coordinate is 32 m is D meters. The value of D is:
[9 Jan. 2020 II]
(a) 32 (b) 50 (c) 60 (d) 40
1
sol. (c) Using 𝑆 = 𝑢𝑡 + 2 𝑎𝑡 2
1
𝑦 = 𝑢𝑦 𝑡 + 2 𝑎𝑦 𝑡 2 (along 𝑦 Axis)
1
⇒ 32 = 0 × 𝑡 + (4)𝑡 2
2
1
⇒ × 4 × 𝑡 2 = 32
2
⇒ 𝑡 = 4𝑠
1
𝑆𝑥 = 𝑢𝑥 𝑡 + 2 𝑎𝑥 𝑡 2 (Along 𝑥 Axis)
1
⇒ 𝑥 = 3 × 4 + × 6 × 42 = 60
2
16. A particle is moving along the 𝑥‐axis with its coordinate with time 𝑡’ given by 𝑥(𝑡) = 10 + 8𝑡 −
3𝑡 2 . Another particle is moving along the 𝑦‐axis with its coordinate as a function of time given by
𝑦(𝑡) = 5 − 8𝑡 3 . At 𝑡 = 1s, the speed of the second particle as measured in the frame of the first
particle is given as √𝑣. Then 𝑣(in m/s) is−
[NA 8 Jan. 2020 I]
sol. (580)
For pariticle ‘A’ For particle ‘B’
𝑋𝐴 = −3𝑡 2 + 8𝑡 + 10 𝑌𝐵 = 5 − 8𝑡 3
⃗ 𝐴 = (8 − 6𝑡)𝑖̂
𝑉 ⃗ 𝐵 = −24𝑡 2 𝑗
𝑉
𝑎𝐴 =‐ 6î 𝑎𝐵 = −48𝑡𝑗̂
At 𝑡 = 1 sec
⃗ 𝐴 = (8 − 6𝑡)𝑖̂ = 2𝑖̂ and⃗⃗𝑣𝐵 = −24𝑗

𝑉
⃗ 𝐵/𝐴 = −𝑣𝐴 + 𝑣𝐵 = −2î‐24j
𝑉
Speed of𝐵w. r. t. 𝐴, √𝑣 = √22 + 242
= √4 + 576 = √580
𝑣 = 580(m/s)
17. A particle moves such that its position vector 𝑟(𝑡) = cos  t î + sin  t 𝑗 where  is a
constant and 𝑡 is time. Then which of the following statements is true for the velocity 𝑣 (t) and
acceleration 𝑎(𝑡) ofthe particle:
[8 Jan. 2020 II]
(a) 𝑣 is perpendicular to 𝑟 and 𝑎 is directed away from the origin
(b) 𝑣 and 𝑎 both are perpendicular to 𝑟
(c) 𝑣 and 𝑎 both are parallel to 𝑟
(d) 𝑣 is perpendicular to 𝑟 and 𝑎 is directed towards the origin
sol. (d) Given, Position vector,
𝑟 = cos 𝑤𝑡𝑖̂ + sin 𝑤𝑡𝑗
𝑑𝑟
Velocity, 𝑣 = = w (− sin wtî+ cos w𝑡𝑗̂)
𝑑𝑡
Acceleration,
𝑑𝑣
𝑎= = −w 2 ( cos w𝑡𝑖̂ + sin w𝑡𝑗̂)
𝑑𝑡
𝑎 = −w 2 𝑟
𝑎 is antiparallel to 𝑟
Also 𝑣. 𝑟 = 0 𝑣 ⊥ 𝑟
Thus, the particle is performing uniform circular motion.

18. ⃗ = 𝑘(𝑦𝑖̂ + 𝑥𝑗̂) , where k is a constant. The general equation
A particle is moving with velocity v
for its path is
[2010]
(a) 𝑦 = 𝑥 2 + constant (b) 𝑦 2 = 𝑥 + constant
(c) 𝑥𝑦 = constant (d) 𝑓 = 𝑥 2 + constant
sol. (d) 𝑣 = 𝑘(𝑦𝑖 + 𝑥𝑗)
𝑣 = 𝑘𝑦𝑖 + 𝑘𝑥𝑗
𝑑𝑥 𝑑𝑦
= 𝑘𝑦, = 𝑘𝑥
𝑑𝑡 𝑑𝑡
𝑑𝑦 𝑑𝑦 𝑑𝑡
= ×
𝑑𝑥 𝑑𝑡 𝑑𝑥
𝑑𝑦 𝑘𝑥
=
𝑑𝑥 𝑘𝑦
𝑦𝑑𝑦 = 𝑥𝑑𝑥 (i) Integrating equation (i)
∫ 𝑦 𝑑𝑦 = ∫ 𝑥 ⋅ 𝑑𝑥
𝑦2 = 𝑥2 + 𝑐
19. A particle has an initial velocity of 3𝑖̂ + 4𝑗̂ and an acceleration of 0.4𝑖̂ + 0.3𝑗̂. Its speed after 10 s
is:
[2009]
(a) 7√2 units (b) 7 units (c) 8.5 units (d) 10units
sol. (a) Given 𝑢

⃗ = 3𝑖̂ + 4𝑗, 𝑎 = 0.4𝑖̂ + 0.3𝑗, 𝑡 = 10s
From 1st equation of motion.

𝑣−𝑢
𝑎=
𝑡
𝑣 = 𝑎𝑡 +u
⇒v=(0.4î+ 0.3j)×10+(3î+4j) ⇒ 4î+3j+3j+4j
⇒ v=7î+7j
⇒ |𝑣⃗ | = √72 + 72 = 7√2 unit.
20. The co‐ordinates of a moving particle at any time ‘t’are given by 𝑥 = 𝛼𝑡 3 and 𝑦 = 𝛽𝑡 3 . The
speed ofthe particle at time 𝑡’ is given by
[2003]
(a) 3𝑡√𝛼 2 + 𝛽 2 (b) 3𝑡 2 √𝛼 2 + 𝛽 2 (c) 𝑡 2 √𝛼 2 + 𝛽 2 (d) √𝛼 2 + 𝛽 2
sol. (b) Coordinates of moving particle at time 𝑡’ are
𝑥 = 𝛼𝑡 3 and 𝑦 = 𝛽𝑡 3
𝑑𝑥 𝑑𝑦
𝑣𝑥 = = 3𝛼𝑡 2 and 𝑣𝑦 = = 3𝛽𝑡 2
𝑑𝑡 𝑑𝑡
𝑣 = √𝑣𝑥2 + 𝑣𝑦2 = √9𝛼 2 𝑡 4 + 9𝛽 2 𝑡 4
= 3𝑡 2 √𝛼 2 + 𝛽 2
𝜋
21. A particle of mass 𝑚 is projected with a speed 𝑢fii om the ground at an angle 𝜃 = w.r.t.
3
horizontal (x‐axis). When it has reached its maximum height, it collides completely in elastically
with another particle of the same mass and velocity 𝑢 î. The horizontal distance covered by the
combined mass before reaching the ground is:
[9 Jan. 2020 II]
3√3𝑢2 3√2𝑢2 5 𝑢2 𝑢2
(a) (b) (c) (d) 2√2
8𝑔 4𝑔 8 𝑔 𝑔
sol. (a) Using principal of conservation of linear momentum for horizontal motion, we have
2𝑚𝑣𝑥 = 𝑚𝑢 + 𝑚𝑢 cos 60∘

3𝑢
𝑣𝑥 =
4
For vertical motion
1 2ℎ
ℎ = 0 + 𝑔𝑇 2 ⇒ 𝑇 = √
2 𝑔
Let 𝑅 is the horizontal distance travelled by the body.

1
𝑅 = 𝑣𝑥 𝑇 + 2 (0)(𝑇)2 (For horizontalmotion)
3𝑢 2ℎ
𝑅 = 𝑣𝑥 𝑇 = ×√
4 𝑔
3√3𝑢2
⇒𝑅=
8𝑔
22. The trajectory of a projectile near the surface of the earth is given as 𝑦 = 2𝑥 − 9𝑥 2 . Ifit were
launched at an angle 𝜃0 with speed 𝑣0 then (𝑔 = 10 ms 2):
[12 April 2019 I]

1
1 5 2 3
(a) 𝜃0 = sin √5 and 𝑣0 = 3 m/s (b) 𝜃0 = cos1 ( ) and 𝑣0 = 5 m/s
√5
1 9 2 3
(c) 𝜃0 = cos1 ( ) and 𝑣0 = 3 m/s (d) 𝜃0 = sin1 ( ) and 𝑣0 = 5 m/s
√5 √5
sol. (c) Given, 𝑦 = 2𝑥 − 9𝑥 2
On comparing with,
𝑔𝑥 2
𝑦 = 𝑥 tan 𝜃 − 2 2 ,
2𝑢 cos 𝜃
We have,
1
tan 𝜃 = 2 or cos 𝜃 =
√5
𝑔 10
and = 9 or 2 =9
2𝑢2 cos2 𝜃 2𝑢2 (1/√5)
𝑢 = 5/3m/s
23. A shell is fired from a fixed artillery gun with an initial speed 𝑢 such that it hits the target on the
ground at a distance R from it. If 𝑡1 and 𝑡2 are the values of the time taken by it to hit the target
in two possible ways, the product 𝑡1 𝑡2 is :
[12 Apri12019 I]
(a) R/4g (b) R/g (c) R/2g (d) 2R/g
sol. (d) 𝑅 will be same for 𝜃 and 90∘ − 𝜃.

Time of flights:
2𝑢 sin 𝜃
𝑡1 = and
𝑔
2𝑢 sin (90∘ − 𝜃) 2𝑢 cos 𝜃

𝑡2 = =
𝑔 𝑔
2𝑢 sin 𝜃 2𝑢 cos 𝜃
Now, 𝑡1 𝑡2 = ( )( )
𝑔 𝑔
2 u 2 sin 2 2𝑅
= =
𝑔 g 𝑔
24. Two particles are projected from the same point with the same speed 𝑢 such that they have the
same range R, but different maximum heights, ℎ1 and ℎ2 . Which of the following is correct?
[12 April 2019 II]
(a) R2 = 4ℎ1 ℎ2 (b) R2 = 16ℎ1 ℎ2 (c) R2 = 2ℎ1 ℎ2 (d) R2 = ℎ1 ℎ2
sol. (b) For same range, the angle of projections are:
𝜃 and 90∘ − 𝜃. So,
𝑢2 sin2 𝜃 𝑢2 sin2 (90o −𝜃) 𝑢2 cos2 𝜃

ℎ1 = and ℎ2 = =
2𝑔 2𝑔 2𝑔
𝑢2 sin 2𝜃
Also, 𝑅 = 𝑔
𝑢2 sin2 𝜃 𝑢2 cos2 𝜃
ℎ1 ℎ2 = ×
2𝑔 2𝑔
𝑢2 𝑢2 (2 sin 𝜃 cos 𝜃)2
=
16𝑔2
𝑅2
=
16
or 𝑅 2 = 16ℎ1 ℎ2
25. A plane is inclined at an angle 𝛼 = 30o with respect to the horizontal. A particle is projected with
a speed u = 2 ms 1, fii om the base ofthe plane, as shown in figure. The distance from the base, at
which the particle hits the plane is close to : (Take g = 10 m/s 2)
[10 April 2019 II]

(a) 20 cm (b) 18 cm (c) 26 cm (d) 14 cm
sol. (a) On an inclined plane, time of flight (T) is given by
𝑇 = 2𝑢 sin 𝜃
𝑔 cos 𝛼
Substituting the values, we get

(2)(2 sin 15∘ ) 4 sin 15∘
𝑇= ∘=
𝑔 cos 30 10 cos 30∘
1
Distance, S = (2 cos 15∘ )𝑇 − 2 𝑔 sin 30∘ (𝑇)2
4 sin 15o 1×10sin30∘ 16sin2 15o

= (2 cos 15∘ ) 1010 cos 30o − ( )
2 100cos2 30o
16√3−16
= = 0.1952m = 20cm
60
26. A body is projected at t = 0 with a velocity 10 ms −1 at an angle of 60∘ with the horizontal. The
radius ofcurvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration
due to gravity g = 10ms −2, the value of R is:
[11 Jan. 2019 I]
(a) 10.3 m (b) 2.8 m (c) 2.5m (d) 5.1m
sol. (b)
Horizontal component of velocity v𝑥 = 10 cos 60∘ = 5m/s

vertical component of velocity vy = 10 cos 30∘ = 5√3m/s
After t = 1 sec.
Horizontal component of velocity v𝑥 = 5m/s
Vertical component of velocity
vy = |(5√3 − 10)|m/s = 10 − 5√3

V2
Centripetal, acceleration a 
R
v2x +v2y 25+100+75−100√3
⇒R= = (i)
an 10 cos 𝜃
From figure (using (i))
10 − 5√3
tan 𝜃 = = 2 − √3 ⇒ 𝜃 = 15∘
5
100(2 − √3)
R= = 2.8m
10 cos 15
27. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a
horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by
the bullets fired by the two guns, on the ground is:
[10 Jan. 2019 I]
(a) 1:16 (b) 1:2 (c) 1:4 (d) 1:8
u2 sin 2𝜃
sol. (a) As we know, range R = g
and, area A = 𝜋R2
A ∝ R2 or, A ∝ u4
A1 u14 14 1
= 4=[ ] =
A2 u2 2 16
28. The initial speed of a bullet fired from a rifle is 630 m/s. The rifle is fired at the center of a target
700 m away at the same level as the target. How far above the centre ofthe target?
[Online April, 2014]

(a) 1.0m (b) 4.2m (c) 6.1m (d) 9.8m
sol. (c) Let 𝑡’ be the time taken by the bullet to hit the target.
700 m = 630ms−1 𝑡
700m 10
⇒𝑡= = sec
630ms−1 9
For vertical motion,
Here, 𝑢 = 0
1
ℎ = 𝑔𝑡 2
2
1 10 2
= × 10 × ( )
2 9
500
= m = 6.1m
81
Therefore, the rifle must be aimed 6.1m above the centre of the target to hit the target.
29. The position of a projectile launched from the origin at 𝑡 = 0 is given by 𝑟 = (40𝑖̂ + 50𝑗̂)m at
𝑡 = 2s. If the projectile was launched at an angle 𝜃 from the horizontal, then 𝜃 is
(take g = 10ms −2) [Online April 9, 2014]
2 3 7 4
(a) tan−1 3 (b) tan−1 2 (c) tan−1 4 (d) tan−1 5
sol. (c) From question,
Horizontal velocity (initial),

40
𝑢𝑥 = = 20m/s
2
1
Vertical velocity (initial), 50 = 𝑢𝑦 𝑡 + 2 𝑔𝑡 2
1
⇒ 𝑢𝑦 × 2 + (−10) × 4
2
or, 50 = 2𝑢𝑦 − 20
70
or, 𝑢𝑦 = = 35m/s
2
𝑢𝑦 35 7
tan 𝜃 = = =
𝑢𝑥 20 4
7
⇒ Angle 𝜃 = tan−1 4
30. A projectile is given an initial velocity of (î+2j) m/s, where î is along the ground and 𝑗 is along
the vertical. If g = 10m/s 2 , the equation of its trajectory is:
[2013]
(a) 𝑦 = 𝑥 − 5𝑥 2 (b) 𝑦 = 2𝑥 − 5𝑥 2 (c) 4𝑦 = 2𝑥 − 5𝑥 2 (d) 4𝑦 = 2𝑥 − 25𝑥 2
sol. (b) From equation, v

⃗ = î + 2𝑗
⇒𝑥=𝑡 (i)
1
𝑦 = 2𝑡 − 2 (10𝑡 2 ) ... (ii)
From(i) and(ii), 𝑦 = 2𝑥 − 5𝑥 2
31. The maximum range of a bullet fired fii om a toy pistol mounted on a car at rest is R 0 = 40m.
What will be the acute angle ofinclination ofthe pistol for maximum range when the car is moving
in the direction of firing with uniform velocity v = 20m/s, on a horizontal surface?
(g = 10m/s 2 ) [Online April 25, 2013]
(a) 30∘ (b) 60∘ (c) 75∘ (d) 45∘
sol. (b)
32. A ball projected from ground at an angle of 45∘ just clears a wall in fiiont. If point of projection
is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of
the wall, the height of the wall is:
(a) 4.4m (b) 2.4m (c) 3.6m (d) 1.6m

sol. (b)
As ball is projected at an angle 45∘ to the horizontal
therefore Range = 4H
10
or 10 = 4H ⇒ H = = 2.5m
4
(Range = 4m + 6m = 10m)
u2 sin2 𝜃
Maximum height, H = 2g
H × 2g 2.5 × 2 × 10
u2 = = = 100
sin2 𝜃 1 2
( )
√2
or, u = √100 = 10ms −1
Height of wall PA
1 g(OA)2
= OA tan 𝜃 − 2 2
2 u cos 𝜃
1 10 × 16
= 4− × = 2.4m
2 10 × 10 × 1 × 1
√2 √2
33. A boy can throw a stone up to a maximum height of l0m. The maximum horizontal distance that
the boy can throw the same stone up to will be
[2012]
(a) 20√2m (b) 10 m (c) 10√2m (d) 20m

𝑢2 sin2 𝜃 𝑢2 sin2 𝜃
sol. (d) 𝑅 = , 𝐻=
𝑔 2𝑔
𝐻 max at 2𝜃 = 90∘
u2
H max 
2g
𝑢2
= 10 ⇒ 𝑢2 = 10𝑔 × 2
2𝑔
𝑢2 sin 2𝜃 𝑢2
𝑅= ⇒ 𝑅 max =
𝑔 𝑔
𝑅 max = 20 metre
34. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the
fountain is 𝑣, the total area around the fountain that gets wet is:
[2011]
𝑣4 𝜋 𝑣4 𝑣2 𝑣2
(a) 𝜋 𝑔2 (b) (c) 𝜋 𝑔2 (d) 𝜋
2 𝑔2 𝑔
sol. (a) Let, total area around fountain
𝐴 = 𝜋𝑅 2max (i)
𝑣 2 sin 2𝜃 𝑣 2 sin 90∘ 𝑣2

Where 𝑅 max = = = (ii)
𝑔 𝑔 𝑔
From equation (i) and(ii)

𝑣4
𝐴=𝜋 2
𝑔
35. A projectile can have the same range ‘R’ for two angles of projection. If 𝑇1 ’ and 𝑇2 ’ to be time of
flights in the two cases, then the product of the two time of flights is directly proportional to.
[2004]
1 1
(a) 𝑅 (b) (c) (d) 𝑅 2
𝑅 𝑅2
sol. (a) A projectile have same range for two angle
Let one angle be 𝜃, then other is 90∘ − 𝜃

2u sin  2u cos 
T1  , T2 
g g
4𝑢2 sin 𝜃 cos 𝜃
then, 𝑇1 𝑇2 = = 2𝑅
𝑔
𝑢2 sin2 𝜃
(𝑅 = )
𝑔
Thus, it is proportional to R. (Range)
36. A ball is thrown from a point with a speed v0 at an elevation angle of 𝜃. From the same point
v0
and at the same instant, a person starts running with a constant speed to catch the ball. Will the
2
person be able to catch the ball? If yes, what should be the angle of projection 𝜃?
[2004]
(a) No (b) Yes, 30∘ (c) Yes, 60∘ (d) Yes, 45∘
sol. (c) Yes, Man will catch the ball, if the horizontal component of velocity becomes equal to the
constant speed of man.

𝑣𝑜
= 𝑣𝑜 cos 𝜃
2
or 𝜃 = 60∘
37. A boy playing on the roof of a 10 m high building throws a ball with a speed of l0m/s at an angle
of 30o with the horizontal. How far from the throwing point will the ball be at the height of 10 m
from the ground?
1 √3
[𝑔 = 10m/s 2 , sin 30o = 2 , cos 30o = ] [2003]
2
(a) 5.20m (b)4.33m (c) 2.60m (d) 8.66m
sol. (d) Horizontal range is required

(10)2 sin (2 × 30o )
𝑅= = 5√3 = 8.66m
10
TOPIC-4 .Relative Velocity in Two Dimensions & Uniform Circular Motion
38. A clock has a continuously moving second’s hand of 0.1m length. The average acceleration of the
tip of the hand (in units of ms −2) is of the order of:
[Sep. 06, 2020 (I)]
(a) 10−3 (b) 10−4 (c) 10−2 (d) 10−1
sol. (a) Here, 𝑅 = 0.1m

2 2
w   0.105 rad / s
T 60
Accelaration of the tip of the clock seconds hand,
a  w2 R  (0.105)2 (0.1)  0.0011  1.1103

3
Hence, average accelaration is of the order of 10
39. When a car is at rest, its driver sees raindrops falling on it vertically. When driving the car with
speed v, he sees that raindrops are coming at an angle 60o from the horizontal. On furter
increasing the speed of the car to (1 + 𝛽)v, this angle changes to 45o . The value of 𝛽 is close to:
[Sep. 06, 2020 (II)]
(a) 0.50 (b) 0.41 (c) 0.37 (d) 0.73
sol. (d) The given situation is shown in the diagram. Here 𝑣𝑟 be the velocity of rain drop.
When car is moving with speed 𝑣,

𝑣𝑟
tan 60∘ = (i)
𝑣
𝑣
When car is moving with speed (1 + 𝛽)𝑣, tan 45∘ = (𝛽+1)𝑣
𝑟
(ii)
Dividing (i) by(ii) we get,
√3𝑣 = (𝛽 + 1)𝑣 ⇒ 𝛽 = √3 − 1 = 0.732.
40. The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4
km/h. What should be the direction of the swimmer with respect to the flow of the river to cross
the river straight?
[9 April 2019 I]
(𝑎) 90∘ (b) 150∘ (c) 120∘ (d) 60∘
𝑢 2 1
sol. (c) sin 𝜃 = 𝑣 = 4 = 2 or 𝜃 = 30∘
with respect to flow, = 90∘ + 30∘ = 120∘
41. Ship A is sailing towards north‐east with velocity km/hr where points east and , north. Ship B is
at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A
will be at minimum distance from B in:
[8 April 2019 I]
(a) 4.2hrs. (b) 2.6hrs. (c) 3.2hrs. (d) 2.2hrs.
sol.
𝑣𝐴 = 30𝑖̂ + 50𝑗 km/ hr
𝑣𝐵 = (−10𝑖̂)km/hr
𝑟𝐵𝐴 =(80î+l50j) km
𝑣𝐵𝐴 = 𝑣𝐵 − 𝑣𝐴 = −10î‐30î‐50î=40î‐50j
|(𝑟𝐵𝐴 ) ⋅ (𝑣𝐵𝐴 )|
𝑡mmimum =
|(𝑣𝐵𝐴 )|2
|(80𝑖̂+150𝑗̂ )(−40𝑖̂−50𝑗̂ )|
= 2
(10√41)
10700 107
𝑡= = = 2.6hrs.
10√41 × 10√41 41
42. Two particles A, B are moving on two concentric circles of radii R1 and R 2 with equal angular
speed w. At t = 0, their positions and direction of motion are shown in the figure:
[12 Jan. 2019 II]
𝜋
The relative velocity vA⃗⃗ − v⃗B⃗ and t = 2𝑤 is given by:
(a) w(R1 + R 2 ) î (b) −w(R1 + R 2 ) î (c) w(R 2 − R1 ) î (d) w(R1 − R 2 ) î

sol. (c) From, 𝜃 = wt =
2
So, both have completed quarter circle
Relative velocity,
vA − vB = 𝑤R1 (−î) − wR 2 (−i) = w(R 2 − R1 )i
43. A particle is moving along a circular path with a constant speed of 10 ms −1. What is the magnitude
of the change in velocity ofthe particle, when it moves through an angle of 60∘ around the centre
of the circle?
(a) 10√3m/s (b) zero (c) 10√2m/s (d) 10 m/s
sol. (d)
Change in velocity, |𝛥v| = √v12 + v22 + 2v1 v2 cos (𝜋 − 𝜃)
𝜃
= 2v sin (⋅.⋅ |v
⃗ 1 | = |v
⃗ 2 |) = v
2
1
= (2 × 10) × sin (30∘ ) = 2 × 10 ×
2
= 10m/s
44. If a body moving in circular path maintains constant speed of 10 ms−1 , then which of the
following correctly describes relation between acceleration and radius?
sol. (c) Speed, 𝑉 = constant (from question)
v2
Centripetal acceleration, a 
r
𝑟𝑎 = constant
Hence graph (c) correctly describes relation between acceleration and radius.
NEWTON’S LAWS OF MOTION
Inertia :
 It is the inability of a body to change its state of rest or of uniform motion or its direction by itself.
 Mass is a measure of inertia in translatory motion
 Heavier the mass, larger the inertia & vice–versa.
Types of inertia: There are three types of inertia. (i) Inertia of rest (ii) Inertia of motion and
(iii)Inertia of direction.
 Inertia of rest: It is the inability of a body to change its state of rest by itself .
Ex:When a bus is at rest and starts suddenly moving forward the passengers inside it will fall back.
Inertia of motion: It is the inability of a body to change its state of uniform motion by itself.
Ex: Passengers in a moving bus fall forward, when brakes are applied suddenly.
Inertia of direction: It is the inability of a body to change its direction of motion by itself .
Ex:When a bus takes a turn, passengers in it experience an outward force.
 A person sitting in a moving train, throws a coin vertically upwards, then
i) it falls behind him, if the train is accelerating
ii) it falls infront of him, if the train is retarding
iii) it falls into the hand of the person, if the train is moving with uniform velocity.
iv) It falls into the hand of the person if the train is at rest
Newton's First Law ( law of Inertia)
 Every body continues to be in its state of rest (or) uniform motion in a straight line unless it is
acted upon by a net external force to change its state
 It defines inertia,force and mechanical equilibrium.
 If the net external force on an object is zero , then acceleration of object is zero.
Linear momentum :
 
 Linear momentum is the product of the mass of a body and its velocity. p  mv
 Linear momentum is a vector.It has the same direction as the direction of velocity of the body. SI
unit: kg m s-1 , CGS unit: g cm s-1
 D.F: MLT-1
Change in momentum of a body in different cases
 
 Consider a body of mass m moving with velocity vi and momentum Pi . Due to a collision (or) due
 
to the action of a force on it suppose its velocity changes to v f and momentum changes to Pf in a
small time interval D t .
  
Change in momentum of body = D P = Pf - Pi
Where Pi  initial momentum
Pf  final momentum
  
D P = mv f - mvi
    
D P = Pf - Pi = Pf2 + Pi 2 - 2 Pf Pi cos q where q = angle between Pf and Pi
Consider a body of mass ‘m’ moving with velocity ‘ v ’along a straight line
 Case (i) : If it hits a wall and comes to rest, Change in momentum of the body
Y
m
v
   
D P = Pf - Pi = 0 - (mv )iˆ
ˆ ; D P = mv , along the normal and away from the wall.
= - mvi
 Case(ii) : If the body rebounds with same speed ‘ v ’ then q = 180 0
m v
     
D P = Pf - Pi = éêë- (mv )iˆùûú- éëê(mv )iˆùúû= - (2mv )iˆ

\ D P = 2mv , along the normal and away from the wall.
 Case (iii) : If the body hits a rigid wall normally with speed v1 and rebounds with speed v2 then   1800 ,
m v1
v2
    
D P = P f - Pi    mv2  iˆ   mv1  iˆ ,

D P = m (v2 + v1 ), along the normal and away from the wall.
 Case (iv) : A body of mass ‘m’ moving with speed ‘ v ’ hits a rigid wall at an angle of incidence q
and rebounds with same speed ‘ v ’

 P is along the normal,away from the wall
 
 P x   mv cos  i  mv cos  i  P y  mv sin  j  mv sin  j
  
 P   P x   P y  2mv cos   i 

D P = 2mv cos q
mv
P 

mv

 Case(v) : In the above case if q is the angle made with wall then D P = 2 mv sin q , along the
normal and away from the wall.
 Case(vi) : Projectile motion :

u
u sin
 ucos 
x
u cos 

usin u
a)In case of projectile motion the change in

momentum of a body between highest point and point of projection is

Pi =  mu cos   iˆ   mu sin   ˆj
 
Pf =  mu cos   iˆ  0 , P = -  mu sin   ˆj
b) The change in momentum of the projectile between the striking point and point of projection is

Pi   mu cos   iˆ   mu sin   ˆj
 
Pf   mu cos   iˆ   mu sin   ˆj  P = -  2mu sin  ˆj
 A particle of mass ‘m’ is moving uniformly with a speed ‘ v ’ along a circular path of radius ‘r’. As it
moves from a point A to another point B, such that the arc AB subtends an angle  at the centre, then
the magnitude of change in momentum is 2 m v sin(  /2) and is directed towards the centre of the
circle.
Pf
B
r

A
Pi
Newton's second law:
 The rate of change of momentum of a body is directly proportional to the resultant (or) net external
force acting on the body and  takes place along the direction of force.


 F net 
dp 
(or) F net 
 
d mv
dt dt 
 dv 
 In a system if only velocity changes and mass remain constant , F net  m  ma
dt
  dm
 In a system, if only mass changes and velocity remians constant F net  v
dt
 Force is a vector and the acceleration produced in the body is in the direction of net force,
 SI unit : newton (N). CGS unit :dyne.
 One newton = 105 dyne.
 D.F=MLT-2
Gravitational units of force: Kilogram weight (kg wt) and gram weight (g wt); 1 kg.wt =
9.8 N, 1 gm.wt= 980 dyne.
 A metallic plate of mass ‘M’ is kept held in mid air by firing ‘n’ bullets in ‘t’ seconds each of mass
‘m’ with a velocity ‘v’ from below.
mnv
(a) If the bullet falls dead after hitting the plate then  Mg
t
2m n v
(b) If the bullet rebounds after hitting the plate with same velocity then  Mg
t
EX.1: A force produces an acceleration16 ms-2 in a mass 0.5 kg and an acceleration 4 ms-2 in an
unknown mass when applied separately. If both the masses are tied together, what will be
the acceleration under same force?
Sol. Force F=ma=0.5 when both masses are joined and same force acts, acceleration is given
F 8
by a 
1
  3.2ms 2
m  m 0.5  (8 / 4)
1
EX.2: When the forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 aree
mutually perpendicular, then the particle remain stationary. If the force F1 is now removed,
then find the acceleration of the particle .
Sol. If mass 'm' is stationary under three forces,
O
90
F2 F3
  
F1  F2  F3  0
  
m 
F1   F2  F3 
F22  F32  F1
F1
F22  F32 F1
Obviously if F1 is removed then the mass will have acceleration, a  (or ) a 
m m
EX.3: A body of mass m=3.513 kg is moving along the x-axis with a speed of 5ms-1.The
magnitude of its momentum is recorded as
Sol. m=3.513kg,v=5ms-1 momentum,
p = mv=3.5135 =17.565kgms-1
EX.4: A very flexible chain of length L and mass M is vertically suspended with its lower end
just touching the table. If it is released so that each link strikes the table and comes to rest.
What force the chain will exert on the table at the moment ‘y’ part of length falls on the table ?
M 
Sol. Since chain is uniform , the mass of ‘y’ part of the chain will be  L y  . When this part reaches the
 
table, its total force exerted must be equal to the weight of y part resting on table + Force due to the
momentum imparted
M 
 dy  2 gy Mg M
F M yg  
L  = y  v. 2 gy
L dt L L
 dy  Mg M My
  v   y 2 gy . 2 gy = 3 g
 dt  L L L
EX.5: A body of mass 8kg is moved by a force F =(3x)N, where x is the distance covered.
Initial position is x  2 m and final position is x =10m. If initially the body is at rest, find the
final speed. [2014E]
dv dv dx
Sol: F=ma  F=m  3 x =m
dt dx dt
dv
3x  8 v  3xdx  8vdv
dx
10 v
10 v
 x2   v2 

2 0

3 xdx  8 vdv   
3
 2 2
 8 2
 0
2 3  96
3[100-4]=8 v  v   36  v =6ms-1
2
8
EX. 6: Sum of magnitudes of the two forces acting at a point is 16 N. If their resultant is
normal to the smaller force, and has a magnitude 8 N, then the forces are ( 2012E)
Sol. F1  F2  16 ——(1) Resultant force is perpendicular to F1, then F22  F12  F 2
F22  F12  82  ( F2  F1 )( F2  F1 )  64
 F2  F1  16  64 ————(2)
Solving(1) &(2),we get F1  6 N , F2  10 N
 b
EX.7: A particle is at rest at x=a. A force F   2 i begins to act on the particle. The particle starts
x
its motion, towards the origin, along X–axis. Find the velocity of the particle, when it reaches
a distance x from the origin.
b d b
Sol. F   x 2  dt  p    x 2
d b dv dx b
 mv    2  m   2
dt x dx dt x
b b
mvdv   2
dx  vdv   2 dx
x mx
v x x
b v2 b  1 
 v dv   
0 a
mx 2
dx  
2 m  x  a
v2 b  1 1  2b  a  x 
  v 
2 m  x a  
m  xa 

EX.8: A particle of mass m is at rest at the origin at time t=0. It is subjected to a force F(t)= F0 e bt
in the X-direction. Its speed V(t) is depicted by which of the following curves.
(AIEEE-2012)
F0 F0b
mb m
1) v(t) 2) v(t)
t t
F0b F0
m mb
3) v(t) 4) v(t)
t t
Sol: As the force is exponentially decreasing, its acceleration, i.e,rate of increase of velocity will decrease
with time.Thus, the graph of velocity will be an increasing curve with decreasing slope with time.
v t
F F0  bt dv F0  bt  dv  F0 e  bt dt
a  e 
m m
 e
dt m
0 0 m
t 0
F  1   F  1  
 v   0   e  bt    0   e bt 
 m  b   0 m b  t
F0 0  bt F

mb
 e  e   0 1  e  bt 
mb
F0
So, velocity increases continuously and attains a maximum value, vmax 
Ans: 3
mb
EX.9: A bus moving on a level road with a velocity v can be stopped at a distance of x ,
by the application of a retarding force F. The load on the bus is increased by 25% by
boarding the passengers. Now, if the bus is moving with the same speed and if the
same retarding force is applied, the distance travelled by the bus before it stops is
[2014E]
Sol :By using equations of motion v2 - u2 = 2as
F F
v 2  u 2  2   s u 2  2   s
m m
Fs 2Fs m1 s1
u2  2  m   m  s m s
m u2 2 2
Given s1 = x ,m1 = m, and

25 m 5m m x 5x
m2 = m + m = m + =  5m / 4  s  s2   1.25x  m
100 4 4 2 4
Applications of variable mass :

 When a machine gun fires ‘n’ bullets each of mass ‘m’ with a velocity v in a time interval ‘t’
nmv
then force needed to hold the gun steadily is F 
t
 When a jet of liquid coming out of a pipe strikes a wall normally and falls dead , then force exerted
by the jet of liquid on the wall is F=Adv2 A = Area of cross section of the pipe v = Velocity of jet
d = density of the liquid
 If the liquid bounces back with the same velocity then the force exerted by the liquid on the wall
is F  2 Adv 2
 If the liquid bounces back with velocity v ' then the force exerted on the wall is F  Adv(v  v)
 When a jet of liquid strikes a wall by making an angle '  ' with the wall with a velocity ‘ v ’ and
rebounds with same velocity then force exerted by the water jet on wall is F  2 Adv 2 sin 
dm
 If gravel is dropped on a conveyor belt at the rate of ,extra force required to keep the belt
dt
 dm 
moving with constant velocity ' u ' is F  u  
 dt 
EX.10:A gardener is watering plants at the rate 0.1litre/sec using a pipe of cross- sectional area
1 cm2. What additional force he has to exert if he desires to increase the rate of watering two
times?
 Av 
2
d
Sol :F = Ad v 2 = . If rate of watering of plant (A v ) is doubled, it means that the amount of water
A
poured/sec is doubled which is possible only if velocity is doubled. Hence, force is to be made 4 times.
 Av 
2
 Additional force = 3 times initial force  3 Adv 2

3 d
A
3  0.1 0.1 103
 4
 3  105 N
10
EX.11: A liquid of density  flows along a horizontal pipe of uniform cross – section A with a
velocity v through a right angled bend as shown in Fig. What force has to be exerted at the
bend to hold the pipe in equilibrium?
Sol :Change in momentum of mass m of liquid as it passes through the bend
v
45°
P  Pf  Pi  2mv
P m
F
t
 
 2 v
t
;  as m   AL 
  . AL  ;
F  2v  as L / t  v 
t
F  2  Av 2
EX.12:A flat plate moves normally with a speed v1 towards a horizontal jet of water of uniform
area of cross section. The jet discharges water at the rate of volume V per second at a speed of
v2 . The density of water is r . Assume that water splashes along the surface of the plate at right
angles to the original motion. The magnitude of the force acting on the plate due to the jet is
dp dm dm V
Sol. Force acting on the plate F   ur Since Av2  V   A(v1  v2 )  (v1  v2 ) 
dt dt dt v2
( ur  v1  v2 = velocity of water coming out of jet w.r.t plate)
V V
F  (v1  v2 ). (v1  v2 )   (v1  v2 ) 2  N
v2 v2

Impulse ( J ) :
 It is the product of impulsive force and time of act ion that produces a finite change in momentum
of body.
 J=Ft = m(v-u) = change in momentum. SI unit: Ns (or) Kg - ms-1; DF: MLT-1
 It is a vector directed along the force
 change in momentum and Impulse are always in the same direction.
 For constant force, J=Ft,
 Impulsive force is a variable, then
 t2
 d p
F J   Fdt
dt , t1
 The area bounded by the force-time graph measures Impulse.

F
t1 t2
t
Application of Impulse :
a) shock absorbers are used in vehicles to reduce the magnitude of impulsive force.
b) A cricketer lowers his hands, while catching the ball to reduce the impulsive force.

EX.13::Find the impulse due to the force F  ai  bt j , where a=2 N and b=4 Ns -1 if this force acts
f r om t i=0 to tf=0.3s
tf
 0.3
 
Sol: J  Fdt  (ai  bt j )dt
ti 0
0.3 0.3 0.3

t2 
J  a  dt i  b  t dt j  a t 0 i  b   j
0.3
0 0  2 0
 0.3  j = 
2
 2  0.3  i  4  0.6i  0.18 j NSec

2

EX.14:A ball falling with velocity v i  (0.65i  0.35 j ) ms -1 is subjected to a net impulse

 
I  0.6i  0.18 j Ns. If the ball has a mass of 0.275kg, calculate its velocity immediately
following the impulse

     I
Sol: mv f  mvi  I ; v f  vi 
m
 0.6iˆ  0.18 ˆj
v f  0.65iˆ  0.35 ˆj 
0.275

v f  0.65iˆ  0.35 ˆj  2.18iˆ  0.655 ˆj


v f  1.53iˆ  0.305 ˆj ms  1 
EX.15:A body of mass 2kg has an initial speed 5 ms 1 . A force acts on it for some time in the
direction of motion. The force–time graph is shown in figure. Find the final speed of the body
F(N)
A B
4
C D
2.5
F G H E
O 2 4 4.5 6.5
t  (sec)
1
Sol. Area of OAF   2  4 = 4
2
Area of ABG F = 2  4 = 8
1
Area of BGHC   4  2.5   0.5 = 1.625
2
Area of CDEH = 2  2.5 = 5
Total area under F-t graph = Change in momentum
 m(v – u) = 18.625
18.625
 v  5  14.25ms 1
2
EX.16:A bullet is fired from a gun. The force on a bullet is, F  600  2  105 t newton. The force
reduces to zero just when the bullet leaves barrel.Find the impulse imparted to the bullet.
Sol. F = 600 – 2 × 105 t , F becomes zero as soon as the bullet leaves the barrel.
0 = 600 – 2 × 105 t  600  2  105 t
t
t = 3 × 10–3 s  Impulse =  Fdt

0
t 3103
 t2 
  600  2 10 t  dt
5
=  600t  2  105 
0  2 0
= 600 × 3 × 10–3 – 105 × 9 × 10–6 = 0.9Ns

Equilibrium: The necessary and sufficient conditions for the translational equilibrium of the rigid
body.
 F  0 ;  Fx  0 ,  Fy  0 ,  Fz  0 For rotational equilibrium
   0 :   x  0,   y  0 ,   z  0
  dv 
 dv 
 
 As for a body , F  0 , ma  0 as Fma
dt
 
 0 ( as m is finite and a  ) =constant or zero
dt v
 If a body is in translatory equilibrium it will be either at rest or in uniform motion.If it is at rest, the
equilibrium is called static,otherwise dynamic.
 If ‘ n ’coplanar forces of equal magnitudes acting simultaneously on a particle at a point, with the
360
angle between any two adjacent forces is ‘  ’ and keep it in equilibrium, then  
n
Lami’s Theorem :
  
 If an object O is in equilibrium under three concurrent forces F1 , F2 and F3 as shown in figure.
F1 F2 F
Then,   3
sin  sin  sin 
F2
 
F1

F3
 If the bob of simple pendulum is held at rest by applying a horizontal force ‘F’ as shown in fig
|||||||||||||||||||||||||||||||||||||||||||||||||||||||
 Tcos 
F
T sin
Mg
If body is in equilibrium
T sin   F , T cos  mg ,
F  mg tan  , F 2  m g   T
2
x l l 2  x2
 
F T mg
EX.17:A mass of 3kg is suspended by a rope of length 2m from the ceiling. A force of 40N in the
horizontal direction is applied at midpoint P of the rope as shown. What is the angle the rope
makes with the vertical in equilibrium and the tension in part of string attached to the ceiling?
(Neglect the mass of the rope, g= 10m/s2)
Sol :Resolving the tension T1 into two mutually perpendicular components, we have
T1 cos   W  30 N T1 sin   40 N
4 1  4 
 tan   (or)   tan    53
0
3  
3
EX.18:A mass M is suspended by a weightless string. The horizontal force required to hold the
mass at 600 with the vertical is (2013E)
Sol :
 l T cos
T 
T sin 
F
X
Mg
F  T sin  - - - - - (1)
Mg  T cos  - - - - - (2)
Dividing Eq.(1) and Eq.(2)
F T sin 

Mg T cos  F  Mg tan 
;
F  Mg tan 600 ; F  3Mg

EX.19:A chain of mass 'm' is attached at two points A and B of two fixed walls as shown in the
figure. Find the tension in the chain near the walls at point A and at the mid point C.
A B
 
C
Sol.
i)
T sin  T sin 
2T sin 
T T
 
T cos   T cos 
  T cos 
T cos 
1
2T sin   mg  T  mg cos ec
2
ii) Tension along horizontal direction is same everywhere
(no external force is acting on it in horizontal direction.)
mg cos  mg cot 
T 1  T cos   
2 sin  2
Newton's third law:
 For every action there is always an equal and opposite reaction
 Action and reaction do not act on the same body and they act on different bodies at same instant of
time
 Action and reaction, known as pair of forces, are equal in magnitude and opposite in directions
acting on different bodies in interaction. So they never cancel each other
 Newton’s third law is not applicable to pseudo forces.
 Newton’s third law defines nature of force and gives the law of conservation of linear momentum.
Examples:
 When we walk on a road we push the road backwards and road applies equal (in magnitude) and
opposite force on us, so that we can move forward.
 When we swim on water we push water backward and water applies equal (in magnitude) and
opposite force on us,so that we can move forward.
 A bird is in a wire cage hanging from a spring balance. When the bird starts flying in the cage, the
reading of the balance decreases.
 If the bird is in a closed cage (or) air tight cage and it hovers in the cage the reading of the spring
balance does not change.
 In the closed cage if the bird accelerates upward the reading of the balance is R  Wbird  ma
Limitations of newton’s third law:-
 Newton’s third law is not strictly applicable for the interaction between two bodies separated by
large distances, of the order of astronomical units.
 It does not apply strictly when the objects move with velocity nearer to that of light
 It does not apply where the gravitational field is strong.
Normal reaction/force : Normal force acts perpendicular to the surfaces in contact when one
body tries to press on the surface of the second body.In this way second body tries to push
away the first body.
A
A A
A N
B B B B
 When the body lies on a horizontal surface N = mg

N
mg
 When the body lies on an inclined surface
N = mg cos θ .
N
 mg sin
mg cos mg

Free Body Diagram:- When several bodies are connected by strings, springs, surfaces of
contact, then all the forces acting on a body are considered and sketched on the body under
consideration by just isolating it . Then the diagram so formed is called Free Body Diagram (FBD).
Some examples:
i) A block is placed on a table and the table is kept on earth.Assuming no other body in the universe
exerts any force on the system,make the FBD of block and table.
m1
m2
FBD of block, N1  m1 g
FBD of table N 2  N1  m2 g  m1 g  m2 g   m1  m2  g
N1
m1
N1
m1g
N2
m 2g
ii) A block of mass M is suspended from the ceiling by means of a uniform string of mass m.Find
the tension in the string at points A,B and C. B is the mid point of string. Also find the tensions A,B
and C if the mass of string is negligible or it is massless.
ii) A block of mass M is suspended from the ceiling by means of a uniform string of mass m.Find
the tension in the string at points A,B and C. B is the mid point of string. Also find the tensions A,B
and C if the mass of string is negligible or it is massless.
A
B m
M
Tension at any point will be weight of the part below it.
m 
So, TA   M  m g,TB    M  g,TC  Mg .
2 
Now if the string is massless: m=0 then TA  TB  TC  Mg . So in a massless string, tension
is the same at every point.
(iii) Find the tension in the massless string
connected to the block accelerating upward.
T
a
T
m m
mg mg
Net force :
Fnet  T  mg Now apply Fnet  ma
 T  mg  ma  T  mg  ma  m  g  a 
Note: If ‘a’ is downward, then replace a with -a; we get T  m  g  a 

In free fall a=g then T=0.
Frames of Reference:
 A system of coordinate axes which defines the position of a particle or an event in two or three
dimensional space is called a frame of reference.
There are two types of frames of reference
a) inertial or unaccelerated frames of reference
b) non-inertial or accelerated frames of reference
Inertial frames of reference :
a) Frames of reference in which Newton’s Laws of motion are applicable are called inertial frame.
b) Inertial frames of reference are either at rest or move with uniform velocity with respect to a
fixed imaginary axis.
c) In inertial frame, acceleration of a body is caused by real forces.
d) Equation of motion of mass ‘m’ moving with acceleration ‘a’ relative to an observer in an

inertial frame is  real  ma
F
Examples:
1) A lift at rest,
2) Lift moving up(or)down with constant velocity,
3) Car moving with constant velocity on a straight road.
Real Force : Force acting on an object due to its interaction with another object is called a real
force.
Ex: Normal force, Tension, weight, spring force, muscular force etc.
a) All fundamental forces of nature are real.
b) Real forces form action, reaction pair.
Non-Inertial frames :
a) Frames of reference in which Newton Laws are not applicable are called non-inertial frames.
b) Accelerated frames move with either uniform acceleration or non uniform acceleration.
c) All the accelerated and rotating frames are non-inertial frames of reference.
d) Examples:
1) Accelerating car on a road.
2) Merry go round.
3) Artificial satellite around the earth.
Pseudo force :
a) In non-inertial frame Newton’s second law is not applicable. In order to make Newton’s second
law applicable in non-inertial frame a pseudo force is introduced.

b) If a is the acceleration of a non-inertial frame, the pseudo force acting on an object of mass m,
 
as measured by an observer in the given non-inertial frame is F Pseudo  ma
i.e. Pseudo force acts on an object opposite to the direction of acceleration of the non-inertial
frame.
c) Pseudo forces exist for observers only in non-inertial frames, such forces have no existence
relative to an inertial frame.
  
d) Equation of motion relative to non-inertial frame is F real 
 F Pseudo  ma
Where a is the acceleration of body as measured in non-inertial frame.

e) Earth is an inertial frame for an observer on the earth but it is an accelerated frame for an
observer at centre of earth (or) in a satellite.
Examples : (i) Centrifugal force and deflection of pendulum relative to accelerating car.(ii) Gain
or loss of weight experienced in an accelerating elevator.
Apparent weight of a body in a moving elevator
Weight of a body on a surface comes due to the reaction of a supporting surface, i.e.,Apparent
weight of a body in a lift
Wapp  Reaction of supporting surface. Consider a person standing on a spring balance , or in a lift.
The following situations are possible:
Case(i) :If lift is at rest or moving with constant velocity then the person will be in translatory
equilibrium. So, R  mg
 Wapp  mg [as Wapp  R ]
or Wapp  W0 [as W0  mg  true weight]
R R R
a a
70 kg 80 kg 60 kg
mg mg mg
(a) (b) (c)
i.e., apparent weight (reading of balance) will be equal to true weight.
Case(ii) : If lift is accelerated up or retarding down with acceleration a from Newton's II law we
have
R  mg  ma or R  m  g  a 
 a  a
or Wapp  m  g  a   mg 1    W0 1   or Wapp W0
 g  g
i.e., apparent weight (reading of balance) will be more than true weight.
Case (iii) : If lift is accelerated down: or retarding up with acceleration ‘a’ mg  R  ma i.e.,
R  m g  a
 a
or Wapp  m  g  a  [as Wapp  R ]  mg 1  g 

 a
i.e., Wapp  W0 1   Wapp W0
 g
i.e., apparent weight (reading of balance) will be lesser than true weight.
Note: If a  g , Wapp will be negative; negative weight will mean that the body is pressed against
the roof of the lift instead of floor (as lift falls more faster than the body) and so the reaction will
be downwards, the direction of apparent weight will be upwards.
Case (iv) : If lift is in freely falling, Then a=g ,
So mg  R  mg i.e., R  0 . So, Wapp  0
v v
Satellite a=g
a=g
Planet
a=g
(a ) (b) (c)
( a ) Freely falling lift
( b ) Satellite motion
( c ) Projectile motion
i.e., apparent weight of a freely falling body is zero.

 This is why the apparent weight of a body is zero, or body is weightless if it is in a (i) lift whose
cable has broken, (ii) orbiting satellite.
EX.20: A mass of 1kg attached to one end of a string is first lifted up with an acceleration 4.9m/
s2 and then lowered with same acceleration. What is the ratio of tension in string in two
cases.
Sol :When mass is lifted up with acceleration 4.9m/s2 T1 = m( g  a ) =1 (9.8 + 4.9)=14.7N
When mass is lowered with same acceleration T2 = m( g  a) =1(9.8 – 4.9)=4.9 N
T1 14.7
   3 :1
T2 4.9
EX.21:The apparent weight of a man in a lift isW1 when lift moves upwards with some
acceleration and is W2 When it is accelerating down with same acceleration. Find the true weight of the
man and acceleration of lift.
Sol :(a) W1  m( g  a),W2  m( g  a )
W1  W2
W1  W2  2mg  W1  W2  2W (W  mg )  W
2
W1 m( g  a ) g  a
(b) W  m( g  a )  g  a
2
g W1  W2  W  W2 
  a  g 1 
a W1  W2  W1  W2 
Connecting Bodies:
 If masses are connected by strings then acceleration of system and tension in the strings on smooth
horizontal surface are
F T T
M1 M2
a)
Free body diagram for M2
T M2 T=M2a .....(1)
F M1 T F-T=M1a ....(2)
from (1) and (2)

F M 2F
a and T 
 M1  M 2   M1  M 2 
T1 T1 T2 T2
b) F M1 M2 M3
F (M 2  M 3 ) F
a ; T1 
M1  M 2  M 3 (M1  M 2  M 3 )
M 3F
T2 
(M 1  M 2  M 3 )
 If masses are connected by a string and suspended from a support then tension in the string when
force F is applied downwards as shown in the figure
T2
M1
T1
M2
F
T1
M2
T1  F  M 2 g ............(1)
F
M2g

T2
M1
T2  T1  M 1 g ............(2)
T1
M1g
From (1) and (2), T2  F  ( M1  M 2 ) g

EX.22: Three blocks connected together by strings are pulled along a horizontal surface by
applying a force F. If F= 36N, What is the tension T2?
27kg
8kg
1kg
T1 T2
F
Sol :Suppose the system slides with acceleration ‘a’.
m1  1kg , m2  8kg , m3  27kg
F  T2  m3a , T2  T1  m2 a , T1  m1a
Solving the above equations ,we get
F 36 36
a    1 ms 2
m1  m2  m3 1  8  27 36
From the above equation, T2  F  m3 a
T2  36  27  1 =9 N
Contact Forces : When two objects are in contact with each other, the molecules at the interface
interact with each other. This interaction results in a net force called contact force. The contact
force can be resolved into two components.
(a) Normal force (N): Component of the contact force along the normal to the interface. Normal
force is independent of nature of the surfaces in contact.
(b) Friction (f): Component of the contact force along the tangent at the interface. Friction
depends on the roughness of the surfaces in contact. This component can be minimised by polishing
the surfaces.
 The tension and contact forces are self adjustable forces. Their magnitude and direction change
when other forces involved in a physical arrangement change.
 Masses are in contact on a smooth horizontal surface:
F f1
f2
M1 M2
f1=f
M2
contact force f1= f2= f = M2a

free body diagram forM1
F f2=f
M1
F-f = M1a .........(1)
free body diagram for M2
f1=f
M2
f=M2a ...........(2)
From (1) and (2)
F M2F
a , contact force, f=
(M1  M 2 ) M1  M 2
 Contact forces are as shown in the figure
F
M1 M2 M3
a) Acceleration of system, F
a
(M1  M 2  M 3 )
b) Contact force between M1 and M 2 f = (M2+M3)a
c) Contact force between M2 and M3, f1 = M3a
Atwood’s Machine :
 Masses M1 and M2 ( M1 > M 2 )are tied to a string , which passes over a frictionless light pulley The
string is light and inextensible.
T T
T T
M2
M1
M1  M 2  g
Acceleration of the system, a  M1  M 2
 2M M 
Tension in the string , T   M  M  g
1 2
 1 2 
 4M M 
Thrust on the pulley , 2T   M  M  g
1 2
 1 2 

 If the pulley begins to move with acceleration a then
 M M   2M M 
i) If the pulley accelerates upward , then anet   M  M  (g  a) and Tnet   M  M  (g  a)
1 2 1 2
 1 2  1 2
 M  M2 
ii) If the pulley accelerates downward, then anet   1  ( g  a) and
 M1  M 2 
 2 M 1M 2 
Tnet    ( g  a)
 M1  M 2 
4 M 1M 2
 Thrust on the pulley when it comes downward with acceleration ‘a’ is T   M  M  ( g  a)
1 2
EX.23:The maximum tension a rope can withstand is 60 kg-wt.The ratio of maximum acceleration
with which two boys of masses 20kg and 30kg can climb up the rope at the same time is
(2011E)
Sol. m1  20kg , m2  30kg , T  60kgwt  600 N
For ‘ m1 ’; T  m1 g  m1a1
600  20  10  20  a1  a1  20ms 2 For ‘ m2 ’; T  m2 g  m2a2
600  30 10  30  a2  a2  10ms 2
a1 : a2  20 :10  2 :1
EX.24:Figure shows three blocks of mass ‘m’ each hanging on a string passing over a pulley.
Calculate the tension in the string connecting A to B and B to C?
Sol. Net pulling force = 2 mg – mg = mg
Total mass = m+ m + m = 3m
T1 T1
B a
a A
T2
T2
mg C a
mg
mg g
Acceleration, a  
3m 3
Considering block A,
T1  mg  ma ; T1  mg  ma
g 4
T1  mg  m    T1  mg
3
  3
T2 T1
a C a A
mg mg
Considering block C,
mg  T2  ma  T2  mg  ma
mg 2
 T2  mg   mg .
3 3
EX.25:A man of mass 60 kg is standing on a weighing machine kept in a box of mass 30 kg as
shown in the diagram. If the man manages to keep the box stationary, find the reading of the
weighing machine.
T T
Sol. we know that Normal reaction = scale reading

For man , T  Mg  R
T
T
m R
R
Mg
mg
For box : T  mg  R
Mg  R  mg  R ; 2 R =(M –m)g
(60 - 30)´ 10
R= = 150 N
2
EX.26:Two unequal masses are connected on twosides of a light string passing over a light and
smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped
for a moment, 1sec after the system is set into motion. Find the time elapsed before the string is
tight again . (g = 10 m/s2)
1kg
2kg
Sol. Net pulling force = 2g – 1g = 10N Mass being pulled = 2 + 1 = 3 kg
10
Acceleration of the system is a = 3
m / s2 Velocity of both the blocks at t = 1 s will be
 10  10
v0  at =  
 3 
(1) = 3
m/s. Now, at this moment velocity of 2kg block becomes zero, while that of
10
1kg block is 3 m/s upwards. Hence, string becomes tight again when displacement of 1 kg block
= displacement of 2 kg block.
1 2 1 2
v0t  gt  gt  gt 2  v0t
2 2
v0 (10 / 3) 1
t   s
g 10 3
EX.27:In the figure, if m1 is at rest, find the relation among m1 , m2 and m3 ?
Sol. m1 is at rest  point B does not move, m2 and m3 move with acceleration
A
1
T
T1 B
T T
m1 m2 m3
 m  m2 
a 3 g ; m3  m2
 m 2  m3 
2m2 m3 g 4m2 m3 g
T = m m ; T 1  2T 
2 3 m2  m3
4m2m2 g 4 1 1
m1 g   
m2  m3 m1 m 2 m 3
 Two blocks are connected by a string passing over a pulley fixed at the edge of a horizontal table
then the acceleration of system and tension in the string  M 2  M 1 
a
T
M1
M2 a
M2g
M 2 g  T  M 2 a and T  M 1a
M2g
a
 1  M2 
M
M 1M 2 g
T  M 1a 
 M1  M 2 
 Acceleration and Tension in the string when bodies are connected as shown in the figure if
M1 > M3.
T2 T1
M2
T2 T1
a M3 M1 a
M 1 g  T1  M 1a ; T1  T2  M 2 a
T2  M 3 g  M 3a
a
 M1  M 3  g
 M1  M 2  M 3 
M 3 g (2 M 1  M 2 ) M 1 g (2 M 3  M 2 )
T2  T1 
M1  M 2  M 3 ; M1  M 2  M 3
 Masses are attached to a string passing through the pulley attached to the edge of an inclined plane,
acceleration of system and tension in the string if M2 moves down
 M  M 1 sin  
a 2 g;
 M1  M 2 
 M M (1  sin  ) 
T  1 2 g
 ( M1  M 2 ) 
T
a
M1 T
 M2
Thrust on the pulley : Resultant Tension =
Tg  T 2  T 2  2T 2 cos(90   )
Tg  2T 2 (1  sin  )  T 2(1  sin  )
EX.28:By what acceleration the boy must go up so that 100 kg block remains stationary on the wedge.
The wedge is fixed and is smooth.(g = 10m/s2)
g
0k
10
M
50 kg
m
53°
Sol :For the block to remain stationary,

4
T  Mg sin  = 100  10  sin 53= 100  10  = 800N
5
For man ; T – mg = ma
T = m(g + a )  800  50 10  a  a = 6m/s2
EX.29:The system as shown in fig is released from rest. Calculate the tension in the strings and
force exerted by the strings on the pulley. Assuming pulleys and strings are massless
3 kg
2 kg 1 kg
Sol: T1-1g=1a — (1)

T2-T1=3a — (2)
2g- T2=2a —(3)
Solving the above equations,
g
we get , a  m / s2
6
P2 T2 T2 T1 T1
3kg P1
T2 T1
T2 F2 F1 T1
2 kg 1 kg
2g 1g
7g 5g
T1  N , T2  N
6 3
7g
Force on pulley P1 is F1  T12  T12  2T1  N
3 2
5 2g
Force on pulley P2 is F2  T22  T22  2T2 
N
3
 If position of masses is interchanged,then the tension in the string and acceleration remains
unchanged.
T T a
a
M1 M2
180(+)
 
 If M2 slides down then M1 moves up on double smooth inclined plane then the acceleration of
 M 2 sin   M1 sin  
system and tension in the string are given by, acceleration , a =  g
 M1  M 2 
M 1M 2 g
Tension , T =  M  M  (sin   sin  )
1 2
Resultant Tension
TR  T 2  T 2  2T 2 cos 180  (   ) 
 2T 2 1  cos     
Note:- If M 2 sin   M 1 sin   a  0
 System does not accelerate
EX.30:In the adjacent fig, masses of A, B and C are 1kg, 3kg and 2kg respectively. Find a) the
acceleration of the system b) tension in the string (g = 10m/s2)
B 90°
C
A
60° 30°
Sol . a) In this case net pulling force

= mA g sin 600  mB g sin 60 0  mC g sin 300
3  3 1
=(1)(10) +(3)(10)  2  -(2)(10)  2  =24.64N
2    
Total mass = 1 + 3 + 2 = 6kg
24.64
 Acceleration of the system a   4.1 m / s 2
6
b) For the tension in the string between A and B.
FBD of Body A
T1
a A
mAg sin 60
mA g sin 60  T1  mA a
T1  mA g sin 60  mA a  mA  g sin 60  a 
 3 
T1  1  10   4.1  4.56 N
 2 
For the tension in the string between B and C
FBD of body C
T2
a
mcg sin 30
T2  mC g sin 300  mC a ; T2  mC  g sin 30  a 

0
 1 
T2  2  10    4.1  18.2 N
 2 
 A force F is applied on the massless pulley as shown in the figure and string is connected to the
block on smooth horizontal surface. Then
T T
T T F
m
F= 2T and T  mablock
 If the block moves a distance ‘x’ the pulley moves x/2 (Total length of the string remains constant)
ablock
Therefore acceleration of the pulley =
2
T F /2 F
=   .
2m 2m 4m
Constrained Motion:
 (a) Constraint : Restriction to the free motion of body in any direction is called constraint.
(b) Constrained Body : A body, whose displacement in space is restricted by other bodies,
either connected to or in contact with it, is called a constrained body.
(c) Kinematic Constraints : These are equations that relate the motion of two or more
particles.
(d) Types of Constraints :
i) General constraints ii) Pulley constraints
iii) Wedge constraints iv) Mixed constraints
General Constraints:
i) A body placed on floor : The floor acting as a constraint restricts the kinematical quantities in
the downward direction such that
m
x
y = 0 ; v y  0 and a y  0 for the body placed on the floor..

ii) Two bodies connected with a string or rod.
inextensible string inextensible rod
A B A B
a b
The string / rod is inextensible.
 Displacements of A and B are equal in horizontal direction  s A  sB
ds A dsB
Differentiating w.r.t time,   v A  vB
dt dt
dv A dvB
Again differentiating   a A  aB
dt dt
iii)Two bodies in contact with each other
 Displacement of A and B are equal in horizontal direction.
A
B
 s A  sB
By differentiating, we will get v A  vB and a A  aB in horizontal direction
String Constraints:
 For example, the motion of block A is downwards along the inclined plane in fig. will cause a
corresponding motion of block B up the other inclined plane.Assuming string AB length is
inextensible, i.e., length of AB is constant.
xA A B xB
A B
 
|||||||||||||||||||||||||||||||||||||||||||||||||||||||
 The displacements of A  x A  and B  xB  are equal  x A  xB

Differentiating w.r.t. time,  v A  vB
Once again differentiating w.r.t. time,  a A  a B
i.e., if one body (A) moves down the inclined plane with certain acceleration, then the other body
will move up inclined plane with an equal acceleration (magnitude).
Alternate Method : First specify the location of the blocks using position co-ordinates SA and
SB.
C D
SA SB
O
A B
 
From the fig. the position co-ordinates are related by the equation s A  lCD  sB  L
where lCD  the length of the string over arc
CD = constant L = total length of the string = constant Differentiating w.r.t. time, we get
ds A dsB
  0  vB   v A
dt dt
The negative sign indicates that when block A has a velocity downward, i.e., in the direction of
positive s A , it causes a corresponding upward velocity of block B, i.e., B moves in the negative
sB direction.
dvB dv
Again differentiating w.r.t. time,   A  a B   a A Similarly
dt dt
a
1)
a
B
x A  xB  v A  v B  a A  a B
2)
A B
x A  xB  v A  v B  a A  a B
Wedge constraints :
For wedges in contact the constriant is that velocity and acceleration along common normal is
same for both bodies
Ex. 30a : Find the relation between velocity of rod and that of wedge at any instant
m1
v1 v2
m2

Sol. Component of velocityalong perpendicular to the contact plane must be equal.
m1

v1
v2
90 0  
m2
v1 cos   v2 cos   90 0   
v1 cos   v2 sin 
v1
tan  
v2
Ex. 30b. Find the relation between a1 and a2
m2
H a2
a1
m1 
L
Sol. Accelaration of rod w.r.t wedge
a1
a2

ar / w  a1iˆ  a2 ˆj
a1

ar / w
a2

ar / w must be along the incline
a2
tan  
a1
Mixed constraints :
Ring sliding on a smooth rod :
 Consider a ring of mass m connected through a string of length L with a block of mass M. If the
ring is moving up with acceleration am and aM is the acceleration of block. As the length of the
string is constant,
L  d 2  y2  x
y

m x
Since, L is constant, differentiating with respect to time t, we get
dL 1 2y  dy  dx
 1   0
dt 2  dt  dt
d 2
y 
2 2
dy dx
Since  vm and  vM and
dt dt
y
cos  so vM  vm cos .
d 2  y2
By differentiating, relation between am and aM can be obtained, however, while doing so remember
that cos is not constant, but it is variable.
Two blocks connected with pulley : If the blocks are connected as shown in fig, then the
length of the string is
d d
y
  x
1 2
L  2 d 2  y2  x
Since, L is constant, differentiating with respect to time t, we get
dL 2 2y  dy  dx
 1   0
dt  dt  dt
2d 2  y 
2 2
 2v1 cos  v2  0 ; v 2   2 v1 cos 

EX.31:A rod of length ‘ l ’is inclined at an angle ‘  ’with the floor against a smooth vertical wall.If
the end A moves instantaneously with velocity v1 ,what is the velocity of end B at the instant
when rod makes ‘  ’ angle with the horizontal.
Sol: Let at any instant,end B and A are at a distance x and y respectively from the point ‘O’.
y
A
v1
l
y

O x B v2 x
Thus we have, x 2  y 2  l 2 .............(1)
Here l is the length of the rod,which is constant. Differentiating eq (1) with respect to time,we get
d 2 d dx dy
( x  y 2 )  (l 2 ) ; 2x  2y 0
dt dt dt dt
dx dy
If  v2 and  v1
dt dt
y
x(v2 )  y (v1 )  0  v2 
v1  v1 tan 
x
EX.32:In the fig, find the acceleration of mass m2
Sol: 1  2 2  cons tan t
on differentiating v1  2v2  0
Again differentiating
a1  2a2  0  a1  2a2
|||||||||||||||||||
T T
T 1
2 T
T
m1
2T
2T
m2
'- ve' sign indicates that the accelerations are in opposite direction.Suppose acceleration of m2 is
a2 downward and then acceleration of m1 will be a1 upwards.
T  m1 g  m1a1
T  m1 g  m1a1
m2 g  2T  m2 a2
a1
T 2T
m1 m2
a2
m1g m 2g
m2 g  2( m1 g  m1a1 )  m2 a2
m2 g  2 m1 g  m2 a2  4 m1a2 ( a1  2 a2 )
'–' sign should not be substituted
 m2  2m1  g
a2  ms 2
4m1  m2
EX.33:In the fig, find the acceleration of m1 and m2
m1 ||||||||||||||
a1
T
1
T 2
T
2T
m2
a2
m2 g
a1
Sol. 1  2 2  cons tan t m1 T
a1  2 a2 ; T  m1a1
m2 g  2T  m2 a2 ; m2 g  2 m1a1  m2 a2
2T
m2 m2 g 2m2 g
m2 g  2m1  2a2   m2 a2 a2  , a1 
a2 4m1  m2 4m1  m2
m2g
EX.34:A pendulum is hanging from the ceiling of a car having an acceleration a0 with respect to
the road. Find the angle made by the string with vertical at equilibrium. Also find the
tension in the string in this position.

a0
T cos 
T

ma0
T sin 
mg
Sol : T sin   ma0 ......  i  ; T cos   mg ......  ii 
a0
dividing (i)&(ii) tan  
g
 a0 
The string is making an angle   tan 1
 
 g
with the vertical at equilibrium
Squaring and adding (i) and (ii)
T 2 sin 2   T 2 cos 2   m 2  a02  g 2 
T  m a02  g 2
EX.35: For what value of ‘a’ the block falls freely?
h
a

x
Sol :In the time the wedge moves a distance ‘x’ towards left with an acceleration a the block falls from
a height ‘h’ with acceleration ‘g’
1 2 1 x a a
x at , h  gt 2   ,  cot    a  g cot 
2 2 h g g
EX.36:A block of mass m is placed on a smooth wedge of inclination  . The whole system is accelerated
horizontally so that the block does not slip on the wedge.Find the i) Acceleration of the wedge ii)
Force to be applied on the wedge iii) Force exerted by the wedge on the block.
Sol. (i). For an observer on the ground : R cos 
R

R sin 
F
mg a
M 
R sin   ma , R cos   mg
 a  g tan 
ii) F = (M + m)a = (M + m) g tan 
iii) Force exerted by the wedge on the block
mg
 R or R  mg sec 
cos 
1
Note :If inclination is given as 1 in x, sin   x
1
tan   xq 1
x2  1
g
 Acceleration a  g tan  
x2  1
EX.37:Two fixed frictionless inclined planes making an angles 300 and 600 with the vertical are
shown in the figure. Two blocks A and B are placed on the two planes. What is the relative
vertical acceleration of A with respect to B? (AIEEE-2010)
A
60° 30°
Sol: mg sin   ma  a  g sin 

where a is along the inclined plane
 vertical component of acceleration is g sin 2 
ar  a AB  a A  aB
g
 2 0 2

 relative vertical acceleration of A with respect to B is g sin 60  sin 30 
0
2
 4.9ms 2
(in vertical direction)

EX.38:For what value of 'a' block slides up the plane with an acceleration 'g' relative to the
inclined plane.
Sol.
Fnet  ma cos   mg sin 
ma   ma cos   mg sin 
If a  g , mg  ma cos   mg sin 
ma cos 
a
ma 
ma sin   mg sin 
mg
mg cos 
 1  sin  
a cos   g  g sin   a  g  
 cos  
 a  g (sec   tan  )
EX.39:A solid sphere of mass 2kg rests inside a cube as shown. The cube is moving with velocity

ˆ
  (5ti + 2tj)ms  1 where 't' is in sec and
' ' is in m/s. What force does sphere exert on cube?
O X

Sol. As given,   5ti  2t j ;
d x d y
 ax   5, a y  2
dt dt
ay
ax
mg
When cube is moving with above accelerations along x and y-axes, the forces that exert on cube are
Fx   max  2  5  10 N
 
Fy   mg  ma y    20  2  2   24 N
 Fx 2   Fy 
2
Net force, F
= 10 2   242  26N

EX.40:A block is placed on an inclined plane moving towards right with an acceleration a0 = g.
The length of the inclined plane is l0 . All the surfaces are smooth. Find the time taken by the
block to reach from bottom to top.
ma cos 
30°
ma0
mg sin 
30°
Sol. ma  ma0 cos300  mg sin 300
ma0 cos300  mg sin300

a
m
3 1 æ 3-
ma0 - mg 1ö÷
2 2 = g ççç ÷
÷
a= çè 2 ø÷
m
1 2 1 2
from s  ut  at ; 0 = at
2 2
1 æ 3- 1ö÷ 2
0 = g ççç ÷
÷t t=
4 0
2 çè 2 ø÷  g ( 3- 1 ) sec
EX.41:A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string
when the trolley rolls up a plane of inclination a with acceleration a0 is
a0
T cos  T 
T sin 
Sol. mg sin 

ma0 cos
mg 
T sin q = ma0 + mg sin a -------------(1)

T cos q = mg cos a -------------(2)
1 a0 + g sin a
2  tan q = g cos a
éa + g sin a ù
q = tan- 1 ê 0 ú
êë g cos a ú û
EX.42:A block slides down from top of a smooth inclined plane of elevation q fixed in an eleva-
tor going up with an acceleration a0.The base of incline has length L. Find the time taken by
the block to reach the bottom.
a0
L
Sol. Let us solve the problem in the elevator frame. The free body diagram is shown. The forces are
N
a0

ma0sin mg
ma0
mgsin

(i) N normal reaction to the plane,
(ii) mg acting vertically downwards,
(iii) ma0 ( pseudo force).acting vertically down
If a is acceleration of the body with respect to inclined plane, taking components of forces parallel
to the inclined plane.
mg sin   ma0 sin   ma  a   g  a0  sin 
This is the acceleration with respect to the elevator
L
The distance travelled is cos  . If ‘t’ is the time for reaching the bottom of inclined plane
L 1
 0   g  a0  sin  .t 2
cos  2
1 1
 2L 2  4L 2
t   
  g  a0  sin  cos     g  a0  sin 2 
Law of conservation of momentum:

 When the resultant external force acting on a system is zero, the total momentum (vector sum) of
the system remains constant. This is called “law of conservation of linear momentum”.
 Newton’s third law of motion leads to the law of conservation of linear momentum.
 Walking, running, swimming, jet propulsion, motion of rockets, rowing of a boat, recoil of a gun
etc., can be explained by Newton’s third law of motion.
 Explosions, disintegration of nuclei, recoil of gun, collisions etc., can be explained on the basis of
the law of conservation of linear momentum.
Applications:
 When a shot is fired from a gun, while the shot moves forwards, the gun moves backwards. This
motion of gun is called recoil of the gun. When a gun of mass ‘M’ fires a bullet of mass ‘m’ with
a muzzle velocity ‘v’, the gun recoils with a velocity ‘V’ given by V = mv/M.
 When a bullet of mass ‘m’ moving with a velocity ‘v’ gets embedded into a block of mass M at rest
and free to move on a smooth horizontal surface, then their common velocity ` V = mv/ (M +
m).
 A boy of mass ‘m’ walks a distance ‘s’ on a boat of mass ‘M’ that is floating on water and initially
at rest. If the boat is free to move, it moves back a distance d = ms / (M + m).
Explosion of Bomb
 A shell of mass ‘M’at rest explodes into two fragments and one of masses ‘m’ moves out with a
velocity ‘ v ’ the other piece of mass (M– m) moves in opposite direction with a velocity of V =
m v / (M – m).
 Suppose a shell of mass m at rest explodes into three pieces of masses m1 , m2 and m3 , moving
  
with velocities v1 , v2 and v3 respectively..
     
m1v1  p1 ; m2 v2  p2 ; m3v3  p3
     
p1  p2  p3  m1 v1  m2 v2  m3 v3  0
  
(as shell is at rest initially)  p3    p1  p2 
So the third piece moves with the same magnitude of the resultant momentum of the other two
pieces but in opposite direction.
P2


P1

P3
 
P3  P12  P22  2 P1 P2 cos    angle between P1 , P2
 
     angle between P3 , P1
P2 sin 
tan  
P1  P2 cos 
Explosion of a shell travelling in a parabolic path at its highest point: (into

two fragments)
 Consider a shell of mass M as a projectile with velocity u and angle of projection  . Suppose the
 
shell breaks into two fragments at maximum height and their initial velocities are v1 and v2
M = m 1 + m2
u (m1 + m2)u cos 
Total momentum of the two parts is constant just before and just after the explosion.
  
 m1  m2  u cos  i  m1v1  m2v2
Case : (i) If the fragments travel in opposite direction after explosion then
  
 m1  m2  u cos  i  m1v1i  m2v2i
Case : (ii) If one fragment retraces its path and falls at the point of projection
  
 m1  m2  u cos  i  m1u cos  i  m2v2
Case:(iii) If one fragment falls freely after explosion
 
 m1  m2  u cos  i  m1 0  m2v2
 
 m1  m2  u cos  i  m2v2
EX.43:A bomb moving with velocity(40i+50j-25k)m/s explodes into two pieces of mass ratio1:4.After
explosion the smaller piece moves away with velocity(200i+70j+15k)m/s.The
velocity of larger piece after explosion is (EAM-2010)
Sol: From Law of conservation of linear momentum
MU  m1v1  m2 v2 ; M  5 x, m1  x, m2  4 x
U  40i  50 j  25kms
 1 ;
v  200i  70 j  15kms
1
 1
here v2 is the velocity of the larger piece

5 x 40i  50 j  25k  
 
x 200i  70 j  15k  4 x  v2 
On simplification, we get v2  45 j  35k

EX.44:A particle of mass 4 m explodes into three pieces of masses m,m and 2m. The equal masses
move along X-axis and Y- axis with velocities 4ms-1 and 6 ms-1 respectively . The magnitude of
the velocity of the heavier mass is (E - 2009)
Sol: M=4 m ,U=0, m1  m, m2  m, m3  2m
v1  4ms 1 , v2  6ms 1 , v3  ?
According to law of conservation of momentum,
  
P1  P 2  P3  0
     
 
P 3   P1  P 2 , P3  P1  P 2
1 2 Cos
P3  P12  P2 2  2 PP
P1 and P2 are perpendicular to each other
P3  P12  P2 2 , m3v3  (m1v1 )2  (m2 v2 )2
2mv3  (m  4)2   m  6 
2
2v3  16  36  v3  13ms 1
EX.45:A rifle of 20kg mass can fire 4 bullets/s. The mass of each bullet is 35  103 kg and its final
velocity is 400ms 1 . Then, what force must be applied on the rifle so that it does not move
backwards while firing the bullets?(2007E)
Sol :Law of conservation of momentum MV + 4mv = 0
4mv 4  35 103  400
V =  = = -2.8 ms 1
M 20
MV 20  2.8
Force applied on the rifle F   = -56 N
t 1
EX.46:All surfaces are smooth.Find the horizontal displacements of the block and the wedge
when the block slides down from top to bottom.
h l
M 
L
Sol : When the block slides down on the smooth wedge, the wedge moves backwards. In the horizontal

direction there is no external force ; Fx  0

 Px =constant
   
Pf  Pi (along x-axis) ; mu  MV  0
x1 = forward distance moved by the block along X-axis.
x2 = backward distance moved by the wedge along X-axis.
 
mu   MV ;
x1 x
m M 2
t t
ML M  cos 
mx1  Mx2 , x1  
M m M m
mL m cos 
x2  
M m M m
EX.47: A bomb of 1 kg is thrown vertically up with speed 100 m/s. After 5 seconds, it explodes
into two parts. One part of mass 400gm goes down with speed 25m/s. What will happen to
the other part just after explosion
Sol : After 5 sec, velocity of the bomb,
v  u  at

v = u j – gt j = (100 – 10 x 5) j = 50 j m/s
m  1kg , m1  0.4kg , m2  0.6kg , v1  25ms 1
According to law of conservation of momentum mv  m1v1  m2 v2
1  50 j   0.4  25 j  0.6 v2
 v2  100 j
 v2  100ms 1 ,vertically upwards
EX.48:A particle of mass 2m is projected at an angle 450 with horizontal with a velocity of 20 2 m/
s. After 1sec, explosion takes place and the particle is broken into two equal pieces. As a
result of explosion one part comes to rest. The maximum height attained by the other part
from the ground is (g = 10m/s2)
Sol : M  2m,  450 , u  20 2ms 1
1
u x  u cos   20 2   20 ms 1
2
1
u y  u sin   20 2   20 ms 1
2
1 2 1
But height attained before explosion , H1 = ut  gt = 20  1   10  12  15 m
2 2
After 1sec, v x  20 ms 1
v y  u y  gt  20  10  10 ms 1
Due to explosion one part comes to rest,
m1  m2  m, v1  0
M ( v x i  v y j )  m1v1  m 2 v 2
2 m (20 i  10 j )  m (0)  m v 2
v 2  40 i  20 j
v y1  20ms 1
Height attained after explosion = H 2 

v 
y
1 2

20  20
 20m
2g 2  10
H TOT  H1  H 2  15  20  35m
Friction:
If we slide or try to slide a body over another surface, the motion of the body is resisted by bonding
between the body and the surface.This resistance is called friction.
 The force of friction is parallel to the contact surfaces and opposite to the direction of intended or
relative motion.
 There are three types of frictional forces
i. Static friction ii. Dynamic friction
iii. Rolling friction
 If a body is at rest and no pulling force is acting on it,force of friction on it is zero.
 If a force is applied to move the body and it does not move,the friction developed is called static
friction, which is equal in magnitude and opposite in direction to the applied force (static friction
friction is self adjusting force).
 If a force is applied to move the body and it moves,the friction developed is called dynamic or
kinetic friction.
 When a body rolls or rotates on the surface of another body friction developed is called as
rolling friction.
 It is due to the deformation at the point of contact and depends on area of contact.
F
A
B
Frictional force
Static friction
State of
t
es
Dynamic
motion
fr
friction
eo
at
St
Pulling
D O C force
Note-i: If you are walking due east the feet slides relatively due west so the frictional force is due
east.
Note-ii: Engine is connected to rear wheels of a car. When the car is accelerated, direction of
frictional force on the rear wheels will be in the direction of motion and on the front wheels in
the opposite direction of motion
Note-iii:In cycling ,the force exerted by rear wheel on the ground makes the force of friction to act
on it in the forward direction. Front wheel moving by itself experience force of friction in backward
direction.
Note-iv:When pedaling is stopped,the frictional force is in backward direction for both the wheels.
Laws of Friction:
 Friction is directly proportional to the normal reaction acting on the body.
 The law of static friction may thus be written as
f s   s N . Where the dimensionless constant  s is called the coefficient of static friction and N is
the magnitude of the normal force.
 fs max  fl  sN ;
f l  Limiting friction
 Coefficient of static friction ( μ s) depends on the nature of the two surfaces in contact and is
independent of the area of contact.
 Static friction is independent of the area of contact between the two surfaces
fk
 Coefficient of kinetic friction ( μ k) = . It is independent of velocity of the body..
N
fR
 Coefficient of rolling friction   R  
N
 Rolling friction depends on the area of the surfaces in contact.
Note : μ S > μ K > μ R
 Friction depends on the nature of the two surfaces in contact i.e., nature of materials, surface
finish, temperature of the two surfaces etc.
Angle of Friction:
 Angle made by the resultant of f and N with the normal reaction N is called angle of friction.
 Friction is parallel component of contact force to the surfaces.
 Normal force is perpendicular component of contact force to the surfaces.
R N
f mg
R f 2  N2
f
When the block is static tan   ;   s
N
s N
When the block is in impending state, tan s   s
N
Where s  maximum angle of friction.
k N
When block is sliding, tan k   k
N
Since  s   k , it follows that s  k .
 s N 
2
FR  fl 2  N 2   N 2  N  s2  1
FR  mg tan 2   1  s  tan  
FR  mg sec 
Motion on a horizontal rough surface: Consider a block of mass m place on a horizontal

surface with normal reaction N.
Case (i) :If applied force F = 0, the force of friction is zero.
F(Applied
force)
f m
|||||||||||||||||||||||||||||||||||||||||||||||||||||||
Case (ii) :If applied force F   f s max , the block does not move and the force of friction is fs  F
Case (iii) : If applied force F =  f s  max block just ready to slide and frictional force  f s max  fl   s N
F  s mg  N  mg  ; (at time t=0 )

Case(iv) : If the above applied force continues to act ( t  0 ) the body gets motion, static friction
converts as kinetic friction and body possesses acceleration
Fext  f k f  fk
a  l   s  k  g
m m
Case (v) :If the applied force is greater than limiting friction the body starts moving and gets
acceleration
F1ext  fk
a Here F 1ext  Fext
m
 If the block slides with an acceleration ‘ a ’under the influence of applied force ‘F’,
FR  F  f k ; ma  F  f k
F  f k F   k mg
a   ( f k   K N   K mg )
m m
N
a
F
fk
|||||||||||||||||||||||||||||||||||||||||||||
mg
Bodies in contact with vertical surfaces:

 A block of mass m is pressed against a wall without falling, by applying minimum horizontal
mg
force F. Then F = 
s
fs
|||||||||||||||||||||||||||||||||||||||||||||||
N
F
mg
f s  mg ; s N  mg
mg
s F  mg  N  F  F 
s
 A block is pressed between two hands without falling, by applying minimum horizontal force
mg
‘F’ by each hand. Then F = 2 
s
fs fs
||||||||||||||||||||||||||||||||||||||||||
||||||||||||||||||||||||||||||||||||||||||
F F
mg
mg
W=2 fs, ; mg  2s F  F  2
s
Sliding block on a horizontal rough
N surface coming to rest :
v=0
u
fk |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
mg
a) The acceleration of the block is a   k g
u2
b) Distance travelled by the block before coming to rest is S
2k g
u
c) Time taken by the block to come to rest is t   g
k
 An insect is crawling in a hemispherical bowl of radius ‘r’. Maximum height upto which it can
crawl is
fs=s N
O

N
r r sin  P
h 
mg cos 
mg sin 
mg
 1 
h = r 1- cos    r  1  
  2
 1 
 s 
Maximum angular displacement upto which it can crawl is ‘ θ ’. Then  s  tan 
 A block is placed on rear horizontal surface of a truck moving along the horizontal with an
acceleration ‘a’. Then
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
1) The maximum acceleration of the truck for which block does not slide on the floor of the truck
is a  s g
2) If a  s g block does not slide and frictional force on the block is f=ma.
3) If a  s g block slips or slides on the floor the acceleration of the block relative to the truck is
a  a  kg
4) If  is the distance of the block from rear side of the truck, time taken by the block to cover a
2
distance  . t=
a - μkg
4) Acceleration of the block relative to ground is a  k g
Body placed in contact with the front surface of accelerated truck:
 When a block of mass ‘m’ is placed in contact with the front face of the vehicle moving with acceleration
a then a pseudo force ‘Fpf ‘ acts on the block in a direction opposite to the direction of motion of the vehicle
fl
a
ma
N
mg
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Under equilibrium, fl = mg; N = ma
g
 S N  mg  S .ma  mg  amin  
s
EX.49: A man of mass 40 kg is at rest between the walls as shown in the figure. If ‘ m ’ between the
man and the walls is 0.8, find the normal reactions exerted by the walls on the man.
Sol. Since man is at rest,
µN1 µN2
F2
mg N2
F1 N1
N1  N 2  0 ( horizontal equilibrium )
 N1  N 2  N , F1  F2  F (say)
 2  N  mg ( vertical equilibrium )
= 2  0.8  N  400  N  250N
EX.50:A 2 kg block is in contact with a vertical wall having coefficient of friction 0.5 between
the surfaces. A horizontal force of 40N is applied on the block at right angles to the wall.
Another force of 15N is applied, on the plane of the wall and at right angles to 40N force.
Find the acceleration of the block.
Sol.
f=20N
F=15N
15N
F=40N F=40N
20N
25N=FR
W=20N
Resultant of W=20N and 15N

FR  202 152 = 25N
frictional force f   N = 0.5 x 40 = 20N
This acts in a direction, opposite to 25N force.
Net force acting on the block, Fnet = 25–20=5
5
acceleration of the block a
2
= 2.5ms–2
EX.51:A block of mass 4 kg is placed on a rough horizontal plane. A time dependent horizontal force F =
kt acts on the block (k = 2 N/s). Find the frictional force between the block and the plane at t = 2
seconds and t = 5 seconds (µ = 0.2)
Sol. Given F = kt
When t = 2sec ; F = 2(2)=4N ..... case (i)
f ms   s mg = 0.2 × 4 × 10 = 8N
Here F < f ms  friction = applied force=4N
When t = 5 sec ; F = 2 (5) = 10N......case(ii)
F > f  frictional force < 8N
EX.52:A block on table shown in figure is just on the edge of slipping. Find the coefficient of
static friction between the block and table
T cos 30
30OT
40N
T sin 30
f
80N
Sol.
f l  T sin 
 mg  T sin 
.........(1)
80  T cos  ...........(2)
T sin 30 0  mg
 ;
T cos 30 0 80
 40 1  2
Tan 300 = 80 ;    1.15
3 2 3
EX.53:When a car of mass 1000 kg is moving with a velocity of 20ms-1 on a rough horizontal
road, its engine is switched off. How far does the car move before it comes to rest if the
coefficient of kinetic friction between the road and tyres of the car is 0.75 ?
Sol. Here v = 20ms 1 , k = 0.75, g = 10ms 2
v2
Stopping distance S   26.67 m
2k g
EX.54:A horizontal conveyor belt moves with a constant velocity V. A small block is projected
with a velocity of 6 m/s on it in a direction opposite to the direction of motion of the belt. The
block comes to rest relative to the belt in a time 4s.   0.3 , g = 10 m/s2. Find V

Sol. V b,c  Vb  Vc  6  V
f   mg  0.3  m  10  3m
F 3m
Retardation a  = m  3m / s 2
m
ur  6 V,Vr  0, t  4 sec , a r   3 ms 2
Vr  u r  ar t , 0 = (6+V) – 3 x 4,V = 6 m/s
EX.55:The rear side of a truck is open. A box of 40 kg mass is placed 5m away from the open end
as shown in figure. The coefficient of friction between the box and the surface is 0.15. On a
straight road, the truck starts from rest and accelerating with 2 m/s2. At what distance from
the starting point does the box fall off the truck? (Ignore the size of the box).
N
Fpf 2
fk a = 2m/s
||||||||||||||||||||||||||||||||||||||||
5m
mg
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Sol :Because of the acceleration of the truck the pseudo force on the box = m x a = 40 x 2 = 80N.
This force acts opposite to the acceleration of the truck. The frictional force on the truck which
acts in the forward direction f k =  N = 0.15 x 40g = 58.8N Since pseudo force is greater than
frictional force, the block will accelerate in backward direction relative to truck with a magnitude
80  58.8
a  0.53 m / s 2
40
The time taken by box to cover the distance 5m is given by
1 2s
s  0  at 2  t   4.34sec
2 a
The distance travelled by truck in this time is , a   2 ms 2
1 1
s  at 2   2  (4.34)2  18.87m
2 2
Sliding of a chain on a horizontal table:
 Consider a uniform chain of mass “m” and length “L” lying on a horizontal table of coefficient of
friction “  s ”. When 1/nth of its length is hanging from the edge of the table, the chain is found
mg
to be about to slide from the table. Weight of the hanging part of the chain =
n
N
L
fs |||||||||||||||||||||||||||||||
 m
m g
 n 
m
g
n
mg  1
Weight of the chain lying on the table = mg 
 mg  1  
n  n
When the chain is about to slide from edge of the table,
The weight of the hanging part of the chain = frictional force between the chain and the table
surface.
mg  1
  s mg 1  
n  n 1
 s 

mg  n 1 
  s mg    n  1
n  n 
L
If l is the length of the hanging part, then n  Substituting this in the above expression we get,
l
l L  1
s   or  n   s
Ll l s
The maximum fractional length of chain hanging from the edge of the table in
 s
equilibrium is L    1
s
L 1
 Fractional length of chain on the table 
L s  1
Connected Bodies :
 A block of mass m 1 placed on a rough horizontal surface, is connected to block of mass m 2
by a string which passes over a smooth pulley.The coefficient of frictionbetween m1 and the
table is  .
a
N
T
m1g T a
m2g
For body of mass m2
m2 g  T  m2 a ————— (i)
For body of mass m1
T  f k  m1a  T   k N  m1a —— (ii)
Solving Eqs (i) and (ii), we get
 m  k m1  m1m2 g
a 2 g; T  (1   )
 m1  m2  m1  m2
EX.56:A block of mass 10kg is pushed by a force F on a horizontal rough plane is moving with
acceleration 5ms 2 . When force is doubled, its acceleration becomes 18ms 2 .Find the
coefficient of friction between the block and rough horizontal plane.  g  10ms 2  .
Sol :
N
a
F
fk
|||||||||||||||||||||||||||||||||||||||||||||
mg
F
On a rough horizontal plane, acceleration of a block of mass ‘m’ is given by a    k g ..........(i)
m
Initially, a = 5ms 2
F
5
10
 k 10  ............  ii   m  10kg 
When force is doubled a = 18ms 2 .
2F
18   k 10  ..........  iii 
10
8
Multiplying Eq(ii) with 2 and subtracting from Eq.(iii) 8   k 10    k   0.8
10
x3
EX.57:A block of mass ‘m’ is placed on a rough surface with a vertical cross section of y
6
. If the
coefficient of friction is 0.5, the maximum height above the ground at which the block can be
placed without slipping is(JEE MAIN -2014)
N
u=0
0
A
mg sin  mg cos
Sol: mg
l h
v 
B
dy d x 3 2
Tan   ( )  Tan  x
dx dx 6 2 x2
At limiting equilibrium,we get   Tan  0.5 
2
x 2  1  x  1
x3
Now putting the values of ‘ x ’in y  , we get
6
1 1
When x  1  y  ; x  1  y  
6 6
So the maximum height above the ground at which the block can be placed without slipping is
1
y m
6
Motion of a body on an inclined plane :
Case (i) :Body sliding down on a smooth inclined plane :
Let us consider a body of mass ‘m’ kept on the plane as shown in fig.
 Normal reaction N  mg cos 
 Acceleration of sliding block a = g sin 
 If l is the length of the inclined plane and h is the height. The time taken to slide down start-
2l 1 2h
ing from rest from the top is t  
g sin sin g
 Sliding block takes more time to reach the bottom than to fall freely in air from the top of the inclined
plane to the ground.
 Velocity of the block at the bottom of the inclined plane is same as the speed attained if block falls
freely from the top of the inclined plane.
V  2 gl sin   2 gh
Case(ii) :Body projected up on a smooth inclined plane :
v=0
B
s h
u
A  C
 If a block is projected up the plane with a velocity u, the acceleration of the block is a   g sin 
u2
 Distance travelled by the block up the plane before its velocity becomes zero is S
2 g sin 
u
 Time of ascent t  g sin 
Case (iii) Motion of a body down the rough inclined plane:
 Let a body of mass m be sliding down a rough inclined plane of angle of inclination θ and
coefficient of kinetic friction μ k.
N
fk

mg sin mg mg cos

Angle of Repose (  ) : Angle of repose is the minimum angle of the rough inclined plane for
which body placed on it may just start sliding down. It is numerically equal to the angle of
friction.
 Let θ be the angle of inclination of a rough inclined plane,  be the angle of repose, m be the
mass of the body and  be the coefficient of friction.
At limiting equilibrium (about to slide)
mg sin  s mg cos  tan  s    tan1  s 
1. When 1   ; the block remains at rest on the inclined plane. Frictional force mg sin 1 (self
adjusting) acceleration a=0
2. When  2   ; the block remains at rest on inclined plane or impending state of motion is achieved.
mg sin  2   s mg cos  2 (at time t=0)
Here  2  1 and f s  f l acceleration a=0
3. When  2   ; and the same inclination is continued the block moves downwards with accleration
a.
mg sin  2  k mg cos  2 acceleration
mg sin  2  k mg cos  2
a
m
 s mg cos  2   k mg cos  2
a  g cos  2 (  s   k )
m
4. When    , the body slides f k  k mg cos 
The resultant force acting on the body down the plane is FR = mgsin -f K ,
FR  mg  sin    k cos  
The acceleration of the body a  g  sin   k cos  

N A
u=0 fk
mg sin
 mg cos
mg
h
B

v
 Velocity of the body at the bottom of the plane V  2 g  sin   k cos   l
 If ‘t’is the time taken to travel the distance ‘l’ with initial velocity u = 0 , at the top of the plane,
2l
t
g (sin   k cos  )
 The time taken by a body to slide down on a rough inclined plane is ‘n’ times the time taken by it
to slide down on a smooth inclined plane of same inclination and length then coefficient of friction
is....
2l
trough g (sin   k cos  )
n  sin 
tsmooth 2l n2 
sin    k cos 
g sin 
 1
 n 2 sin   n 2 k cos   sin    k  Tan 1  2 
 n 
Body projected up a rough inclined plane:
If a body is projected with an initial velocity ‘u’to slide up the plane, the kinetic frictional force
acts down the plane and the body suffers retardation due to a resultant force
FR =(mg sin  +fk )
 acceleration a = - g(sin θ + μ k cos θ )
2l
 Time taken to stop after travelling a distance l along the plane, t = g (sin    k cos  )
 Force required to drag with an acceleration ‘a’ is F = ( μ k mg cos θ + mg sin θ + m a )

EX.58:A body is moving down a long inclined plane of angle of inclination ‘  ’for which the
coefficient of friction varies with distance x as   x   kx , where k is a constant. Here x is the
distance moved by the body down the plane. The net force on the body will be zero at a
distance x0 is given by
f
N
Sol : 
mg sin mg cos
mg

F  mg sin   f
N  mg cos  ; f   N   mg cos 
F  mg sin    mg cos 
F  mg  sin   kx cos  
tan 
If F = 0 ; sin   kx0 cos   0  x0 
k
EX.59:A body of mass ‘m’ slides down a smooth inclined plane having an inclination of 450 with the
horizontal. It takes 2s to reach the bottom. If the body is placed on a similar plane having
coefficient of friction 0.5 Then what is the time taken for it to reach the bottom?
Sol : Mass = m,  =450,  =0.5 Time taken by the body to reach the bottom without friction is
2l
T1 
g sin  = 2sec
Time taken with friction is
2l T1 sin    cos 
T2   
g  sin    cos   T2 sin 
sin  sin 450

T2  T1 2
sin    cos  sin 450  (0.5) cos 450
(1/ 2)
2  2  2  2.828s
(1/ 2)  (0.5)(1/ 2)
EX.60:Two blocks of masses 4 kg and 2 kg are in contact with each other on an inclined plane of
inclination 300 as shown in the figure. The coefficient of friction between 4 kg mass and the
inclined plane is 0.3, where as between 2 kg mass and the plane is 0.2. Find the contact force
between the blocks.
g
2k
g
4k
0
30
Sol :The acceleration of 4 kg mass,

If   300 , k  0.3
1 3
a4  g (sin   k cos  )  10   0.3    2.6ms
2
 2 2 
The acceleration of 2 kg mas
1 3 2
a2  10   0.2    3.27 ms
 2 2 
 a2  a4
Thus, there will be contact force between the blocks and they move together. If ‘a’ is the common
acceleration,
(m1  m2 )a  (m1  m2 ) gsin   ( 1 m1  2 m2 ) g cos 
1 3
6a  6  10   (0.3  4  0.2  2) 10 
2 2
6a  30  13.856  a  2.7 ms 2
For, 4 kg mass; mg sin   f contact  f friction  ma
1 3
4´ 10´ + f c - 0.3´ 10´ 10´ = 4 x 2.7
2 2
fc = 10.8 + 10.4 – 20 Þ f c = 1.2N
EX.61:A 30kg block is to be moved up an inclined plane at an angle 300 to the horizontal with a
velocity of 5ms–1. If the frictional force retarding the motion is 150N, find the horizontal
force required to move the block up the plane. (g=10ms–2.)
Sol.
mg
0
30
The force required to move a body up an inclined plane is F = mg sin q + f k
= 30(10) sin 300 +150 = 300N.
If P is the horizontal force, F = P cos q
F 300 300´ 2
P= = 0
= = 200 3 = 346 N
cos q cos 30 3
EX.62: A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction
is twice that of resultant downward force along the inclined plane, then find the angle between
the inclined plane and the horizontal
Sol :   0.5 , N  mg cos 
N  2 F , F  mg (sin    cos  ) N  2 mg  sin    cos  
mg cos   2mg  sin    cos  

c os   2 cos   tan    
1 1
 tan    tan   1    450
2 2
EX.63:In the given figure, the wedge is acted upon by a constant horizontal force 'F'. The wedge is
moving on a smooth horizontal surface. A ball of mass 'm' is at rest relative to the wedge. The ratio
of forces exerted on 'm' by the wedge when 'F' is acting and 'F' is withdrawn assuming no friction
between the edge and the ball,is equal to
Sol.

|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
When Force F is applied
Fcos
N1

F = ma
F sin
mg sin
mg cos 
N1 = mgcos  + Fsin 
( F cos   mg sin   F  mg tan  )
N1 F sin 
If F=0 ; N2 = mgcos  , N  1  mg cos 
2
N1 mg tan  sin 
 1  1  tan 2   sec 2 
N2 mg cos 
 Two blocks of mass m and M are placed on a rough inclined plane as shown, when    
i)Minimum value of M for which m slides upwards is
M  m  sin   s cos  
m
mg sin
M

Mg
mg
ii) Maximum value of M for which m slides downwards: M  m  sin   s cos  

 A body is released from rest from the top of an inclined plane of length ‘L’ and angle of
L
inclination ‘  ’. The top of plane of length  n  1 is smooth and the remaining part is
n
rough. If the body comes to rest on reaching the bottom of the plane then find the value of
coefficient of friction of rough
N
u=0
mg sin
 mg cos
L
mg h
n
 n 1 
 L
  n 
v
For smooth part :

L
Using v 2  u 2  2as ; V  2a1 ,
2
n
a1  g sin  , a2  g (sin    cos  )
 n 1  L  n 1 
For rough part 0  V 2  2 a2  L 2a1  2 a 2  L
 n  n  n 
 n 
g sin    g sin    cos    n  1   Tan 
 n  1 
 A body is pushed down with velocity ‘u’ from the top of an inclined plane of length ‘L’ and angle
L
of inclination ‘  ’. The top of plane of length  n  1 is rough and the remaining part is smooth. If
n
the body reaches the bottom of the plane with a velocity equal to the initial velocity ‘u’, then
the value of coefficient of friction of rough plane is  K  n  tan  
Pushing & Pulling of a Lawn Roller :
i) A Roller on Horizontal Surface Pushed by an Inclined Force :
N
F
 F cos 
F
fr ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
F sin 
mg
 When a lawn roller is pushed by a force ‘F’,which makes an angle  with the horizontal, then
 Normal reaction N = mg + F sin  .
 Frictional force f r  r N  r  mg  F sin  
The net horizontal pushing force is given by F1= F(cos  – µr sin  ) – µrmg
ii)A Lawn Roller on a Horizontal Surface Pulled by an Inclined Force
N
F sin  F

F cos 
|||||||||||||||||||||||||||||||||||||||||||||||||||||||
fr
mg
 Let a lawn roller be pulled on a horizontal road by a force ‘F’, which makes an angle  with the
horizontal.
 Normal reaction N =mg – Fsin 
 Frictional force fr = µrN = µr(mg – Fsin  )
 The net horizontal pulling force is F2 = F (cos + r sin ) - r mg Pulling is easier thanPushing.
Applying an Inclined Pulling Force :
Let an inclined force F be applied on the body so as to pull it on the horizontal surface as shown
in the figure.
N
F sin  F

F cos 
|||||||||||||||||||||||||||||||||||||||||||||||||||||||
fr
mg
The body is in contact with the surface,and just ready to move
N + F sin =mg  N = mg - F sin 
frictional force f r  F cos 
Fcos =  r N ,Fcos =  r (mg - F sin )
r mg mgsin
cos      Tan  r 
F F
 cos  r sin 
For F to be minimum cos     should be
maximum  cos      1      0,   
  angle of friction.
2r  1
r
r  r mg
 Fmin  mgsin  mgsin From the figure, sin   2  1 , Fmin   2  1
r r
Minimum horizontal pulling force, when   0

cos  0     cos 
mg sin 
F  mg tan 
cos 
Applying an Inclined Pushing Force :
Let an inclined force F is applied on the body so as to push it on the horizontal surface as shown
in the figure.
N
Fsin F

f F cos 
||||||||||||||||||||||||||||||||||||||||||
mg
The body is in contact with the surface, and just ready to move, N=mg+Fsin
frictional force f l = Fcos 
Fcos   s N  Fcos   s  mg  Fsin 
 s mg
F
 cos  s sin 
mgsin
F Tan  s 
cos    
For F to be minimum   0
mg sin 
 Fmin   mg tan    mg
cos  s
(since  s  tan  )
EX.64: A block of mass m kg is pushed up against a wall by a force P that makes an angle ‘  ’ with the
horizontal as shown in figure. The coefficient of static friction between the block and the wall is
 . The minimum value of P that allows the block to remain stationary is
P sin 
P
f

N P cos 
Sol : 
mg
At equilibrium, P sin   f  mg , N  P cos 

f   mg  P sin  ; N   mg  P sin 
P cos   mg  P sin  ; P  sin    cos    mg
mg
P
 sin    cos 
Block on Block:
 Case I: Bottom block is pulled and there is no friction between bottom block and the horizontal
surface.
mu
f
mB F
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
 When the bottom block is pulled upper block is accelerated by the force of friction acting upon it.
 The maximum acceleration of the system of two blocks to move together without slipping
is amax   s g , where s is the coefficient of static friction between the two blocks. The maximum
applied force for which both the blocks move together is Fmax  s g  mu  m B 
 If a   s g blocks move together and applied force is F  (m B  mu )a In this case frictional force
between the two block f  mu a .
 If a   s g , blocks slip relative to each other and have different accelerations. The acceleration of
F   k mu g
the upper block is au  k g and that for the bottom block is aB  mB
 Case - II:Upper block pulled and there is no
friction between bottom block and the horizontal surface.
mu F
f
mB
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
 When the upper block is pulled, bottom block is accelerated by the force of friction acting on it.
 The maximum acceleration of the system of two blocks to move together without slipping is
mu
amax   s g
mB where , s =coefficient of static friction between the two blocks The maximum
mu
force for which both blocks move together is Fmax  s m g  m u  m B 
B
 If a  amax , blocks move together and frictional force between the two blocks is f  mB a The
applied force on the upper block is F  (m B  mu )a
 If a  amax blocks slide relative to each other and hence they have different accelerations.The
mu
acceleration of the bottom block is aB   K m g and the acceleration of the upper block is
B
F   k mu g
au 
mu
 A number of blocks of identical masses m each are placed one above the other. Force required
to pull out Nth block from the top is
F = (2N–1)  mg
EX.65:A block of mass 4kg is placed on another block of mass 5kg, and the block B rests on a
smooth horizontal table, for sliding the block A on B, a horizontal force 12N is required
to be applied on it .How much maximum horizontal force can be applied on ‘B’ so that both
A and B move together? Also find out the acceleration produced by this force.
M1 A
F M2 B
Sol: Here M 1 =4kg and M 2 =5kg
L imiting friction between the blocks is f lim Acceleration of system is
F F F
a   m / s2
M1  M 2 4  5 9
Because of this acceleration the block A experiences a pseudo force of magnitude
F
Fpseudo  M 1a  4 
9
M1 a
M1 A
12N
F B
M2
As block A move together with B, Fpseudo  f lim For maximum value of applied force
4F
Fpseudo  f lim ;  12  F  27 N
9
27
The acceleration of blocks = 3m / s 2
9
EX.66:Two blocks of masses ‘m’ and ‘M’ are arranged as shown in the figure. The coefficient
of friction between the two blocks is ‘  ’, where as between the lower block and the horizontal
surface is zero. Find the force ‘F’ to be applied on the upper block, for the system to be
under equilibrium?
F
m
M
F m T
Sol :
mg
m T
mg
On the upper block,
F=T+f=T+  N; F = T +  mg ……(1)
On the lower block T =  mg ……… (2)
from (1) and (2), we get, F = 2  mg
EX.67 : Block A weighs 4N and block B weighs 8N.The coefficient of kinetic friction is 0.25 for all
surfaces.Find the force F to slide B at a constant speed when (a) A rests on B and moves with
it (b) A is held at rest and (c) A and B are connected by a light cord passing over a smooth
pulley as shown in fig (a),(b) and (c) respectively
A A
R2 R2
F f2
F B R1 f 1 B R1 f1
S S
(a) (b)
A
T
f2 R2
f2
F
B R1 T
S f1
(C)
Sol :(a) When A moves with B the force opposing the motion is the only force of friction between B and
S the horizontal and as velocity of system is constant,
F  f1   R1  0.2  4  8   3N
(b)When A is held stationary, the friction opposing the motion is between A and B. So
F   R1   R2  3  0.25  4   4 N
(c)In this situation for dynamic equilibrium of B
F   R1   R2  T ...........(i)
While for the uniform motion of A,
T   R2 ............(ii)
Substituting T from eqn (ii) in (i) we get F   R1  2  R2  3  2  1  5 N
EX.68 : The apparent weight of a person inside a lift is W1 when lift moves up with certain
acceleration and is W2 when lift moves down with same acceleration. The weight of person
when lift moves up with constant speed is
W1  W2 W1  W2
1) 2) 3) 2W1 4) 2 W2
2 2
Sol : key-1
w1  m( g  a), w2  m( g  a ) ,
w3  mg
EX.69 : A rope of length 10m and linear density 0.5kg/m is lying length wise on a smooth horizontal
floor. It is pulled by a force of 25 N. The tension in the rope at a point 6m away from the
point of application is
1) 20 N 2) 15 N 3) 10 N 4) 5 N
F
Sol : key-3. F  ma ,For one unit  L
F  l
For l units  l , T  F  1  
L  L
EX.70 : Three blocks of masses m1 , m2 and m3 are connected by a massless string as shown in figure
on a frictionless table. They are pulled with a force T3 = 40 N. If m1  10kg , m2  6kg and
m3  4kg , then tension T2 will be
1) 10 N 2) 20 N 3) 32 N 4) 40 N
Sol : key-3.. Fnet  ma , T3  T2  m3 a , T2  T1  m2 a , T1  m1a
solving the above equations,we get
EX.71 : A horizontal force F pushes a 4 kg block (A) which pushes against a 2 kg block (B) as
shown. The blocks have an acceleration of 3m/s 2 to the right. There is no friction between
the blocks and the surfaces on which they slide. What is the net force B exerts on A?
1) 6 N to the right 2) 12 N to the right 3) 6 N to the left 4) 12 N to the left
Sol : key-3. F  ma , Fnet  F  f , f  ma

T3
a
m1  m2  m3 , T2  (m1  m2 )a
EX.72 : The momentum of a body in two perpendicular directions at any time't' are given by
3t 2
PX  2t  6 and PY 
2
 3 .The force acting on the body at t = 2 sec is
2
1) 5 units 2) 2 units 3) 10 units 4) 15 units
dP dP
Sol : key-3. Fx  x , Fy  y , F  Fx 2  Fy 2
dt dt
EX.73 : A particle of mass m, initially at rest is acted upon by a variable force F for a brief
interval of time T. It begins to move with a velocity u after the force stops acting. F is shown
in the graph as a function of time. The curve is a semicircle.
 F02 T 2  F0T  F0T

1) u  2) u  3) u  4) u 
2m 8m 4m 2m
Sol : key-3. Impulse = Area of semi circle
EX.74 :A balloon of mass M is descending at a constant acceleration  . When a mass m is released
from the balloon it starts rising with the same acceleration  . Assuming that its volume does
not change, what is the value of m?
 2 g g
1) M 2) M 3) M 4) M
g g  2
Sol : key-2. While descending, Mg  FB  M 
While ascending FB   M  m  g   M  m  
Where ‘ FB ’is the buoyancy force
EX.75 :In the following figure, the pulley is massless and frictionless. There is no friction between
the body and the floor. The acceleration produced in the body when it is displaced through a
certain distance with force ‘P’ will be
P P P P
1) 2) 3) 4)
M 2M 3M 4M
T
Sol : key-2. F  ma , 2T  p, a  m
EX.76 :The pulley arrangements shown in figure are identical, the mass of the rope being negligible.
In case I,the mass m is lifted by attaching a mass 2m to the other end of rope with a constant
downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass m
in case I is
F = 2mg
m
m
2m
1) zero 2) more than that in case II
3)less than that in case II 4)equal to that in case II
 m1  m2 
Sol : key-3. F  ma , a  g   ,F –T = 0 and
 m1  m2 
T = 2mg also T–mg = ma1
Finally a < a1
EX.77 :The string between blocks of masses ‘m’ and ‘2m’ is massless and inextensible.The system
is suspended by a massless spring as shown. If the string is cut, the magnitudes of accelerations
of masses 2m and m (immediately after cutting)
g g g g
1) g , g 2) g , 3) ,g 4) ,
2 2 2 2
Sol : key-3. For m1 , F  ma  m1 g ;For m2 , T  m2 g  m2 a1

 
EX.78 :Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 respec-
 
tively at time t  0 . They collide at time t0 . Their velocities become v11 and v21 at time 2t0
while still moving in air. The value of  m1v1  m2 v2    m1v1  m2 v2  is

1 1  
1) zero 2)  m1  m2  gt0
1
3) 2  m1  m2  gt0 4)  m1  m2  gt0
2
Sol : key-3. mv1  m(u  at )
m1v11  m2v21  m  v1  2 gt0   m  v2  2 gt0 
EX.79 :In order to raise a block of mass 100kg a man of mass 60kg fastens a rope to it and passes
5g
the rope over a smooth pulley. He climbs the rope with an acceleration relative to rope.
4
The tension in the rope is  g  10ms 2 
1) 1432N 2) 928 N 3) 1218N 4) 642N
Sol : key-3. mm  60kg , mB  100kg
‘a’ be acceleration of rope.
5g
arel  , arel  am  a ; am  arel  a
4
5g
am  a
4
T  mB g  mB a  1
T  mm g  m2 am   2 
solving (1) & (2), we get
T - 100g = 100a -------(3)
 5g 
T - 60g = 60   a  ------(4)
 4 
(3 ) - (4) , 40 g  100a  75 g  60a
35 2
160a  350  a  ms
16
35
T = 100g + 100a = 1000 + 100   1218 N
16
EX.80 :In the pulley-block arrangement shown in figure.Find the relation between acceleration
of block A and B.
1) aB=-3aA 2) aB=-aA 3) aB=-2aA 4) aB=-4aA
Sol : key-1
x1  x3  1
differentiating with respect to time,
we get v1  v3  0
Again differentiating w.r.to to time,
a1  a3  0  a1   a3 , a3   a1
 x1  x3    x2  x3    2 x1  x2  2 x3   2
differentiating w.r.to time, v1  v2  2v3  0
Again differentiating w.r.to time, a1  a2  2a3  0
a1  a2  2a1  0 ; 3a1  a2  0
a2  3a1 ; aB  3a A
EX.81 :Three equal weights A, B and C of mass 2 kg each are hanging on a string passing over a
fixed frictionless pulley as shown in the fig. The tension in the string connecting weights B
and C is
1) zero 2) 13 N 3) 3.3 N 4) 19.6 N

Sol : key-2. For A, T1  m1a  m1 g
For B, (m 2  m3 ) g  T2  (m 2  m3 )a
Fnet
a
m1  m2  m3 ;For C , m3 g  T2  m3 a
EX.82 :In the figure shown a3 = 6m/s2 (downwards) and a2 = 4m/s2 (upwards). Find acceleration
of 1.
1) 1m/sec2 upwards 2) 2m/sec2 upwards

3) 1m/sec2 downwards 42m/sec2 downwards
Sol : key-1
Since the points 1,2,3 and are movable, so let their displacements are x1 , x2 , x3 and x4 We observe
that the length of the strings between 1 and 4 and 2 and 3 are constants.
x1  x4  1 ;  x2  x4    x3  x4    2
Differentiating twice w.r.t time, we get
a1  a 4  0  a1   a 4  1
a2  a4  a3  a4  0  a 2  a3  2 a 4  0   2 
a2  4ms 2 , a3  6ms 2 ; 4  6  2a4  0
2a4  2  a4  1ms 2 ; From (1),
a1   a4 ; a1  1ms 2 ; a1  1ms 2 upwards
EX.83 :A man of mass m stands on a platform of equal mass m and pulls himself by two ropes
passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his
weight, his upward acceleration would be
g g
1) 2) 3) g 4) zero
2 4
 mg 
Sol : key-4. F  4   upwards
 2 
W  2mg  downwards
F = W , a  0
EX.84 :Two masses each equal to m are lying on X-axis at (-a, 0) and (+a, 0), respectively, as
shown in fig. They are connected by a light string. A force F is applied at the origin along
vertical direction. As a result, the masses move towards each other without loosing contact
with ground. What is the acceleration of each mass? Assume the instantaneous position of
the masses as (-x, 0) and (x, 0), respectively
y
-a, 0 a, 0
-x x
m m
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
2F a 2
 x2  2F x F x F x
1)
m x
2) m
a 2
 x2  3) 2m
a 2
 x2  4) m
a 2
 x2 
Sol : key-3. F  2T sin  , ma1  T cos 
EX.85 :Two blocks A and B are separated by some distance and tied by a string as shown in the
figure. The force of friction in both the blocks at t = 2s is.
1) 4N(  ), 5N(  ) 2) 2N(  ), 5N(  )

3) 0N(  ), 10N(  ) 4) 1N(  ), 10N(  )
Sol : key-4. At t = 2s, F1 = 4N
f1  1m1 g  0.6  1 10  6 N
f 2  2 m2 g  0.5  2  10  10 N
Fnet  F2  F1  15  4  11N
As, Fnet  f1  f 2 .
The system will remain at rest and the values of frictional forces on the blocks will be,
T  4  f1 and T  15  f 2 ; 4  f1  15  f 2
f1  f 2  11N  (1)
4th option,
f1  1N , f 2  10 N ; f1  f 2  11N
EX.86 :The car A is used to pull a load B with the pulley arrangement shown. If A has a forward
velocity v A determine an expression for the upward velocity vB , of the load in terms of VA and
 .  is angle between string and horizontal
h l
B
A
1 1
1) VA cos  2) VA sin  3) VA cos  4) VA tan 
2 2
Sol : key-1. Let  angle between string and horizontal T is tension in string
T
2T

VB VA
 T .V  0 ; T cos  VA  2TVB  0
1
VB  VA cos 
2
EX.87 :The force acting on the block of mass 1kg is given by F = 5 -2t. The frictional force acting
on the block after time t = 2 seconds will be    0.2 
  0.2 F  (5  2t)N
1) 2N 2) 3N 3) 1N 4) Zero
Sol : key-2.
N
(5-2t)
f
10
f max  10  0.2  2 N
Initial force = 5N > 2N
 block will move with acceleration
5  2t  f
a  5  2t  2
1
dv
 3  2t v  3t  t 2
dt
 at t  0, v  0  v  0  t  0,3sec
 at t = 2sec block is moving
 f max will act i.e., frictional force acting = 2N
EX.88 :A body of mass 2kg travels according to the law x(t) = pt + qt2 + rt3 where, q = 4ms-2, p =
3ms-1 and r = 5ms-3. The force acting on the body at t = 2 seconds is
a) 136 N b) 134 N c) 158 N d) 68 N
Sol: (a) Given, mass = 2kg
x(t) =pt + qt2 + rt3
dx
v  p + 2qt + 3rt 2
dt
dv
a  0 + 2qt + 6rt
dt
at t = 2s; a = 2q + 6 x 2 xr
= 2q +12r
= 2 x 4 + 12 x 5
= 8 + 60 = 68 m/s
Force = F = ma = 2 x 68 = 136N
  
EX.89 :A body with mass 5 kg is acted upon by a force F   3 i  4 j  N . If its initial velocity at t
 
  
 1
= 0 is v   6 i  12 j  ms , the time at which it will just have a velocity along they- axis is
 
a) never b) 10 s c) 2 s d) 15 s
Sol: (b) Given, mass = m = 5kg
  
A cting force = F   3 i  4 j  N
 
  
 1
Initial velocity at t = 0, v   6 i  12 j  ms
 
  

F  3i 4 j 
Retardation, a  m    5  5  m / s
2
 
As final velocity is along Y-axis only, its x- component must be zero.
From v = u + at, for X-component only,
EX.90 :The motion of a particle of mass m is given by x = 0 for t < 0s, x(t) = Asin 4  t for 0< t<(1/
4) s (A > 0), and x = 0 for t >(1/4) s. Which of the following statements is true?
a) The force at t = (1/8) s on the particle is -m16  2 A.
b) The particle is acted upon by an impulse of magnitude m47  2 A at t = 0 s and t = (1/4) s.
c) The particle is not acted upon by any force.
d) The particle is not acted upon by a constant force.
e)There is no impulse acting on the particle.
Sol: (a,b,d)
Given, x = 0 for t < 0 s
1
x(t) = A sin 4  t; for 0 < t < s
4
1
x =0; for t > s
4
1
For, 0 < t < s
4
dx dv  t 
v (t) = =4 Acos 4  t
dt dt
dv  t 
a(t) = = -16  2 a sin4  t
dt
1 1
At = s, a(t) = -16  2 Asin4  x = -16  2A
8 8
F = ma(t) = -16  2 A x m = -16  2 mA
Impulse = Change in linear momentum
1
I = Fxt = (-16  2 Am) x = -4  2Am
4
the impulse (Change in linear momentum)
1
at t = 0 is same as, t = s
4
Clearly, force depends upon A which is not constant. Hence, force is also not constant.
EX.91 :In figure a body A of mass m slides on plane inclined at angle 1 ,to the horizontal and  is
the coefficent of friction between Aand the plane. A is connected by a light string passing
over a frictioness pulley to another body B, also of mass m, sliding on a frictionless plane
inclined at angle 2 to the horizontal. Which of the following statements are true?
B
A m
m
1 2
a) A will never move up the plane.
sin 2  sin 1
b) A will just start moving up the plane when   cos 1
c) For A to move up the plane, 2 must always be greater than 1 .
d) B will always slide down with constant speed.
Sol: (b,c) Let A moves up the plane frictional force on A will be downward as shown.
N1
A B N2
s1
1 g co mg cos
sin m 2
2
mg
mg
f mg mg 2
sin
1
2
When A just starts moving up

mg sin 1  f  mg sin 2
 mg sin 1  mg cos 1  mg sin 2
sin 2  sin 1

cos 1
When A moves upwards
f = mg sin 2 - mg sin 1 > 0
 sin 2 > sin 1  2 > 1
EX.92 :Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as
shown in figure. Calculate T1 and T2 when whole system is going upwards with acceleration
= 2m/s2 (use g = 9.8 ms-2).
T1
5 kg
T2
3 kg
a) T1 = 5N, T2 = 38N b) T1 = 35.4N, T2 = 94.4N

c) T1 = 94.4N, T2 = 35.4N d) T1 = 0N, T2 = 35.4N
Sol:Given, m1 = 5kg, m2 = 3kg
g = 9.8m/s2 and a=2m/s2
T1
5kg T2
T1
T2
a a
a 5g T2
3kg 3g
T2
For the upper block T1 - T2 - 5g = 5a
 T1 - T2 = 5 (g + a)
For the lower block T2 - 3g = 3a
 T2 = 3(g + a) = 3(9.8+2) = 35.4N
From Eq. (i) T1 = T2 + 5 (g + a)
= 35.4 + 5 (9.8 + 2) = 94.4N
    1
EX.93 :A cricket ball of mass 150 g has an initial velocity u   3 i  4 j  ms and a final velocity
 
  
v    3 i  4 j  ms 1 after being hit. The change in momentum (final momentum- initial
 
momentum) is (in kg ms-1)
  
   
  
a) zero b)   0.45 i  0.6 j  c)   0.9 i  1.2 j  d) 5  i  j  i
     
 
Sol: (c) Given, u = 3iˆ  4ˆj m/s and
 
v = - 3iˆ  4ˆj m/s
mass ofthe ball = 150 g = 0.15 kg
 P= mv-mu
 ˆ ˆ  
ˆ ˆ 
 P = m (v - w) = -(0.15)  3i  4 j  3i  4 j  
= (0.15)  6iˆ  8ˆj
Hence,  p =  0.9iˆ  1.2ˆj

EX.94 :In figure the co-efficient of friction between the floor and the body B is 0.1. The co-
efficient of friction between the bodies B and A is 0.2. A force F is applied as shown B. The
mass of A is m/2 and of B is m. Which of the following statements are true?
B F
a) The bodies will move together if F= 0.25 mg.
b) The body A will slip with respect to B if F=0.5 mg.
c) The bodies will move together if F= 0.5 mg.
d) The bodies will be at rest if F=0.1 mg.
Sol: (a,b,d)
Consider the adjacent diagram Frictonal force on
B(f1) and frictional force on A(t2) will be as shown.
Let A and B are moving together
F  f1 F  f1 2  F  f1 
a common   
mA  mB  m / 2   m 3m
Pseudo force on A = (mA) x acommon
2  F  f1  m 2  F  f1   F  f1 
 mA    
3m 2 3m 3m
A f2
f2
f1 B F
The force (F) will be maximum when

Pseudo force on A = Frictional force on A
F f
 max 1  m A g
3
F f m
 max 1  0.2   g  0.1mg
3 2
 Fmax  0.3mg  f1
3
 0.3mg   0.1 mg  0.45mg
2
 Hence, maximum force upto which bodies will move together is Fmax = 0.45mg
a) Hence, for F = 0.25 mg < Fmax bodies will move together
b) For F = 0.5mg > Fmax, body A will slip with respect to B
c) For F = 0.5mg > Fmax, bodies slip
3
 f1 max  m Bg   0.1  m  g  0.15mg
2
m
 f 2 max  m A g   0.2    g  0.1mg
2
Hence, minimum force required for movement of the system (A + B)
fmin= (f1)max+ (f2)max
= 0.15mg + 0.1mg = 0.25mg
d) Ginve, force F = 0. lmg < Fmin
Hence, the bodies will be at rest
EX.95 :A body of mass 10kg is acted upon by two perpendicular forces, 6N and 8N. The resultant
acceleration of the body is
a) 1 ms-2 at an angle of tan-1  4 / 3 6N force.
b) 0.2 ms-2 at an angle of tan-1  4 / 3 w.r.t. 6N force.
c) 1 ms-2 at an angle of tan-1  4 / 3 w.r.t.8N force.
d) 0.2 ms-2 at an angle of tan-1  4 / 3 w.r.t.8N force.
Sol: (a) Consider the adjacent diagram
Given, mass = m = 10 kg
F1 = 6N, F2 = 8N
Resultant force= F  F12  F22  36  64 =10N
F2
F
PN
2
1
F1
F 10
a   1m / s 2 ; along R
m 10
Let 1 , be angle between R and F1
8 4
tan 1  
6 3
4
1  tan 1   w.r.t. F = 6N
3 1
Let 2 be angle between F and F2

6 3
tan 2  
8 4
3
2  tan 1   w.r.t F = 8N
4 2
EX.96 :Mass m1 moves of a slope making an angle q with the horizontal and is attached to mass m2
by a string passing over a frictionless pulley as shown in figure. The coefficient of friction
between m1 and the sloping surface is  . Which of the following statements are true ?
a) If m2 > m1 sin , the body will move up the plane
b)If m2 > m1 (sin  +  cos  ), the body will move up the plane
c) If m2 < m1 (sin  +  cos  ), the body will move up the plane
d) If m2 < m1 (sin  -  cos  ),the body will move down the plane
Sol: (b,d)
Let m1, moves up the plane.
Different forces involved are shown in the diagram
N = Normal reaction
f = Frictional force
T = Tension in the string
f =  N =  m1g cos 
N T T

sin  mg cos
mg f
m1g
For the system (m1 + m 2) to move up
m2g - (m1g sin  + f) > 0
 m2g - (m1g sin  +  m1g cos  ) > 0
 m2 > m1(sin  +  cos  )
Hence, option (b) is corrected
Let the body moves down the plane, in this case facts up the plane.
Hence, m1g sin  - f > m2g
 m1g sin -  m1g cos  > m2g
 m1 (sin  -  cos  ) > m2
 m2 < m1 (sin  -  cos  )
Hence, option (d) is correct.
EX.97 :When a body slides down from rest along a smooth inclined plane making an angle of 450
with the horizontal, it takes timeT. When the same body slides down from rest along a rough
inclined plane making the same angle and through the same distance, it is seen to take time pT,
where p is some number greater than 1. Calculate the coefficient of friction between the body
and the rough plane.
a
 =450
 1   1  1 1
a)  1  2  b)  1  2  c) 2 d) - 2
 P   P  P P
Sol: (a) consider the diagram where a body slides down from along an inclined plane of inclination
 (= 450)
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane
Here a = gsin  ,  = 450

g
 a = gsin450 =
2
Let the travelled distance be s.
1
Using equation of motion, s = ut + at2, weget
2
1 g 2 gT 2
s = 0.t + T or s 
2 2 2 2
On rough inclined plane Acceleration fo the body
a = g(sin  -  cos  )
g 1   

2
 1 
 As, sin 45  cos 45 
0 0

 2
Again using equation of motion,
1
s = ut + at2, we get
2
1 g 1   
s  0(pT)  (pT) 2
2 2
g 1    p 2 T 2
or s 
2 2
From Eqs. (i) and (ii), we get
gT 2 g 1    p T
2 2

2 2 2 2
or 1    p 2  1
1
or 1    
p2
 1 
or   1  2 
 p 
TOPIC-1 1ST,2ND & 3RD Laws of Motion
1. A particle moving in the 𝑥𝑦 plane experiences a velocity dependent force 𝐹⃗ = 𝑘(𝑣𝑦 𝑖̂ + 𝑣𝑥 𝑗̂) ,
where 𝑣𝑥 and 𝑣𝑦 are 𝑥 and 𝑦 components of its velocity 𝑣⃗. If 𝑎⃗ is the acceleration of the
particle, then which of the following statements is true for the particle? [Sep. 06, 2020 (II)]
(a) Quantity 𝑣⃗ × 𝑎⃗ is constant in time
(b) 𝐹⃗ Arises due to a magnetic field
(c) Kinetic energy of particle is constant in time
(d) Quantity 𝑣⃗ ⋅ 𝑎⃗ is constant in time
sol. (a) Given
𝐹⃗ = 𝑘(𝑣𝑦 𝑖̂ + 𝑣𝑋 𝑗)
𝐹𝑥 = 𝑘𝑣𝑦 𝑖̂, 𝐹𝑦 = 𝑘𝑣𝑥 𝑗
𝑚𝑑𝑣𝑥 𝑑𝑣𝑥 𝑘
= 𝑘𝑣𝑦 ⇒ = 𝑣𝑦
𝑑𝑡 𝑑𝑡 𝑚
𝑑𝑣𝑦 𝑘
Similarly, = 𝑚 𝑣𝑥
𝑑𝑡
dv y vx
  v y dv y  vx dvx
dvx vy
𝑣𝑦2 = 𝑣𝑋2 + 𝐶
𝑣𝑦2 − 𝑣𝑋2 = constant
𝑘
𝑣⃗ × 𝑎⃗ = (𝑣𝑥 𝑖̂ + 𝑣𝑦 𝑗̂) × (𝑣 𝑖̂ + 𝑣𝑥 𝑗̂)
𝑚 𝑦
𝑘 𝑘
= (𝑣𝑥2 𝑘̂ − 𝑣𝑦2 𝑘̂ ) 𝑚 = (𝑣𝑥2 − 𝑣𝑦2 ) 𝑚 𝑘̂ = constant
2. A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate
𝑑𝑀(𝑡)
= 𝑏𝑣 2 (𝑡) , where 𝑣(𝑡) is its instantaneous velocity. The instantaneous acceleration of the
𝑑𝑡
satellite is : [Sep. 05, 2020 (II)]

𝑏𝑣 3 2𝑏𝑣 3 𝑏𝑣 3
(a) −𝑏𝑣 3 (𝑡) (b) − 𝑀(𝑡) (c) − 𝑀(𝑡) (d) − 2𝑀(𝑡)
sol. (b) From the Newton’s second law,

𝑑𝑝 𝑑(𝑚𝑣) 𝑑𝑚
𝐹= = = 𝑣 ( 𝑑𝑡 ) (i)
𝑑𝑡 𝑑𝑡
𝑑𝑀(𝑡)
We have given, = 𝑏𝑣 2 (𝑡) (ii)
𝑑𝑡
Thrust on the satellite,

𝑑𝑚
𝐹 = −𝑣 ( 𝑑𝑡 ) = −𝑣(𝑏𝑣 2 ) = −𝑏𝑣 3 [Using (i) and (ii)]
−𝑏𝑣 3
⇒ 𝐹 = 𝑀(𝑡)𝑎 = −𝑏𝑣 3 ⇒ 𝑎 =
𝑀(𝑡)
3. A small ball of mass 𝑚 is thrown upward with velocity 𝑢 from the ground. The ball experiences a
resistive force 𝑚𝑘𝑦 2 where 𝑣 is its speed. The maximum height attained by the ball is:
[Sep. 04, 2020 (II)]
1 𝑘𝑢2 1 𝑘𝑢2 1 𝑘𝑢2 1 𝑘𝑢2
(a) tan−1 (b) ln ((+ )) (c) tan−1 (d) ln ((+ ))
2𝑘 𝑔 𝑘 2𝑔 𝑘 2𝑔 2𝑘 𝑔
sol.
𝐹⃗ = 𝑚𝑘𝑣 2 − 𝑚𝑔 (⋅.⋅mg and 𝑚𝑘𝑦 2 act opposite to each other)

𝑎⃗ = −[𝑘𝑣 2 + 𝑔]
𝑑𝑣 𝑑𝑣
⇒ 𝑣 ⋅ 𝑑ℎ = −[𝑘𝑦 2 + 𝑔] (⋅.⋅ 𝑎 = 𝑣 𝑑ℎ)
0 ℎ
𝑣 ⋅ 𝑑𝑣
⇒∫ 2
= ∫ 𝑑ℎ
𝑢 𝑘𝑦 + 𝑔 0
1
⇒ ln [𝑘𝑣 2 + 𝑔]0𝑢 = −ℎ
2𝑘
1 𝑘𝑢2 + 𝑔
⇒ ln [ ]=ℎ
2𝑘 𝑔
4. A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the
ball is affected by a drag force equal to m𝛾v 2 (where m is mass ofthe ball, v is its instantaneous
velocity and 𝛾 is a constant). Time taken by the ball to rise to its zenith is: [10 April 2019 I]
1 𝛾 1 𝛾
(a) tan−1 (√g V0 ) (b) sin−1 (√g V0 )
√𝛾g √𝛾g
1 𝛾 1 2𝛾
(c) 𝑙n (1 + √g V0 ) (d) tan−1 (√ g V0 )
√𝛾g √2𝛾g
sol. (a) Net acceleration

dv
= a = −(g + 𝛾v 2 )
dt
Let time t required to rise to its zenith (v = 0) so,
0 −dv 𝑡
∫𝑣0 g+𝛾v2 = ∫0 d t [for H max , v = 0]
l   v0 
t tan 1 
g  g 
 
5. A ball is thrown vertically up (taken as + z‐axis) from the ground. The correct momentum‐height
(p‐h) diagram is: [9 April 2019 I]
sol. (d) v 2 = u2 − 2gh

or v = √𝑢2 − 2𝑔ℎ
Momentum, P = mv = 𝑚√𝑢2 − 2𝑔ℎ
u2
At h = 0, P = mu and at h  , P  0
f
upward direction is positive and downward direction is negative.
6. A particle of mass m is moving in a straight line with momentump. Starting at timet = 0, a force
𝐹 = 𝑘t acts in the same direction on the moving particle during time interval T so that its
momentum changes from p to 3p. Here 𝑘 is a constant. The value of T is: [11 Jan. 2019 II]
𝑘 p 2𝑘 2𝑝
(a) 2√p (b) 2√𝑘 (c) √ p (d) √ k
sol. (b) From Newton’s second law

dp
= 𝐹 = kt
dt
Integrating both sides we get,
3p T T
t2
∫ d p = ∫ k tdt ⇒ [p]3p
p = k[ ]
p 0 2 0
kT 2 p
⇒ 2p = ⇒ T = 2√
2 k
R
7. A particle of mass m is acted upon by a force 𝐹 given by the empirical law = t2 v(t) . If this law
is to be tested experimentally by observing the motion starting from rest, the best way is to plot:
1
(a) log v(t) against (b) v(t) against t 2
t
1
(c) log v(t) against (d) log v(t) against t
t2
R dv R
sol. (a) From 𝐹 = t2 v(t) ⇒ m dt = t2 v(t)
dv Rdt
Integrating both sides  dt   mt 2
R
In v = − mt
1
. . ln v ∝ t
8. A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth
rigid horizontal rod of length L(L >> r) and are at rest at random positions. The rod is mounted
between two rigid supports (see figure). If one of the beads is now given a speed v, the average
force experienced by each support after a long time is (assume all collisions are elastic):
mv2 mv2 mv2

(a) (b) (c) (d) zero
2(L−nr) L−2nr L−nr
sol. (b) Space between the supports for motion of beads is L − 2nr
2mV mV 2
Average force experienced by each support, F  
2  L  2nr  L  2nr
V
9. A body of mass 5 kg under the action of constant force 𝐹⃗ = 𝐹x î + 𝐹y ĵ has velocity at t = 0s as

⃗⃗ = +6ĵm/s. The force 𝐹⃗ is: [Online April 11, 2014]
⃗⃗ = (6î − 2ĵ)m/s and at t = 10s as v
v
3 4 3 4
(a) (−3ĵ + 4j)N (b) (− 5 î + 5 ĵ) N (c) (3î‐4j)N (d) (5 î‐ 5 ĵ) N
sol. (a) From question,

Mass of body, 𝑚 = 5 kg
Velocity at 𝑡 = 0, 𝑢 = (6î‐2j) m/s
Velocity at 𝑡 = 10s, 𝑣 = +6𝑗m/s
Force, 𝐹 =?
𝑣−𝑢 6𝑗̂ −(6𝑖̂−2𝑗̂ ) −3𝑖̂+4𝑗̂
Acceleration, 𝑎 = = = m/s2
𝑡 10 5
(−3𝑖̂+4𝑗̂ )
Force, 𝐹 = 𝑚𝑎 =5× = (−3𝑖̂ + 4𝑗̂)𝑁
5
10. A particle of mass m is at rest at the origin at time 𝑡 = 0. It is subjected to a force 𝐹(𝑡) = 𝐹0 e−𝑏𝑡
in the 𝑥 direction. Its speed v(𝑡) is depicted by which of the following curves? [2012]
sol. (c) Given that 𝐹(𝑡) = 𝐹0 𝑒 −𝑏𝑡

𝑑𝑣
⇒ 𝑚 𝑑𝑡 = 𝐹0 𝑒 −𝑏𝑡
𝑑𝑣 𝐹0 −𝑏𝑡
= 𝑒
𝑑𝑡 𝑚
𝑣
𝐹0 𝑡
∫ 𝑑 𝑣 = ∫ 𝑒 −𝑏𝑡 𝑑𝑡
0 𝑚 0
𝑡
𝐹0 𝑒 −𝑏𝑡 𝐹0
𝑣= [ ] = [−(𝑒 −𝑏𝑡 − 𝑒 −0 )]
𝑚 −𝑏 0 𝑚𝑏
𝐹0
⇒𝑣= [1 − 𝑒 −𝑏𝑡 ]
𝑚𝑏
11. This question has Statement 1 and Statement 2. Of the four choices given after the Statements,
choose the one that best describes the two Statements.
Statement 1: If you push on a cart being pulled by a horse so that it does not move, the cart pushes
you back with an equal and opposite force.
Statement 2: The cart does not move because the force described in statement 1 cancel each other.
(a) Statement 1 is true, Statement2 is true, Statement 2 is the correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement2 is true, Statement 2 is not the correct explanation of Statement 1.
sol. (a) According to newton third law of motion i.e. every action is associated with equal and opposite
reaction.
12. Two fixed frictionless inclined planes making an angle 30O and 60∘ with the vertical are shown in
the figure. Two blocks A and B are placed on the two planes. What is the relative vertical
acceleration of A with respect to B? [2010]
(a) 4.9ms −2 in horizontal direction (b) 9.8ms−2 in vertical direction

(c) Zero (d) 4.9ms−2 in vertical direction
sol. (d) mg sin 𝜃 = 𝑚𝑎
𝑎 = 𝑔 sin 𝜃
Vertical component of acceleration = 𝑔sin2 𝜃
Relative vertical acceleration ofA with respect to B is
𝑔(sin2 60 − sin2 30]
3 1 𝑔
= 𝑔 (4 − 4) = = 4.9m/s 2 in vertical direction
2
13. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves
0.2m while applying the force and the ball goes upto 2 m height further, find the magnitude of the
force. (Consider g = 10m/s2 ). [2006]
(a) 4N (b) 16N (c) 20N (d) 22N
sol. (d) For the motion of ball, just after the throwing
𝑣 = 0, 𝑠 = 2m, 𝑎 = −𝑔 = −10ms−2
𝑣 2 − 𝑢2 = 2𝑎𝑠 for upwardjourney
⇒ −𝑢2 = 2(−10) × 2 ⇒ 𝑢2 = 40
When the ball is in the hands of the thrower
𝑢 = 0, 𝑣 = √40ms −1
𝑠 = 0.2m
𝑣 − 𝑢2 = 2𝑎𝑠
2
⇒ 40 − 0 = 2(𝑎)0.2 ⇒ 𝑎 = 100m/s2
𝐹 = 𝑚𝑎 = 0.2 × 100 = 20𝑁
⇒ 𝑁 − 𝑚𝑔 = 20 ⇒ 𝑁 = 20 + 2 = 22𝑁
Note :
𝑊ℎ𝑎𝑛𝑑 + 𝑊𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝛥𝐾
⇒ 𝐹(0.2) + (0.2)(10)(2.2) = 0 ⇒ 𝐹 = 22𝑁
14. A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is
completed in 0.1s, the force ofthe blow exerted by the ball on the hand of the player is equal to
[2006]
(a) 150N (b) 3 N (c) 30N (d) 300N
sol. (c) Given, mass of cricket ball, 𝑚 = 150𝑔 = 0.15 kg
Initial velocity, 𝑢 = 20m/s
𝑚(𝑣−𝑢) 0.15(0−20)
Force, 𝐹 = = = 30𝑁
𝑡 0.1
15. A particle of mass 0.3 kg subject to a force 𝐹 = −𝑘𝑥 with 𝑘 = 15N/m. What will be its initial
acceleration if it is released from a point 20 cm away from the origin? [2005]
(a) 15m/s 2 (b) 3m/s 2 (c) 10m/s2 (d) 5m/s2
sol. (c) Mass (𝑚) = 0.3 kg
Force, 𝐹 = 𝑚. 𝑎 = −𝑘𝑥
⇒ 𝑚𝑎 = −15𝑥
⇒ 0.3𝑎 = −15𝑥
15 −150
⇒𝑎=− 𝑥= 𝑥 = −50𝑥
0.3 3
𝑎 = −50 × 0.2 = 10m/s2
16. A block is kept on a frictionless inclined surface with angle of inclination𝛼’. The incline is given an
acceleration ‘a’ to keep the block stationary. Then a is equal to [2005]
(a) 𝑔 cosec 𝛼 (b) 𝑔/ tan 𝛼 (c) 𝑔 tan 𝛼 (d) 𝑔

sol. (c) When the incline is given an acceleration a towards the right, the block receives a reaction 𝑚𝑎
towards left.
For block to remain stationary, Net force along the incline should be zero.
mg sin 𝛼 = 𝑚𝑎 cos 𝛼 ⇒ 𝑎 = 𝑔 tan 𝛼
17. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards with an initial accelerationof 10m/s 2 .
Then the initial thrust of the blast is [2003]
(a) 3.5 × 105 N (b) 7.0 × 105 N (c) 14.0 × 105 N (d) 1.75 × 105 N
sol. (b) In the absence of air resistance, if Thrust (F) the rocket moves up with an acceleration a, then
thrust
𝐹 = 𝑚𝑔 + 𝑚𝑎
𝐹 = 𝑚(𝑔 + 𝑎) = 3.5 × 104 (10+10)
= 7 × 105 N
18. Three forces start acting simultaneously on a particle moving with velocity, v
⃗⃗. These forces are
represented in maguitude and direction by the three sides of a triangle ABC. The particle will now
move with velocity [2003]
(a) less than 𝑣⃗ (b) greater than 𝑣⃗

(c) |𝑣⃗| in the direction of the largest force BC (d) 𝑣⃗ , remaining unchanged
sol. (d) Resultant force is zero, as three forces are represented by the sides of a triangle taken in the same
order. From Newton’s second law, 𝐹⃗𝑛𝑒𝑡 = 𝑚𝑎⃗.
Therefore, acceleration is also zero i.e., velocity remains unchanged.
19. A solid sphere, a hollow sphere and a ring are released from top of an inclined plane (frictionless) so
that they slide down the plane. Then maximum acceleration down the plane is for (no rolling) [2002]
(a) solid sphere (b) hollow sphere (c) ring (d) all same
sol. (d) This is a case of sliding (if plane is friction less) and therefore the acceleration of all the bodies is
same.
TOPIC 2, Motion of Connected Bodies, Pulley& Equilibrium of Forces
20. A mass of10 kg is suspended by a rope of length 4 m, from the ceiling. A force 𝐹 is applied
horizontally at the midpoint ofthe rope such that the top half of the rope makes an angle of 45∘ with
the vertical. Then 𝐹 equals: (Take 𝑔 = 10 m/s2 and the rope to be massless) [7 Jan. 2020 II]
(a) 100 N (b) 90 N (c) 70N (d) 75N
sol. (a) From the free body diagram
10 kg
𝑇 cos 45∘ = 100𝑁 (i)
𝑇 sin 45∘ = 𝐹 (ii)
On dividing (i) by(ii) we get
𝑇 cos 45∘ 100
=
𝑇 sin 45∘ 𝐹
⇒ 𝐹 = 100N
21. An elevator in a building can carry a maximum of 10 persons, with the average mass of each person
being 68 kg. The mass ofthe elevator itself is 920 kg and it moves with a constant speed of 3 m/s.
The frictional force opposing the motion is 6000 N. Ifthe elevator is moving up with its full capacity,
the power delivered by the motor to the elevator (𝑔 = 10m/s 2 ) must be at least: [7 Jan. 2020 II]
(a) 56300 W (b) 62360 W (c) 48000 W (d) 66000 W
sol. (d) Net force on the elevator = force on elevator + frictional force
⇒ 𝐹 = (10𝑚 + 𝑀)𝑔 + 𝑓
where, 𝑚 = mass ofperson, 𝑀 = mass ofelevator, 𝑓 = fiictional force
⇒ 𝐹 = (10 × 68 + 920) × 9.8 + 600
⇒ 𝐹 = 22000𝑁
⇒ 𝑃 = 𝐹𝑉 = 22000 × 3 = 66000𝑊
22. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied
on the rope at some point, the rope deviated at an angle of 45o at the roof point. If the suspended
mass is at equilibrium, the magnitude of the force applied is (g = 10ms−2 ) [9 Jan. 2019 II]
(a) 200 N (b) 140 N (c) 70 N (d) 100 N
sol.
At equilibrium,
mg 100
tan 45∘ = =
𝐹 𝐹
𝐹 = 100N
23. A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m
and radius R. Ifthe string does not slip on the cylinder, with what acceleration will the mass fall or
release? [2014]
2g g 5g
(a) b) (c) (d) g
3 2 6
sol. (b) From figure,
Acceleration 𝑎 = 𝑅𝛼 … (i)
And 𝑚𝑔 − 𝑇 = 𝑚𝑎 …(ii)
From equation (i) and (ii)
𝑎
𝑇 × 𝑅 = 𝑚𝑅 2 𝛼 = 𝑚𝑅 2 ( )
𝑅
or 𝑇 = 𝑚𝑎
⇒ 𝑚𝑔 − 𝑚𝑎 = 𝑚𝑎
𝑔
⇒ 𝑎= 2
24. Two blocks of mass M1 = 20 kg and M2 = 12 kg are connected by a metal rod of mass 8 kg. The
system is pulled vertically up by applying a force of 480 N as shown. The tension at the mid‐point
of the rod is: [Online April 22, 2013]
(a) 144 N (b) 96N (c) 240 N (d) 192N
𝐹
sol. (d) Acceleration produced in upward direction a = M
1 +M2 +Massofmetalrod
480
= 20+12+8 = 12 ms −2
Tension at the mid point

Mass of rod
T = (M2 + ) a = (12 + 4) × 12 = 192N
2
25. A uniform sphere of weight W and radius 5 cm is being held by a string as shown in the figure. The
tension in the string will be: [Online April 9, 2013]
𝑊 𝑊 𝑊 𝑊
(a) 12 (b) 5 12 (c) 13 (d) 13 12
5 5
sol. (d)
PQ = √oP 2 + OQ2
= √132 + 52 = 12
13
Tension in the string T = w cos 𝜃 = 12 W
26. A spring is compressed between two blocks of masses 𝑚1 and 𝑚2 placed on a horizontal
fiictionless surface as shown in the figure. When the blocks are released, they have initial velocity of
𝑣1 and 𝑣2 as shown. The blocks travel distances 𝑥1 and 𝑥2 respectively before coming to rest.
x 
The ratio  1  is [Online May 12, 2012]
 x2 
𝑚2 𝑚1 𝑚 𝑚
(a) (b) (c) √𝑚2 (d) √𝑚1
𝑚1 𝑚2 1 2
sol. (a)
27. A block of mass 𝑚 is connected to another block ofmass 𝑀 by a spring (massless) of spring
constant𝑘. The block is kept on a smooth horizontal plane. Initially the blocks are at rest and the
spring is unstretched. Then a constant force 𝐹 starts acting on the block of mass 𝑀 to pull it. Find
the force of the block of mass 𝑚. [2007]
𝑀𝐹 𝑚𝐹 (𝑀+𝑚)𝐹 𝑚𝐹
(a) (𝑚+𝑀)
(b) (c) (d) (𝑚+𝑀)
𝑀 𝑚
sol. (d) Writing free body‐diagrams for 𝑚 &M,
we get 𝑇 = 𝑚𝑎 and 𝐹 − 𝑇 = 𝑀𝑎 where 𝑇 is force due to spring

⇒ 𝐹‐ 𝑚𝑎 = 𝑀𝑎 or, 𝐹 = 𝑀𝑎 + 𝑚𝑎
𝐹
Acceleration of the system 𝑎 = 𝑀+𝑚.
𝐹 𝑚𝐹
Now, force acting on the block of mass m is 𝑚𝑎 = 𝑚 (𝑀+𝑚) = 𝑚+𝑀
28. Two masses 𝑚1 = 5g and 𝑚2 = 4.8 kg tied to a string are hanging over a light frictionless pulley.
What is the acceleration of the masses when left free to move? (g = 10ms−2 ) [2004]
(a) 5 m/s2 (b) 9.8m/s 2 (c) 0.2m/s2 (d) 4.8m/s 2
sol. (c)Here,𝑚1 = 5kgand𝑚2 = 48kg
If a is the acceleration of the masses,
𝑚1 𝑎 = 𝑚1 𝑔 − 𝑇 (i)
𝑚2 𝑎 = 𝑇 − 𝑚2 𝑔 (ii)
Solving (i) and (ii) we get
𝑚1 − 𝑚2
𝑎=( )𝑔
𝑚1 + 𝑚2
(5 − 4.8) × 9.8
⇒𝑎= m/s2 = 0.2m/s2
(5 + 4.8)
29. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring
reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s 2 ,
the reading of the spring balance Will be [2003]
(a) 24N (b) 74N (c) 15N (d) 49N
sol. (a) When lift is stationary, 𝑊1 = 𝑚𝑔 (i)
When the lift descends with acceleration a, 𝑊2 = 𝑚(𝑔 − 𝑎)
49
𝑊2 = (10 − 5) = 24.5𝑁
10
30. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force
𝑃 is applied at the free end ofthe rope, the force exerted by the rope on the block is [2003]
𝑃𝑚 𝑃𝑚 𝑃𝑀
(a) (b) (c) 𝑃 (d)
𝑀+𝑚 𝑀−𝑚 𝑀+𝑚
sol. (d) Taking the rope and the block as a system

we get 𝑃 = (𝑚 + 𝑀)𝑎
𝑃
Acceleration produced, 𝑎 = 𝑚+𝑀
Taking the block as a system,

𝑀𝑃
Force on the block, 𝐹 = 𝑀𝑎 𝐹 = 𝑚+𝑀
31. A light spring balance hangs from the hook of the other light spring balance and a block of mass 𝑀
kg hangs liom the former one. Then the true statement about the scale reading is [2003]
(a) both the scales read 𝑀 kg each
(b) the scale of the lower one reads 𝑀 kg and of the upper one zero
(c) the reading of the two scales can be anything but the sum of the reading will be 𝑀 kg
(d) both the scales read M/2 kg each
sol. (a) The Earth exerts a pulling force Mg. The block in turn exerts a reaction force Mg on the spring of
spring balance 𝑆1 which therefore shows a reading of M kgf.
As both the springs are massless. Therefore, it exerts a force of Mg on the spring of spring balance
𝑆2 which shows the reading of 𝑀𝑘𝑔𝑓.
32. A lift is moving down with acceleration 𝑎. A man in the lift drops a ball inside the lift. The
acceleration of the ball as observed by the man in the lift and a man standing stationary on the
ground are respectively [2002]
(a) 𝑔, 𝑔 (b) 𝑔 − 𝑎, 𝑔 − 𝑎 (c) 𝑔 − 𝑎, 𝑔 (d) 𝑎, 𝑔
sol. (c) Case ‐ I: For the man standing in the lift, the acceleration of the ball
𝑎⃗𝑏𝑚 = 𝑎⃗𝑏 − 𝑎⃗𝑚 ⇒ 𝑎𝑏𝑚 = 𝑔 − 𝑎
Case‐ II: The man standing on the ground, the acceleration of the ball
𝑎⃗𝑏𝑚 = 𝑎⃗𝑏 − 𝑎⃗𝑚 ⇒ 𝑎𝑏𝑚 = 𝑔 − 0 = 𝑔
33. When forces 𝐹1 , 𝐹2 , 𝐹3 are acting on a particle ofmass 𝑚 such that 𝐹2 and 𝐹3 are mutually
perpendicular, then the particle remains stationary. Ifthe force 𝐹1 is nowremoved then the
acceleration of the particle is [2002]
(a) 𝐹1 /𝑚 (b) 𝐹2 𝐹3 /𝑚𝐹1 (c) (𝐹2 − 𝐹3 )/𝑚 (d) 𝐹2 /𝑚.
sol. (a) When forces 𝐹1 , 𝐹2 and 𝐹3 are acting on the particle, it remains in equilibrium. Force 𝐹2 and
𝐹3 are perpendicular to each other,
𝐹1 = 𝐹2 + 𝐹3
𝐹1 = √𝐹22 + 𝐹32
The force 𝐹1 is now removed, so, resultant of 𝐹2 and 𝐹3 will now make the particle move with
𝐹
force equal to 𝐹1 . Then, acceleration, 𝑎 = 𝑚1
34. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is
perpendicular to the smaller force. Then the magnitudes of the forces are [2002]
(a) 12N, 6N (b) 13N, 5N (c) 10 N, 8N (d) 16N, 2N.
sol. (b) Let the two forces be 𝐹1 and 𝐹2 and let 𝐹2 < 𝐹1 . 𝑅 is the resultant force.
Given 𝐹1 + 𝐹2 = 18 (i)
From the figure 𝐹22 + 𝑅 2 = 𝐹12
𝐹12 − 𝐹22 = 𝑅 2
35. A light string passing over a smooth light pulley connects two blocks of masses 𝑚1 and 𝑚2
(vertically). If the acceleration of the system is 𝑔/8, then the ratio of the masses is [2002]
(a) 8: 1 (b) 9: 7 (c) 4: 3 (d) 5: 3
sol. (b) For mass m1 𝑚1 𝑔 − 𝑇 = 𝑚1 𝑎 (i)
For mass 𝑚2 𝑇 − 𝑚2 𝑔 = 𝑚2 𝑎 (ii)
(𝑚1 −𝑚2 )𝑔
Adding the equations we get 𝑎= 𝑚1 +𝑚2
𝑔
Here 𝑎 = 8
m1
1
1 m2 m m m 9
  1 1  8 1  8  1 
8 m m2 m2 m2 7
1
1
m2
36. Three identical blocks of masses 𝑚 = 2 kg are drawn by a force 𝐹 = 10.2N with an acceleration
of 0.6ms−2 on a fiictionless surface, then what is the tension (in N) in the string between the blocks
𝐵 and 𝐶? [2002]
(a) 9.2 (b) 3.4 (c) 4 (d) 9.8

sol. (b) 𝛤orce =mass × acceleration
𝐹 = (𝑚 + 𝑚 + 𝑚) × 𝑎
𝐹 = 3m × a
𝐹
𝑎=
3𝑚
10.2
𝑎= m/s2
6
10.2
𝑇2 = 𝑚𝑎 = 2 × = 3.4N
6
37. One end of a massless rope, which passes over a massless and frictionless pulley 𝑃 is tied to a hook
𝐶 while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of
maximum safe acceleration (in ms−2 ) can a man of 60 kg climb on the rope? [2002]
(a) 16 (b) 6 (c) 4 (d) 8

sol. (c) Tension, 𝑇 = 360N
Mass of a man m = 60 kg
𝑚𝑔 − 𝑇 = 𝑚𝑎
𝑇
𝑎=𝑔−
𝑚
360
= 10 − = 4𝑚/𝑠 2
60
TOPIC-3 Friction
38. An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts
slipping after it is at height ℎ from the bottom. If the coefficient of friction between the ground and
the insect is 0.75, then ℎ is: (g = 10ms−2 ) [Sep. 06, 2020 (I)]
(a) 0.20 m (b) 0.45 m (c) 0.60m (d) 0.80m
sol. (a) For balancing, mgsin 𝜃 = 𝑓 = 𝜇𝑚𝑔 cos 𝜃
3
⇒ tan 𝜃 = 𝜇 = = 0.75
4
4 𝑅
ℎ = 𝑅 − 𝑅 cos 𝜃 = 𝑅 − 𝑅 ( ) =
5 5
𝑅
ℎ= = 0.2m [radius, 𝑅 = 1m]
5
39. A block starts moving up an inclined plane of inclination 30∘ with an initial velocity of 𝑣0 . It
𝑣0
comes back to its initial position with velocity . The value of the coefficient of kinetic friction
2
𝐼
between the block and the inclined plane is close to . The nearest integer to 𝐼 is
1000
[NA Sep. 03, 2020 (II)]

sol. (346)
Acceleration of block while moving up an inclined plane,
𝑎1 = 𝑔 sin 𝜃 + 𝜇𝑔 cos 𝜃
⇒ 𝑎1 = 𝑔 sin 30∘ + 𝜇𝑔 cos 30∘
𝑔 𝜇𝑔√3
=2+ (i) (𝜃 = 30o )
2
Using 𝑣 2 − 𝑢2 = 2𝑎(𝑠)
⇒ 𝑣02 − 02 = 2𝑎1 (𝑠) (u = 0)
⇒ 𝑣02 − 2𝑎1 (𝑠) = 0
𝑣2
⇒ 𝑠 = 𝑎0 (ii)
1
Acceleration while moving down an inclined plane

𝑎2 = 𝑔 sin 𝜃 − 𝜇𝑔 cos 𝜃
⇒ 𝑎2 = 𝑔 sin 30∘ − 𝜇𝑔 cos 30∘
𝑔 𝜇√3
⇒ 𝑎2 = 2 − 𝑔 (iii)
2
Using again 𝑣 2 − 𝑢2 = 2𝑎𝑠 for downward motion

𝑣 2 𝑣2
⇒ ( 20 ) = 2𝑎2 (𝑠) ⇒ 𝑠 = 4𝑎0 (iv)
2
𝑣02 𝑣2
Equating equation (ii) and (iv) = 4𝑎0 ⇒ 𝑎1 = 4𝑎2
𝑎1 2
𝑔 𝜇𝑔√3 𝑔 𝜇√3
⇒ + = 4( − )
2 2 2 2
⇒ 5 + 5√3𝜇 = 4(5 − 5√3𝜇) (Substituting, 𝑔 = 10m/s2 )
⇒ 5 + 5√3𝜇 = 20 − 20√3𝜇 ⇒ 25√3𝜇 = 15
√3 346
⇒𝜇= = 0.346 =
5 1000
𝑙 346
So, = 1000
1000
40. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20N,
making an angle of 30o with the horizontal, as shown in the figures. The coefficient of friction
between the block and floor is 𝜇 = 0.2. The difference between the accelerations of the block, in
case (B) and case (A) will be: (𝑔 = 10 ms−2 ) [12 April 2019 II]
(a) 0.4 ms −2 (b) 3.2 ms −2 (c) 0.8 ms −2 (d) 0 ms −2

sol. (c) A: 𝑁 = 5g + 20 sin 30∘
20 sin 30∘
𝐹−𝑓 20 cos 30∘ −𝜇𝑁
Accelaration, 𝑎1 = =
𝑚 5
√3
20 × 2 − 0.2 × 60
=[ ] = 1.06m/s 2
5
B: 𝑁 = 5g − 20 sin 30∘
1
= 50 − 20 × = 40N
2
𝐹−𝑓 20 cos 30∘ − 0.2 × 40
𝑎2 = =[ ] = 1.86m/s 2
𝑚 5
2
Now 𝑎2 − 𝑎1 = 1.86 − 1.06 = 0.8m/s
41. Two blocks A and B masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure.
The coefficient offiiction between A and B is 0.2 and between B and the surface of the table is
also 0.2. The maximum force 𝐹 that can be applied on B horizontally, so that the block A does not
slide over the block B is: [Take g = 10m/s 2 ] [10 April 2019 II]
(a) 8N (b) 16N (c) 40N (d) 12N

sol. (b) Taking (A + B) as system
𝐹 − 𝜇(𝑀 + 𝑚)𝑔 = (𝑀 + 𝑚)𝑎
𝐹 − 𝜇(𝑀 + 𝑚)𝑔
⇒𝑎=
(𝑀 + 𝑚)
𝐹−(0.2)4×10 𝐹−8
𝑎= =( ) (i)
4 4
But, 𝑎 max = 𝜇g = 0.2 × 10 = 2

𝐹−8
=2
4
⇒ 𝐹 = 16N
42. A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum
force 2 N down the inclined plane. The maximum external force up the inclined plane that does not
move the block is 10 N. The coefficient of static friction between the block and the plane is:
[Take g = 10m/s 2 ] [12 Jan. 2019 II]
√3 √3 1 2
(a) ` (b) (c) (d)
2 4 2 3
sol. (a) From figure,

2 + mg sin 30∘ = 𝜇mg cos 30∘ and 10 = mg sin 30∘ + 𝜇 mg cos 30∘
= 2𝜇mg cos 30∘ − 2
⇒ 6 = 𝜇mg cos 30∘ and 4 = mg cos 30∘
3
By dividing above two ⇒ 2 = 𝜇 × √3
√3
Coefficient of friction, 𝜇 = 2
43. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is
applied on the block. The coefficient of static friction between the plane and the block is 0.6. What
should be the minimum value of force P, such that the block does not move downward?
(take g = 10ms 2 ) [9 Jan. 2019 I]
(a) 32N (b) 18N (c) 23 N (d) 25 N

sol. (a)
100
mg sin 45∘ = = 50√2
√2
[.⋅ m = 10kg, g = 9.8ms −2 ]
1
𝜇mg cos 𝜃 = 0.6 × mg × = 0. 6′ 50√2
√2
3 + mg sin 𝜃 = P + 𝜇mg cos 𝜃
3 + 50√2 = P + 30√2
P = 31.28 = 32N
44. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless
pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15.
The minimum weight m that should be put on top of m2 to stop the motion is [2018]
(a) 18.3 kg (b) 27.3 kg (c) 43.3 kg (d) 10.3 kg
sol. (b) Given: m1 = 5kg; m2 = 10kg; 𝜇 = 0.15
FBD for m1 , m1 g − T = m1 a ⇒ 50 − T = 5 × a
and, T − 0.15(m + 10)g = (10 + m)a
For rest 𝑎 = 0
or, 50 = 0.15(m + 10)10
3
⇒ 5 = 20 (m + 10)
100
= m + 10 m = 23.3kg; close to option (b)
3
45. A given object takes 𝑛 times more time to slide down a 45∘ rough inclined plane as it takes to
slide down a perfectly smooth 45O incline. The coefficient of kinetic friction between the object
and the incline is: [Online Apri115, 2018]
1 1 1 1
(a) √1 − 𝑛2 (b) 1 − 𝑛2 (c) (d) √1−𝑛2
2−𝑛2
sol. (b) The coefficients of kinetic friction between the object and the incline
1 1
𝜇 = tan 𝜃 (1 − 2
) ⇒ 𝜇 = 1 − 2 (𝜃 = 45∘ )
n n
46. A body of mass 2kg slides down with an acceleration of 3m/s2 on a rough inclined plane having
a slope of 30∘ . The external force required to take the same body up the plane with the same
acceleration will be: (g = 10m/s 2 ) [Online April 15, 2018]
(a) 4N (b) 14N (c) 6N (d) 20N
sol. (d) Equation of motion when the mass slides down Mg sin 𝜃 − 𝑓 = 𝑀𝑎
⇒ 10 − 𝑓 = 6(𝑀 = 2kg, 𝑎 = 3m/s2 , 𝜃 = 30∘ given)
𝑓 = 4N
Equation of motion when the block is pushed up
Let the external force required to take the block up the plane acceleration be 𝐹
F‐Mg sin 𝜃 − 𝑓 = 𝑀𝑎
⇒ 𝐹 − 10 − 4 = 6
𝐹 = 20N
47. A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational
acceleration. On an inclined plane inside the rocket, making an angle 𝜃 with the horizontal, a point
object of mass m is kept. The minimum coefficient offiiction 𝜇m ln between the mass and the
inclined surface such that the mass does not move is: [Online April 9, 2016]
(a) tan 2𝜃 (b) tan 𝜃 (c) 3 tan 𝜃 (d) 2 tan 𝜃
sol. (b) Let 𝜇 be the minimum coefficient of friction
At equilibrium, mass does not move so,

3mg sin 𝜃 = 𝜇3mg cos 𝜃
. . 𝜇 min = tan 𝜃
48. Given in the figure are two blocks A and B ofweight 20 N and 100 N, respectively. These are
being pressed against a wall by a force 𝛤 as shown. If the coefficient of fiiction between the blocks
is 0.1 and between block B and the wall is 0.15, the fiictional force applied by the wall on block
B is: [2015]
(a) 120N (b) 150N (c) 100N (d) 80N

sol. (a)
Assuming both the blocks are stationary N = 𝐹
𝑓1 = 20N
𝑓2 = 100 + 20 = 120N
Considering the two blocks as one system and due to equilibrium f= 120N
49. A block of mass m = 10 kg rests on a horizontal table. The coefficient of fiiction between the
block and the table is 0.05. When hit by a bullet ofmass 50 g moving with speed v, that gets
embedded in it, the block moves and comes to stop after moving a distance of2 m on the table. If a
v
freely falling object were to acquire speed after being dropped from height H, then neglecting
10
energy losses and taking g = 10ms−2 , the value of H is close to: [Online April 10, 2015]
(a) 0.05 km (b) 0.02 km (c) 0.03km (d) 0.04km
sol. (d) 𝑓 = 𝜇(𝑀 + 𝑚)𝑔
𝑓 𝜇(𝑀 + 𝑚)𝑔
𝑎= = = 𝜇𝑔
𝑀+𝑚 (𝑀 + 𝑚)
= 0.05 × 10 = 0.5ms−2
Initia1momenmm 0.0.5𝑉
𝑉0 = =
(𝑀 + 𝑚) 1005
𝑣 2 − 𝑢2 = 2𝑎𝑠
0 − 𝑢2 = 2𝑎𝑠
𝑢2 = 2𝑎𝑠
0.05𝑣 2
( ) = 2 × 0.5 × 2
10.05
Solving we get 𝑣 = 201√2
Object falling from height 𝐻.
𝑉
= √2𝑔𝐻
10
201√2
= √2 × 10 × 𝐻
10
𝐻 = 40m = 0.04 km
x3
50. A block of mass m is placed on a surface with a vertical cross section given by y  . If the
6
coefficient of friction is 0.5, the maximum height above the ground at which the block can be
placed without slipping is: [2014]
1 2 1 1
(a) m (b) m (c) m (d) m
6 3 3 2
sol. (a) At limiting equilibrium,

𝜇 = tan 𝜃
𝑑𝑦 𝑥2
tan 𝜃 = 𝜇 = 𝑑𝑥 = (from question)
2
Coefficient of friction 𝜇 = 0.5

x2
0.5 
2
⇒ 𝑥 = ±1
𝑥3 1
Now, 𝑦 = = 6𝑚
6
51. Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with
acceleration ‘a’ perpendicular to the axis of the cylinder. What is 𝐹fiiction at point P? It is assumed
that the cylinder does not slip. [Online Apri119, 2014]
Ma Ma
(a) Mg (b) Ma (c) (d)
2 3
sol. (d) Force of friction at point P,

1
𝛤fiiction = 3 Ma sin 𝜃
1
= Ma sin 90∘ [here 𝜃 = 90∘ ]
3
Ma
=
3
52. A heavy box is to dragged along a rough horizontal floor. To do so, person A pushes it at an angle
30∘ from the horizontal and requires a minimum force 𝐹A , while person B pulls the box at an
angle 60o from the horizontal and needs minimum force 𝐹B . If the coefficient of friction between
√3 F
the box and the floor is 5 , the ratio A is [Online Apri119, 2014]
FB
5 3 2
(a) √3 (b) (c) √2 (d)
√3 √3
sol.
𝜇mg
𝐹A =
sin 𝜃 − 𝜇 cos 𝜃
Similarly,
𝜇mg
𝐹B =
sin 𝜃 + 𝜇 cos 𝜃
 mg
FA sin    cos  √3
 [𝜇 = given]
FB  mg 5
sin    cos 
√3
sin 30∘ + cos 30∘
= 5
√3
sin 60∘ − cos 60∘
5
1 √3 √3
+ × 2
=2 5
√3 √3 1
2 − 5 ×2
1 3 1 8
1   
2 5
  2 5
3 1 34
1  
5  5 10
8 2
= =
√3 × 4 √3
53. A small ball of mass m starts at a point A with speed vo and moves along a fiictionless track AB as
shown. The track BC has coefficient of friction 𝜇. The ball comes to stop at C after travelling a
distance L which is: [Online April 11, 2014]
B←L←C
2h v2
o h v2o h v2 h v2
(a) + 2𝜇g (b) + 2𝜇g (c) + 𝜇go (d) o
+ 2𝜇g
𝜇 𝜇 2𝜇 2𝜇
sol. b) Initial speed at point A, 𝑢 = 𝑣0

Speed at point B, 𝑣 =?
𝑣 2 − 𝑢2 = 2𝑔ℎ
𝑣 2 = 𝑣02 + 2𝑔ℎ
Let ball travels distance ‘S’ before coming to rest
𝑣2 𝑣02 + 2𝑔ℎ
𝑆= =
2𝜇𝑔 2𝜇𝑔
𝑣02 2𝑔ℎ ℎ 𝑣02
= + = +
2𝜇𝑔 2𝜇𝑔 𝜇 2𝜇𝑔
54. A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a
smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move
together is 12 N, the maximum force that can be applied to B for the blocks to move together will
be: [Online April 9, 2014]
(a) 30N (b) 25 N (c) 27N (d) 48N
sol. (c) Minimum force on 𝐴
= frictional force between the surfaces = 12𝑁
12N
Therefore maximum acceleration 𝑎 max = = 3m/s2
4kg
Hence maximum force, 𝐹max = total mass × 𝑎 max

= 9 × 3 = 27N
55. A block is placed on a rough horizontal plane. A time dependent horizontal force 𝐹 = kt acts on the
block, where k is a positive constant. The acceleration‐ time graph of the block is :
sol. (b) Graph (b) correctly depicts the acceleration‐time graph of the block.
56. A body starts from rest on a long inclined plane of slope 45∘ . The coefficient of fiiction between the
body and the plane varies as 𝜇 = 0.3𝑥, where 𝑥 is distance travelled down the plane. The body will
have maximum speed (for 𝑔 = 10m/s2 ) when x= [Online April 22, 2013]
(a) 9.8m (b) 27m (c) 12m (d) 3.33m
sol. (d) When the body has maximum speed then
𝜇 = 0.3x = tan 45∘
x = 3.33m
57. A block of weight 𝑊 rests on a horizontal floor with coefficient of static friction 𝜇. It is desired to
make the block move by applying minimum amount of force. The angle 𝜃 from the horizontal at
which the force should be applied and magnitude of the force 𝐹 are respectively.
𝜇𝑊 1 𝜇𝑊
(a) 𝜃 = tan−1 (𝜇), 𝐹 = (b) 𝜃 = tan−1 (𝜇) , 𝐹 =
√1+𝜇2 √1+𝜇 2
𝜇 𝜇𝑊
(c) 𝜃 = 0, 𝐹 = 𝜇𝑊 (d) 𝜃 = tan−1 (1+𝜇) , 𝐹 = 1+𝜇
sol. (a) Let the force 𝐹 is applied at an angle 𝜃 with the horizontal.
For horizontal equilibrium,

𝐹 cos 𝜃 = 𝜇𝑅 (i)
For vertical equilibrium,
𝑅 + 𝐹 sin 𝜃 = mg
or, 𝑅 = 𝑚𝑔 − 𝐹 sin 𝜃 (ii)
Substituting this value of 𝑅 in eq. (i), we get 𝐹 cos 𝜃 = 𝜇(mg − 𝐹 sine)
= 𝜇mg − 𝜇𝐹 sin 𝜃
or, 𝐹( cos 𝜃 + 𝜇 sin 𝜃) = 𝜇mg
𝜇mg
or, 𝛤 = (iii)
cos 𝜃+𝜇 sin 𝜃
For 𝐹 to be minimum, the denominator ( cos 𝜃 + 𝜇 sine) should be maximum.

𝑑
( cos 𝜃 + 𝜇sin𝜃) = 0 or, —sine +𝜇 cos 𝜃 = 0 or, tan 𝜃 = 𝜇
𝑑𝜃
or, 𝜃 = tan−1 (𝜇)

𝜇
Then, sin 𝜃 = and
√1+𝜇2
1
cos 𝜃 =
√1 + 𝜇 2
Hence, 𝐹 min
𝜇𝑤 𝜇𝑤
= 2 =
1 𝜇 √1 + 𝜇 2
+
√1 + 𝜇 2 √1 + 𝜇 2
58. An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the
insect and the surface is 1/3. If the line joining the center of the hemispherical surface to the insect
makes an angle 𝛼 with the vertical, the maximum possible value of 𝛼 so that the insect does not
slip is given by [Online May 12, 2012]
(a) cot 𝛼 = 3 (b) sec 𝛼 = 3 (c) cosec 𝛼 = 3 (d) cos 𝛼 = 3
sol.
The insect crawls up the bowl upto a certain height h only till the component ofits weight along the
bowl is balanced by limiting frictional force.
For limiting condition at point A
𝑅 = 𝑚𝑔 cos 𝛼 (i)
𝐹1 = 𝑚𝑔 sin 𝛼 (ii)
Dividing eq. (ii) by(i)
1 𝐹1
tan 𝛼 = = = 𝜇[𝐴𝑠 𝐹1 = 𝜇𝑅]
cot 𝛼 𝑅
1 1
⇒ tan 𝛼 = 𝜇 = 3[⋅.⋅ 𝜇 = 3(Given)]
cot   3
59. The minimum force required to start pushing a body up rough (frictional coefficient 𝜇) inclined
plane is 𝐹1 while the minimum force needed to prevent it from sliding down is 𝐹2 . If the inclined
F
plane makes an angle 𝜃 from the horizontal such that tan 𝜃 = 2𝜇 then the ratio 1 is
F2
[2011 RS]
(a) 1 (b) 2 (c) 3 (d) 4
sol. (c)
When the body slides up the inclined plane, then mg sin 𝜃 + 𝑓1 = 𝐹1

or, 𝛤1 = mg sin 𝜃 + 𝜇mg cos 𝜃
When the body slides down the inclined plane, then mg sin 𝜃 − 𝑓2 = 𝐹2
or 𝐹2 = mg sin 𝜃 − 𝜇mg cos 𝜃
𝐹1 sin 𝜃 + 𝜇 cos 𝜃
=
𝐹2 sin 𝜃 − 𝜇 cos 𝜃
𝐹1 tan 𝜃 + 𝜇 2𝜇 + 𝜇 3𝜇
⇒ = = = =3
𝐹2 tan 𝜃 − 𝜇 2𝜇 − 𝜇 𝜇
60. If a spring of stiffness ‘k’ is cut into parts ’ and ‘B’ of length ℓ𝐴 : ℓ𝐵 = 2: 3, then the stiffness of
spring 𝐴 ’ is given by [2011 RS]
3𝑘 2𝑘 5𝑘
(a) (b) (c) 𝑘 (d)
5 5 2
sol. (d) It is given ℓ𝐴 : ℓ𝐵 = 2: 3

2ℓ 3𝑙
ℓ𝐴 = , 𝑝𝐵 = ( )
5 5
1
We know that 𝑘 ∝ ℓ
If initial spring constant is 𝑘, then

𝑘ℓ = 𝑘𝐴 ℓ𝐴 = 𝑘𝐵 ℓ𝐵
2ℓ 5𝑘
𝑘ℓ = 𝑘𝐴 ( 5 ) 𝑘𝐴 = 2
61. A smooth block is released at rest on a 45o incline and then slides a distance 𝑑’. The time taken to
slide is 𝑛’ times as much to slide on rough incline than on a smooth incline. The coefficient of
friction is [2005]
1 1 1 1
(a) 𝜇𝑘 = √1 − 𝑛2 (b) 𝜇𝑘 = 1 − 𝑛2 (c) 𝜇𝑠 = √1 − 𝑛2 (d) 𝜇𝑠 = 1 − 𝑛2
sol.
Smooth rough
On smooth inclined plane, acceleration of the body= 𝑔 sin 𝜃. Let 𝑑 be the distance travelled
1
𝑑= (𝑔 sin 𝜃)𝑡12 ,
2
2𝑑
𝑡1 = √ ,
𝑔 sin 𝜃
On rough inclined plane,

𝑚𝑔 sin 𝜃 − 𝜇𝑅
𝑎=
𝑚
⇒ 𝑎 = 𝑚𝑔 sin 𝜃 − 𝜇𝑚𝑔 cos 𝜃
𝑚
⇒ 𝑎 = 𝑔 sin 𝜃 − 𝜇𝑘 𝑔 cos 𝜃
1
𝑑= (𝑔 sin 𝜃 − 𝜇𝑘̂𝑔 cos 𝜃)𝑡22
2
2𝑑
𝑡2 = √
𝑔 sin 𝜃 − 𝜇𝑘̂ 𝑔 cos 𝜃
According to question, 𝑡2 = 𝑛𝑡1
2𝑑 2𝑑
𝑛√ =√
𝑔 sin 𝜃 𝑔 sin 𝜃 − 𝜇𝑘̂𝑔 cos 𝜃
Here, 𝜇 is coefficient of kinetic friction as the block moves over the inclined plane.
sin 𝜃 = ( sin 𝜃 − 𝜇𝑘̂ cos 𝜃)𝑛2
1 1
⇒𝑛= ⇒ 𝑛2 =
√1−𝜇𝑘 1−𝜇𝑘
1
⇒ 𝜇𝑘 = 1 − 𝑛 2
62. The upper half of an inclined plane with inclination  is perfectly smooth while the lower half is
rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of
fiiction for the lower half is given by [2005]
(a) 2 cos  (b) 2 sin  (c) tan  (d) 2 tan 
sol. (d) For first half
acceleration = 𝑔 sin 𝜑;
For second half
acceleration = −(𝑔 sin 𝜑 − 𝜇𝑔 cos 𝜑)
For the block to come to rest at the bottom, acceleration in I half = retardation in II half.
𝑔 sin 𝜑 = −(𝑔 sin 𝜑 − 𝜇𝑔 cos 𝜑)
⇒ 𝜇 = 2 tan 𝜑
NOTE
According to work‐energy theorem, 𝑊 = 𝛥𝐾 = 0
(Since initial and final speeds are zero) Work done by fii iction + Work done by gravity = 0𝑖. 𝑒.,
ℓ
−(𝜇 𝑚𝑔 cos (|)) 2 + 𝑚𝑔ℓ sin 𝜑 = 0
𝜇
or 2
cos 𝜑 = sin 𝜑 or 𝜇 = 2 tan 𝜑
63. Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be
stopped is [𝜇k = 0.5] [2005]
(a) 1000 m (b) 800 m (c) 400 m (d) 100 m
sol. (a) Given, initial velocity, 𝑢 = 100m/s. Final velocity, 𝑣 = 0.
Acceleration, 𝑎 = 𝜇𝑘 𝑔 = 0.5 × 10
𝑣 2 − 𝑢2 = 2𝑎𝑠 or
⇒ 02 − 𝑢2 = 2(−𝜇𝑘 𝑔)𝑠
1
⇒ −1002 = 2 × − × 10 × 𝑠
2
⇒ 𝑠 = 1000m
64. A block rests on a rough inclined plane making an angle of 30∘ with the horizontal. The coefficient
of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N,
the mass ofthe block (in kg) is (take 𝑔 = 10m/s2 ) [2004]
(a) 1.6 (b) 4.0 (c) 2.0 (d) 2.5
sol.
Since the body is at rest on the inclined plane, mg sin 30∘ = 𝛤orce of fiiction
⇒ 𝑚 × 10 × sin 30∘ = 10
⇒ 𝑚 × 5 = 10 ⇒ 𝑚 = 2.0 kg
65. A horizontal force of 10 N is necessarytojust hold a block stationary against a wall. The coefficient
offiiction between the block and the wall is 0.2. The weight of the block is [2003]
(a) 20N (b) 50N (c) 100N (d) 2N

sol. (d) Horizontal force, 𝑁 = 10 N.
Coefficient of friction 𝜇 = 0.2.
The block will be stationary so long as Force of friction = weight of block

𝜇𝑁 = 𝑊
⇒ 0.2 × 10 = 𝑊
⇒ 𝑊 = 2𝑁
66. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in
10 s. Then the coefficient of friction is [2003]
(a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06
𝑣−𝑢
sol. (d) 𝑢 = 6m/s, 𝑣 = 0, 𝑡 = 10𝑠, 𝑎 =? Acceleration 𝑎 = 𝑡
0−6
⇒𝑎=
10
−6
⇒ 𝑎= = −0.6m/s2
10
f = 𝜇N ↑ N
The retardation is due to the frictional force
𝑓 = 𝑚𝑎 ⇒ 𝜇𝑁 = 𝑚𝑎
𝑚𝑎
⇒ 𝜇𝑚𝑔 = 𝑚𝑎 ⇒ 𝜇 = 𝑚𝑔
𝑎 0.6
⇒𝜇= = = 0.06
𝑔 10
TOPIC-4, Circular Motion, Banking of Road
67. A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per
second. A coin placed at a distance of 1.25cm from the axis of rotation remains at rest on the disc.
The coefficient of friction between the coin and the disc is (g = 10m/s 2 ) [Online Apri115, 2018]
(a) 0.5 (b) 0.7 (c) 0.3 (d) 0.6
mv 2
sol. (d) Using,  mg   mrw2
r
w = 2𝜋𝑛 = 2𝜋 × 3.5 = 7𝜋rad/ sec
Radius, 𝑟 = 1.25 cm = 1.25 × 10−2 m Coefficient of friction, 𝜇 =?
𝑚𝑟𝑤 2
𝜇𝑚𝑔 = (𝑣 = 𝑟w))
𝑟
22 2
𝑟(𝑗)2 1.25×10−2 ×(7× )
7
⇒ 𝜇= =
𝑔 10
1.25 × 10−2 × 222

= = 0.6
10
68. A conical pendulum of length 1 m makes an angle 𝜃 = 45∘ w.r.t. Z‐axis and moves in a circle in
the XY plane. The radius of the circle is 0.4m and its centre is vertically below O. The speed of the
pendulum, in its circular path, will be: (g = 10m/s 2 ) [Online April 9, 2017]
(a) 0.4 m/s (b) 4 m/s (c) 0.2 m/s (d) 2 m/s
∘ 2
sol. (d) Given, 𝜃 = 45 , r = 0.4m, g = 10m/s
mv 2
T sin   (i)
r
T cos 𝜃 = mg (ii)
v2
From equation (i) & (ii) we have, tan  
rg
v 2 = rg 𝜃 = 45∘
Hence, speed of the pendulum in its circular path,
v = √rg = √04 × 10 = 2m/s
69. A particle is released on a vertical smooth semicircular track from point X so that OX makes angle
𝜃 from the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y
where OY makes angle 𝜑 with the horizontal. Then: [Online April 19, 2014]
1 2 3
(a) sin 𝜑 = cos 𝜑 (b) sin 𝜑 = 2 cos 𝜃 (c) sin 𝜑 = 3 cos 𝜃 (d) sin 𝜑 = 4 cos 𝜃
sol. (c)
70. A body of mass ‘m’ is tied to one end of a spring and whirled round in a horizontal plane with a
constant angular velocity. The elongation in the spring is 1 cm. If the angular velocity is doubled, the
elongation in the spring is 5 cm. The original length of the spring is: [Online April 23, 2013]
(a) 15 cm (b) 12 cm (c) 16 cm (d) 10 cm
sol. (a)
71. A point P moves in counter‐clockwise direction on a circular path as shown in the figure. The
movement of ‘P’ is such that it sweeps out a length 𝑠 = 𝑡 3 + 5, where s is in metres and 𝑡 is
in seconds. The radius of the path is 20 m. The acceleration of’P’ when 𝑡 = 2s is nearly. [2010]
(a) 13m/s 2 (b) 12 m/s2 (c) 7.2ms 2 (d) 14m/s 2

sol. (d) 𝑠 = 𝑡 3 + 5
𝑑𝑠
⇒ velocity, 𝑣 = 𝑑𝑡 = 3𝑡 2
𝑑𝑣
Tangential acceleration at = = 6𝑡
𝑑𝑡
𝑣2 9𝑡 4
Radial acceleration ac = =
𝑅 𝑅
At 𝑡 = 2𝑠, 𝑎𝑡 = 6 × 2 = 12m/s2
9 × 16
𝑎𝑐 = = 7.2m/s2
20
Resultant acceleration
= √𝑎𝑡2 + 𝑎𝑐2
= √(12)2 + (72)2 = √144 + 51.84

= √19584 = 14m/s2
72. For a particle in uniform circular motion, the acceleration 𝑎⃗ at a point P(R, 𝜃) on the circle
ofradius R is (Here 𝜃 is measured fii om the x‐axis) [2010]
v2 v2 v2 v2
(a) − cos 𝜃 î+ sin 𝜃𝑗 (b) − sin 𝜃 î+ cos 𝜃𝑗
𝑅 𝑅 𝑅 𝑅
v2 v2 v2 v2
(c) − cos 𝜃 î  sin 𝜃𝑗 (d) î+ 𝑗
𝑅 𝑅 𝑅 𝑅
−𝑣 2 𝑣2
sol. (c) Clearly a  ac cos (‐ î )  ac sin  (‐ j) = cos 𝜃î − sin 𝜃𝑗̂
𝑅 𝑅
73. An annular ring with inner and outer radii 𝑅1 and 𝑅2 is rolling without slipping with a uniform
angular speed. The ratio of the forces experienced by the two particles situated on the inner and
F
outer parts of the ring , 1 is [2005]
F2
2
R  𝑅2 R1
(a)  1  (b) (c) (d) 1
𝑅1
 R2  R2
sol.
Let 𝑚 is the mass of each particle and w is the angular speed of the annular ring.
v12 w2 R12
a1    R1w2
R1 R1
𝑣22
𝑎2 = = w2 𝑅2
𝑅2
Taking particle masses equal
𝐹1 𝑚𝑎1 𝑚𝑅120) 𝑅1
= = =
𝐹2 𝑚𝑎2 𝑚𝑅20)2 𝑅2
NOTE:
The force experienced by any particle is only along radial direction.
Force experienced by the particle, 𝐹 = 𝑚 w2 𝑅
𝐹1 𝑅1
=
𝐹2 𝑅2
74. Which of the following statements is FALSE for a particle moving in a circle with a constant angular
speed? [2004]
(a) The acceleration vector points to the centre of the circle
(b) The acceleration vector is tangent to the circle
(c) The velocity vector is tangent to the circle
(d) The velocity and acceleration vectors are perpendicular to each other.
sol. (b) Only option (b) is false.
since acceleration vector is always radial (i.e. towards the center) for uniform circular motion.
75. The minimum velocity (in ms −1 ) withwhich acardriver must traverse a flat curve ofradius 150 m
and coefficient offriction 0.6 to avoid skidding is [2002]
(a) 60 (b) 30 (c) 15 (d) 25
sol. (b) The maximum velocity of the car is
𝑣 max = √𝜇𝑟𝑔
Here 𝜇 = 0.6, 𝑟 = 150m, 𝑔 = 9.8
𝑣 max = √06 × 150 × 9.8 ≃ 30m/s
WORK ENERGY AND POWER
Work :
Work is said to be done on a body, only when energy of the body changes (Mechanical energy or Thermal
energy)
A force or a Torque is responsible for work done on a body.
Workdone by a force :
Work is said to be done on a body by a force, if it displaces the point of application of force in its direaction
or opposite to its direction.
Here force may be constant or variable i.e function of time or function of position
Work done by constant force:
 
When a constant force F acts on a particle and the particle moves through a displacement S ,
then the force is said to do work W on the particle.
 
W  F .S
 S cos 
F  F
F
  
  
S S F cos S
 
The scalar (dot) product of F and S , can be evaluated as
 
W  F .S =FS cos 
Where

F is the magnitude of F ,

S is the magnitude of S
 
 is the angle between F and S .
 W=magnitude of the force x component of displacement in the direction of force
W  FS cos  F  S cos 
 W =component of the force in the direction of displacement x magnitude of the displacement
W   F cos   S
 Work is a scalar quantity.
 SI Unit is Nm or joule (J).
 CGS unit is erg.
 1J  IN  1m; 1erg  1 dyne  1cm
Dimensional formula of work is  ML T  .
2 2

 Relation between joule and erg: 1 joule = 107 erg
Other units of work:
Electron Volt  eV   1.6 1019 J
Kilowatt hour  3.6  106 J
Work done by multiple forces:
If a number of forces act on a body or particle then:
W  W1  W2  W3  .......
 
W   F1.ds   F2 .ds  ......
  

W   F1  F2  ....... .ds 
   
W   FR .ds  as FR   F 
Work done in displacing a particle under the action of a number of forces is equal to the work
done by the resultant force.
Nature of Work:
Work done by a force may be positive or negative or zero.
Ex:
(a) If we lift a body from rest to a height h
F
mg
Work done by lifting force F , W1  Fh cos 00  Fh  ve 

Work done by gravitational force , W2  mgh cos 1800  mgh  ve 
So, net work W  W1  W2  Fh  mgh   F  mg  h
Now, if the body is in equilibrium F=mg, W=0
(b) If a body is pulled on a rough horizontal road through a displacement S
N
S
F
f
mg
Work done by normal reaction and gravity W1  0 as force is  to S

Work done by pulling force F, W2  FS cos 00  FS  ve 
Work done by frictional force f , W3  fs cos 1800    mg  ve 
Net work W  W1  W2  W3  0  FS  fS   F  f  S
Now, if the body is in dynamic equilibrium f  F ; So, W=0
Zero Work:
Work done is zero if
1. Force and displacement are perpendicular.
2. Displacement of point of application of forced is zero.
3. Net force acting on the body is zero.
 
‘ As W   F .d s

so, if d s  0, W  0
i.e., if the displacement of a particle or body is zero whatever be the force, work done is zero
(except non-conservative force)
‘ (a) When a person tries to displace a wall or stone by applying a force and it (actually its centre
of mass) does not move, the work done is zero.
(b) A weight lifter does work in lifting the weight from the ground but does not work in holding it
up.
As W   Fds cos  , so W  0, if   90 , i.e.,
0
if force is always perpendicular to motion, work done by the force will be zero though neither
forced nor displacement is zero. This is why:
(a) When a porter moves with a suitcase on his head on a horizontal level road, the work done by
the lifting force or force of gravity is zero.
(b) When the bob of a simple pendulum swings, the work done by tension in the string is zero.
Work done by Variable Force:
ds

F
When the magnitude and direction of a force varies with position,

then the work done by such a force for an infinitesimal displacement ds is given by
 
dW  F .ds
B   B
‘ The total work done in going from A to B is W AB      F cos   ds
F
A
.ds
A
 ^ ^ ^   ^ ^ ^
In terms of rectangular components F  Fx i  Fy j  Fz k ; ds  dx i  dy j  dz k
B B B
W   Fx dx   FY dy   Fz dz
A A A
Graphical representation of work done:
The area enclosed by the F-S graph and displacement axis gives the amount of work done by the force.
F
P Q
O R S
Work = FS = Area of OPQR
Work done by variable force
xi dx xf S
For a small displacement dx the work done will be the area of the strip of width dx
Xf Xf
W  dw   F dx
Xi Xi
 If area lies above X-axis work done is +ve

if the area enclosed below X-axis work done is –ve
F
xi xf
O
‘
Negative work
Applications on work:
If force is changing linearly from F1 to F2 over a displacement S then work done is
 F  F2 
W  1 s
 2 
 
If force displaces the particle from its initial position ri to final position r f then displacement vector
  
S  rf  ri
y

F
 
rf S

ri
X
    
W  F .S  F . rf  ri  
Work done in pulling the bob of mass m of a simple pendulum of length L through an angle  to vertical by
means of a horizontal force F.
O
L-h  T
L
C
A F
h mg
B
Lh h h
cos    1  ;  1  cos 
L L L
h  L 1  cos  
•Work done by gravitational force W  mgh  mgL 1  cos  

•Work done by horizontal force F is W=FL sin 
•Work done by tension T in the string is zero.
Work done by gravitational force in pulling a uniform rod of m ass m and length l through an angle  is
given by
‘  C.G
h
C.G
l
W  mg 1  cos   ,
2
l
Where is the distance of centre of mass from the support.
2
A ladder of mass ‘m’ and length ‘L’ resting on a level floor is lifted and held against a wall at an angle 
with the floor
L /2

L
Work done by gravitational force is Wg   mgh  mg   sin 
2
A bucket full of water of total mass M is pulled by using a uniform rope of mass m and length l .
l
Work done by pulling force. W  mgl  mg
2
A block of mass m is suspended vertically using a rope of negligible mass.
If the rope is used to lift the block vertically up with uniform acceleration ‘a’,
work done by tension in the rope is
‘
W  m g  a h  h  height 
If block is lowered with acceleration ‘a’, then
W  m  g  a  h
1
A uniform chain of mass M and length L is kept on smooth horizontal table such that of its length is
nth
hanging over the edge of the table.
The work done by the pulling force to bring the hanging part onto the table is
L
2n L
C.G--
n
M   M   L  MgL
W    gh    g    2
 n   n   2n  2n
M
Mass of hanging part is
n
1
A uniform chain of mass M and Length L rests on a smooth horizontal table with nth part of its length is
1
hanging from the edge of the table.

1
Work done in pulling the chain practically such that nth part is hanging from the edge of the
2
MgL  1 1 
table is given by
W   
2  n12 n22 
A uniform chain of mass ‘M’ and length L is suspended vertically. The lower end of the chain is lifted upto
point of suspension
G2
l/4
G1 G1
l/4
G2
L L L
h   =raise in centre of mass of lower half of the chain.
4 4 2
M L MgL
Work done by gravitational force is Wg   g 
2 2 4
The work done in lifting a body of mass ‘m’ having density ' d1 ' inside a liquid of density ' d 2 ' through a
height ‘h’ is
h
FB
mg
 d 
W  mg h  mgh 1  2 
‘  d1 
A body of mass ' m ' is placed on a frictionless horizontal surface.
A force F acts on the body parallel to the surface such that it moves with an acceleration ‘a’,
through a displacement ‘S’.
The work done by the force is W  FS  maS   0 
0
A body of mass ' m ' is placed on a rough horizontal surface of coefficient of friction  .
A force F acts on the body parallel to the surface such that it moves with an acceleration ‘a’,
through a displacement ‘S’.
The work done by the friction of the force is f  umg cos  ; but  00
 f  umg cos 00  umg
W   f  ma  S   umg  ma  S  m  ug  a  S
If the body moves with uniform velocity then W  f S  umg S
A body of mass m is sliding down on a smooth inclined plane of inclination  .
If L is length of inclined plane then work done by gravitational force is
N
L
mg sin  
mg sin 
mg

Wg  FS  mg sin  L
A body of mass ' m ' is moved up the smooth inclined plane of inclination  and length L by a constant
horizontal force F then work done by the resultant force is
W   F cos   mg sin   L
N F cos 
F
mg sin   mg cos
mg F sin 

A body of mass ' m ' is sliding down on rough inclined plane of inclination  .
If L is the length of incline and  K is the coefficient of kinetic friction then work done by the
resultant force on the body is
N
fx
mg sin   mg cos
mg

W   mg sin   f k  L   mg sin   k mg cos   L
 mgL  sin    k cos  

‘
A uniform solid cylinder of mass m , length l and radius r is lying on ground with curved surface in contact
with ground.
If it is turned such that its circular face is in contact with ground then work done be applied
force is
l
l/2
r
l  l 
W  mgh  mg   r   h   r 
2  2 
Two blocks of masses m1 and m2  m1  m2  connected by an inextensible string are passing over a
smooth, massless pulley.
The two blocks are released from the same level. At any instant ‘t’, if ‘x’ is the displacement of
each block then
T T
m2
x
m2 g
x
m1
m1 g
Work done by gravity on block m1 , W1   m1 gx ‘

Work done by gravity on block m2 ,W2  m2 gx
Work done by gravitational force on the system, Wg  m1 gx  m2 gx
1 
Wg   m1  m2  gx   m1  m2  g  at 2   v 2  u 2  2as 
2   
 m1  m2  g 2t 2   m1  m2  g 
2
Wg   a  
2  m1  m2   m1  m2 
Note:
In this work done on the two blocks by tension is zero.
W  T  x  T  x  0
Work Done in Conservative and Non-conservative Field:
 In conservative field, work done by the force (line integral of the force
i.e.  F.d l ) is independent of the path followed between any two points.
W A B  W AB  W AB
Path I Path II Path III
 F.d l   F.d l   F.d l

or
Path I Path II Path III
‘
I
A B
II
III
 In conservative field work done by the force (line integral of the force
i.e.  F.d l ) over a closed path/loop is zero.
W AB  WB A  0
 
or  F.d l  0
A B
Conservative force :
The forces of these type of fields are known as conservative forces.
Example : Electrostatic forces, gravitational forces, elastic forces, magnetic forces etc and all the central
forces are conservative in nature.
If a body of mass m lifted to height h from the ground level by different path as shown in the figure
B B B B
‘
I II III IV
l h
h3 ‘
q h2
h1
A A A A
Work done through different paths

W I  F. s  mg  h  mgh
h
WII  F. s  mg sin   l  mg sin    mgh
sin 
WIII  mgh 1  0  mgh 2  0  mgh 3  0  mgh 4
 mg (h1  h2  h3  h4 )  mgh
WIV   F. d s  mgh
It is clear that WI  WII  WIII  WIV  mgh .

Further if the body is brought back to its initial position A, similar amount of work (energy) is
released from the system, it means W AB  mgh and W BA  mgh .
Hence the net work done against gravity over a round trip is zero.
W Net  W AB  W BA  mgh  (mgh )  0
i.e. the gravitational force is conservative in nature.
NOTE :
  ‘’
 dU
Under conservative force F  where U is Potential Energy.. U   dU    F .dr
dr
 ^ ^ ^
( F  Fx i  Fy k  Fz k
 ^ ^ ^
dr  dx i  dy j  dz k )
  u ^ u ^ u ^ 
F  i j k 
 x y z 
Non-conservative forces :
A force is said to be non-conservative if work done by or against the force in moving a body from one
position to another, depends on the path followed between these two positions and for complete cycle this work
done can never be zero.
Example: Frictional force, Viscous force, Airdrag etc.
If a body is moved from position A to another position B on a rough table, work done against frictional force
shall depend on the length of the path between A and B and not only on the position A and B.
W AB  mgs
Further if the body is brought back to its initial position A, work has to be done against the frictional force,
which opposes the motion. Hence the net work done against the friction over a round trip is not zero.
R
s
W BA  mgs .
W Net  W AB  W BA  mgs  mgs  2 mgs  0 .
i.e. the friction is a non-conservative force.
‘

::PROBLEMS::

1.   
A body is displaced from rA  2iˆ  4 ˆj  6kˆ to rB  6iˆ  4 ˆj  2kˆ under a constant force

ce 
ˆ ˆ ˆ 
F  2i  3 j  k . Find the work done.
SOLUTION :
    
Work done W  F .S ;W  F . rB  rA  
   
W  2iˆ  3 ˆj  kˆ  6iˆ  4 ˆj  2kˆ  2iˆ  4 ˆj  6kˆ 
  
 ^ ^

^ ^
^ ^
W   2 i  3 j  k  . 4 i  8 j  8 K 
  
W  8  24  8  24 units
 ^ ^ ^
2. A force F  2 x i  2 j  3z 2 k N is acting on a particle. Find the work done by the force in displacing
the body from (1,2,3) m to (3,6,1) m..
SOLUTION :
x2 y2 z2
Work done W   F dx   F dy   F dz
x1
x
y1
y
z1
z
3 6 1
W   2 xdx   2dy   3z 2 dz
1 2 3
3 1
 x2   z3 
W  2    2  y 2  3    10 J
6
 2 1  3 3
3. If W1 , W2 and W3 represent the work done in moving a particle from A to B along three different
paths 1,2 and 3 respectively As shown) in the gravitational field of a point mass m, find the
correct relation between W1 , W2 and W3 [IIT-2003]
B
m
2
1
A) W1  W2  W3 B) W1  W2  W3 C) W1  W2  W3 D) W2  W1  W3
SOLUTION :
Work done will be same in all the cases because gravitational field is a conservative field.
Thus work done is independent of the point B,
therefore W1  W2  W3
4. The force acting on an object varies with the distance travelled by the object as shown in the figure.
Find the work done by the force in moving the object from x=0m to x=14m.
F(N)
O 6 10 14 S
SOLUTION :
. Work done=Area under F- S curve.
1  1 
W    6  4    4  4       4   36 J
2  2 

5.  
A body is displaced from  0, 0  to 1m,1m  along the path x  y by a force F  x j  yi N . The
2
work done by this force will be

4 5 3 7
A) J B) J C) J D) J
3 6 2 5
SOLUTION :
1,1  
W  F .ds
 0,0 

Here ds  dxi  dy j  dzk
1,1
W 
 0,0 
 x dy  ydx   as x  y 
2
1,1
 y3 x2  5
W     J
 3 2   0,0 6
6. The displacment x (in m), of a particle of mass m (in kg) is related to the time t (in second) by
t  x  3 . Find the work done in first six second. (in mJ)
SOLUTION :
x   t  3   t 2  6t  9
2
.
dx
v  2t  6
dt
at t  0, v  6 ;
at t  6, v  6
1
m  6   18m
2
initial KE 
2
1
m  6   18m
2
final KE 
2
7. When a rubber band is stretched by a distance ‘x’, it exerts a restoring force of magnitude
F  ax  bx 2 , where a and b are constants. Find the work done in stretching the unstretched
rubber band by ‘L’. (JEE MAIN 2014)
SOLUTION :
The restoring force exerted by the rubber band when it is stretched by a distance ‘x’ is F  ax  bx 2 .
The small amount of work done on the rubber band in stretching through a small distance ‘dx’ is
dW  Fdx   ax  bx 2  dx
The total work done in stretching the unstretched rubber band by ‘L’ is
L L L L
W   Fdx    ax  bx dx   axdx   bx 2 dx
2
0 0 0 0
L L
 x2   x3  aL2 bL3
W  a  b   
 2 0  3 0 2 3
 
 x y ˆj 
8. The work done on a particle of mass m by a force k  2 iˆ 
 x  y 
2 3/ 2
 x 2
 y 
2 3/ 2  (k being a
 
constant of appropriate dimensions) when the particle is taken from the point  a, 0  along a circular
path of radius a bout the origin in the x-y plane is [IIT-2013]
2k k k
A) B) C) D) zero
a a 2a
SOLUTION :
 
W=  F .dr
  rB
xiˆ yjˆ
rB
xdx ydy
 k    ˆ
dxi 
 ˆ
dyj  k  
rA  x  y 

rA  x  y     x2  y2 
3/ 2 3/ 2
 2 2 3/ 2
x 2
 y 2 3/ 2  2 2
 
rB
1   x2   y 2 
rB
1
 k d    d    k  2  x2  y2 
x  y2  rA  x  y 
3/2
2 2 2 3/2
rA
2
    
rB r r r
1 2rdr dr  1 1 1
 k  3 d r2   k 
B B B
rA
2r rA
2r 3
 k 
rA
r 2
 k   
 r  rA
k  
 rA rB 
But rA  a and rB  a ; W  0
9. Forces acting on a particle moving in a straight line varies with the velocity of the particle as

F where  is constant. The work done by this force in time interval t is:

1
A) t B)  t C) 2t D)  2 t
2
SOLUTION :

F

d 
m    m d    dt
dt 
 m 2 
   t
 2 
; KE  t  work done
10. A particle of mass ‘m’ is projected at an angle  to the horizontal with an initial velocity u. Find
the workdone by gravity during the time it reaches the highest point.
SOLUTION :

O x
X
^    u 2 sin 2   ^
Fy  mg j; ry  H max   j
 2g 
   ^
  u sin   ^
2 2
W  F y r y    mg j  . j
   2g 
1
W   mu 2 sin 2  
2
11. A 10 kg block is pulled along a frictionless surface in the form of an arc of a circle of radius 10 m.
The applied force is 200 N. Find the work done by (a) applied force and (b) gravitational force in
displacing thr ough an angle 600
SOLUTION :
600 r
r F
Work done by applied force W  Fr sin 

3
W  200  10  sin 600  200 10  1732 J
2
work done by gravitational force W  mgr 1  cos  
W  10  9.8 10 1  cos 600 
 1
W  98  10 1    490 J
 2
12. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the
edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire
chain back onto the table?
SOLUTION :
M=4 kg, L=2m, l  0.6m , g  10m / s 2
l M  l
Work done W  mg   l g
2  L  2
4 0.6
W     0.6 10   3.6 J
2 2
13. Find the work done in lifting a body of mass 20 kg and specific gravity 3.2 to a height of 8 m in
water? (g=10m/s2)
SOLUTION :

Given specific gravity   3.2
b
b  3.2   w  3.2  1000  3200

 w   1000 
Workdone W= mgh 1     20 10  8  1  
 3200 
 b 
 2200 
W  20  10  8    1100 J
 3200 
14. A block of mass ‘m’ is lowered with the help of a rope of negligible mass through a distance ‘d’
with an acceleration of g/3. Find the work done by the rope on he block?
SOLUTION :
During lowering a block, tension in rope is T  m  g  a  andS  d
work done W  m  g  a  d
 g 2mgd
W  m  g   d  
 3 3
15. If the system shown is released from rest. Find the net workdone by tension in first one second
(g=10m/s2)
3 kg B
A 2 kg
SOLUTION :
 m  m1   3 2 
a 2 g   10  2m / s
2
m
 1  m2   2  3 
2m1m2 g 2  2  3 10
T  24 N
m1  m2 23
1 2 1
for each blocks S  at   2 1  1m
2 2
Wnet  W1  W2  TS  TS  0
Energy:
Energy is the ability or capacity to do work. Greater the amount of energy possessed by the body, greater
the work it will be able to do.
Energy is cause for doing work and work is effect of energy
Energy is a scalar.
Energy and work have same units and dimensions.
The different forms of energy are
• Mechanical energy,
•Light energy,
•Heat energy,
•Sound energy,
•Electrical energy,
•Nuclear energy .....etc.
Mechanical energy is of two types
1) Potential Energy
2) Kinetic Energy
Potential energy (U):
Potential energy of body is the energy possessed by a body by virtue of its position or configuration in the
field.
Potential energy is defined only for conservative forces. It does not exist for non-conservative forces.
In case of conservative forces.
 dU    u2 r2
 
F  
 dr


 dU   F .dr  
u1
dU   r F .dr
1
 
r2
U 2  u1    F .dr  W
r1
r  

r
If 1   , U 1  0 U   F .dr  W

P.E can be +ve or -ve or can be zero.

P.E depends on frame of reference.
EX-
1.Water stored in a dam,
2.A stretched bow,
3.A loaded spring etc., possesses P.E
In case of conservative force (field) potential energy is equal to negative of work done in shifting the body
from some reference position to given position.
Potential energy can be defined only for conservative forces. It does not exist for non-conservative forces.
A moving body may or may not have potential energy.
Potential energy should be considered to be a property of the entire system, rather than assigning it to any
specific particle.
Kinetic energy:
Kinetic energy is the energy possessed by a body by virtue of its motion.
1 2
Kinetic energy of a body of mass ‘m’ moving with a velocity ‘v’, KE  mv
2
Kinetic energy is a scalar quantity.
The kinetic energy of an object is a measure of the work an object can do by the virtue of its motion.
Examples for bodies having K.E
1) A vehicle in motion
2) Water flowing along a river
3) A bullet fired from a gun
Kinetic energy depends on frame of reference.
EX-kinetic energy of a person of mass m sitting in a train moving with speed v is zero in the frame of train but
1 2
mv in the frame of earth.
2
Relation between KE and Linear Momentum:
1 2 P2 1
 KE  mv   Pv  P  mv 
2 2m 2
If two bodies of different masses have same momentum then lighter body will have greater KE
 1
 KE 
 m 
When a bullet is fired from a gun the momentum of the bullet and gun are equal and opposite.
KEbullet M gun
i.e 
KEgun M bullet
Hence, the KE of the bullet is greater than that of the gun
A body can have energy without momentum. But it can not have momentum without energy.
A bullet of mass ‘m’ moving with velocity ‘v’ stops in wooden block after penetrating through a distance
‘x’. If F is resistance offered by the block to the bullet
(Assuming F is constant inside the block)
1 2 mv 2
mv  Fx; F   v 2 ax
2 2x
For a given body
The graph between KE and P is a parabola.
KE
The graph between KE and P is a straight line passing through the origin.
1
Its slope=
2m
P
1
The graph between KE and P is a rectangular hyperbola.
1
P
A particle is projected up from a point at an angle ' ' with the horizontal. At any time ‘t’ if ‘P’ is linear
momentum, ‘y’ is vertical displacement and ‘x’ is horizontal displacement, then nature of the curves drawn
for KE of the particle (K) against these parameters are
i) K-y graph:
K  K i  mgy;
It is a straight line
K
ii) K-t graph:

 1 
K  K i  mg  u y t  gt 2 
 2 
1 2
 y  uyt  gt ;
2
It is a parabola
K
iii) K - x graph:
 gx 2 
K  K i  mg  x tan   2 
 2u x 
 g 
 y   tan   x   2  x 2 ;
 2u x 
It is also parabola
K
iv) K -P2 graph:

1
It is a straight line passing through origin and slope=
2m
P2 = 2mK
P2
P2  K
Restoring force and spring constant : When a spring is stretched or compressed from its normal position
(x = 0) by a small distance x, then a restoring force is produced in the spring to bring it to the
normal position.
According to Hooke’s law this restoring force is proportional to the displacement x and its
direction is always opposite to the displacement.
x=0
F
m Fext
+x
F
m Fext
–x
Fig. 6.20
i.e. F  x
…(i)
F  k x
where k is called spring constant.
If x = 1, F = k (Numerically)
k=F
Hence spring constant is numerically equal to force required to produce unit displacement (compression or
extension) in the spring. If required force is more, then spring is said to be more stiff and vice-versa.
Actually k is a measure of the stiffness/softness of the spring.
F
Dimension : As k 
x
[F] [MLT 2 ]
\ [k ]    [MT 2 ]
[x ] L
Units : S.I. unit Newton/metre, C.G.S unit Dyne/cm.
Spring force is an example of a variable force which is conservative.

In an ideal spring, the spring force Fs is directly proportional to ‘x’ .
Where x is the displacement of the block from equilibrium position.
i.e., Fs  Kx .
The constant K is called spring constant.
The work done on the block by the spring force as the block moves from undeformed position
x=0 to x=x1
 
dW  F .dx   Kxdx
x1
1 1
W   dW    Kxdx   K  x 2    Kx12
x1
0
2 0 2
x2
If the block moves from x=x1 to x=x2 the work done by spring force is W    Kxdx
x1
1 1 1
W   K  x12  x22   Kx12  Kx22
2 2 2
Potential energy stored in a spring:
The change in potential energy of a system corresponding to a conservative internal force is
x  
du    F .dx,
0
dU = - (work done by the spring force)

  Kx 2  1 2
dU     ;U f  U i  Kx
 2  2
since Ui is zero when spring is at its natural length
1 2
U f  Kx
2
Work - energy theorem:
Work done by all forces acting on a body is equal to change in its kinetic energy.
1 2 1
i.e., W  K f  K i
mv  mu 2
2 2
Where Kf and Ki are the final and initial kinetic energies of the body.
Work energy theorem is applicable not only for a single particle but also for a system of particles.
When it is applied to a system of two or more particles change in kinetic energy of the system is
equal to work done on the system by the external as well as internal forces.
Work - energy theorem can also be applied to a system under the action of variable forces,
pseudo forces, conservative as well as non-conservative forces.
Application of work-energy theorem:
A body of mass m starting from rest acquire a velocity ‘v’ due to constant force F.
Neglecting air resistance.
1 2
Work done = change in Kinetic energy mv
2
A particle of mass ‘m’ is thrown vertically up with a speed ‘u’. Neglecting the air friction, the
work done by gravitational force, as particle reaches maximum height is
Wg   K  K f  K i
1 1 1
Wg   mu 2  m  0   mu 2
2 2 2
A particles of mass ‘m’ falls freely from a height ‘h’ in air medium onto the ground. If ‘v’ is the
velocity with which it reaches the ground, the work done by air friction is W f and work done by
gravitational force Wg then,
1 2 1
Wg  W f  mv  0  mv 2
2 2
A block of mass ‘m’ slides down a frictionless incline of inclination ' ' to the horizontal.
If h is the height of incline, the velocity with which body reaches the bottom of incline is
1
Wg  K ; mgh  mv 2  0
2
1
mgh  mv 2 ; v  2 gh
2
A body of mass ‘m’ starts from rest from the top of a rough inclined plane of inclination ' ' and
length ' l ' .
The velocity ‘v’ with which it reaches the bottom of incline if k is the coefficient of kinetic
friction is
Wg  W f  k
1 2
 mg sin   l    k mg cos   l  mv  0
2
v  2 gl  sin    k cos  
A bob of mass m suspended from a string of length l is given a speed u at its lowest position then
the speed of the bob v when it makes an angle  with the vertical is
1
Wg  WT  K  mgl 1  cos    0  m v2  u2 
2
v  u 2  2 gl 1  cos  
A bullet of mass ‘m’ moving with velocity ‘v’ stops in a wooden block after penetrating through
a distance x.
If ‘f’ is the resistance offered by the block to the bullet.
W f  K f  K i ;  fx  0  KEi
KEi mv 2 P2
i.e., stopping distance x   
f 2f 2mf
A block of mass ‘m’ attached to a spring of spring constant ‘K’ oscillates on a smooth horizontal
table. The other end of the spring is fixed to a wall. It has a speed ‘v’ when the spring is at
natural length. The distance it moves on a table before it comes to rest is calculated as below
WS . F  Wg  WN  K  S .F  spring force 
Let the mass be oscillating with amplitude ‘x’,
1 2
On compressing the spring WS .F   Kx
2
Wg =FS cos 90 = 0; WN=NS cos 900 =0
0
1 1 m
WS .F  K f  K i   Kx 2  0  mv 2  x  v
2 2 K
A pile driver of mass ‘m’ is dropped from a height ‘h’ above the ground. On reaching the ground
it pierces through a distance ‘s’ and then stops finally. If R is the average resistance offered by
ground then
1 1
Wg  WR  K f  K i  mu 2  mv 2
2 2
 h
mg  h  s     Rs   0; R  mg  1  
 s
u= 0
air
h
v1  2 gh
s soil
v=0
Here time of penetration is given by impulse equation (R-mg) t = 0  m 2 gh
A body of mass ‘m’ is initially at rest. By the application of a constant force, its velocity changes
to v0 in time t0 the kinetic energy of the body at time ‘t’ is
W  K  K f  K i   K i  0
1  1
K f  W  mas  ma  at 2   ma 2t 2
2  2
2
vo 1  v0  2
Since a  ; K f  m   t
to 2  t0 
Types of Equilibrium:

A body is said to be in translatory equilibrium, if net force acting on the body is zero, i.e., F net  0
dU
If the forces are conservative F  
dr
and for equilibrium F = 0,
dU dU
so   0 or  0, At equilibrium position
dr dr
slope of U-r graph is zero or the potential energy is optimum (maximum or minimum or constant)
There are three types of equilibrium
(i) Stable equilibrium (ii) Unstable equilibrium
(iii) Neutral equilibrium
Stable equilibrium:
1.Net forced is Zero
dU
2.  0 or slope of U-r graph is zero
dr
3.When displaced from its equilibrium position, a net retarding forces starts acting on the body, which has a
tendency to bring the body back to its equilibrium position
d 2U
4.PE in equilibrium position is minimum as compared to its neighboring points as is positive
dr 2
5.When displaced from equilibrium position the centre of gravity of the body comes down
Unstable equilibrium:
1.Net force is zero
dU
2.  0 (or) slope of U-r graph is zero
dr
3.When displaced from its equilibrium position, a net force starts acting on the body which moves the body
in the direction of displacement or away from the equilibrium position
d 2U
4.PE in equilibrium position is maximum as compared to other positions as is negative
dr 2
5.When displaced from equilibrium position the centre of gravity of the body goes up
Neutral equilibrium:
1.Net force is zero
dU
2.  0 (or) slope of U-r graph is zero
dr
3.When displaced from its equilibrium position the body has neither the tendency to come back nor move
away from the original position.
d 2U
4.PE remains constant even if the body is moving to neighbouring points 0
dr 2
5.When displaced from equilibrium position the centre of gravity of the body remains constant
Potential energy curve :

A graph plotted between the potential energy of a particle and its displacement from the centre
of forceis called potential energy curve.
Figure shows a graph of potential energy function U(x) for one dimensional motion.
As we know that negative gradient of the potential energy gives force.
dU
 F
dx
U(x)
C D
A
B
O x
Nature of force :
(i) Attractive force :
On increasing x, if U increases,
dU
dx
 positive ,
then F is in negative direction

i.e. force is attractive in nature.
In graph this is represented in region BC.
(ii) Repulsive force :
On increasing x, if U decreases,
dU
dx
 negative ,
then F is in positive direction

i.e. force is repulsive in nature.
In graph this is represented in region AB.
(iii) Zero force :
On increasing x, if U does not change,
dU
0
dx
then F is zero
i.e. no force works on the particle.
Point B, C and D represents the point of zero force or these points can be termed as position of equilibrium.
Law of conservation of Mechanical energy:

Total mechanical energy of a system remains constant, if only conservative forces are acting on
a system of particles and the work done by all other forces is zero.
U f  U i  W
From work energy theorem W  k f  ki
U f  U i    k f  k i 
U f  k f  U i  ki  U  K  cons tan t
The sum of potential energy and kinetic energy remains constant in any state.
A body is projected vertically up from the ground. When it is at height ‘h’ above the ground,
its potential and kinetic energies are in the ration x : y .
If H is the maximum height reached by the body, then
x h h x
 or 
y H h H x y
:: PROBLEMS : :
1. Two spheres whose radii are in the ratio 1 : 2 are moving with velocities in the ratio 3 : 4. If their
densities are in the ratio 3 : 2, then find the ratio of their kinetic energies.
r1 1 v1 3 1 3
 ,  , 
r2 2 v2 4  2 2
1 2 1 14 
K .E  mv  V   v 2    r 3   v 2
2 2 23 
3 2 3 2
KE1 1  r1   v1  3 1 3
         
KE2  2  r2   v2  2 2 4
KE1 3 1 9 27
   
KE2 2 8 16 256
2. An engine is pumping water continuously. The water passes through a nozzle with a velocity  .
As water leaves the nozzle, the mass per unit length of the water jet is m0 . Find the rate at
which kinetic energy is imparted to the water.
1 3 1 2 1 32 1 12
A) m0 v B) m0 v C) m0 v D) m0 v
2 2 2 2
SOLUTION :
dm
m0 
dx
d d 1  1  dm 
 KE   dt  2 mv 2   2  dt  v 2
dn    
1  dm   dx  2 1
     v  m0 v
3
2  dx   dt  2
3. A particle is projected at 60 to the horizontal with a kinetic energy ‘K’ . Find the kinetic energy
0
at the highest point? (JEE MAIN 2007)

SOLUTION :
1
Initial kinetic energy is K  mu 2
2
The velocity at highest point vx  u cos  .
Kinetic energy of a particle at highest point
1 2 1 K
KH  mvx  mu 2 cos 2   K cos 2 600 
2 2 4
4. An athlete in the Olympic games covers a distance of 100 m in 10s. His kinetic energy can be
estimated to be in the range. (JEE MAIN 2008)
1) 200J-500J 2) 2  10 J  3  10 J
5 5 3) 20,000J-50,000J 4) 2,000J - 5,000J
SOLUTION :
Approximate mass of the athlete = 60kg
Average velocity = 10m/s
1 1
Approximate K .E. mv   60  10  3000 J
2 2
2 2
Range of KE = 2000 J to 5000 J
5. Kinetic energy of a particle moving along a circle of radius ‘r’ depends on the distance as KE =
cs2, (c is constant, s is displacement). Find the force acting on the particle
SOLUTION :
1 2  2c 
KE  mv  cs 2  v    s
2  m
dv 2c ds 2c
t    v
dt m dt m
2c  2c  2c
Ft  m t  mv  m s  2cs
m  m  m
2
 mv 2 
Total force F  Ft  F   2cs   
2 2 2
c 
 r 
s2
F  2cs 1  2
r
6. A r ectangular plank of mass m 1 and height ‘a’ is on a horizontal surface. On the top of it another
rectangular plank of mass m2 and height ‘b’ is placed. Find the potential energy of the system?
m2 b
a
m1
SOLUTION :
Total potential energy of system U = U1 + U2
a  b   m   b 
 m1 g  m2 g  a    1  m2  a  m2    g
2  2   2   2 
7. A ring ‘A’ of mass ‘m’ is attached to a stretched spring of force constant K, which is fixed at C on a
smooth vertical circular track of radius R. Points A and C are diametrically opposite. When the ring
slips from rest on the track to point B, making an angle of 30 with AC. ( ACB  30 ) spring
becomes unstretched. Find the velocity of the ring at B
A(m)
D B
30
C
1 1
 KR 2 2  KR 2 2
   
2 2
A)  2 3  gR 3  B)  2 3  gR 
 2m   m 
1 1
 2 KR 2 2  KR 2 2
   
2 2
C)  2 3  gR 3  D)  2 1  gR 
 m   2m 
SOLUTION :
Decrease in elastic PE + Decrease in PE = Increase in KE
1 2 1
Kx  mg  AD   mv 2
2 2
x  AC  CB  2 R  2 R cos 30  R 2  3    As CBA  90 

R
AD  AB cos 60   AC sin 30  cos 60  2
1
  R 1 2
2
So, KR 2 2  3  mg  mv
2 2 2
1
 KR 2 2
 
2
v 2 3  gR 
 m 
8. A rod of mass m and length L is held vertical. Find its gravitational potential energy with respect
to zero potential energy at the lower end?
dy
y
SOLUTION :
Choose a small element of length dy, then
m
mass of the element dm    dy .
L
The potential energy of the element dU = (dm) g ( y ) Potential energy of the entire rod
L L L
m m
U    dm  gy     dy  .gy  g  ydy
0 0
L L 0
L
m  y2  mgL
U  g  
L  2 0 2
12 12
 2E   2E 
9. The total mechanical energy of a particle is E. The speed of the particle at x    is   .
 K   m 
Find the potential energy of the particle at x :
1 2 1 2 2 2
A) zero B) Kx C) Kx D) Kx
2 4 5
SOLUTION :
12 12
 2E   2E 
At x    , v 
 K   m 
2E 1 2
or v  or mv  E
2
m 2
12
 2E 
So, at x    , kinetic energy is equal to total mechanical energy..
 K 
12
 2E 
Hence, PE at x    is zero
 K 
So, U  x   0
10. A chain of length  and mass ‘m’ lies on the surface of a smooth hemisphere of radius R >  with
one end tied to top of the hemisphere. Find the gravitational potential energy of the chain?
SOLUTION :
(Rd  )
R  d
y=R cos 
Consider a small element of chain of width d  at angle  from the vertical

m
The mass of the element dm    Rd

The gravitational potential energy of the element du = (dm)gy
The gravitational potential energy of total chain
l l
R R
m 
U    dm  gy    Rd g  R cos  
0 0
l 
mgR 2 l
mgR 2 l 
U sin  0R  sin  
l l R
11. A spring of force constant ‘k’ is stretched by a small length ‘x’. Find work done in stretching it
further by a small length ‘y’?
SOLUTION :
1 2
Initial potential energy U i kx
2
1
k x  y
2
Final potential energy U f 
2
1 1
Work done W  U f  U i  k  x  y   kx
2 2
2 2
1
W ky  2 x  y 
2
12. Under the action of force 2kg body moves such that its position ‘x’ varies as a function of time t
t3
given by x  , x is in meter and t in second. Calculate the workdone by the force in first two
3
seconds.
SOLUTION :
From work-energy theorem W  KE
t3 dx 2
x , Velocity v  t
3 dt
At t = 0, v1=0, At t= 2s, v2= 4 m/s
1 1
W m  v22  v12    2  42  0   16 J
2 2
13. A uniform chain of length ‘l’ and ,as ‘M’ is on a smooth horizontal table, with (1/n)th part of its
length hanging from the edge of the table. Find the kinetic energy of the chain as it completely
slips off the table.
L
n
SOLUTION :
Work done W  U i  U f  K f  K i
Mgl Mgl 1  1
 2  Mv 2 ; v  gl 1  2 
2 2n 2  n 
3
14. A particle moves move on the rough horizontal ground with some initial velocity V0 . If of its
4
kinetic energy lost due to friction in time t0 . The coefficient of friction between the particle and the
ground is
V0 V0 3V0 V0
A) 2 gt B) 4 gt C) 4 gt D) gt
0 0 0 0
SOLUTION :
1
K .Ei  mV02 ;
2
11
K .E f  mV02
42
V  V0   gt0
15. Two blocks having masses 8 kg and 16 kg are connected to the two ends of a light spring. The
system is placed on a smooth horizontal floor. An inextensible string also connects B with ceiling
as shown in figure at the initial moment. Initially the spring has its natural length. A constant
horizontal force F is applied to the heavier block as shown. What is the maximum possible value
of F so that lighter block doesn’t loose contact with ground.
4m
5m A
B 8 kg 16 kg F
SOLUTION :
Draw FBD of B to get extension in spring. When block B just looses contact with ground
resultant force on it is zero.
T N

kx
mg
Kx
Kx  T cos   0  T  ; T sin   N  mg  0
cos 
Kx
When N = 0 then T sin   mg  sin   mg
cos 
mg 80 60
x  
K tan  K   4 / 3 K
If spring has to just extend till this value then from work energy theorem we get
1 2
Fx 
Kx  F  30 N
2
16. A 0.5 kg block slides from the point A (see figure) on a horizontal track with an initial speed of 3m/s
towards a weightless horizontal spring of length 1 m and force constant 2 Newton/m.The part AB of
the track is frictionlessand the part BC has the coefficients of static and kinetic friction as 0.22 and
0.2 respectively. If the distances AB and BD are 2m and 2.14 m respectively find the total distance
through which the block moves before it comes to rest completely (Take g = 10 m/s2).
A B D C
A) 4.20 m B) 4.14 m C) 4.24 m D) 4.26 m
SOLUTION :
From A to B, there will be no loss of energy.
Now let block compresses the spring by an amount x and comes momentarily to rest.
Then, loss of energy will be equal to the work done against friction.
Therefore,
x
kx
A B C v=0 C f
1 1
k mg (BD  x )  mv 2  kx 2
2 2
1 1
Substituting the values (0.2) (0.5) (10) (2.14 + x) = (0.5)(3)2  (2)( x )2
2 2
Solving this equation, we get x = 0.1m
17. A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring
and compresses it till the block is motionless. The kinetic frictional force is 15 N and spring
constant is 10,000Nm -1. Find the compression in the spring? (JEE MAIN 2007)
SOLUTION :
1 2 1
KE  mv  W friction  Kx 2
2 2
1 1
  2  42  15 x   10000  x 2
2 2
 5000 x 2  15 x  16  0
 x  0.055m or x  5.5 cm
18. A block of mass 1 kg kept over a smooth surface is given velocity 2 m/s towards a spring of spring
constant 1 N/m at a distance of 10m. Find after what time block will be passing through P again
P k
m V
10 m
A)  20  2  sec B) 10sec C) 10  2  sec D) 10    sec
SOLUTION :
s m s
t   
v k v
19. A particle is suspended vertically from a point O by an inextensible massless string of length L. A
vertical line AB is is at a distance of L/8 from O as shown. The object is given a horizontal velocity
u. At some point, its motion ceases to be circular and eventually the object passed through the line
AB. At the instant of crossing AB, its velocity is horizontal. Find u.
[1999]
A
L
8
SOLUTION :
C v
90-
D Q

mg

L + Lsin
L
. L 8
P u
Lcos
Now, we have following equations
mv 2
1) TQ  0 Therefore, mg sin   ...... (1)
L
2) v 2  u 2  2gh  u 2  2gL(1  sin ) ...... (2)
1
3) QD  (range ) ........ (3)
2
 3 3
u  gL  2  
 2 

Comprehension-I
The potential energy U (in J) of a particle is given by  ax  by  , where a and b are constants. The mass
of the particle is 1 kg and x and y are the coordinates of the particle in metre. The particle is at rest at
 4a, 2b  at time t  0 .
20. Find the speed of the particle when it crosses x-axis
A) 2 a  b B) C)
1 2
a  b2 D)
a 2
 b2 
2 2
a b
2 2
2 2
SOLUTION :
  1  U  U  
a F /m 
m  X
i
Y 
j    ai  bj since, m  1 kg  
21. Find the speed of the particle when it crosses y-axis
A) 4 a 2  b 2 B) 2 2  a 2  b 2  C) 2  a 2  b2  D) a 2
 b2 
SOLUTION :
 
acceleration ax   a, a y  b

acceleration a  a x  a y  a  b
2 2 2 2
u x   at , v y  bt
1 a
X  4a  ax t 2  4a  t 2
2 2
1 b
Y  2b  a y t 2  2b  t 2
2 2
particle crosses x-axis, when y = 0
22. Find the acceleration of the particle
A) 4 a 2
 b2  B) 2 2  a 2  b 2  C) 2  a 2  b2  D) a 2
 b2 
SOLUTION :
Particle crosses y-axis, when x = 0
23. Find the coordinates of the particle at t = 1 second
A)  3.5a,1.5b  B)  3a, 2b  C)  3a,3b  D)  3a, 4b 
SOLUTION :
Coordinate at t  1 sec will be  3.5a,1.5b 
24. In the below figure, what constant force ‘P’is required to bring the 50 kg body, which starts from
rest to a velocity of 10m/s in 7m? (Neglect friction)
p
300
50 kg
300
SOLUTION :
Work done by force P in displacing the block by 7 m, W1  F cos   S 
7 3
W1  P cos 30 0  7  PJ
2
W2=mgh = 50 x 9.8 x 7 sin 300 = 17
25. Three springs A,B and C each of force constant K, are connected at O. The other ends of B and C
can slide on smooth sliders. A pan is hanging from other end of the spring A. When a block of mass
m is placed int he pan, find the amount of workdone by the gravity on block system after it stops
vibrating. The spring C does not sag:
3m2g2 m2g2 2m2g2 m2g2

A) B) C) D)
2K K K 2K
SOLUTION :
The system will adjust in such a way by sliding the spring C remains unstretched and spring A and B
remains vertical.
Thus, effective force constant is giveny by
1 1 1
  ;
K' K K
K
K'
2
there is no effect of spring C.
K ' x  mg ;
2mg
x ,
K
2mg
W  mg .x  mg .
k
26. Figure shows a spring fixed at the bottom end of an incline of inclination 370. A small block of mass
2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses
the spring by 20 cm, stops momentarily and then rebounds through a distance 1m up and incline.
Find (i) the friction coefficient between the plane and the block and (ii) the spring constant of the
spring . (g= 10 ms-2)
37 0
SOLUTION :
Applying work energy theorem for downward motion of the body W=  KE
1 2
mg sin   x  d   f  l1  Kx  KE
2
1
20sin 37 0  5     20 cos 37 0  5  K  0.2   0
2
2
80  0.02 K  60  1
For the upward motion of the body
1 2
 mg sin  l2   f  l2   Kx  KE
2
1
K  0.2   0
2
2  10 sin 370 1    20 cos 37 0  1 
2
16  0.02 K  12   2 
Adding equations (1) and (2), we get
96   48    0.5
Now, use the value of  in equation (1), we get K=1000 N/m
27. Figure shows a light, inextensible string attached to a cart that can slide along a frictionless horizontal
rail aligned along an x axis. The left end of the string is pulled over a pulley, of negligible mass and
friction and fixed at height h = 3m from the ground level. The cart slides from x 1 = 3 3 m to x2 = 4
m and during the move, tension in the string is kept constant 50 N. Find change in kinetic energy of
the cart in joules. (Use 3 = 1.7) in form of 10 x n, where n =
A
h = 3m
x2 x1
SOLUTION :
Change in kinetic energy = Work done by the force; so W  50  1 Along the string);
so W  50 Joule
a b
28. In a molecule, the potential energy between two atoms is given by U  x   12
 6 . Where ‘a’ and ‘b’
x x
are positive constants and ‘x’ is the distance between atoms. Find the value of ‘x’ at which force is zero
and minimum P.E at that point. (JEE MAIN 2010)
SOLUTION :
dU
Force is zero  0
dx
i.e., a  12  x 13  b  6  x 7  0
12a 6b 12a 6b
13
 7  0  13  7
x x x x
1
2a  2a  6
x 6
x   
b  b 
Substituting the value of x
12 6
 b 6  b 6
 U min  a   b 
 2a   2a 
 b2   b 2  b 2
U min        U min 
 4a   2a  4a
29. In the figures A) and B) AC, DG and GF are fixed inclined planes, BC = EF = x and AB = DE = y. A
small block of mass M is released from the point A. It slides down AC and reaches C with a speed
VC. The small block is released from rest from the point D. It slides down DGF and reaches the
point F with speed VF. The coefficients of kinetic frictions between the block and both the surfaces
AC and DGF are m. Calculate VC and VF.
A D
B C E F
(a) (b)
A) 1.7 m/s B) 2.7 m/s C) 3.7 m/s D) 0.7 m/s
SOLUTION :
In both the cases work done by friction will be Mgx
1 1
 MVC2  MVF2  Mgy  Mgx
2 2
 VC  VF  2gy  2gx
30. Two balls of same mass are projected as shown, by compressing equally (say x) the springs of different
force constants K1 and K 2 by equal magnitude. The first ball is projected upwards along smooth wall
and the other on the rough horizontal floor with coefficient of friction  . If the first ball goes up by
height h , then the distance covered by the second ball will be :
K2
K1
2hK 2 hK1 3hK 2 hK 2

A)  K B) 2  K C) 2  K D)  K
1 2 1 1
SOLUTION :
1
K1 x 2  mgh
2
2mgh
x 
K1
1
K 2 x 2   mgx0
2
2  mgx0
x2 
K2
h K2
equating x0   . K
1
31. A particle is acted by a force F=kx, where k is a +ve constant. Its potential energy at x=0 is zero.
which curve correctly represents the variation of potential energy of the block with respect to x ?
[IIT-2004]
U
U
X X
A) B)
U
U
X X
C) D)
SOLUTION :
1
U   fFdx   fkxdx   kx 2
2
32. A massless platform is kept on a light elastic spring as shown in figure. When a sand particle of
0.1 kg mass is dropped on the pan from a height of 0.24m, the particle strikes the pan and the
spring compresses by 0.01 m. From what height should particle be dropped to cause a compression
of 0.04m.
0.1 kg
SOLUTION :
By conservation of mechanical energy
1 2
mg  h  y   Ky
2
h=height of particle
y=compression of the spring
As here particle and spring remain same
2 2
h1  y1  y1  0.24  0.01  0.01 
  ;   ; h2  3.96m
h2  y2  y2  h2  0.04  0.04 
33. Block A of mass 1kg is placed on the rough surface of block B of mass 3 kg. Block B is placed on
smooth horizontal surfac. Blocks are given the velocities as shown. Find net work done by the
frictional force. [ in (-) ve J]
8m/sec
A
4m/sec
B
SOLUTION :
1  3 v  1 8   3 4  20 ; v  5m / sec
1 39
for block A, W f  1  52  82    J
2 2
1 27
for block B, W f   3  52  4 2    J
2 2
net work done by friction = -6 J
34. A small mass ‘m’ is sliding down on a smooth curved incline from a height ‘h’ and finally moves
through a horizontal smooth surface. A light spring of force constant K is fixed with a vertical rigid
stand on the horizontal surface, as shown in the figure. Find the value for the maximum compression
in the spring if mass ‘m’ is released from rest from height ‘h’ and hits the spring on the horizontal
surface.
A
m
h
C K
SOLUTION :
Conservation of energy b/w positions A and C
 PE A  block  KEA   PEC  spring  KEC
1 2 1 2mgh
mgh  0  Kx  0; mgh  Kx 2 ; x 
2 2 K
35. A vehicle of mass 15 quintal climbs up a hill 200 m high. It then moves on a level road with a speed
of 30ms-1. Calculate the potential energy gained by it and its total mechanical energy while running
on the top of the hill.
SOLUTION :
m= 15 quintal = 1500kg, g=9.8ms-2, h=200m
P.E. gained,
U=mgh = 1500 x 9.8 x 200=2.94 x 106J
1 2 1
mv   1500   30   0.675  10 6 J
2
K .E. 
2 2
Total mechanical energy
E  K  U   0.675  2.94  106  3.615 6 J
36. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to
its lower end. The mass is released with the spring initially unstretched Then,the maxumum extension
in the spring is [IIT-2002]
4Mg 2Mg Mg Mg
A) B) C) D)
k k k 2k
SOLUTION :
Loss in P.E = Gain in K.E + P.E stored in spring
1
MgX max  0  kX 2 max ;
2
2Mg
X max 
k
37. A block of mass 0.18kg is attached to a spring of force-constant 2 N/m. The coefficient of friction
between the block and the floor is 0.1 Initially the block is at rest and the spring is un-stretched. An
impulse is given to the block as shown in the figure.The block slides a distance of 0.06m and
comes to rest for the first time . The initial velocity of the block in m/s is V  N /10 . Then N is:
[IIT-2011]
SOLUTION :
Decrease in mechanical energy = work done against friction
1 1
m 2  kx 2    mg  x
2 2
2 gx  k
v
m
Putting m  0.18kg , x  0.06m, k  2 Nm 1 ,
  01 we get
4
  0.4m / s  m/s
10
N  4
38. A particle is released from height H. At certain height from the ground its kinetic energy is twice
its gravitational potential energy. Find the height and speed of particle at that height
SOLUTION :
K.E= 2PE
But KE=TE - PE
mg  H  h   2mgh; mgH  3mgh
H
h ; Also K .E  2 P.E ,
3
1 2 H  gH
mv  2mg  v2
2  3  3
39. A heavy flexible uniform chain of length  r and mass  r lies in a smooth semicircular tube AB of
radius ‘r’. Assuming a slight disturbance to start the chain in motion, find the velocity v with which it
will emerge from the end of the tube?
C.G 2r / 
B A
Reference level
SOLUTION :
B A
r
2
C.G
2r
Centre of gravity of a semicircular arc is at a distance from the centre.

 2r 
Initial potential energy U i    r  g  
 
  r 
Final potential energy U f    r  g  
 2 
When the chain is completely slipped off the tube, all the links of the chain have the same velocity v.
1 1
Kinetic energy of chain k  mv 2    r  v 2
2 2
From conservation of energy,
 2r    r  1
 rg      r  g      r  v
2
   2  2
2 
On solving we get, v  2rg   
 2 
40. An observer and a vehicle, both start moving together from rest with accelerations 5 m/s2 and 2 m/
s2, respectively. There is a 2 kg block on the floor of the vehicle, and   0.3 between their surfaces.
Find the work done by frictional force on the 2 kg block as observed by the running observer,
during first 2 seconds of the motion.
2
2m/s
5m/s2
2kg
SOLUTION :
FBD of the block,
Fpseudo
f
f L  6 N , Fpseudo  4 N
 f  4N
acc. of the block with respect to observes  2  5  3 m / s 2
1
 displacement of the block w.r.to observes   3  4  6 m
2
 work done by friction w.r. to observers  24 Joule
 x4 x2 
41. The potential energy of 1 kg particle free to move along X-axis is given by    4  2  J. The
U x 
 
total mechanical energy of the particle is 2 J. Find the maximum speed of the particle.
SOLUTION :
dU
For maximum value of U,  0.
dx
4 x3 2 x
   0 or x  0, x  1.
4 2
d 2U d 2U
At x  0, 2  1 and At x  x  1, 2  2
dx dx
Hence U is minimum at x  x  1 with value
1 1 1
U min    J
4 2 4
1 9
K max  U min  E or K max   2 or K max 
4 4
1 2 9 3
 mv   vmax  , ms 1
2 4 2
42.. A ball leaves the track at B which is at 3m height from bottom most point of the track. The ball
further rises upto 4m height from the bottom most point before falling down. Find h ( in m), if the
track at B makes an angle 300 with horizontal.
B
h 3m
SOLUTION :
u B 2 sin 2  2 g  h  hB  sin 2 300
h0  3   3
2g 2g
4  given   3  h sin 2 300  hB sin 2 300
h  3
 3  ;
4 4
7 h
  h 7m
4 4
43. Two blocks A and B are placed one over other. Block B is acted upon by a force of 20 N which
displaces it through 5 m. Find work done by frictional force on block A
 = 0.5
A 2kg
20N
B 3kg
Frictionless
SOLUTION :
Limiting friction between the blocks f L  10 N
10
 amax A   5 m / s2
2
i.e., for shipping between A, B, Bmust move with 5 m / s 2
5m/s2
10N
F
 F  25 N
but given F  20 N
44. A and B are smooth light hinges equidistant from C, which can slide on ABC. The spring of force
constant K is fixed at its one end C and connected to light rods AD and BD at point D. A block of
mass m is suspended at D. Find the velocity of the block, when CAD changes from 30 to 45 .
AD  BD  L
A C B
D
1 1
   2
   
2
KL 2 2 KL2 2
A)  gL  2 1  B)  gL 2  2 1 
 2m   2m 
1
 2
1
 K L2  2
C)  gL  2  1   
2
KL
D)  gL 
2
2 1  
 4m   2m 
SOLUTION :
L
Inititally, CD  L sin 30 
2
L
Finally, CD  L sin 45 
2
Increase in elastic PE + Increase in KE = Decrease in PE
2
1  L L 1  L L
K    mv2  mg   
2  2 2 2  2 2
1
 2
  K L2
 
2
On solving, v   gL 2 1  2 1 
 4m 
45. Figure shows a particle sliding on a frictionless track which terminates in a straight horizontal
section. If the particle starts slipping from the point A, how far away from the track will the
particle hit the ground?
A
1.0 m
0.5 m
SOLUTION :
Applying the law of conservation of mechanical energy for the points A and B,
1 2
mgH  mv  mgh
2
v2 g
g  or v 2  g  v  g  3.1ms 1
2 2
After point B the particle exhibits projectile motion with   00 and y=-0.5 m
Horizontal distance travelled by the body
2h 2  0.5
Ru  3.1  1m
g 9.8
Power:
The rate of doing work is called power.
Power or average power is given by
work done
Pavg  ,
time
Power is a scalar
SI Unit: watt (W) (or) J/s,
CGS Unit: erg/sec
Other units: kilo watt, mega watt and horse power
One horse power (H.P)=746 watt.
Instantaneous Power:
 W 
P  Lt  
t  0
 t 
 
It is also calculated by P  FV cos   F .V
Relation Between Pavg and Pins:
W mv 2 1  v  1 1  
Pave    mv    mav  F .V
t 2t 2 t 2 2
1
Pave 
Pinst
2
The area under P - t graph gives work done
dW
P W   P.dt
dt
dW
The slope of W - t curve gives instantaneous power P  dt  tan 
power
Work

O time O time
Applications on power:
The power of a machine gun firing ‘n’ bullets each of mass ‘m’ with a velocity ‘v’ in a time interval ‘t’ is
given by
1 
n  mv 2  2
P 
2   nmv
t 2t
A crane lifts a body of mass ‘m’ with a constant velocity v from the ground, its power is
P = Fv = m g v
1
Power of lungs of a body blowing a whistle is P  (mass of air blown per sec) (velocity)2
2
Power of a heart pumping blood = (pressure) (volume of blood pumped per sec)
dm
A conveyor belt is moving with a constant speed ‘v’ horizontally and gravel is falling on it at a rate of .
dt
dm
Then additional force required to maintain speed v is F  v
dt
dm
additional power required to drive the belt is, P  Fv  v
2
dt
When a liquid of density '  ' coming out of a hose pipe of area of cross section ‘A’ with a velocity ‘v’
strikes the wall normally and stops dead.
1 mv2 1
The npower exerted by the liquid is P    Av3
2 t 2
( mass = density x volume = m=   A  l )
A vehicle of mass ‘m’ is driven with constant acceleration along a straight level road against a constant external
resistance ‘R’ when the velocity is ‘v’, power of engine is
P= Fv = (R + ma)v
If P is a rated power of a device and if its efficiency is x%, useful power is (output power)
x
P1 P
100
If a motor lifts water from a well of depth ‘h’ and delivers with a velocity ‘v’ in a time t then power of the
motor
1
mgh  mv 2
P 2
t
If a body of mass ‘m’ starts from rest and accelerated uniformly to a velocity v0 in a time t0 , then the work
done on the body in a time ‘t’ is given by
2
1 1 v t v v 
W  mv 2  m  o  ; a  0 ; v  at   0  t
2 2  to  to  to 
Instantaneous power, P = Fv = m a v
v0  v0  v02
P  m   t  m t
to  to  t02
A motor pump is used to deliver water at a certain rate from a given pipe. To obtain ‘n’ times water from
the same pipe in the same time by what among of (a) force and (b) power of the motor should be increased.
If a liquid of density '  ' is flowing through a pipe of cross section ‘A’ at speed ‘v’ the mass
dm
coming out per second will be  Av  .
dt
To get ‘n’ times water in the same time
1
 dm   dm 
   n   A ' v '  '  n  Av  
 dt   dt 
As the pipe and liquid are not changed,
    ; A '  A & v '  nv
1
 dm 
 nv   n 
dm
v
dm F '  dt   dt   n 2
as F  v  
dt F  dm   dm 
v  v 
 dt   dt 
as P  Fv 
P ' F ' v '  n F   nv 

2
   n3
P Fv Fv
 F '  n2 F  P '  n3P
To get ‘n’ times of water force must be increased n 2 times while power n3 times.
Position and velocity of an automobile w.r.t. time:
An automobile of mass ‘m’ accelerates starting from rest, while the engine supplies constant
power, its position and velocity changes w.r.t. time as
Velocity:
As Fv = P = constant
dv  dv 
i.e. m v p F  m 
dt  dt 
P v2 P
or  vdv   dt on integrating we get  t  C1
m 2 m
As initially the body is at rest,
ie. v = 0 at t =0  C1  0;
1/ 2
 2Pt 
v   v t 1/2
 m 
Position:
From the above expression
1/2 1/ 2
 2 Pt  ds  2 Pt 
v   or   
 m  dt  m 
1
1/ 2 1
 2 Pt   2 p 2 2
  m 
ds    dt  
 m 
  t dt
1/ 2
 2P  2 3/ 2
integrating on both sides we get S    .t  C2
 m  3
Now at t = 0, S = 0  C2=0
1/ 2
 8 P  3/2
 t , S t
3/ 2
S
 9m 
:: PROBLEMS ::
1. An automobile is moving at 100 kmph and is exerting attractive force of 3920 N. What horse
power must the engine develop, if 20% of the power developed is wasted?
SOLUTION :
5
Velocity = 100 kmph 100  m/s
18
Force= 3920 N;; Useful power = 80%
W F .S 80 5
Power =  F .v  P  3920  100 
t t 100 18
100 5
P  3920 100   13.16 104 W  182.5hp
80 18
2. The coefficient of friction between a particle moving with some velcoity V0 and the rough horizontal
 V0 
surface is   . Find how much kinetic energy is lost in time t0 due to friction:
 2 gt0 
A) 1/4 B) 1/2 C) 3/4 D) 2/3
SOLUTION :
v1  v0    g  t0
 v 
 v0   0  gt0  v0 / 2
 2 gt0 
velocity left  v0 / 2
2
1  v0  11 
K.E left  m     mv0 2 
2 2 42 
1
 of initial K.E
4

3. The velocity of a particle is v  ati  bt 2 j , where t is the time in second. Match the columns for
t  1 second:
Column-I
A) Acceleration of particle is B) Tangential acceleration is
C) Radial acceleration is D) Radius of curvature of path is
Column-II
p) less than  a 2  b 2 
3/2
q) less than ab
r) less than  a  b 
2 2
s) greater than 2b
SOLUTION :
A-s; B-r; C-q; D-p

v  ati  bt 2 j  ai  b j for t  1 second

a  ai  2bt j

a   a 2  4b 2   2b
1/ 2
1
  a 2  2b 2  
2 2
  a  4b  
2 2 


 a 2
 b 2
 
ab
  ab
a  b2 
2 1/ 2
v2  a  b 
2 2 3/ 2
  a 2  b2 
3/2
R 
an ab
4. The 50 N collar starts from rest at A and is lifted with a constant speed of 0.6 m/s along the
smooth rod. Determine the power developed by the force F at the instant shown.
0.9 m
 T F
T
1.2 m
A
SOLUTION :
Since the collar is lifted with a constant speed
T cos   mg  0  T cos   mg  5  10
 
Now, P  F .v  T cos   v; HereT  F
P=50 x v = 50 x 0.6 = 30W
5. What is the minimum value of the mass M so that the block is lifted off the table at the instant
shown in the diagram ? Assume that the blocks are initially at rest.
m m
A) 0 B) C) m sin 600 D) none of these
sin 60 tan 600
Frictionless
Table
m
60
SOLUTION :
The accelerations of the blocks along the string are equal;
now apply F = ma for both the blocks.
6. A machine delivers power to a body which is directly proportional to velocity of the body. If the body
starts with a velocity which is almost negligible, find the distance covered by the body in attaining a
velocity v.
SOLUTION :
 dv 
Power P  Fv cos 0  Fv  m   v v
 dt 
dv
mv  K 0 v, Where K 0  cons tan t
dt
dv  dv  dx
m  K0 ; m    K0
dt  dx  dt
dv K 
mv  K 0 ; vdv   0  dx
dx  m
v x
 K0 
Integrating  vdv    m dx;
0 0 
v2  K0  1 mv 2
  x  x 
2  m 2 K0
Passage
Rod AO3 of length L can rotate about A. Initially rod was at potition AO2 , when spring OB of force
ce
constant K, attached to block B of mass m was at position OA with unstretched length L. The
smooth block B can slide on rod when pulled by the block D of mass m through a massless spring
and smooth pulley at O1 .
7. Find the velocity of the block B, when the rod and spring at B make an angle of 30 with their
respective initial positions : B is the middle point of the block)
A C O2
B
O3
 O1
D
1 1 1
   
1
  
10mgL  KL2 2  3 2  2

 2mgL  KL2  
2 1  2  5mgL  KL2

2 1  2

 6mgL  KL2

2 1  2

A)  8m  B)  4m

 C)  4m  D)  4m 
       
SOLUTION :
ABO  90
(Since, AOB  30 and OAB  60 )
L 3 L
OB  L cos 30  ; AB 
2 2
L
BC  BA sin 30 
4
L
Distance by which B ahs gone down  BC 
4
Distance by which D has gone down  AB  L sin 30  L
Decrease in PE = Increase in KE + Increase in elastic PE
2
L 1 1  3
mgL  mg   2m  v 2  K  L  L 
4 2 2  2 
1

10mgL  KL2 2  3 2  2
  
on solving, v   8m 
 
8. Find the work done by the frictional force (if slider is rough) at the instant when rod and the spring
attached at block B make an anlge of 30 with their respective initial positions.
1 2
    mgL
2 2
A) KL 2  3  mgL B) KL 2  3
2

2 4
1 2
  5 1 2
 
2 2
C)KL 2  3  mgL D) KL 2 1
8 4 2
SOLUTION :
W  KE  0
W f  WN  WS  Wg  0
WN  Work done by normal reaction = 0 Acts perpendicular to displacement)
WS  Work done by spring force
2
1  L 3
 0  K  L  
2  2 
5
Wg  Work done by force of gravity  mgL
4
 1   5
2

2 2 3
W f    0  KL    mgL 
 2  2  4 
 
1
  5
2
 KL2 2  3  mgL
8 4
9. Find the power of an engine which can draw a train of 400 metric ton up the inclined plane of 1 in
98 at the rate 10 ms-1. The resistance due to friction acting on the train is 10 N per ton.
SOLUTION :
1
Given sin   ; m  400  103 kg
98
frictional force f = 10  400 = 4000N;
velocity v = 10 ms-1
 Power P   mg sin   f  v
 1  
 P   400  103  9.8    4000   10
 98  
=440000W = 440 KW
10. A block A of mass m slides on a smooth slider in the system as shown. A block c of same mass
hanging from a pulley pulls block A. When the block A was at position B, the spring was unstretched.
Find the speed of the block A when AB  OB  L
(m)
A
B
90
Spring constant = K
C
O
1 1 1
1
 gL KL2 2  2  22  gL KL2 2  2
KL2
 

2 2
A)  
m 
 B)  gL  2  1  C)  gL 
2 KL2
 2  1  D)   
 2  2m   m   2 m 
SOLUTION :
Decrease in PE = Increase in KE + Increase in elastic PE
1 1
mgL    2m  v 2  Kx 2
2 2
s  mv  KL  2  1
2 2 1 2
2
1
 2
KL2
 
2
v   gL  2 1 
 2m 
11. A hose pipe has a diameter of 2.5 cm and is required to direct a jet of water to a height of atleast
40m. Find the minimum power of the pump needed for this hose.
SOLUTION :
Volume of water ejected per sec
2
d
Av      2 ghm3 / s;  v  2 gh
2
1
Mass ejected per sec is M   d 2  2 gh  Kg / s
4
1 2 1 2 3
Kinetic energy of water leaving hose/sec K .E  mv   d   2 gh  2  
2 8
1 3
 3.14   2.5  102    2  9.8  40  2  1000  21.5 KJ
2
8
12. A body of mass m accelerates uniformly from rest to velocity v0 in time t0 , find the instantaneous
v0
power delivered to body when velocity is .
2
SOLUTION :
v0 mv0
Acceleration a  t ; ForceF  t
0 0
v0  mv0  v0 mv02
Instantaneous power P=F.    
2  t0  2 2t0
13. A particle of 500 gm mass moves along a horizontal circle of radius 16m such that normal acceleration
of particle varies with time as an  9t 2
Column - I Column - II
A) Tangential force on particle at t = 1 second (in newton) p) 72
B) Total force on particle at t = 1 second ( in newton) q) 36
C) Power delivered by total force at t = 1 sec ( in watt) r) 75
D) Averge power developed by total force over first one second (in watt) s) 6
SOLUTION :
A-s,B-r,C-p,D-q
14. A wind -powered genrator converts wind energy into electrical energy. Assume that the generator
converts a fixed fraction of the wind energy intercepted by its blades into electrical energy . For
wind speed  , the electrical powwer output will be proportional to
[IIT-2008]
A)  B)  2 C)  3 D)  4
SOLUTION :
dp d
F   mv   v  Apv   Apv 2
dt dt
 P  Fv   APv  v  APv 3  P v 3
2
Comprehension
A block of mass m sits at rest on a frictionless table in a rail car that is moving with speed vc along a straight
horizontal track (fig.) Aperson riding in the car pushes on the block with a net horizontal force F for a time t in
the direction of the car’s motion.
Ground Train
m F
S1
S
15. What is the final speed of the block according to a person in the car?
Ft 2Ft Ft
A) B) C)  D) zero
m m m
SOLUTION :
Ft
V  u  at  0 
m
16. According to a person standing on the ground outside the train?

Ft 2Ft Ft
A) Vc + B) Vc – C) – Vc D) zero
m m m
SOLUTION :
Ft
Vg  V0 
m
17. How much did K.E of the block change according to the person in the car?
F2 t 2 F2 t 2 2F 2 t 2
A) B) C) D) none of these
2m m m
SOLUTION :
1 1 F 2t 2 1 F 2t 2
mv 2  
2 2 m2 2 m
18. In terms of F, m, & t, how far did the force displace the object according to the person in car?
Ft 2 Ft 2 2Ft 2 4Ft 2
A) B) C) D)
m 2m m m
SOLUTION :
1F 2
S train  0t  t
2m
19. According to the person on the ground. The displacement of block is

Ft 2 Ft 2 Ft 2 Ft 2
A)  2v c t B)  vct C)  vct D)  vct
2m 2m m 2m
SOLUTION :
S ground  VC t  Strain ;
1F 2
 VC t  .t
2m
Vertical circular motion
Vertical circular motion with variable speed:
O
r
V1
A
Consider a body of mass ‘m’ tied at one end of a string of length ‘r’ and is whirled in a vertical circle by
fixing the other end at ‘O’.
Let V 1 be the velocity of the body at the lowest point.
V2
mg Vhor
TH
O Thor

T
mV2
TL P
r
mg sin
 mg cos
V1 mg
A
mg
Velocity of the body at any point on the vertical circle:
1 1
TE A  TEP ; mV12  0  mV2  mgh
2 2
V2  V12  2 gh, but h  r 1  cos  
V2  V12  2 gr 1  cos   ;V  V12  2 gr 1  cos  
If V2 is the velocity of the body at highest point   180 

0
V2  V12  2 gr 1  1  V12  4 gr
Tension in the string at any point:
Let T be the tension in the string when the string makes an angle  with vertical.
mV2
T  mg cos 
r
1) At the lowest point  = 00 tension in the string is
mV12
TL   mg  max imum.
r
2) At the highest point   1800 .
mV22
The tension in the string is TH   mg (min)
r
3) When the string is horizontal,   90 ,
2
mVhorz
tension in the string at this position is T hor  
r
4) The difference in maximum and minimum tension in the string is
mV12 mV22
Tmax  Tmin   mg   mg
r r
m
 V12  V22   2mg
r
m
 4 gr   2mg  4mg  2mg  6mg
r
5) Ratio of maximum tension to minimum tension in the string is
mV12
 mg
Tmax V 2  rg
 r 2  12
Tmin mV2 V2  rg
 mg
r
When the particle is at ‘P’
a) Tangential force acting on the particle is Ft= mg sin  .
Tangential acceleration at  g sin 
b) Centripetal force acting on the particle is
 mV 2 
Fe     T  mg cos  .
 r 
V2
Centripetal acceleration ac 
r
c) Net acceleration of the particle at the point ‘P’ is a  at2  ac2 .
d) The net force acting on the particle at point ‘P’ is F  Ft 2  Fc2
Ft a t
Angle made by net force or net acceleration with centripetal component is  and tan  = F a
c c
Condition for vertical circular motion of a body

B
V2
mg
T2
O
T1
V1
A mg
mV22
We know that T2   mg
r
The body will complete the vertical circular path when tension at highest point is such that
mV22
T2  0, mg  0;V2 min  gr
r
Hence the minimum speed at highest point to just complete the vertical circle is gr
From the law of conservation of mechanical energy total energy at lowest point A=total energy at highest
point B
U A  KE A  U B  KEB
1 1
O  mV12  mg  2r   mV22
2 2
1 1
mV12  2mgr  mgr V2  gr 
2 2
5
= mgr  V1  5 gr
2
For the body to continue along a circular path the critical velocity at lowest point is 5gr
Critical velocity at any point on the vertical circle:
O
T V
Rh
B P

mg
A V1
From the Law of conservation of energy total energy at point ‘A’= total energy at point P
U A  KE A  U P  KEP
1 1
O  mV12  mgh  mV2
2 2
1 1
m  5 gR   mgR 1  cos    mV2
2 2
5 gmR 1
 mgR  mgR cos   mV2
2 2
5 gmR 1
 mgR  mgR cos   mV2
2 2
mgR 1
3  2 cos    mV2
2 2
V  gR  3  2cos  
Minimum tension in the string to just complete vertical circle:
Let T be the tension in the string when the string is making an angle  from lowest point
O T V
 P
 mg cos
mg sin  mg
mv2 m
T  mg cos    mg cos   gR  3  2 cos  
R R
 mg cos   3mg  2mg cos 
 3mg cos   3mg  3mg 1  cos  
In case of non uniform circular motion in a vertical plane if velocity of the body at the lowest point is less than
5gr , the particle will not complete the circle in vertical plane, the particle may either oscillate about the
lowest point or it leaves the circle with out looping.
Condition for oscillating about the lowest position:
1)If 0  VL  2 gr . in this case, velocity becomes zero before tension vanishes and the particle oscillates
about its lowest position with angular amplitude 00    900
2)If velocity of the body at the lowest point VL  2 gr , then the maximum height reached by the body just
VL2
before its velocity becomes zero is given by h 
2g
VL2
3) The angle made by the string with the vertical when its velocity becomes zero is given by cos   1 
2 gr
Note: If 0  VL  2 gr then the particle oscillates such that 00    900
Condition for leaving the circular path without looping:
 If 2 gr  VL  5 gr .
the particle is not able to complete the vertical circle,
it goes to certain height and leaves the circular path  900    1800 
while leaving the circular path T = 0 but V  0
The angle made by the string with downward vertical when the tension in the string becomes zero is given
2 VL2
by cos   
3 3gr
VL2  gr
The heigth at which the tension in the string becomes zero is given by h 
3g
When car moves on a concave bridge of radius
N

 v
mg cos
Concave Bridge
mg
mv 2
Centripetal force  N  mg cos  
r
mv 2
and normal reaction N  mg cos  
r
When car moves on a convex-bridge of radius r
N
mg cos

mg
mv 2
Centripetal for = mg cos   N 
r
mv 2
and normal reaction N  mg cos  
r
A ball of mass ‘M’ is suspended vertically by a string of length ‘L’.
A bullet of mass ‘m’ is fired horizontally with a velocity ‘u’ onto the ball, sticks to it.
For the system to complete the vertical circle, the minimum value of ‘u’ is given by
u
 M  m 5 gL
m
0
L
u m
M
A nail is fixed at a certain distance ‘x’ vertically below the point of suspension of a simple pendulum of
length L.
The bob is released when the string makes an angle  with vertical.
The bob reaches the lowest position then describes a vertical circle whose centre coincides with the nail.
Then
L  3  2 cos  
xmin 
5
x
L
o
nail
L-x
A body of mass ‘m’ is allowed to slide down from rest, from the top of a smooth incline of height ‘h’ .
For the body to move in loop of radius ‘r’ on arriving at the bottom.
h
r
 5r 
a) Minimum height of smooth incline h   
 2
b)’h’ is independent of mass of the body
A small body is freely sliding down from the top of a smooth convex-hemisphere of radius r, placed on a
table with its flat face on the table then
a) Normal reaction on the body is zero at the instant the body leaves the hemisphere.
b) The vertical height from table a which the body leaves the hemisphere is h = 2r/3
h r

If the position vector of the body with respect to the centre of curvature makes an angle  with
vertical when the body leaves the hemisphere, then cos  =2/3
2 gr
d) velocity of block at that instant is V 
3
e) If the block is given a horizontal velocity ‘u’ from the top of the smooth convex-hemisphere
2 u2
then the angle  with vertical at which the block leaves hemisphere is cos   
3 3gr
: : PROBLEMS : :
1. A nail is located at certain distance vertically below the point of suspension of a simple pendulum. The
pendulum bob is released from the position where the string makes an angle 600 from the vertical.
Calculate the distance of the nail from the point of suspension such that the bob will just perform
revolutions with the nail as the centre. Assume the length of the pendulum to be 1m.
60 0
SOLUTION :
Velocity of bob at lowest position
V  2 g  1  cos  

 2 g   1  cos 600   2 g  g  ... 1
2
Let ‘d’ be the distance of nail from the point of suspension. The bob will have to complete the circle of
radius r = l  d
To complete vertical circle
Vmin  5 gr  5 g 1  d    2 
Equating, equations (1) and (2) , we get
4l 4
gl  5 g 1  d   d    0.80m
5 5
PASSAGE
A bead of mass m is threaded on a smooth circular wire centre O, radius a, which is fixed in vertical plane.
3mg
A light string of natural length ‘a’, elastic constant = and breaking strength 3mg connects the bead to
a
the lowest point A of the wire. The other end of the string is fixed to ring at point B near point A. The string
is slaked initially. The bead is projected from A with speed u.
P
A
u
B
2. The smallest value u0 of u for which the bead will make complete revolutions of the wire will be
A) u0  5 ga B) u0  6 ga C) u0  7 ga D) u0  2 ga
SOLUTION :
When particle is at highest position, the elastic force is downwards
3mg
Fl   2a  a   3mg
a
it v is velocity at height point at B
mu0 2
 Fl  mg  N
a
1
If V = 0, then KE at lowest point A will be mu02   Elastic energy  gPE  at B
2
1  3mg  2
=   a  mg 2a
2 a 
u 02  7 ga
3. A bob of mass M is suspended by a massless string of length L. The horizontal velocity  at just
sufficient ot make it reach the point B.The angle  at which the speed of the bob is half of that A,
satisfies [IIT-2008]
B
L 
P
    3 3
A)   B)   C)   D)  
4 4 2 2 4 4
SOLUTION :
As the body just reaches the topmost point B, therefore ,  A  5 gL and  B  gL
Let at point P having angular displacement  the speed becomes half of the initial value A .
Using the law of conservation of energy
Energy at A=Energy at B
1 1
m 2 A  m 2 P  mgL 1  cos  
2 2
1
m  A 2   P 2   mgL 1  cos  
2
1  5 gL 
m  5 gL    mgL 1  cos  
2  4 
15 7
 1  cos   ; cos   
8 8
3
Which means  
4
4. If v  2u0 , the tension T in the elastic string when the bead is at the highest point B of the wire is
3mu02  4u02 
A) B) 4mg C) 2mg D)   g m
a  a 
SOLUTION :
When v  2v0  2 7 ga , For po int B
m  2v0 
2
 3mg  mg  N
a
4mv0 2
 4mg  N
a
N   4  28  mg =-24 mg
negative sign denote it acts downwards and adds to tension total tension in string T = 3mg+N
 4v 2 
T   0  gm
 a 
5. The elastic energy stored in the string when the bead is at the highest point B will be
3mga 2mga
A) B) 2mga C) 4mga D)
2 2
SOLUTION :
. Elastic PE stored in the string
1  3mg  2 3
  a  mga
2 a  2
6. A body slides without friction from a height H=60 cm and then loops the loop of radius R=20cm at
the bottom of an incline. Find the ratio of forces exerted on the body by the track at the positions A,B
and C (g = 10 ms-2)
C
H B
R
A
SOLUTION :
From data H=3R
Velocity at A, VA  2 gH  2 g  3R   6 gR
Velocity at B, VB  4 gR
Velocity at C, VC  2 gR
mVA2
Reaction force at A= R1   mg cos  00 
R
m  6 gR
  mg  7 mg
R
mVB2 m  4 gR
Reaction force at B  R2   mg cos  900    0  4mg
R R
mVC2 m  2 gR
Reaction force at C  R3   mg cos 1800    mg  mg
R R
 R1 : R2 : R3  7 : 4 :1
7. A small body A starts sliding from the height h down an inclined groove passing into a half - circle of
radius h/2 (see figure). Assuming the friction to be negligible, find the velocity of the body at the
highest point of its trajectory After breaking off the groove).
9 8 27 10
A) gh B) gh C) gh D) gh
27 27 8 27
SOLUTION :
v 2 At end of track) = 2gh.
mu 2
Let body break at angle  the  mg cos  . (1)
h
2u

h
u 2  v 2  2g 1  cos  
2
2 2
solving cos  = &u  gh .
3 3
v at highest pt is
2 2 8
u cos   gh   gh
3 3 27
8. A hemispherical vessel of radius R moving with a constant velocity v0 and containing a ball, is
suddenly haulted. Find the height by which ball will rise in the vessel, provided the surface is smooth:
v0 2 2v0 2 v0 2
A) B) C) D) none of these
2g g g
SOLUTION :
1
mv0 2  mgR 1  cos  
2
v0 2
 R  R cos   required height
2g
9. A particle of mass m initially at rest starts moving from point A on the surface of a fixed smooth
hemisphere of radius r as shown. The particle looses its contact with hemisphere at point B. C is
centre of the hemisphere. The equation relating  and  ' is
A
B
r 
'
C
A) 3sin   2 cos  ' B) 2sin   3cos  ' C) 3sin  '  2cos  D) 2sin   3cos  '
SOLUTION :
mv 2
Let v be the speed of particle at B when it is about to loose contact  mg sin  '
r
Applying conservation of energy
1 2
mv  mg  r cos   r sin  '
2
3sin  '  2cos 
10. A heavy particle hanging from a fixed point by a light inextensible string of length  is projected
horizontally with a speed of g . Find the speed of the particle and the inclination of the string to the
vertical at the instant of the motion, when the tension in the string is equal to the weight of the particle.
SOLUTION :
T
 B
mg cos
hu mg
A mg sin
  gl
Let T=mg at an angle ' ' as shown in the
figure and h   1  cos     i 
Applying law of conservation of mechanical energy between the points A and B,
1
we get m  u 2  v 2   mgh
2
Here u 2  g    ii 
and v 2  u 2  2 gh   iii 
mv 2
Further T  mg cos   T  mg 

v 2  g  1  cos     iv 
From equations (i), (iii) and (iv)
we have g  1  cos    g   2 g  1  cos  
2 2
cos   ;   cos 1  
3  
3
g
From equation (iv) v 
3
11. A bob of mass m is suspended from a fixed support with a light string and the system with bob and
support is moving with a uniform horizontal acceleration. The breaking strength of the string is
mg 2 . Find the workdone by the tension in the string in the first one second:
mg 2 mg 2
A) 2mg 2
B) C) D) mg 2 2
2 2
SOLUTION :
T sin   ma and T cos   mg
So, a  g tan 
mg
Now, T   mg 2 (given)
cos 
1
cos   , or   450
2
and a  g tan   g tan 450  g
1 2
workdone, W  T.S  T  at  2
2 
 1

 mg 2    g 1 2 = m g 2 / 2
 2
2
12. A bob attached to one end of a string, other end of which is fixed at peg A. The bob is taken to a
position where string makes an angle of 300 with the horizontal. On the circular path of the bob in
vertical plane there is a peg ‘B’ at a symmetrical position with respect to the position of release as
shown in the figure. If Vc and Va be the minimum speeds in clockwise and anticlock wise directions
respectively, given to the bob in order to hit the peg ‘B’ then ratio Vc : Va is equal to
PegB Bob
30° 30°
PegA
A) 1:1 B) 1: 2 C)1: 2 D) 1:4

SOLUTION :
For complete circular motion speed at highest point in gR
Apply conservation energy va  2 gR
For clock wise motion
mvc2
T  mg cos 60 
R
vC to minimum T=0 ;
gR
vC 
2
13. A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric
spheres A and B as shown. The sphere A has radius R and sphere B has a radius  R  d  . All
surfaces are smooth. The diameter of ball is slightly less than d. The ball is given a gentle push so
that angle made by radius vector of the ball with vertical is  . N A and N B are the magnitudes of
normal reaciton forces on the ball exerted by spheres A and B respectively:
h

R A
d B
Match the columns:

Column-I
1  2  1  3  1  3  1  2 
A)   cos   B)   cos   C)   cos   D)   cos  
3 4 4 3
Column-II
p) N B =0 and N A = mg  3cos   2  q) N B =0 and N A = mg  4cos   2 
r) N A =0 and N B = mg  2  3cos   s) none of these

SOLUTION :
A-P, B-S, C-S, D-R
 d
h   R   1  cos   v 2  2 gh
 2
or N  mg  3cos   2 
Ball will lose contact with sphere A, when N = 0 3cos   2  0
PASSAGE-
A particle of mass M attached to an inextensible string is moving in a vertical circle of radius. R
about fixed point O. It is imparted a velocity u in horizontal direction at lowest position as shown in
figure.
P h
u
Following information is being given

i) Velocity at a height h can be calculated by using formula v 2  u 2  2 gh
ii) Particle will complete the circle if u  5 gR
iii) Particle will oscillates in lower half  0    90  if 0  u  2 gR

0 0
M 2
iv) The magnitude of tension at a height ‘h’ is calculated by using formula T  u   gR  3gh
R
14. If R  2m, M  2kg and u  12m / s . Then value of tension at lowest position is
A) 120 N B) 164 N C) 264 N D) zero
SOLUTION :
Put h = 0 T = 164 N
15. Tension at highest point of its trajectory in above question will be

A) 100 N B) 44 N C) 144 N D) 264 N
SOLUTION :
Put h = 2R T= 144 N
16. If M  2kg , R  2m and u  10m / s . Then velocity of particle when   600 is

A) 2 5 m / s B) 4 5 m / s C) 5 2 m / s D) 5 m / s
SOLUTION :
R
At   60 h  R  R cos 60  2
0
R
Put h  in v 2  u 2  2 gh
2
Collisions
The strong interaction among bodies involving exchange of momentum in a short interval of time is called
collision.
During collision bodies may or may not come into physical contact.
Ex:
In the collision of  particle with nucleus, due to coulombic repulsive forces  particle is
scattered away without any physical contact.
Based on the direction of motion of colliding bodies, collisions are classified into
(i) Head on or one dimensional collision
(ii) oblique collision
Head on (or) one dimensional collision
A u B u A v1 B v
1 2 2
Before Collision After Collision

It is the collision in which the velocities of the colliding bodies are confined to same straight line before and
after collision.
Oblique Collision:
It is the collision in which the velocities of the colliding bodies are not confined to same straight line before
and after collision.
Oblique collision may be two dimensional or three dimensional.
When a particle hits elastically and obliquely another stationary particle of same mass, then they move
perpendicular to each other after collision.
Types of Collision: Based on conservation of kinetic energy collisions are classified into
(i) Elastic Collision
(ii) Inelastic collision
Elastic Collision:
It is the collision in which both momentum and kinetic energy are conserved. Forces involved
during collision are conservative in nature
Ex:
1. Collision between atomic particles.
2. Collision between two smooth billiard balls.
3. Collision of  particle with nucleus.
Inelastic collision:
It is the collision in which momentum is conserved but not kinetic energy. Some or all the forces
involved during collision are non conservative.
Ex:
Collision between two vehicles.
Perfectly inelastic collision:
It is the collision in which the colliding bodies stick together and move as a single body after
collision
In perfectly inelastic collision the momentum remains conserved but the loss of kinetic energy is maximum.
Ex:
A bullet is fired into a wooden block and remains embedded in it.
Line of impact:
The line passing through the common normal to the surfaces in contact during impact is called
line of impact.
The force during collision acts along this line on both bodies.
EX-1:
Two balls A and B are approaching each other such that their centres are moving along line CD.
Line of impact and
line of motion
A B
C D
A B
Head on Collision
EX-2:
Two balls A and B are approaching each other such that their centres are moving
along dotted lines as shown in figure.
B
Line of motion Line of motion
of ball B of ball A
A
Line of impact
Oblique collision
Ex3:
Ball is falling on a stationary wedge Line of motion
of ball
Line of impact
Oblique Collision
Elastic collision in one dimension:
When two particles of masses m1 and m2 are moving along the line joining their centers with
velocities u1 and u2 (u1 > u2) before collision. Then v1 and v2 are their velocities after collision
u1 u2 m1 m2
v1 v2
m1 m2 During After collision
Before collision Collision
   
From the conservation of linear momentum m1 u1  v1  m2 v2  u2   
1 1 1 1
From Law of conservation of K.E m1u12  m2u22  m1v12  m2 v22
2 2 2 2
   
 u1  u2  v2  v1
i.e Relative velocity of approach before collision = Relative velocity of separation after collision
Velocities after collision are
  m  m    2m  
 1   u1  
1 2 2
v  u2
m
 1  m2  m
 1  m 2 
  2m    m  m  
 v2    u1  
1 2 1
 u2
 m1  m2   m1  m2 
Special cases:
1)If colliding particles have equal masses
i.e m1= m2 = m ;
   
v1  u2 , v2  u1
2) If two bodies are of equal masses and the second body is at rest
 
ie., m1  m2  m and u2  0
   
then v1  0 ; v2  u1
 
3)A lighter particle collides with heavier particle which is at rest m1 <<< m2, u2  0
  
v1  u1 ; v2  0
4)A heavier body collides with lighter body at rest
 
m1 >>> m2 , u2  0 ;
   
v1  u1 ; v2  2u1
Applications:
A body of mass m1 moving with a velocity v1 collides elastically with a stationary mass m2
  m  m  
1)Velocity of first body after collision v1   m  m  u1
1 2
 1 2 
  2m1  
2)Velocity of second body after collision v2   m  m  u1
 1 2 
3)KE of first body after collision (or) KE retained by first body

2
1 1  m1  m2  2
K.E1= m1v1  m1 
2
 u1
2 2  m1  m2 
2 2
1  m  m2   m1  m2 
K .Enet  m1u12  1   KEi  
2  m1  m2   m1  m2 
2
K .Eret  m1  m2 
4)Fraction of KE retained by 1 body st  
K .Ei  m1  m2 
5)KE of second body after collision (or) KE transferred to the second body
2
1 1  2m1  2
KE2  m2 v22  m2   u1
2 2  m1  m2 
 4m1m2   1 
KE2   2 
m1u12 
 (m1  m2 )   2 
 4m m 
KEtra   1 2
 KEi
  m  m 2 
 1 2 
6)Fraction of KE transferred form 1st body to second body (or) Fraction of KE lost by 1st body is
KEtra 4m1m2

KE i  m1  m2 2
P1 m1v1 m1  m2
7)Fraction of momentum retained by m1 P  m u  m  m
i 1 1 1 2
8)Fraction of momentum transferred from 1st body to second body

P2 Pi  P1 P  m  m2  2m2
 1 1  1  1 
Pi Pi Pi  m1  m2  m1  m2
Coefficient of restitution
Newton introduced a dimensionless parameter called the coefficient of restitution (e) to measure
the elasticity of collision.
It is defined as the ratio of the relative velocity of separation to the relative velocity of approach
of the two colliding bodies
 
Re lative velocity of separation v2  v1
e   
Re lative velocity of approach u1  u2
This formula is applied along the line of impact.
Here the velocities mentioned in the expression should be taken along the line of impact.
For a perfectly elastic collision e = 1
For an inelastic collision 0 < e < 1
For completely inelastic collision e = 0
A body dropped freely from a height ‘h’ strikes the floor the rebounds to a height h1
h1
e
h
after nth rebound hn  e 2n h
t t1
t2
h t3
h1
h2
h3
When a freely falling ball strikes the ground with a velocity ‘v’ and rebounds with a velocity v1 then
v1
e
v
after nth rebound Vn  e nV
Total distance travelled by the ball before it stops bouncing
d  h  2h1  2h2  2h3  .......
 h  2e 2 h  2e 4 h  2e 6 h  .......
 h  2e 2 h 1  e 2  e 4  ......
1  e 2 
d  h 2 
1  e 
Total time taken by the ball to stop bouncing
T  t  2t1  2t2  2t3  .......
2h 2h1 2h2 2h3

 2 2 2  ........
g g g g
2h 2h
  2e 1  e  e 2  .......
g g 
2h 2h
  2e 1  e  e 2  .......
g g 
Average speed of the ball during its entire journey is given by
Total dis tan ce travelled
Average speed 
Total time taken
1  e 2 
h
1  e2  gh 1  e 
2
  
2 h 1  e  2 1  e 2
g 1  e 
Average velocity of the ball during its entire journey is given by
Net displacement
Average velocity 
Total time taken
h gh 1  e 
 
2h 1  e  2 1  e 
g 1  e 
Change in momentum in 1s t collision
 mv1   mu    mv1  mu 
 meu  mu  mu 1  e 
Change in momentum in 2nd collision
 m  v2  v1   m  e 2u  eu   meu 1  e 
Total change in momentum before it stops is
p  mu 1  2e  2e 2  ........ u  2 gh 
 
1  e  1  e 
 mu    m 2 gh 
1  e  1  e 
Distance travelled before second impact is d 2  h  2h1  h 1  2e 
2
Distance travelled before third impact is d 3  h  2 h1  2h2  h 1  2e  2e 

2 4
2h
Time taken for second impact is t2  t  2t1  1  2e 
g
2h
Time taken for third impact is t3  t  2t1  2t2  g
1  2e  2e2 
Application
A particle of mass m moving with a speed u strikes a smooth horizontal surface at an angle  . The particle
rebounds at an angle  with a speed v. The coefficient of restitution is ‘e’.
m
m
u v


 
Since no external impulse acts in the horizontal direction, momentum of the ball is conserved in
the horizontal direction.
mu cos   mv cos 
u cos   v cos  .......... 1
By def of coefficient of restitution we get
eu sin   v sin    2
tan 
from (1) and (2), tan   e tan  tan   e
On squaring eq (1) and (2) and adding we get
v 2  u 2  cos 2   e 2 sin 2  
v  u cos 2   e 2 sin 2 
A ball is projected with an initial velocity u at an angle  to the horizontal surface.
If ‘e’ is the coefficient of restitution between the ball and the surface then
Y
u

x
1st 2 nd 3rd
2u sin 
1)Time taken for 1st collision, T 
g
2)Time interval between 1st and 2nd collisions,
2v1 sin 
T1   v1  eu 
g
2  eu  sin 
T1  eT
g
3)Time interval between 2nd and 3rd collisions,
2v2 sin g 2  e u  sin 

2
T2 
g

g
 e 2T  v
2  e2u 
4)The total time of flight is

T 1  T  T1  T2  ........  T  eT  e2T  e3T  .......
 T 1  e  e 2  e3  .......
T
T1 
1 e
If collision is elastic, e = 1 the T1 = 
5)The horizontal distance covered by the ball before 1st collision is
u 2 sin 2
R  u cos   T
g
6)The horizontal distance covered by it between 1st and 2nd collisions, R1  u cos   eT  eR
7)horizontal distance covered between 2nd and 3rd collisions, R2  u cos   e 2T  e 2 R
8)Total horizontal distance covered by the ball is
R1  R0  R1  R 2  R3  ........
 R  eR  e 2 R  .........  R 1  e  e 2  .....
R
R1 
1 e
For perfectly elastic collision e = 1 and R1 = 
9)The maximum height reached by the ball before 1st collision
u 2 sin 2  (u sin  ) 2
H 
2g 2g
10)Maximum height it reaches between 1st and 2nd collisions is
 eu sin  
2
H 1
  e2 H
2g
11)The sum of maximum heights reached by the ball is
H 1  H  H1  H 2  ........  H  e 2 H  e 4 H  ...........
H
 H 1  e 2  e 4  ............. , H1 
1  e2
If the collision is elastic e = 1 and H’= 
Head on inelastic collision
   
Two bodies of masses m1 and m2 moving with initial velocities u1 and u2 u1  u2 collide.  
 
After collision two bodies will move with velocities V1 and V2 .
   

From Law of conservation of linear momentum m1 u1  v1  m2 v 2  u 2   
   
By the definition of coefficient of restitution v 2  v1  e u1  u 2  
  m  em    1  e  m2  
v1   1 2
 u1   u2
 m1  m2   m1  m2 
  1  e  m1    m  em  
v2    u1  
2 1
 u2
 1m  m 2  m
 1  m2 
If m1  m2  m, u2  0 then
u1 u
v1  1  e  ; v2  1  e  1
2 2
v1 1  e

v2 1  e
Loss of kinetic energy of the system:
KE  KEI  LE F
1 mm    2

KE   1 2  u1  u 2 1  e 2 
2  m1  m2 

In case of perfectly in-elastic collision, e = 0
 loss in KE of system is
1 mm    2
KE   1 2  u1  u 2
2  m1  m2 
 
If two bodies are approaching each other then loss in KE of the system is maximum
1 mm 
KEmax   1 2   u1  u2 
2
2  m1  m2 
Ballistic pendulum:
It is an arrangement used to determine the velocities of bullets.
A log of wood of mass ‘M’ is suspended by a string of length ' l ' as shown in the figure. A bullet of mass ‘m’
is fired horizontally into the wooden block with a velocity ‘u’
Case I:
Let the bullet gets embedded in the block and system rises to a height ‘h’ as shown in the figure.
m
M
h
m u M
From the law of conservation of linear momentum
m1u1  m2u2   m1  m2  v
mu
mu  0   m  M  v  v  .......... 1
mM
KE of the system after collision is given by
1
KE   m  M  v2
2
PE at highest point = (m + M) gh
1
From LCE,  m  M  v 2   m  M  gh
2
v 2  2 gh  or  v  2 gh ...........  2 
From (1) and (2) velocity of the bullet
M m M m
u 2 gh  2 gl 1  cos  
m m
Loss in KE of the system = KE1 - KE2
1 1
KE  mu 2   m  M  v 2
2 2
1 2 m 2u 2 
KE   mu   m  M  2 
2   m  M  
1  mM  2
KE  u
2  m  M 
Case II:
If the bullet emerges out of the block with velocity ‘v’ then
mu  mv  MV Where V  2 gh
Collisions in two dimensions (oblique collisions)
1.A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of
individual particles changes along common normal direction.
2.No component of impulse acts along common tangent direction. Hence, linear momentum (or) linear
velocity of individual particles remains unchanged along this direction.
3.Net impulse on both the particles is zero during collision. Hence, net momentum of both the particles
remain conserved before and after collision in any direction.
4.Definition of coefficient of restitution can be applied along common normal direction.
u1 u2
1 2
m1 m1
Y
Tangential axis
X
m1 m2
Normal axis
v1 v2
1 2
m1 m2
From law of conservation of linear momentum along x - axis:

m1u1 cos 1  m2u2 cos  2  m1v1 cos 1  m2v2 cos  2
Along y - axis:
m1u1 sin 1  m2u2 sin  2  m1v1 sin 1  m2 v2 sin  2
v1 cos 1  v2 cos  2
Coefficient of restitution e   u cos   u cos 
1 1 2 2
: : PROBLEMS : :
1. A bullet of mass ‘m’ moving at a speed ‘v’ hits a ball of mass ‘M’ kept at rest. A small part having
mass m1 breaks from the ball and sticks to the bullet. The remaining ball is found to move at a
speed v2 in the direction of the bullet. Find the velocity of the bullet after the collision.
SOLUTION :
Mass of bullet = m and speed = v.
M ass of the ball M and fractional mass of the ball m1 According to law of conservation of linear
momentum
mv + 0 = (m + m1) v1 + (M - m1 ) v2
Where v1 = final velocity of the
(bullet + fractional mass)
mv   M  m1 
v1  v
 m  m1  2
2. Two bodies of masses m1 and m2 are moving with velocities 1ms-1 and 3ms-1 respectively in opposite
directions. If the bodies undergo one dimensional elastic collision, the body of mass m1 comes to
rest. Find the ratio of m1 and m2
SOLUTION :
u1  1m / s, u2  3m / s, v1  0
 m  m2   2m2 
v1   1  u1    u2
 m1  m2   m1  m2 
 m  m2   2m2 
0 1 1     3
 m1  m2   m1  m2 
m1 7
m1  m2  6m2 ; m1  7 m2 ; 
m2 1
3. Two identical balls A and B are released from the positions as shown in the figure. They collide
elastically on the horizontal portion. The ratio of heights attained by A and B after collision (neglect
friction
A
B
4h
h
450 600
SOLUTION :
As mass of two balls are equal, they exchange their velocities after collision.
u A2
u A  2 gh, uB  2 g  4h   8 gh ; hA   h;
2g
vB2 sin 2 600 9h 13h

hB  h   h 
2g 4 4
  v 
B
2
 u B2  2 gh  VB2  u B2  2 gh  vB2  6 gh 
hA 4

hB 13
4. n elastic balls are placed at rest on a smooth horizontal plane which is circular at the end with
m m m
radius ‘r’ as shown in the figure. The masses of the balls are m, , 2 ......... n 1 respectively..
2 2 2
Find the minimum velocity that should be imparted to the first ball of mass ‘m’ such that the ‘nth’
ball will complete the vertical circle.
r
1 2 n
SOLUTION :
Let speed to be imparted to the first ball be v0. Consider the impact between the first two balls and v1 and v2
be the velocities of balls 1 and 2 after the impact respectively.
m
According to law of conservation of linear momentum mv0  mv1  v2  1
2
1 2 1 2 1m 2
According to law of conservation of kinetic energy mv0  mv1    v2   2 
2 2 2 2 
4
Solving equations (1) and (2) , we get v2  v0
3
n 1
4
Similarly, for n ball vn   
th v0   3
3
For the nth ball to complete the vertical circular motion vn  5 gr   4 
From equations (3) and (4) , we have
n 1 n 1
4 3
  V0  5 gr ;V0   5 gr
3 4
5. Ball 1 collides with an another identical ball 2 at rest as shown in the figure. For what value of
coefficient of restitution e, the velocity of second ball become two time that of first ball after
collision?
1 2
SOLUTION :
Here m1  m2 and u2  0
 1 e   1 e 
After collision, v2    u & v1   u
 2   2 
 1 e   1 e 
Given v2  2v1 ;  u  2 u
 2   2 
1
1+e = 2 - 2e ; 3e = 1; e 
3
6. A body ‘A’ with a momentum ‘P’ collides with another identical stationary body ‘B’ one
dimensionally. During the collision, ‘B’ gives an impulse ‘J’ to the body ‘A’. Then the coefficient
of restitution is
SOLUTION :
From the law of conservation of linear momentum,
m1u1  m2u2  m1v1  m2 v2
mu  m  0   mv1  mv2
 P  P1  P2 where P2  J ,  given 
v2  v1 mv2  mv1 P2  P1
e   
u1  u2 mu  0 P
P2   P  P2 
2 P2  P 2 J  P 2 J
    1
P P P P
7. A ball of mass m collides with the ground at an angle  with the vertical. If the collision lasts for time
t, the average force exerted by the ground on the ball is: (e = coefficient of restitution between the
ball and the ground)
u

SOLUTION :
Impulse = change in linear momentum.
eu sin
u sin  u sin 
u cos  u cos 
(Before Collision) (After Collision)
mu cos  1  e 
 Ft  m  eu cos   u cos   or F 
t
8. A ball strikes a horizontal floor at an angle   45 with the normal to floor. The coefficient of
0
restitution between the ball and the floor is e = 1/2 . The fraction of its kinetic energy lost in the
collision is
SOLUTION :
Let ‘u’ be the velocity of ball before collision. Speed of the ball after collision will become
v  u 2 sin 2   e 2u 2 cos 2
2 2
 u   u  5
      8u
 2  2 2
 Fraction of KE lost in collision
1 1
mu 2  mv 2 2
v 5 3
2 2  1    1 
1 u 8 8
mu 2
2
9. Two equal spheres A and B lie on a smooth horizontal circular groove at opposite ends of a diameter. At
time t = 0, A is projected along the groove and it first impinges on B at time t = T1 and again at time t =
T2
T2. If ‘e’ is the coefficient of restitution, find the ratio of T1
v1
t T1
u1 u2  0
A
A
B B v2
SOLUTION :
R
T1  ........ 1
u1
v2  v1
 e  v2  v1  eu1
u1
Time taken for A to collide with B again is
2 R 2 R
T2  T1   T2  T1  .......  2 
v2  v1 eu1
T2 2  e
from (1) and (2) , T  e
1
10. After perfectly inelastic collision between two identical particles moving with same speed in different
directions, the speed of the combined particle becomes half the initial speed of either particle. The
angle between the velocities of the two before collision is
SOLUTION :
In perfectly inelastic collision between two particles, linear momentum is conserved. Let  be the angle
between the velocities of the two particles before collision. Then
P 2  P12  P22  2 P1 P2 cos  or
2
 v 1
 2m    mv    mv   2  mv  mv  cos  or 1  1  1  2 cos  or cos    2 ;  or   120
2 2 0
 2 
11. A bullet of mass ‘m’ moving with velocity ‘u’ passes through a wooden block of mass M= nm as
shown in figure. The block is resting on a smooth horizontal floor. After passing through the
block, velocity of the bullet becomes ‘v’. Its velocity relative to the block is
m
u M=nm
SOLUTION :
Let v be the velocity of block. Then from conservation of linear momentum.
uv
mu  mv  mnv  or  v   
 n 
 Velocity of bullet relative to block will be
 u  v  1  n  v  u
vt  v  v '  v   
 n  n
12. A block of mass 0.50 Kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass
of 1.00 kg and then they move together as a single body. Find the energy loss during the collision (JEE
MAIN 2008)
SOLUTION :
From LCLM, m1u1  m2u2   m1  m2  v
2 1
0.50  2  1  0   0.5  1 v  v  ms
3
1 1
 Energy loss KE  m1u12   m1  m2  v 2
2 2
2
1 1 2
KE   0.5  2   1.5     0.67 J
2
2 2 3
13. Consider a rubber ball freely falling from a height h=4.9 m on a horizontal elastic plate. Assume
that the duration of collision is negligible and the collision with the plate is totally elastic. Then the
velocity as a function of time and the height as a function of time will be; (JEE MAIN 2009)
V Y
Y
V
1)
h
2) V1 h
V1
O t
t
t t1 V1
O O
Y V Y
V
V1
h h
3) V1 4)
t1 3 t1
t
O t1 2 t1 3 t1 t
t O 2 t1 4 t1 t
V1
V1
SOLUTION :
When ball strikes the surface its velocity will be reversed so correct option is (3)
14. A pendulum consists of a wooden bob of mass ‘ m’ and of length l. A bullet of mass m1 is fired towards
v1
the pendulum with a speed v1 and it emerges out of the bob with a speed . Find the initial speed of
3
the bullet if the bob just completes the vertical circle.
SOLUTION :
From the Law of conservation of momentum
 v  m 2v
mv  m1  v1  1  or v  1  1
 3 m 3
To describe a vertical circle v  5 gl
m1 2v1 m 3 5 gl
hence 5gl =   v1  
m 3 m1 2
15. Two billiard balls of same size (radius r) and same mass are in contact on a billiard table. A third ball
also of he same size and mass strikes them symmetrically and remains at rest after the impact. The
coefficient of restitution between the balls is
SOLUTION :
v

u
r 1
sin    ;   300
2r 2
From conservation of linear momentum
u
mu  2mv cos300 or v
3
relative velocity of separation
Now e 
relative velocity of approach
in common normal direction
v u/ 3 2
Hence, e  0
 
u cos 30 u 3/2 3
: : THEORY BITS : :
1. A perfectly elastic ball P1 of mass ‘m’ moving with velocity v collides elastically with three exactly
similar balls P1 , P2 , P3 lying on a smooth table, Velocity of the four balls after collision aree
P1 P2 P3 P4
1) 0,0,0,0 2) v, v, v, v 3) v, v, v,0 4) 0, 0, 0, v
KEY:4
2. A bucket full of water is drawn up by a person. In this case the work done by the gravitational
force is
1) negative because the force and displacement are in opposite directions
2) positive because the force and displacement are in the same direction
3) negative because the force and displacement are in the same direction
4) positive because the force and displacement are in opposite directions
KEY:1
3. A motor car of mass m travels with a uniform speed v on a convex bridge of radius r. When the car
is at the middle point of the bridge, then the force exterted by the car on the bridge is
mv 2 mv 2 mv 2
1) mg 2) mg  3) mg  4) mg 
r r r
KEY:3
4. A man is rowing a boat upstream and inspite of that the boat is found to be not moving with respect
to the bank. The work done by the man is
1) zero 2) positive 3) negative 4) may be +ve or –ve
KEY:1
5. A ball of mass ‘m’ moving with a speed u undergoes a head – on elastic collision with a ball of mass
‘nm’ initially at rest. Find the fraction of the incident energy transferred to the heavier ball.
n n 2n 4n
1) 2)  n  12 3) 1  n 2 4) 1  n 2
n 1
KEY:4
6. A ball is thrown vertically upwards from the ground. Work done by air resistance during its time of
flight is
1) positive during ascent and negative during descent 2) positive during ascent and descent
3) negative during ascent and positive during 4) negative during ascent and descent
KEY:4
7. Workdone by force of friction
1) can be zero 2) can be positive 3) can be negative 4) any of the above
KEY:4
8. Identify the non-conservative force in the following
1) weight of a body 2) force between two ions
3) magnetic force 4) air resistance
KEY:4
9. If x, F and U denote the displacement, force acting on and potential energy of particle, then
dU dU 
1 dU 
1) U= F 2) F  3) F  4) F  x  dx 
dx dx  
KEY:3
10. In the case of conservative force
1) work done is independent of the path 2) work done in a closed loop is zero
3) work done against conservative force is stored is the form of potential energy
4) all the above
KEY:4
11. A body of mass ‘m’ moving with a constant velocity V hits another body of the same mass moving
with the same velocity V but in opposite direction and sticks to it. The velocity of the compound
body after the collision is
1) 2V 2) V 3) V/2 4) zero
KEY:4
12. When the momentum of a body is doubled, the kinetic energy is
1) doubled 2) halved
3) becomes four times 4) becomes three times
KEY:3
13. Internal forces can change
1) Kinetic Energy 2) Mechanical energy 3) Momentum 4) 1 and 2
KEY:4
14. If the momentum of a particle is plotted on X-axis and its kinetic energy on the Y-axis, the graph
is a
1) straight line 2) parabola 3) rectangular hyperbola 4) circle
KEY:2
15. A 2 kg mass moving on a smooth frictionless surface with a velocity of 10ms 1 hits another 2kg mass
kept at rest, in an inelastic collision. After collision, if they move together
1) they travel with a velocity of 5ms 1 in the same direction
2) they travel with a velocity of 10ms 1 in the same direction
3) they travel with a velocity of 10ms 1 in opposite direction
4) they travel with a velocity of 5ms 1 in opposite direction
KEY:1
16. When two identical balls are moving with equal speeds in opposite direction, which of the following is true?
For the system of two bodies
1) momentum is zero, kinetic energy is zero 2) momentum is not zero, kinetic energy is zero
3) momentum is zero, kinetic energy is not zero 4) momentum is not zero, kinetic energy is not zero
KEY:3
17. About a collision which are not correct
1) physical contact is must 2) colliding particles cannot change their direction of motion
3) the effect of the external force is not considered 4) linear momentum does not conserved
KEY:1
18. The product of linear momentum and velocity of a body represents
1) half of the kinetic energy of the body 2) kinetic energy of the body
3) twice of the kinetic energy of the body 4) mass of the body
KEY:3
19. The K.E of a freely falling body
1) is directly proportional to height of its fall 2) is inversely proportional to height of its fall
3) is directly proportional to square of time of its fall 4) 1 and 3 are true
KEY:4
20. Choose the false statement
1) In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation
2) In an inelastic collision the relative velocity of approach is less than the relative velocity of separation
3) In an inelastic collision the relative velocity of separation is less than the relative velocity of approach
4) In perfect inelastic collision relative velocity of separation is zero
KEY:2
21. Consider the following statements
A) Linear momentum of a system of particles is zero
B) Kinetic energy of a system of particles is zero then
1) A does not imply B & B does not imply A 2) A implies B and B does not imply A
3) A does not imply B but B implies A 4) A implies B and B implies A
KEY:3
22. A small sphere of mass ‘m’ is attached to a cord and rotates in a vertical plane about a point O . If
the averages speed of the sphere is increased, the cord is most likely to break at the orientation
when the mass is at :
A
m
l
C O D
1) bottom point B 2) the point C 3) the point D 4) top point A
23. If force acting on a body is inversely proportional to its speed, then its kinetic energy is
1) Linearly related to time 2) Inversely proportional to time
3) Inversely proportional to the square of time 4) A constant
KEY:1
24. In one-dimensional elastic collision, the relative velocity of approach before collision is equal to
1) relative velocity of separation after collision
2) ‘e’ times relative velocity of separation after collision
3) ‘1/e’ times relative velocity of separation after collision
4) sum of the velocities after collision
KEY:1
25. Which of the following graphs depicts the variation of K.E. of a ball bouncing on a horizontal floor
with height? (Neglect air resistances)
K K
1) 2)
h h
K
3) 4) None of these
h
KEY:1
26. Which of the following statement is correct?
1) KE of a system cannot be changed without changing its momentum
2) KE of a system cannot be changed without changing its momentum
3) Momentum of a system cannot be changed without changing its KE
4) A system cannot have energy without having momentum
KEY:1
27. Six steel balls of identical size are lined up along a straight frictionless groove. Two similar balls
moving with speed v along the groove collide with this row on the extreme left end. Then
1) one ball from the right end will move on with speed v
2) two balls from the extreme right end will move on with speed v and the remaining balls will be at rest
3) all the balls will start moving to the right with speed v/8
4) all the six balls originally at rest will move on with speed v/6 and the incident balls will come to rest
KEY:2
28. Two bodies of masses m 1 and m2 have equal momentum. Their K.E. are in the ratio
1) m2 : m1 2) m1:m2 3) m2:m1 4) m12 : m22
KEY:3
29. A lighter body moving with a velocity v collides with a heavier body at rest. Then
1) the lighter body rebounced with twice the velocity of bigger body
2) the lighter body retraces its path with the same velocity in magnitude
3) the heavier body does not move practically
4) both (2) and (3)
KEY:4
30. A body can have
1) changing momentum and finite kinetic energy 2) zero kinetic energy and finite momentum
3) zero acceleration and increasing kinetic energy 4) finite acceleration and zero kinetic energy
KEY:1
31. These diagrams represent the potential energy U of a diatomic molecule as a function of the inter-
atomic distance r. The diagram corresponds to stable molecule found in nature is
U U
1) 2)
r
r
U U
3) 4)
r r
KEY:1
32. A man pushes a wall and fails to displace it. He does
1) negative work 2) positive but not maximum work
3) maximum work 4) no work at all
KEY:4
33. In the fig. the potential energy U of a particle plotted against its position x from origin. Which of
the following statement is correct?
U
O x1 x2 x3 x
1) at x1 is in stable equilibrium 2) at x2 is in stable equilibrium

3) at x3 is in stable equilibrium 4) at x1, x2 and x3 particle is in unstable equilibrium
KEY:4
34. Two identical bodies moving in opposite direction with same speed, collided with each other. If the
collision is perfectly elastic then
1) after the collision both comes to rest
2) after the collision first comes to rest and second moves in the opposite direction with same speed.
3) after collision they recoil with same speed
4) both and 1 and 2
KEY:3
35. When a spring is wound, a certain amount of PE is stored in it. If this wound spring is dissolved in
acid the stored energy
1) In completely lost
2) Appears in the form of electromagnetic waves
3) Appears in the form of heat raising the temperature of the acid
4) Appears in the form of KE by splashing acid drops
KEY:3
36. A bottle of soda water is rotated in a vertical circle with the neck held in hand. The air bubbles are
collected
1) near the neck 2) near the bottom
3) at the middle 4) uniformly in the bottle
KEY:1
37. Two springs have their force constants k1 and k2 and they are stretched to the same extension. If
K2>K1 work done is
1) same in both the springs 2) more in spring K1
3) more in spring K2 4) independent of spring constant K
KEY:3
38. Two springs have their force constants K1 and K2 (K2>K1). When they are stretched by the same
force, work done is
1) same in both the springs 2) more in spring K1
3) more in spring K2 4) independent of spring constant K
KEY:2
39. A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding
force. then
1) Lorry will come to rest in a shorter distance 2) Car will come to rest in a shorter distance
3) Both come to rest in same distance 4) any of above
KEY:3
40. A cricket ball and a ping-pong ball are dropped from the same height in a vacuum chamber from
same height. When they have fallen half way down, they have the same
1) velocity 2) potential energy 3) kinetic energy 4) mechanical energy
KEY:1
41. A cyclist free-wheels from the top of a hill, gathers speed going down the hill, applies his brakes
and eventually comes to rest at the bottom of the hill. Which one of the following energy changes
take place.
1) Potential to kinetic to heat energy 2) Kinetic to potential to heat energy
3) chemical to heat to potential energy 4) Kinetic to heat to chemical energy
KEY:1
42. If ‘E’ represents total mechanical energy of a system while ‘U’ represents the potential energy,
then E-U is
1) always zero 2) negative 3) either positive or negative 4) positive
KEY:4
43. For a body thrown vertically upwards, its direction of motion changes at the point where its total
mechanical energy is
1) greater than the potential energy 2) less than the potential energy
3) equal to the potential energy 4) zero
KEY:3
44. Negative of work done by the conservation forces on a system is equal to
1) the change in kinetic energy of the system 2) the change in potential energy of the system
3) the change in total mechanical energy of the system 4) the change in the momentum of the system
KEY:2
45. Which of the following statements is wrong?
1) KE of a body is independent of the direction of motion
2) In an elastic collision of two bodies, the momentum and energy of each body is conserved
3) If two protons are brought towards each other, the P.E. of the system increases
4) A body can have energy without momentum
KEY:2
46. When a body falls from an aeroplane there is increase in its:
1) acceleration 2) potential energy 3) kinetic energy 4) mass
KEY:3
47. An agent is moving a positively charged body towards another fixed positive charge. The work
done by the agent is
1) positive 2) negative 3) zero 4) may be positive or negative
KEY:1
48. A body is moved along a straight line by a machine delivering constant power. The distance moved
by the body in time t is proportional to
1) t1/2 2) t3/4 3) t3/2 4) t2
KEY:3
49. Potential energy is defined for
1) non-conservative forces only 2) conservative forces only
3) both conservative & non-conservative forces 4) neither conservative nor non-conservative forces
KEY:2
50. A particle is projected at t = 0 from a point on the ground with certain velocity at an angle with the
horizontal. The power of gravitational force is plotted against time. Which of the following is the
best representation?
P P P
1) 2) 3) t 4) P t
t t
KEY:3
51. A body starts form rest and acquires a velocity V in time T. The instantaneous power delivered to
the body in time ‘t’ is proportional to
V V2 2 V2 V2 2
1) t 2) t 3) 2 t 4) 2 t
T T T T
KEY:3
52. Two bodies of different masses have same linear momentum. The one having more K.E. is
1) Lighter body 2) Heavier body 3) both 4) none
KEY:1
53. A car drives along a straight level frictionless road by an engine delivering constant power. Then
velocity is directly proportional to
1
1) t 2) 3) t 4) t2
t
KEY:3
54. The change in kinetic energy per unit ‘space’ (distance) is equal to
1) power 2) momentum 3) force 4) pressure
KEY:3
55. A particle is projected with a velocity u making an angle  with the horizontal. The instantaneous
power of the gravitational force
1) varies linearly with time 2) Is constant throughout the path
3) Is negative for complete path 4) varies inversly with time
KEY:1
56. A gramphone record is revolving with an angular velocity  . A coin is placed at a distance R
from the centre of the record. the static coefficient of friction is  . The coin will revolve with the
record if
g g g g
1) R  2) R  only 3) R  4) R 
2 2 2 2
KEY:4
57. A rock of mass m is dropped to the ground from a height h. A second rock with mass 2m is dropped
from the same height. When second rock strikes the ground, its kinetic energy will be
1) twice that of the first rock 2) four times that of the first rock
3) the same as that of the first rock 4) half that of the first rock
KEY:1
58. A car is moving up with uniform speed along a fly over bridge which is part of a vertical circle. The true
statement from the following is
1) Normal reaction on the car gradually decreases and becomes minimum at highest position of bridge
2) Normal reaction on the car gradually increases and becomes maximum at highest position
3) Normal reaction on car does not change
4) Normal reaction on the car gradually decreases and becomes zero at highest position
KEY:2
59. A vehicle is moving with uniform speed along horizontal, concave and convex surface roads. The
surface on which, the normal reaction on the vehicle is maximum is
1) Concave 2) Convex 3) Horizontal 4) Same at all surfaces
KEY:1
 
60. A ball with initial momentum P collides with rigid wall elastically. If P1 be it’s momentum after
collision then
       
1) P1  P 2) P1   P 3) P1  2 P 4) P1  2 P
KEY:2
61. Internal force can change
1) Linear momentum as well as kinetic energy 2) linear momentum but not the Kinetic energy
3) the kinetic energy but not linear momentum 4) neither the linear momentum nor the kinetic energy
KEY:3
62. Which of the following forces is called a conservative force?

1) Frictional force 2) Air resistance 3) Electrostatic force 4) Viscous force
KEY:3
63. Two particles of different masses collide head on. Then for the system
1) loss of KE is zero, if it were perfect elastic collision
2) If it were perfect inelastic collision, the loss of KE of the bodies moving in opposite directions is more than
that of the bodies moving in the same direction
3) loss of momentum is zero for both elastic and inelastic collision
4) 1, 2 and 3 are correct
KEY:4
64. For the same kinetic energy, the momentum shall be maximum for which of the following particle?
1) Electron 2) Proton 3) Deuteron 4) Alpha particle
KEY:4
65. In an inelastic collision, the kinetic energy after collision
1) is same as before collision 2) is always less than the before collision
3) is always greater than that before collision 4) may be less or greater than that before collision
KEY:2
66. A ball hits the floor and rebounds after an inelastic collision. In this case
1) the momentum of the ball just after the collision is same as that just before the collision
2) The mechanical energy of the ball remains the same in the collision
3) The total momentum of the ball and the earth is conserved
4) the total kinetic energy of the ball and the earth is conserved
KEY:3
67. A body of mass ‘m’ moving with certain velocity collides with another identical body at rest. If the
collision is perfectly elastic and after the collision, if both the bodies moves
1) in the same direction 2) in opposite direction
3) in perpendicular direction 4) at 45° to each other
KEY:3
68. A heaver body moving with certain velocity collides head on elastically with a lighter body at rest,
then
1) smaller body continues to be in the same state of rest
2) smaller body starts to move in the same direction with same velocity as that of bigger body
3) the smaller body start to move with twice the velocity of the bigger body in the same direction
4) the bigger body comes to rest
KEY:3
69. In which of the following, the work done by the mentioned force is negative? The work done by
1) the tension of the cable while the lift is ascending
2) the gravitational force when a body slides down an inclined place
3) the applied force to maintain uniform motion of a block on a rough horizontal surface
4) the gravitational force when a body is thrown up
KEY:4
70. Two bodies P and Q of masses m1 and m2 (m2>m1) are moving with velocity v1 and v2 respectively,
collided with each other. Then the force exerted by P on Q during the collision is
1) greater that the force exerted by Q on
2) less than the force exerted by Q on
3) same as the force exerted by Q on P
4) same as the force exerted by P on Q but opposite in direction
KEY:4
71. The coefficient of restitution (e) for a perfectly elastic collision is
1) 1 2) 0 3)  4)1
KEY:4
72. A ball of mass M moving with a velocity V collides head on elastically with another of same mass
but moving with a velocity v in the opposite direction. After collision.
1) the velocities are exchanged between the two balls
2) both the balls come to rest
3) both of them move at right angles to the original line of motion
4) one ball comes to rest and another ball travels back with velocity 2V
KEY:2
73. A small bob of a simple pendulum released from 30° to the vertical hits another bob of the same
mass and size lying at rest on the table vertically below the point of suspension. After elastic
collision, the angular amplitude of the bob will be
1) 30° 2) 60° 3) 15° 4) zero
KEY:4
74. Two spheres ‘X’ and ‘Y’ collide. After collision, the momentum of X is doubled . Then
1) the initial momentum of X and Y are equal
2) the initial momentum of X is greater then that of Y
3) the initial momentum of Y is double that of X
4) the loss in momentum of Y is equal to the initial momentum of X
KEY:4
75. A shell is fired into air at an angle  with the horizontal from the ground. On reaching the maximum
height,
1) its kinetic energy is not equal to zero 2) its kinetic energy is equal to zero
3) its potential energy is equal to zero 4) both its potential and kinetic energies are zero
KEY:1
76. A bullet is fired into a wooden block. If the bullet gets embedded in wooden block, then
1) momentum alone is conserved
2) kinetic energy alone is conserved
3) both momentum and kinetic energy are conserved
4) neither momentum nor kinetic energy are conserved
KEY:1
77. During collision, which of the following statements is wrong?
1) three is a change in momentum of individual bodies
2) the change in total momentum of the system of colliding particle is zero
3) the change in total energy is zero
4) they law of conservation of momentum is not valid
KEY:4
: : PRACTICE BITS : :
1. A man weighing 80 kg climbs a staircase carrying a 20 kg load. The staircase has 40 steps, each
of 25 cm height. If he takes 20 seconds to climb, the work done is
1) 9800 J 2) 490 J 3) 98 x 105 J 4) 7840 J
KEY:1
HINT:
 
W  F .S =FS= (M+m) g  n  heach step 
2. A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius
0.75m such that it has speed is 5ms–1 when the string is horizontal. Tension in string when it is
horizontal on other side is (g-10ms–2) [2007M]
1) 30N 2) 26N 3) 20N 4) 6N
KEY:3
HINT:
mvH2
T where vH  3 gr
r
3. A body of mass 10 kg moving with a velocity of 5ms-1 hits a body of 1 gm at rest. The velocity of
the second body after collision assuming it to be perfectly elastic is
1) 10 ms 1 2) 5 ms 1 3)15 ms 1 4) 0.10 ms 1
KEY:1
HINT:
 2m1 
v2   u1
 m1  m2 

4. A force F  (6i  8 j ) N , acts on a particle and displaces it over 4 m along the X-axis and 6 m along
the Y-axis. The total work done during the two displacements is
1) 72 J 2) 24 J 3) - 24 J 4) zero
KEY:3
HINT:
 ^  ^
W  Wx  Wy Wx  F .x i , Wy  F . y j
5. In the above problem the ratio of distance travelled in two consecutive rebounds
1) 1 : e 2) e : 1 3) 1 : e² 4) e² : 1
KEY:3
HINT:
hn e 2n h
 
6.    
An object has a displacement from position vector r1  2iˆ  3 ˆj m to r 2  4iˆ  6 ˆj m under a

 
force F  3x iˆ  2 yjˆ N , then work done by the force is
2
1) 24 J 2) 33 J 3) 83 J 4) 45 J
KEY:3
HINT:
x2 y2
W   dw  Fx dx   Fy dy
x1 y1
7. A body starts from rest and is acted on by a constant force. The ratio of kinetic energy gained by
it in the first five seconds to that gained in the next five second is
1) 2 : 1 2) 1 : 1 3) 3 : 1 4) 1 : 3
KEY:4
HINT:
1 1
KE  mv 2  m  gt   v  gt 
2
2 2
K1 t2
 2 1 2 where t1  5sec and t2  10 sec
K 2 t2  t1
8. A body starts from rest and moves with uniform acceleration. What is the ratio of kinetic energies
at the end of 1st, 2nd and 3rd seconds of its journey?
1) 1 : 8 : 27 2) 1 : 2 : 3 3) 1 : 4 : 9 4) 3 : 2 : 1
KEY:3
HINT:
1 2 1
KE  mv  mg 2t 2  v  gt 
2 2
9. A liquid of specific gravity 0.8 is flowing in a pipe line with a speed of 2 m/s. The K.E. per cubic
meter of it is
1) 160 J 2) 1600 J 3) 160.5 J 4) 1.6 J
KEY:2
HINT:
1 2 1 K .E 1 2
KE  mv   V  v 2 ;  v
2 2 V 2
10. A 60 kg boy lying on a surface of negligible friction throws horizontally a stone of mass 1 kg with
a speed of 12 m/s away from him. As a result with what kinetic energy he moves back?
1) 2.4 J 2) 72 J 3) 1.2 J 4) 36 J
KEY:3
HINT:
1
m1v1  m2 v2 ; KE  m2 v22
2
11. Two stones of masses m and 2 m are projected vertically upwards so as to reach the same height.
The ratio of the kinetic energies of their projection is
1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4
KEY:2
HINT:
1 2
KE  mv , When two bodies reach the same
2
KE1 m1

height, v1  v2 ; KE  m  v  2 gh
2 2

12. A tank of size 10 m  10 m  10 m is full of water and built on the ground. If g = 10 ms-2, the
potential energy of the water in the tank is
1) 5  107 J 2) 1  108 J 3) 5  104 J 4) 5  105 J
KEY:1
HINT:
h
P.E  mgh1 ; here h 1  and m    V
2
13. A spring when compressed by 4 cm has 2 J energy stored in it. The force required to extend it by
8 cm will be
1) 20 N 2) 2 N 3) 200 N 4) 2000 N
KEY:3
HINT:
1 2U
U  Kx12  K  2 and F  Kx2
2 x1
14. The elastic potential energy of a stretched spring is given by E = 50x 2 where x is the displacement
in meter and E is in joule, then the force constant of the spring is
1) 50 Nm 2) 100 N m-1 3) 100 N/m² 4) 100 Nm
KEY:2
HINT:
1
U  Kx 2   1 , U  50 x 2  (2), Compare equation (1) and (2) to find K
2
15. A body of mass 2 kg is projected with an initial velocity of 5 m s -1 along a rough horizontal table.
The work done on the body by the frictional forces before it is brought to rest is
1) 250 J 2) 2.5 J 3) -250 J 4) 25 J
KEY:4
HINT:
1 1 
W f  mv 2f  mvi2 
2 2 
16. A body of mass 20 gms is moving with a certain velocity. It collides with another body of mass 80 gm at
rest. The collision is perfectly inelastic. The ratio of the kinetic energies before and after collision of
the system is
1) 2 : 1 2) 4 : 1 3) 5 : 1 4) 3 : 2
KEY:3
HINT:
m1u1   m1  m2  v
1 1
KEi  m1u12 ; KE f   m1  m2  v 2
2 2
17. An object is acted on by a retarding force of 10 N and at a particular instant its kinetic energy is 6
J. The object will come to rest after it has travelled a distance of
1) 3/5 m 2) 5/3 m 3) 4 m 4) 16 m
KEY:1
HINT:
According to work energy theorem
1 1 
W f  mv 2f  mvi2 
2 2 
1
18. A bead of mass kg starts from rest from “A” to move in a vertical plane along a smooth fixed
2
quarter ring of radius 5m, under the action of a constant horizontal for F=5 N as shown. The
speed of bead as it reaches point “B” is
F A
R=5m
B
1) 14.14 m/s 2) 7.07 m/s 3) 5 m/s 4) 25 m/s
KEY:1
HINT:
Applying the work - energy theorem, we get
1
 mv 2  0  W1  W2
2
=Horizontal force  displacement + Vertical force  displacement.
= F  R + mg  R
 ^ ^ ^  ^ ^ ^
19. If F  2 i  3 j  4 k acts on a body and displaces it by S  3 i  2 j  5 k then the work done by the
force is
1) 12 J 2) 20 J 3) 32 J 4) 64 J
KEY:3
HINT:
 
W  F .S
20. A cradle is ‘h’ meters above the ground at the lowest position and ‘H’ meters when it is at the
highest point. If ‘v’ is the maximum speed of the swing of total mass ‘m’ the relation between ‘h’
and ‘H’ is
1) ½ mv2 + h = H 2) (v2/2g) + h = H 3) (v2/g) + 2h = H 4) (v2/2g) + H = h
KEY:2
HINT:
K. E at mean = P. E at extreme position
.
21. AB is a frictionless inclined surface making an angle of 300 with horizontal. A is 6.3 m above the ground
while B is 3.8 m above the ground. A block slides down from A, initially starting from rest. Its velocity on
reaching B is
A
6.3m
300 B
3.8 m
1) 7 m s- 1
2)14 m s-1 3) 7.4 m s-1 4) 4.9 m s-1
KEY:1
HINT:
Gain in K.E = Loss of P.E = mg (h1-h2)
22. A shot is fired at 30° with the vertical from a point on the ground with kinetic energy K. If air
resistance is ignored, the kinetic energy at the top of the trajectory is
1) 3K/4 2) K/2 3) K 4) K/4
KEY:4
HINT:
1
At maximum height, K  mu cos 
1 2 2
1
At projection, K  mu 2
2
23. A stone of mass “m” initially at rest and dropped from a height “h” strikes the surface of the earth with a
velocity “v”. If the gravitational force acting on the stone is W, then which of the following identities is
correct?
1) mv - mh = 0 2) ½ mv2 - Wh2 = 0 3) ½ mv2 - Wh = 0 4) ½ mv2 - mh = 0
KEY:3
HINT:
Gain in K.E = Loss of P.E
24. A ball is dropped on to a horizontal floor. It reaches a height of 144 cm on the first bounce and 81
cm on the second bounce. The height it attains on the third bounce is
1) 45.6 cm 2) 81 cm 3) 144 cm 4) 0 cm
KEY:1
HINT:
hn  e 2n h

25.  
A motor boat is going in a river with a velocity V  4iˆ  2 ˆj  kˆ ms . If the resisting force due to
1

 
stream is F  5iˆ  10  6kˆ N, then the power of the motor boat is
1) 100W 2) 50 W 3) 46W 4) 23W
KEY:3
HINT:
 
P  F .V
26. An 8 gm bullet is fired horizontally into a 9 kg block of wood and sticks in it. The block which is
free to move, has a velocity of 40 cm/s after impact. The initial velocity of the bullet is
1) 450 m/s 2) 450 cm/s 3) 220 m/s 4) 220 cm/s
KEY:1
HINT:
mu   m  M  v
27. Two riffles fire the same number of bullets in a given interval of time. The second fires bullets of
mass twice that fired by the first and with a velocity that is half that of the first. The ratio of their
power is
1) 1 : 4 2) 4 : 1 3) 1 : 2 4) 2 : 1
KEY:4
HINT:
P1  m1  v12 
   
P2  m2  v22 
28. A stationary body explodes into two fragments of masses m1 and m2. If momentum of one fragment
is P, the energy of explosion is.
p2 p2 p 2  m1  m2  p2
1) 2 m  m 2) 3) 4) 2 m  m
 1 2 2 m1m2 2m1m2  1 2
KEY:3
HINT:
p12 p2
E  E1  E2   22
2m1 2m
29. A force of 1200 N acting on a stone by means of a rope slides the stone through a distance of 10 m
in a direction inclined at 600 to the force. The work done by the force is
1) 6000 3J 2) 6000 J 3) 12000 J 4) 8000 J
KEY:2
HINT:
W=FS cos 
30. A crane can lift up 10,000 kg of coal in 1 hour from a mine of 180 m depth. If the efficiency of the
crane is 80 %, its input power must be (g = 10 m s-2)
1) 5 kW 2) 6.25 kW 3) 50 kW 4) 62.5 kW
KEY:2
HINT:
Pout W mgh
 , wherePout  
Pin t t
31. A man carries a load of 50 kg through a height of 40 m in 25 seconds. If the power of the man is
1568 W, his mass is
1) 5 kg 2) 1000 kg 3) 200 kg 4) 50 kg
KEY:4
HINT:
W  m  M  gh
P 
t t
32. A body dropped freely from a height h on to a horizontal plane, bounces up and down and finally
comes to rest. The coefficient of restitution is e. The ratio of velocities at the beginning and after
two rebounds is
1) 1 : e 2) e : 1 3) 1 : e² 4) e² : 1
KEY:3
HINT:
vn  e n v
33. An electric motor creates a tension of 4500 newton in a hoisting cable and reels it in at the rate of 2m/s.
What is the power of the motor?
1) 15 kW 2) 9 kW 3) 225 W 4) 9000 kW
KEY:2
HINT:
 
Pinst  F .V  FV cos 
34. The mass of a simple pendulum bob is 100 gm. The length of the pendulum is 1 m. The bob is
drawn aside from the equilibrium position so that the string makes an angle of 60° with the vertical
and let go. The kinetic energy of the bob while crossing its equilibrium position will be
1) 0.49 J 2) 0.94 J 3) 1 J 4) 1.2 J
KEY:1
HINT:
K .Emean  P.Eextreme  mg 1  cos  
35. A juggler throws continuously balls at the rate of three in each second each with a velocity of 10
m s-1. If the mass of each ball is 0.05 kg his power is
1) 2 W 2) 50 W 3) 0.5 W 4) 7.5 W
KEY:4
HINT:
1 
n  mv 2 
W  2 
P 
t t
36. A body of mass 2 kg attached at one end of light string is rotated along a vertical circle of radius 2m.
If he string can withstand a maximum tension of 140.6 N, the maximum speed with which the stone
can be rotated is
1) 22 m/s 2) 44 m/s 3) 33 m/s 4) 11 m/s
KEY:4
HINT:
 v2 
Tmax  m   g 
 r 

37. The work done by a force F  3iˆ  4 ˆj  5kˆ displaces the body from a point (3,4,6) to a point (7,2,5)
is
1) 15 units 2) 25 units 3) 20 units 4) 10 units
KEY:1
HINT:
    

W  F .S  F . r2  r1 
38. A lawn roller is pulled along a horizontal surface through a distance of 20 m by a rope with a force
of 200 N. If the rope makes an angle of 60° with the vertical while pulling, the amount of work
done by the pulling force is
1) 4000 J 2) 1000 J 3) 2000 3 J 4) 2000 J
KEY:3
HINT:
 
W  F .S  FS cos 
39. A pilot of mass m can bear a maximum apparent weight 7 times of mg. The aeroplane is moving in a
vertical circle. If the velocity of aeroplane is 210 m/s while diving up from the lowest point of vertical
circle, the minimum radius of vertical circle should be
1) 375 m 2) 420 m 3) 750 m 4) 840 m
KEY:3
HINT:
At lowest point of vertical circle,
mv 2
Tmax   mg
rmin
40. A simple pendulum is oscillating with an angular amplitude 60 0. If mass of bob is 50 gram, the
tension in the string at mean position is (g = 10ms–2)
1) 0.5 N 2) 1 N 3) 1.5 N 4) 2N
KEY:2
HINT:
mv 2 m
T  mg   mg   2 gl 1  cos  
r l
41. A body is moving in a vertical circle such that the velocities of body at different points are critical.
The ratio of velocities of body at angular displacements 60 0, 1200 from lowest point is
1) 5: 2 2) 3: 2 3) 3:1 4) 2 : 1
KEY:4
HINT:
v1 3  2 cos 1
v  gR  3  2 cos    
v2 3  2 cos  2
42. A 6 kg mass travelling at 2.5 ms 1 collides head on with a stationary 4 kg mass. After the collision
the 6 kg mass travels in its original direction with a speed of 1 ms 1 . The final velocity of 4 kg
mass is
1) 1 m/s 2) 2.25ms 1 3) 2 ms 1 4) 0 ms 1
KEY:2
HINT:
According to law of conservation of linear momentum, m1u1  m2u2  m1v1  m2 v2
43. A block of mass 1 kg moving with a speed of 4ms-1, collides with another block of mass 2 kg which
is at rest. The lighter block comes to rest after collision. The loss in the K.E of the system.
1) 8 J 2) 4  10 7 J 3) 4 J 4) 0 J
KEY:3
HINT:
m1u1  m2u2  m1v1  m2 v2

1 1
KE  m1u12  m2 v22
2 2
44. A bolt of mass 0.3kg falls from the ceiling of an elevator moving down with an uniform speed of
7m/s. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is
the heat produced by impact.
1)8.82J 2)7.72J 3)6.62J 4)5.52J
KEY:1
HINT:
Heat produced = loss of potential energy = mgh
45. A marble going at a speed of 2ms-1 hits another marble of equal mass at rest. If the collision is
perfectly elastic, thn the velocity of the first after collision is
1) 4 ms-1 2) 0ms-1 3) 2ms-1 4) 3ms-1
KEY:2
HINT:
  m  m    2m  
v1   1 2
 u1  
2
 u2
m
 1  m2  m
 1  m 2 
46. By applying the brakes without causing a skid, the driver of a car is able to stop his car in a
distance of 5 m, if it is going at 36 kmph. If the car were going at 72 kmph, using the same brakes,
he can stop the car over a distance of
1) 10 m 2) 2.5 m 3) 20 m 4) 40 m
KEY:3
HINT:
KE1 W1 FS1
W  KE ; 
KE2 W2 FS 2
47. A massive ball moving with a speed v collides head on with a fine ball having mass very much smaller than
the mass of the first ball at rest. The collision is elastic and then immediately after the impact, the second
ball will move with a speed approximately equal to
1) V 2) 2V 3) v/3 4) infinite
KEY:2
HINT:
m2  m1 and
  2m    m  m  
v2   1
 u1  
2 1
 u2
m
 1  m 2  m
 1  m2 
48. A 1 kg ball moving at 12 m/s collides head on with a 2 kg ball moving in the opposite direction at 24
m/s. The velocity of each ball after the impact, if the coefficient of restitution if 2/3 is
1) -28 m/s; -4 m/s 2) 28 m/s; -4 m/s 3) 20 m/s; 24 m/s 4) -20m/s; -4 m/s
KEY:1
HINT:
  m  em    m (1  e)  
v1   1 2
 u1  
2
 u2
 m1  m2   m1  m2 
  m (1  e)    m2  em1  
v2    1  u1   u2
 m1  m2   m1  m2 
49. A body of mass m moving at a constant velocity v hits another body of the same mass moving at
the same velocity v/s but in the opposite direction and sticks to it. The common velocity after
collision is
1) v 2) v/4 3) 2v 4) v/2
KEY:2
HINT:
m1u1  m2u2   m1  m2  v
50. A neutron, one of the constituents of a nucleus, is found to pass two points 60 meters apart in a
time interval of 1.8  10-4 sec. The mass of a neutron is 1.7  10-27 kg. Assuming that the speed is
constant, its kinetic energy is
1) 9.3  10-17 joule 2) 9.3  10-14 joule 3) 9.3  10-21 joule 4) 9.3  10-11 joule
KEY:1
HINT:
2
1 1 s
KE  mv 2  m  
2 2 t
51. A block of wood of mass 9.8 kg is suspended by a string. A bullet of mass 200 gm strikes horizontally
with a velocity of 100 ms 1 and gets imbedded in it. The maximum height attained by the block is
g  10 ms 
2
1) 0.1 m 2) 0.2 m 3) 0.3 m 4) 0

KEY:2
HINT:
v2
mu   m  M  v andh 
2g
52. A 15 gm bullet is fired horizontally into a 3 kg block of wood suspended by a string. The bullet sticks in the
block, and the impact causes the block to swing 10 cm above the initial level. The velocity of the bullet
nearly is (in ms-1)
1) 281 2) 326 3) 184 4) 58
KEY:1
HINT:
v2
mu   m  M  v and h 
2g
53. A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph.
If g = 10 m s-2, the power of its engine is
1) 4 kW 2) 50 kW 3) 625 kW 4) 25 kW
KEY:1
HINT:
 
P  F .v  Fv  mg sin  v
54. A rubber ball drops from a height h and after rebounding twice from the ground, it rises to h/2. The
co - efficient of restitution is
1/ 2 1/ 4 1/ 6
1 1 1 1
1) 2)   3)   4)  
2 2 2 2
KEY:3
HINT: hn  e 2n h
55. A bullet fired into a trunk of a tree loses 1/4 of its kinetic energy in traveling a distance of 5 cm.
Before stopping it travels a further distance of
1) 150 cm 2) 1.5 cm 3) 1.25 cm 4) 15 cm
KEY:4
HINT:
  S KE1
W  F .S  KE ; 1 
S 2 KE2
56. In the above problem, the ratio of times of two consecutive rebounds
1) 1 : e 2) e : 1 3) 1 : e² 4) e² : 1
KEY:1
HINT:
tn  e n t
57. A ball is dropped on to a horizontal floor. It reaches a height of 144 cm on the first bounce and 81
cm on the second bounce. The coefficient of restitution is
1) 0 2) 0.75 3) 81/144 4) 1
KEY:2
HINT:
h2
e
h1
58. A ball is dropped from height 'H' on to a horizontal surface. If the coefficient of restitution is 'e'
then the total time after which it comes to rest is
2H  1  e  2H  1  e  2H  1  e 2  2H  1  e 2 
1)   2)   3)   4)  
g 1 e  g 1 e  g  1  e 2  g  1  e 2 
KEY:2
HINT:
2h 2h1 2h2
.t   2    and hn  e 2 n h
g g g
59. The length of a ballistic pendulum is 1 m and mass of its block is 0.98 kg. A bullet of mass 20 gram
strikes the block along horizontal direction and gets embedded in the block. If block + bullet
completes vertical circle of radius 1m, the striking velocity of bullet is
1) 280 m/s 2) 350 m/s 3) 420 m/s 4) 490 m/s
KEY:2
HINT:
According to law of conservation of linear momentum mu = (M + m) v
u
 M  m 5 gr
m
60. A 6 kg mass collides with a body at rest. After the collision, they travel together with a velocity
one third the velocity of 6kg mass. The mass of the second body is
1) 6 kg 2) 3 kg 3) 12 kg 4) 18 kg
KEY:3
HINT:
m1u1  m2u2   m1  m2  v
JEE MAINS PREVIOUS QUESTIONS AND SOLUTIONS
WORK ,ENERGY AND POWER
Work :
1. A person pushes a box on a rough horizontal platform surface. He applies a force of 200
𝐍 over a distance of 15 𝐦. Thereafter, he gets progressively tired and his applied force
reduces linearly with distance to 100 N. The total distance through which the box has
been moved is 30 𝐦. What is the work done by the person during the total movement
ofthe box? [4 Sep. 2020 (II)]
(a) 3280 𝐉 (b) 2780 𝐉 (c) 5690 𝐉 (d) 5250 𝐉
SOLUTION : . (d)
The given situation can be drawn graphically as shown in figure.
Work done = Area under F‐x graph
= Area ofrectangle 𝑨𝑩𝑪𝑫 + Area oftrapezium BCFE
𝟏
𝑾 = 𝟐𝟎𝟎 × 𝟏𝟓 + 𝟏𝟎𝟎 + 𝟐𝟎𝟎 × 𝟏𝟓 = 𝟑𝟎𝟎𝟎 + 𝟐𝟐𝟓𝟎
𝟐
⇒ 𝑾 = 𝟓𝟐𝟓𝟎𝐉
2. A small block starts slipping down from a point 𝑩 on an inclined plane 𝑨𝑩, which is
making an angle 𝜽 with the horizontal section 𝑩𝑪 is smooth and the remaining section
𝑪𝑨 is rough with a coefficient offriction 𝝁. It is found that the block comes to rest as it
reaches the bottom (point 𝑨) ofthe inclinedplane. IfBC = 𝟐𝑨𝑪, the coefficient offriction
is given by 𝝁 = 𝒌 𝐭𝐚𝐧 𝜽. The value of 𝒌 is [NA 2 Sep. 2020 (I)]
SOLUTION : (3)
𝐈𝐟𝑨𝑪 = 𝒍 then according to question, 𝑩𝑪 = 𝟐𝒍 and 𝑨𝑩 = 𝟑𝒍.
′ 𝐬𝐢𝐧 𝜽
Here, work done by all the forces is zero. Wffiction +𝑾𝒎𝒈 = 𝟎
𝒎𝒈 𝟑𝒍 𝐬𝐢𝐧 𝜽 − 𝝁𝒎𝒈 𝐜𝐨𝐬 𝜽 𝒍 = 𝟎
⇒ 𝝁𝒎𝒈 𝐜𝐨𝐬 𝜽𝒍 = 𝟑𝒎𝒈𝒍 𝐬𝐢𝐧 𝜽
⇒ 𝝁 = 𝟑 𝐭𝐚𝐧 𝜽 = 𝒌 𝐭𝐚𝐧 𝜽
𝒌=𝟑
3. Consider a force 𝑭 = −xî+yj. The work done by this force in moving a particle from point
𝐀(1, 𝟎) to 𝐁 𝟎, 𝟏 along the line segment is: (all quantities are in SI units)
[9 Jan. 2020 I]
𝟏 𝟑
(a) 𝟐𝐉 (b) 𝐉 (c) 1J (d) 𝐉
𝟐 𝟐
SOLUTION : (c)
Work done, 𝑾 = 𝑭 ⋅ 𝒅𝒔
= −𝒙𝒊 + 𝒚𝒋 ⋅ 𝒅 × 𝒊 + 𝒅𝒚𝒋
𝟎 𝟏
⇒𝑾=− 𝒙 𝒅𝒙 + 𝒚 𝒅𝒚
𝟏 𝟎
𝟏 𝟏
= 𝟎+ + = 𝟏𝑱
𝟐 𝟐
4. A block of mass 𝐦 is kept on a platform which starts from rest with constant
acceleration 𝐠/𝟐 upward, as shown in fig. work done by normal reaction on block in
time 𝐭 is: [10 Jan. 2019 I]
𝐦𝐠 𝟐 𝐭 𝟐 𝐦𝐠 𝟐 𝐭 𝟐 𝟑𝐦𝐠 𝟐 𝐭 𝟐
(a) − (b) (c) 𝟎 (d)
𝟖 𝟖 𝟖
SOLUTION : . (d)
𝐦𝐠 𝟑𝐦𝐠
Here, 𝐍 − 𝐦𝐠 = 𝐦𝐚 = ⇒𝐍= 𝐍 = normal reaction
𝟐 𝟐
𝟑𝐦𝐠 𝟏
Now, work done by normal reaction ‘N’ on block in time 𝐭, 𝐖 = 𝐍𝐒 = 𝐠/𝟐 𝐭 𝟐
𝟐 𝟐
𝟑𝐦𝐠 𝟐 𝐭 𝟐
or, 𝐖 = 𝟖
5. A body of mass starts moving 𝐟𝐢𝐢𝐎 𝐦 rest along 𝐱‐axis so that its velocity varies as
𝐯 = 𝐚 𝐬 where a is a constant 𝐬 and is the distance covered by the body. The total work
done by all the forces acting on the body in the first second after the start ofthe motion
is: [Online Apri116, 2018]
𝟏 𝟏
(a) 𝟖 𝐦𝐚𝟒 𝐭 𝟐 (b) 𝟒𝐦𝐚𝟒 𝐭 𝟐 (c) 𝟖𝐦𝐚𝟒 𝐭 𝟐 (d) 𝟒 𝐦𝐚𝟒 𝐭 𝟐
SOLUTION : (a)
𝐝𝐬
From question, 𝐯 = 𝐚 𝐬 = 𝐝𝐭
𝐚𝟐 𝐭 𝟐
or, 𝟐 𝐬 = 𝐚𝐭 ⇒ 𝐒 = 𝟒
𝐚𝟐
𝜞=𝐦×
𝟐
𝐦𝐚𝟐 𝐚𝟐 𝐭 𝟐 𝟏
Work done = × = 𝟖 𝐦𝐚𝟒 𝐭 𝟐
𝟐 𝟒
6. When a rubber‐band is stretched by a distance 𝐱, it exerts restoring force

ofmaguitude 𝜞 = 𝐚𝐱 + 𝐛𝐱 𝟐 where a and 𝐛 are constants. The work done in stretching
the unstretched rubber‐band by 𝐋 is: [2014]
𝟏 𝐚𝐋𝟐 𝐛𝐋𝟑 𝟏 𝐚𝐋𝟐 𝐛𝐋𝟑
(a) 𝐚𝐋𝟐 + 𝐛𝐋𝟑 (b) 𝟐 𝐚𝐋𝟐 + 𝐛𝐋𝟑 (c) + (d) 𝟐 +
𝟐 𝟑 𝟐 𝟑
SOLUTION : . (c)
Work done in stretching the rubber‐band by a distance 𝒅𝒙 is
𝒅𝑾 = 𝑭𝒅𝒙 = 𝒂𝒙 + 𝒃𝒙𝟐 𝒅𝒙
Integrating both sides,
𝑳 𝑳
𝒂𝑳𝟐 𝒃𝑳𝟑
𝑾= 𝒂 𝒙𝒅𝒙 + 𝒃 𝒙𝟐 𝒅𝒙 = +
𝟎 𝟎 𝟐 𝟑
7. A uniform chain oflength 2 𝐦 is kept on a table such that a length of60 cm hangs freely
from the edge ofthe table. The total mass ofthe chain is 4 𝐤𝐠. What is the work done in
pulling the entire chain on the table? [2004]
(a) 12 𝐉 (b) 𝟑. 𝟔𝐉 (c) 𝟕. 𝟐𝐉 (d) 1200 𝐉
SOLUTION : . (b)
Mass of over hanging part of the chain
𝟒
𝒎′ = 𝟐 × 𝟎. 𝟔 𝐤𝐠 = 𝟏. 𝟐 kg
Weight ofhanging part ofthe chain

= 𝟏. 𝟐 × 𝟏𝟎 = 𝟏𝟐𝐍
C.M. ofhanging part = 𝟎. 𝟑𝐦 below the table
Workdone in putting the entire cha in on the table = 𝟏𝟐 × 𝟎. 𝟑𝟎 = 𝟑. 𝟔 J.
8 . A force 𝑭 = 𝟓𝒊 + 𝟑𝒋 + 𝟐𝒌 𝑵 is applied over a particle which displaces it from its origin
to the point 𝒓 = 𝟐𝒊 − 𝒋 𝒎. The work done on the particle injoules is [2004]
(a) +𝟏𝟎 (b) +𝟕 (c) −𝟕 (d) +𝟏𝟑
SOLUTION : (b)
Given, Force, 𝑭 =(5î+3j+2k )
Displacement, x=(2î‐j)
Work done,
𝑾 = 𝑭 ⋅ 𝒙 =(5î+3j+2k ). (2î‐j)
= 𝟏𝟎 − 𝟑 = 𝟕joules
9. Aspring ofspring constant 𝟓 × 𝟏𝟎𝟑 𝐍/𝐦 is stretched initially by 𝟓𝐜𝐦 from the
unstretched position. Then the work required to stretch it further by another 5 cm
[2003]
(a) 𝟏𝟐. 𝟓𝟎 N‐m (b) 𝟏𝟖. 𝟕𝟓 N‐m (c) 𝟐𝟓. 𝟎𝟎 N‐m (d) 𝟔. 𝟐𝟓 N‐m
SOLUTION : . (b)
Spring constant, 𝒌 = 𝟓 × 𝟏𝟎𝟑 𝐍/𝐦
Let 𝒙𝟏 and 𝒙𝟐 be the initial and fmal stretched position of the spring, then
𝟏
Work done, 𝑾 = 𝟐 𝒌 𝒙𝟐𝟐 − 𝒙𝟐𝟏
𝟏
= × 𝟓 × 𝟏𝟎𝟑 𝟎. 𝟏 𝟐
− 𝟎. 𝟎𝟓 𝟐
𝟐
𝟓𝟎𝟎𝟎
= × 𝟎. 𝟏𝟓 × 𝟎. 𝟎𝟓 = 𝟏𝟖. 𝟕𝟓 Nm
𝟐
10. Asprin 𝐠𝐨fforce constant 8𝟎𝟎𝐍/𝐦𝐡𝐚𝐬𝐚𝐧 extension of5 cm. The work done in extending
it 𝐟𝐢𝐢 𝐨𝐦𝟓 cm to 15 cm is [2002]
(a) 16 𝐉 (b) 𝟖𝐉 (c) 32 𝐉 (d) 24 𝐉
SOLUTION : (b)
Small amount ofwork done in extending the spring by 𝒅𝒙 is
𝒅𝑾 = 𝒌𝒙𝒅𝒙
𝟎.𝟏𝟓
𝑾=𝒌 𝒙 𝒅𝒙
𝟎.𝟎𝟓
𝟖𝟎𝟎
= 𝟎. 𝟏𝟓 𝟐 − 𝟎. 𝟎𝟓 𝟐
𝟐
= 𝟒𝟎𝟎 𝟎. 𝟏𝟓 + 𝟎. 𝟎𝟓 𝟎. 𝟏𝟓 − 𝟎. 𝟎𝟓
= 𝟒𝟎𝟎 × 𝟎. 𝟐 × 𝟎. 𝟏 = 𝟖𝐉
ENERGY
11. A cricket ball ofmass 𝟎. 𝟏𝟓 kg is thrown verticallyup by a bowling machine so that it
rises to a maximum height of20 𝐦 after leaving the machine. Ifthe part pushing the
ball applies a constant force 𝑭 on the ball and moves horizontally a distance 𝐨𝐟𝟎. 𝟐𝐦
while launching the ball, the value ofF (in N) is 𝒈 = 𝟏𝟎𝐦𝐬 −𝟐
[NA 3 Sep. 2020 (I)]
SOLUTION : . (𝟏𝟓𝟎. 𝟎𝟎)
From work energy theorem,
𝟏
𝑾 = 𝑭 ⋅ 𝒔 = 𝜟𝑲𝑬 = 𝒎𝒗𝟐
𝟐
Here 𝑽𝟐 = 𝟐𝒈𝒉
𝟐 𝟏 𝟏𝟓
𝑭⋅𝒔=𝑭× = × × 𝟐 × 𝟏𝟎 × 𝟐𝟎
𝟏𝟎 𝟐 𝟏𝟎𝟎
𝑭 = 𝟏𝟓𝟎 N.
12. A particle 𝒎 = 𝟏𝐤𝐠 slides down a fiictionless track (AOC) starting from rest at a
point 𝑨 (height 2 m). After reaching 𝑪, the particle continues to move fieely in air as a
projectile. When it reaching its highest point 𝑷 (height 1 m), the kinetic energy ofthe
particle (in 𝑱) is: (Figure drawn is schematic and not to scale; take 𝒈 = 𝟏𝟎ms 2)—.
[NA 7 Jan. 2020 I]
SOLUTION :
(𝟏𝟎. 𝟎𝟎) Kinetic energy = change in potential energy of the particle,
𝐊𝐄 = 𝐦𝐠𝜟𝐡
Given, 𝒎 = 𝟏𝐤𝐠,
𝜟𝒉 = 𝐡𝟐 − 𝐡𝟏 = 𝟐 − 𝟏 = 𝟏𝒎
𝑲𝑬 = 𝟏 × 𝟏𝟎 × 𝟏 = 𝟏𝟎𝑱
13. A particle moves in one dimension from rest under the influence ofa force that varies
with the distance travelled bythe particle as shown in the figure. The kinetic energy of
the particle after it has travelled 3 𝐦 is: [7 Jan. 2020 II]
Distance (in m) →
(a) 4 𝐉 (b) 𝟐. 𝟓𝐉 (c) 𝟔. 𝟓 𝐉 (d) 5 𝐉
SOLUTION : . (c)
We know area under F‐x graph gives the work done by the body
𝟏
𝑾= × 𝟑 + 𝟐 × 𝟑 − 𝟐 + 𝟐 × 𝟐 = 𝟐. 𝟓 + 𝟒 = 𝟔. 𝟓𝐉
𝟐
Using work energy theorem,
𝜟 K. 𝐄 = work done
𝜟 K.E = 𝟔. 𝟓𝐉
14. A spring whose unstretched length is 𝒍 has a force constant 𝒌. The spring is cut into
two pieces of unstretched lengths 𝟏𝟏 and 𝒍𝟐 where, 𝒍𝟏 = 𝒏𝒍𝟐 and 𝒏 is an integer. The
ratio 𝒌𝟏 𝒍𝒌𝟐 of the corresponding force constants, 𝒌𝟏 and 𝒌𝟐 will be:
[12 April 2019 II]
𝟏 𝟏
(a) 𝒏 (b) 𝒏𝟐 (c) 𝒏 (d) 𝒏𝟐
SOLUTION : (c)
𝒍𝟏 + 𝒍𝟐 = 𝒍 and 𝒍𝟏 = 𝒏𝒍𝟐
𝒏𝒍 𝒍
𝒍𝟏 = 𝒏+𝟏 and 𝒍𝟐 = 𝒏+𝟏
𝟏
As 𝒌 ∝ 𝒍 ,
𝒌𝟏 𝒍/ 𝒏 + 𝟏 𝟏
= =
𝒌𝟐 𝒏𝒍 / 𝒏 + 𝟏 𝒏
15. Abodyofmass 1 kg falls fieely from aheight of l𝟎𝟎𝐦, on a platform of mass 3 kg which is
mounted on a spring having spring constant 𝐤 = 𝟏. 𝟐𝟓 × 𝟏𝟎𝟔 𝐍/𝐦. The body sticks to the
platform and the spring’s maximum compression is found to be 𝐱. Given that
𝐠 = 𝟏𝟎𝐦𝐬 −𝟐 , the value ofx will beclose to: [11 April 2019 I]
(a) 40 cm (b) 4 cm (c) 80 cm (d) 8 cm
SOLUTION : (b)
Velocity of 1 kg blockjust before it collides with 3 kg block = 𝟐𝐠𝐡 = 𝟐𝟎𝟎𝟎𝐧𝒚𝐒
Using principle of conservation oflinear momentumjust before andjust after collision,
we get
𝟐𝟎𝟎𝟎
𝟏 × 𝟐𝟎𝟎𝟎 = 𝟒𝐯 ⇒ 𝐯 = 𝐦/𝐬 𝐯
𝟒
Initial compression ofspring

𝟏. 𝟐𝟓 × 𝟏𝟎𝟔 𝐱 𝟎 = 𝟑𝟎 ⇒ 𝐱 𝟎 ≈ 𝟎
using work energy theorem,
𝐖𝐠 + 𝐖𝐬𝐩 = 𝜟𝐊𝐄
𝟏
⇒ 𝟒𝟎 × 𝐱 + × 𝟏. 𝟐𝟓 × 𝟏𝟎𝟔 𝟎𝟐 − 𝐱 𝟐
𝟐
𝟏
× 𝟒 × 𝐯𝟐
=𝟎−
𝟐
solving 𝐱 ≈ 𝟒 cm
16. Auniform cable ofmass 𝐌’ and length 𝐋’ is placed on a horizontal surface such that
𝟏 𝐭𝐡
its part is hanging below the edge ofthe surface. To lift the hanging part of the
𝒏
cable upto the surface, the work done should be: [9 April 2019 I]
𝑴𝒈𝑳 𝑴𝒈𝑳 𝟐𝑴𝒈𝑳
(a) (b) (c) (d) nMgL
𝟐𝒏𝟐 𝒏𝟐 𝒏𝟐
SOLUTION : . (a)
𝐖 = 𝐮𝐟 − 𝐮𝒋
𝒎𝒈 𝑳 𝑴𝒈𝑳
=𝟎− − × = .
𝒏 𝟐𝒏 𝟐𝒏𝟐
17 . A wedge ofmass 𝐌 = 𝟒𝐦 lies on a fiictionless plane. 𝐀 particle of mass 𝒎 approaches

the wedge with speed 𝒗. There is no friction between the particle and the plane or
between the particle and the wedge. The maximum height climbed by the particle on
the wedge is given by: [9 April 2019 II]
𝒗𝟐 𝟐𝒗𝟐 𝟐𝒗𝟐 𝒗𝟐
(a) (b) (c) (d) 𝟐𝒈
𝒈 𝟕𝒈 𝟓𝒈
SOLUTION : (c)
𝒎𝒗 = 𝒎 + 𝑴 𝑽’
𝒎𝒗 𝒎𝒗 𝒗
or 𝒗 = 𝒎+𝑴 = 𝒎+𝟒𝒎 = 𝟓
Using conservation ofME, we have

𝟏 𝟏 𝒗 𝟐
𝒎𝒗𝟐 = 𝒎 + 𝟒𝒎 + 𝒎𝒈𝒉
𝟐 𝟐 𝟓
or 𝒉 = 𝟐𝒗 𝟐
𝟓𝒈
18. A particle moves in one dimension from rest under the influence ofa force that varies
with the distance travelled bythe particle as shown in the figure. The kinetic energy of
the particle after it has travelled 3 𝐦 is: [8 April 2019 I]
𝐃𝐢𝐬𝐭 𝑪 𝐦𝐜𝐞 (in 𝐦) →
(a) 4 𝐉 (b) 𝟐. 𝟓𝐉 (c) 𝟔. 𝟓𝐉 (d) 5 𝐉

SOLUTION : (c)
We know area under 𝜞 − 𝐱 graph gives the work done by the body
𝟏
𝑾= × 𝟑+𝟐 × 𝟑−𝟐 +𝟐×𝟐
𝟐
= 𝟐. 𝟓 + 𝟒
= 𝟔. 𝟓𝐉
Using work energy theorem,
𝜟 K.E = work done
𝜟 K.E = 𝟔. 𝟓𝐉
19. A particle which is experiencing a force, given by 𝜞 = 𝟑𝐢 − 𝟏𝟐𝐣, undergoes a
displacement of 𝐝 = 𝟒𝐢. If the particle had a kinetic energy of 3 𝐉 at the beginning ofthe
displacement, what is its kinetic energy at the endof the displacement?
[10 Jan. 2019 II]
(a) 9 𝐉 (b) 12 𝐉 (c) 10 𝐉 (d) 15 𝐉
SOLUTION : . (d)
Work done = 𝜞. 𝐝 = 𝟑𝐢 − 𝟏𝟐𝐉 . 𝟒𝐢 = 𝟏𝟐𝐉
From work energy theorem,
𝐰𝐧𝐞𝐭 = 𝜟𝐊. 𝐄. = 𝐤 𝐟 − 𝐤 𝐢
⇒ 𝟏𝟐 = 𝐤 𝐟 − 𝟑
𝐊 𝐟 = 𝟏𝟓𝐉
20. A block ofmass 𝐦, lying on a smooth horizontal surface, is attached to a spring (of
negligible mass) of spring constant 𝐤. The other end of the spring is fixed, as shown in
the figure. The block is initally at rest in its equilibrium position. Ifnow the block is
pulled with a constant force 𝜞, the maximum speed ofthe block is: [9 Jan. 2019 I]
𝟐𝜞 𝜞 𝝅𝜞 𝜞
(a) (b) 𝝅 (c) (d)
𝐦𝐤 𝐦𝐤 𝐦𝐤 𝐧𝐤
SOLUTION : . (d)
Maximum speed is at mean position or equilibrium At equilibrium Position
𝜞
𝜞 = 𝐤𝐱 ⇒ 𝐱 =
𝐤
From work‐energy theorem,
𝐖𝜞 + 𝐖𝐬𝐩 = 𝜟𝐊𝐄
𝟏 𝟏
𝜞 𝐱 − 𝐤𝐱 𝟐 = 𝐦𝐯 𝟐 − 𝟎
𝟐 𝟐
𝟐
𝜞 𝟏 𝜞 𝟏
𝜞 − 𝐤 = 𝐦𝐯 𝟐
𝐤 𝟐 𝐤 𝟐
𝟏 𝜞𝟐 𝟏
⇒ = 𝐦𝐯 𝟐
𝟐𝐊 𝟐
𝜞
or, 𝐯 𝐦𝐚𝐱 = 𝐦𝐤
21. A force acts on a 2 kg object so that its position is given as a function of time as
𝒙 = 𝟑𝐭 𝟐 + 𝟓. What is the work done by this force in first 5 seconds? [9 Jan. 2019 II]
(a) 850 𝐉 (b) 950 𝐉 (c) 875 𝐉 (d) 900 𝐉
SOLUTION : (d)
Position, 𝐱 = 𝟑𝐭 𝟐 + 𝟓
𝐝𝐱 𝐝 𝟑𝐭 𝟐 +𝟓
Velocity, 𝐯 = ⇒𝐯=
𝐝𝐭 𝐝𝐭
⇒ 𝐯 = 𝟔𝐭 + 𝟎
At 𝐭 = 𝟎 𝐯 = 𝟎
And, at 𝐭 = 𝟓 𝐬𝐞𝐜 𝐯 = 𝟑𝟎𝐦/𝐬
𝟏
𝟏
According to work‐energy theorem, 𝐰 = 𝜟𝐊𝐄 o𝐫𝐖=𝟐 mv2‐0 = 𝟐 𝟐 (30)2 = 𝟗𝟎𝟎𝐉
22 . Aparticle is moving in a circular path ofradius a under the action ofan attractive
𝒌
potential 𝑼 = − 𝟐𝒓𝟐 . Its total energyis: [2018]
𝒌 𝒌 𝟑 𝒌
a) − 𝟒𝐚𝟐 (b) 𝟐𝐚𝟐 (c) zero (d) − 𝟐 𝐚𝟐
SOLUTION : . (c)
𝝏𝐮 𝐊
𝜞=−𝐫 = 𝟑𝐫
𝝏𝐫 𝐫
Since particle is moving in circular path
𝐦𝐯 𝟐 𝐊 𝐊
𝜞= = 𝟑 ⇒ 𝐦𝐯 𝟐 = 𝟐
𝐫 𝐫 𝐫
𝟏 𝐊
K.E. = 𝐦𝐯 𝟐 =
𝟐 𝟐𝐫 𝟐
Total energy = 𝐏. 𝐄. +𝐊. 𝐄.

𝐊 𝐊 𝐊
= − 𝟐𝐫 𝟐 + 𝟐𝐫 𝟐 = Zero (⋅.⋅ P.E. = − 𝟐𝐫 𝟐 given)
23. Two particles ofthe same mass 𝐦 are moving in circular orbits because of force, given
by 𝜞 𝐫 = −𝐫 𝟑 −𝟏𝟔
𝐫 The first particle is at a distance = 𝟏 , and the second, at 𝐫 = 𝟒. The best estimate for
the ratio of kinetic energies of the first and the second particle is closest to
(a) 𝟏𝟎−𝟏 (b) 𝟔 × 𝟏𝟎−𝟐 (c) 𝟔 × 𝟏𝟎𝟐 (d) 𝟑 × 𝟏𝟎−𝟑
SOLUTION : . (b)
As the particles moving in circular orbits, So
𝐦𝐯 𝟐 16 2
−= − + 𝐫
𝐫𝐫
𝟏 𝟏
Kinetic energy, 𝐊𝐄𝟎 = 𝟐 𝐦𝐯 𝟐 = 𝟐 𝟏𝟔 + 𝐫 𝟒
𝟏
For first particle, = 𝟏 , 𝐊 𝟏 = 𝟐 𝐦 𝟏𝟔 + 𝟏
𝟏
Similarly, for second particle, 𝐫 = 𝟒, 𝐊 𝟐 = 𝟐 𝐦 𝟏𝟔 + 𝟐𝟓𝟔
𝐊𝟏 𝟏𝟔 + 𝟐𝟓𝟔 𝟏𝟔 + 𝟏 𝟏𝟕
= = = 𝟔 × 𝟏𝟎−𝟐
𝐊𝟐 𝟐 𝟐 𝟐𝟕𝟐
24. A body ofmass 𝐦 = 𝟏𝟎−𝟐 kg is moving in a medium and experiences a frictional forc
𝟏
𝜞 = −𝐤𝐯 𝟐 . Its intial speed is 𝐯𝟎 = 𝟏𝟎𝐦𝐬−𝟏 . If, after 10 𝐬, its energyis 𝟖 𝒎𝐯𝟎𝟐 , the value
ofk will be : [2017]

(a) 𝟏𝟎−𝟒 kg 𝐦−𝟏 (b) 𝟏𝟎−𝟏 kg 𝐦−𝟏 𝐬 −𝟏
(c) 𝟏𝟎−𝟑 𝐤𝐠𝐦−𝟏 (d) 𝟏𝟎−𝟑 kg 𝐬 −𝟏
SOLUTION : (a)
Let 𝑽𝒇 is the fmal speed ofthe body.
From questions,
𝟏 𝟏 𝑽𝟎
𝒎𝑽𝟐𝒇 = 𝟖 𝒎𝑽𝟐𝟎 ⇒ 𝑽𝒇 = = 𝟓 mls
𝟐 𝟐
𝒅𝑽 𝒅𝑽
𝑭=𝒎 = −𝒌𝑽𝟐 𝟏𝟎−𝟐 = −𝒌𝑽𝟐
𝒅𝒕 𝒅𝒕
𝟓 𝟏𝟎
𝒅𝑽
𝟐
= −𝟏𝟎𝟎𝑲 𝒅𝒕
𝟏𝟎 𝑽 𝟎
𝟏 𝟏
− 𝟏𝟎 = 𝟏𝟎𝟎𝑲 𝟏𝟎 or, 𝑲 = 𝟏𝟎−𝟒 𝒌𝒈𝒎−𝟏
𝟓
25. An object is dropped 𝐟𝐢𝐢 𝐨𝐦 a height 𝐡𝐟𝐢𝐢𝐎 𝐦 the ground. Every time it hits the ground it
looses 50% of its kinetic energy. The total distance covered as 𝐭 → ∞ is
𝟓 𝟖
(a) 3 𝐡 (b) ∞ (c) 𝟑 𝐡 (d) 𝟑 𝐡
SOLUTION : . (a)
(K.E.)’ = 𝟓𝟎% ofK.E. after hit i.e.,
𝟏 𝟓𝟎 𝟏 𝐯
𝐦𝐯 ′𝟐 = × 𝐦𝐯 𝟐 ⇒ 𝐯 ′ =
𝟐 𝟏𝟎𝟎 𝟐 𝟐
𝟏
Coefficient ofrestitution = 𝟐
Now, total distance travelled by object is ( 1)
𝟏 + −𝟐
𝐇=𝐡 =𝐡 = 𝟑𝐡
𝟏
𝟏−𝟐
26. A time dependent force 𝜞 = 𝟔𝐭 acts on a particle ofmass 1 𝐤𝐠. Ifthe particle starts 𝐟𝐢: 𝐨𝐦
rest, the work done by the force during the first 1 second will be [2017]
(a) 9 𝐉 (b) 18 𝐉 (c) 𝟒. 𝟓𝐉 (d) 22 𝐉
SOLUTION : . (c)
𝒅𝑽
Using, 𝜞 = 𝐦𝐚 = 𝒎 𝒅𝒕
𝒅𝑽
𝟔𝒕 = 𝟏. 𝒅𝒕 [𝒎 = 𝟏 kg given]
𝒗 𝟏
𝒕𝟐
𝒅𝑽 = 𝟔 𝒕𝒅𝒕𝑽 = 𝟔 = 𝟑𝐦𝐬 −𝟏
𝟎 𝟐 𝟎
[𝒕 = 𝟏 𝐬𝐞𝐜 given] From work‐energy theorem,

𝟏 𝟏
𝐖 = 𝜟𝐊𝐄 = 𝒎 𝑽𝟐 − 𝒖 𝟐 = × 𝟏 × 𝟗
𝟐 𝟐
= 𝟒. 𝟓𝐉
27. Velocity‐time graph for a body ofmass 10kg is shown in figure. Work‐done on the
body in first two seconds ofthe motion is: [Online Apri110, 2016]
(a) – 𝟗𝟑𝟎𝟎𝐉 (b) 12000 𝐉 (c) – 𝟒𝟓𝟎𝟎𝐉 (d) – 𝟏𝟐𝟎𝟎𝟎𝐉

SOLUTION : . (c)
𝐯−𝟒 𝟎−𝟓𝟎
Acceleration (a) = = = −𝟓𝐦/𝐬𝟐
𝐭 𝟏𝟎−𝟎
𝐮 = 𝟓𝟎𝐦/𝐬
𝐯 = 𝐮 + 𝐚𝐭 = 𝟓𝟎 − 𝟓𝐭
Veocity in first two seconds 𝐭 = 𝟐
𝐯 𝐚𝐭𝐭=𝟐 = 𝟒𝟎𝐦/𝐬
From work‐energy theorem,
𝟏
△ 𝐊. 𝐄. = 𝐖 = 𝟒𝟎𝟐 − 𝟓𝟎𝟐 × 𝟏𝟎 = −𝟒𝟓𝟎𝟎𝐉
𝟐
28. A point particle of mass 𝐦, moves long the uniformly rough track PQR as shown in the
figure. The coefficient of friction, between the particle and the rough track equals 𝝁.
The particle is released, 𝐟𝐢𝐢𝐎 𝐦 rest 𝐟𝐢𝐢 𝐨𝐦 the point 𝐏 and it comes to rest at a point R.
The energies, lost by the ball, over the parts, PQ and QR, ofthe track, are equal to each
other, and no energy is lost when particle changes direction from PQ to QR.
The value ofthe coefficient of friction 𝝁 and the distance = 𝐐𝐑 , are, respectively close
to: [2016]
𝐇𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 → 𝐐
Surface
(a) 𝟎. 𝟐𝟗𝐚𝐧𝐝𝟑. 𝟓𝐦 (b) 𝟎. 𝟐𝟗𝐚𝐧𝐝𝟔. 𝟓𝐦
(c) 𝟎. 𝟐 and 𝟔. 𝟓𝐦 (d) 𝟎. 𝟐 and 𝟑. 𝟓𝐦
SOLUTION : (a)
Work done by fiiiction at 𝐐𝐑 = 𝝁𝐦𝐠𝒙
𝟏 𝟐
In triangle, 𝐬𝐢𝐧 𝟑𝟎∘ = 𝟐 = 𝑷𝑸
⇒ 𝑷𝑸 = 𝟒𝒎
Work done by friction at 𝐏𝐐 = 𝝁𝐦𝐠 × Cos 𝟑𝟎∘ × 𝟒
𝟑
= 𝝁𝐦𝐠 ×
× 𝟒 = 𝟐 𝟑𝝁𝐦𝐠
𝟐
Since work done by fiiction on parts 𝑷𝑸 and 𝑸𝑹 are equal, 𝝁𝐦𝐠𝒙=𝟐 𝟑 umg
⇒ 𝑿=𝟐 𝟑 ≅ 𝟑. 𝟓𝒎
Usingwork energytheorem mg 𝐬i𝐧𝟑𝟎∘ × 𝟒 = 𝟐 𝟑𝝁𝐦𝐠 + 𝝁𝐦𝓧
⇒ 𝟐 = 𝟒 𝟑𝝁
⇒ 𝝁 = 𝟎. 𝟐𝟗
29. A person trying to lose weight byburning fat lifts a mass of10 kg upto a height ofl 𝐦
1000 times. Assume that the potential energy lost each time he lowers the mass is
dissipated. How much fat will he use up considering the work done only when the
weight is lifted up? Fat supplies 𝟑. 𝟖 × 𝟏𝟎𝟕 𝐉 of energy per kg which is converted to
mechanical energy with a 20% efficiency rate. Take 𝐠 = 𝟗. 𝟖𝐦𝐬−𝟐 : [2016]
(a) 𝟗. 𝟖𝟗 × 𝟏𝟎 kg
−𝟑
(b) 𝟏𝟐. 𝟖𝟗 × 𝟏𝟎 kg
−𝟑
(c) 𝟐. 𝟒𝟓 × 𝟏𝟎−𝟑 kg (d) 𝟔. 𝟒𝟓 × 𝟏𝟎−𝟑 kg

SOLUTION : (b)
𝐖 𝐦𝐠𝐡×𝟏𝟎𝟎𝟎 𝟏𝟎×𝟗.𝟖×𝟏×𝟏𝟎𝟎𝟎 𝟗𝟖𝟎𝟎𝟎
𝐧 = 𝐢𝐧𝐩𝐮𝐭 = = Input = = 𝟒𝟗 × 𝟏𝟎𝟒 𝐉
𝐢𝐧𝐩𝐮𝐭 𝐢𝐧𝐩𝐮𝐭 𝟎.𝟐
𝟒𝟗×𝟏𝟎𝟒
Fat used = 𝟑.𝟖×𝟏𝟎𝟕 = 𝟏𝟐. 𝟖𝟗 × 𝟏𝟎−𝟑 𝐤𝐠.
30. A particle is moving in a circle ofradius 𝐫 under the action of a force 𝜞 = 𝜶𝐫 𝟐 which is
directed towards centre of the circle. Total mechanical energy (kinetic energy +
potential energy) ofthe particle is (take 𝝁)tential energy = 𝟎 for 𝐫 = 𝟎):
𝟏 𝟓 𝟒
(a) 𝟐 𝜶𝐫 𝟑 (b) 𝟔 𝜶𝐫 𝟑 (c) 𝟑 𝜶𝐫 𝟑 (d) ff
SOLUTION : (b)
As we know, 𝐝𝐔 = 𝜞. 𝐝𝐫
𝒓 𝒂𝒓𝟑
𝑼= 𝟎
𝜶 𝒓𝟐 𝒅𝒓 = (i) As, 𝒎𝒗𝟐 =𝜶𝒓𝟐
𝟑
𝒓
𝒎 𝒗 = 𝒎𝜶𝒓𝟑
𝟐 𝟐
𝟏
or, 𝟐𝐦 𝐊𝐄 = 𝟐 𝜶𝒓𝟑 (ii)
Total energy = Potential energy + kinetic energy Now, 𝐟𝐢𝐢 𝐨𝐦 eqn (i) and (ii)
Total energy = 𝐊. 𝐄. +𝐏. 𝐄.
𝜶𝒓𝟑 𝜶𝒓𝟑 𝟓 𝟑
= + = 𝜶𝒓
𝟑 𝟐 𝟔
31. A block ofmass 𝒎 = 𝟎. 𝟏 kg is connected to a spring of unknown spring constant 𝐤. It is
compressed to a distance 𝒙 from its equilibrium position and released from rest. After
𝐱
approaching half the distance from equilibrium position, it hits another block and
𝟐
comes to rest momentarily, while the other block moves with a velocity 𝟑𝐦𝐬 −𝟏 .
The total initial energy ofthe spring is: [Online April 10, 2015]
(a) 𝟎. 𝟑 𝐉 (b) 𝟎. 𝟔 𝐉 (c) 𝟎. 𝟖𝐉 (d) 𝟏. 𝟓𝐉
SOLUTION : . (b)
Applying momentum conservation
𝒎𝟏 𝒖𝟏 + 𝒎𝟐 𝒖𝟐 = 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐
𝟎. 𝟏𝒖 + 𝒎 𝟎 = 𝟎. 𝟏 𝟎 + 𝒎 𝟑
𝟎. 𝟏𝒖 = 𝟑𝒎
𝟏 𝟏
𝟎. 𝟏𝒖𝟐 = 𝒎 𝟑 𝟐
𝟐 𝟐
Solving we get, 𝒖 = 𝟑
𝟏 𝟏 𝒙 𝟐 𝟏
𝒌𝒙𝟐 = 𝟐 𝑲 + 𝟐(𝟎. 𝟏)32
𝟐 𝟐
𝟑
⇒ 𝟒 𝒌𝒙𝟐 = 𝟎. 𝟗
𝟑 𝟏
⇒ × 𝒌𝒙𝟐 = 𝟎. 𝟗
𝟐 𝟐
𝟏
𝑲𝜿𝟐 = 𝟎. 𝟔𝐉 (total initial energy of the spring)
𝟐
𝟏 𝐭𝐡
32. A bullet looses of its velocity passing through one plank. The number of such
𝐧
planks that are required to stop the bullet can be: [Online Apri119, 2014]
𝐧𝟐 𝟐𝐧𝟐
(a) 𝟐𝐧−𝟏 (b) 𝐧−𝟏 (c) infinite (d) 𝐧
SOLUTION : (a)
Let 𝐮 be the initial velocity ofthe bullet ofmass 𝐦. After passing through a plank of width
𝐱, its velocity decreases to 𝐯.
𝟒 𝟒 𝐮 𝐧−𝟏
𝐮 − 𝐯 = 𝐧 𝐨𝐫, 𝐯 = 𝐮 − 𝐧 = 𝐧
If 𝜞 be the retarding force applied by each plank, then using work‐ energy theorem,
𝟐
𝟏 𝟏 𝟏 𝟏 𝐧−𝟏
𝜞𝐱 = 𝐦𝐮𝟐 − 𝐦𝐯 𝟐 = 𝐦𝐮𝟐 − 𝐦𝐮𝟐
𝟐 𝟐 𝟐 𝟐 𝐧𝟐
𝟏 𝟏− 𝐧−𝟏 𝟐
= 𝐦𝐮𝟐
𝟐 𝐧𝟐
𝟏 𝟐𝐧 − 𝟏
𝜞𝐱 = 𝐦𝐮𝟐
𝟐 𝐧𝟐
Let 𝐏 be the number of planks required to stop the bullet.
Total distance travelled by the bullet before coming to rest = 𝐏𝐱
Using work‐energy theorem again,
𝟏
𝜞 𝐏𝐱 = 𝟐 mu2‐0
𝟏 𝟐𝐧−𝟏 𝟏
or, 𝐏 𝜞𝐱 = 𝐏 𝐦𝐮𝟐 = 𝟐 𝐦𝐮𝟐
𝟐 𝐧𝟐
𝐧𝟐
𝐏=
𝟐𝐧 − 𝟏
33. A spring ofunstretched length 1 has a mass 𝐦 with one end fixed to a rigid support.
Assuming spring to be made ofa uniform wire, the kinetic energypossessed by it ifits
free end is pulled with uniform velocity 𝐯 is: [Online Apri112, 2014]
𝟏 𝟏 𝟏
(a) 𝟐 𝐦𝐯 𝟐 (b) 𝐦𝐯 𝟐 (c) 𝟑 𝐦𝐯 𝟐 (d) 𝟔 𝐦𝐯 𝟐
SOLUTION : . (d)
34. Two springs of force constants 300 𝐍/𝐦 (Spring A) and 400 𝐍/𝐦(Spring B) arejoined
together in series. The combination is compressed by 𝟖. 𝟕𝟓 cm. The ratio of energy
𝑬 𝑬
stored in A and 𝐁 is 𝑬𝑨 . Then 𝑬𝑨 is equalto: [Online April 9, 2014]
𝑩 𝑩
𝟒 𝟏𝟔 𝟑 𝟗
(a) 𝟑 (b) (c) 𝟒 (d) 𝟏𝟔
𝟗
SOLUTION : . (a)
Given : 𝐤 𝐀 = 𝟑𝟎𝟎𝐍/𝐦, 𝐤 𝐁 = 𝟒𝟎𝟎𝐍/𝐦
Let when the combination of springs is compressed by force F. Spring A is compressed by 𝐱.
Therefore compression in spring 𝐁
𝐱 𝐁 = 𝟖. 𝟕𝟓 − 𝐱 cm
𝜞 = 𝟑𝟎𝟎 × 𝐱 = 𝟒𝟎𝟎 𝟖. 𝟕𝟓 − 𝐱
Solving we get, 𝐱 = 𝟓 cm
𝐱 𝐁 = 𝟖. 𝟕𝟓 − 𝟓 = 𝟑. 𝟕𝟓𝐜𝐦
𝟏 𝟐
𝐄𝐀 𝟐 𝐤 𝐀 𝐱 𝐀 𝟑𝟎𝟎 × . 𝟓 𝟐 𝟒
= = =
𝐄𝐁 𝟏 𝐤 𝐱 𝟐 𝟒𝟎𝟎 × 𝟑𝟕𝟓 𝟐 𝟑
𝟐 𝐁 𝐁
35. The force 𝑭 = 𝑭î on a particle ofmass 2 𝐤𝐠, moving along the 𝒙‐axis is given in the
figure as a function ofits position 𝒙. The particle is moving with a velocity of5 𝐦/𝐬 along
the 𝒙‐axis at 𝒙 = 𝟎. What is the kinetic energy ofthe particle at𝒙 = 𝟖𝐦?
𝟎 𝟏𝟐𝟑𝟒𝟓𝟔𝟕𝟖
(a) 34 𝐉 (b) 𝟑𝟒. 𝟓𝐉 (c) 𝟒. 𝟓𝐉 (d) 𝟐𝟗. 𝟒𝐉
SOLUTION : (d)
36. A particle gets displaced by

𝜟𝒓 = 𝟐𝒊 + 𝟑𝒋 + 𝟒𝒌 𝐦 under the action of a force 𝑭 = 𝟕𝒊 + 𝟒𝒋 + 𝟑𝒌 . The change in its
kinetic energy is [Online May 7, 2012]
(a) 38 𝐉 (b) 70 𝐉 (c) 𝟓𝟐. 𝟓𝐉 (d) 126 𝐉
SOLUTION : (a)
According to work‐energy theorem,
Change in kinetic energy = work done =→ 𝑭. 𝜟 → 𝒓 =(7î+4j+3k ).(2î+3j+4k )
= 𝟏𝟒 + 𝟏𝟐 + 𝟏𝟐 = 𝟑𝟖𝐉
37. At time 𝒕 = 𝟎 a particle starts moving along the 𝒙‐axis. If its kinetic energy increases
uniformly with time ‘t’, the net force acting on it must be proportional to [2011 𝐑𝐒]
𝟏
(a) constant (b) 𝒕 (c) (d) 𝒕
𝒕
SOLUTION : (c)
K.E. ∝ 𝒕
K.E. = 𝒄𝒕 [Here, 𝒄 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭]
𝟏
⇒ 𝟐 𝒎𝒗𝟐 = 𝒄𝒕
𝒎𝒗 𝟐
⇒ = 𝒄𝒕
𝟐𝒎
𝒑𝟐
⇒ 𝟐𝒎 = 𝒄𝒕 ⋅.⋅ 𝒑 = 𝒎𝒗
⇒ 𝒑 = 𝟐𝐜𝐭𝐦
𝒅𝒑 𝒅 𝟐𝒄𝒕𝒎
⇒𝑭= =
𝒅𝒕 𝒅𝒕
𝟏
⇒ 𝑭 = 𝟐𝐜𝐦 × 𝟐 𝒕
𝟏
⇒𝑭∝ 𝒕
38. The potential energy function for the force between two atoms in a diatomic molecule
𝒂 𝒃
is approximately given by 𝑼 𝒙 = − , where a and 𝐛 are constants and 𝐱 is the
𝒙𝟏𝟐 𝒙𝟔
distance between the atoms. Ifthe dissociation energy of the molecule is

𝑫 =[𝑼 𝒙 = ∞ − 𝑼𝐚𝐭 equilibrium] , 𝑫 is [2010]
𝒃𝟐 𝒃𝟐 𝒃𝟐 𝒃𝟐
(a) 𝟐𝒂 (b) 𝟏𝟐𝒂 (c) 𝟒𝒂 (d) 𝟔𝒂
SOLUTION : . (d)
−𝒅𝑼 𝒙
At equilibrium: 𝑭 = 𝒅𝒙
−𝒅 𝒂 𝒃
⇒𝑭= − 𝒙𝟔
𝒅𝒙 𝒙𝟏𝟐
𝟏𝟐𝒂 𝟔𝒃
⇒𝑭=− + 𝒙𝟕
𝒙𝟏𝟑
𝟏
𝟏𝟐𝒂 𝟔𝒃 𝟐𝒂 𝟔
⇒ 𝒙𝟏𝟑 = ⇒𝒙=
𝒙𝟕 𝒃
𝒂 𝒃 𝒃𝟐
𝑼𝐚𝐭 equilibrium = 𝟐𝒂 𝟐
− 𝟐𝒂 = − 𝟒𝒂 and 𝑼 𝒙=∞ = 𝟎
𝒃 𝒃
𝒃𝟐 𝒃𝟐
𝑫=𝟎− − =
𝟒𝒂 𝟒𝒂
39. An athlete in the olympic games covers a distance of100 𝐦 in 10 𝐬. His kinetic energy
can be estimated to be in therange [2008]
(a) 200 J‐ 𝟓𝟎𝟎𝐉 (b) 𝟐 × 𝟏𝟎𝟓 𝐉 − 𝟑 × 𝟏𝟎𝟓 𝐉
(c) 20, 000 J‐50,000 𝐉 (d) 2,000 J‐5, 000 𝐉
SOLUTION : . (d)
The average speed ofthe athelete
𝟓 𝟏𝟎𝟎
𝒗= = = 𝟏𝟎𝐦/𝐬
𝒕 𝟏𝟎
𝟏
𝒎𝒗𝟐
𝐊. 𝐄. =
𝟐
Assuming the mass of athelet to 40 kg his average K.E would be
𝟏
× 𝟒𝟎 × 𝟏𝟎 𝟐 = 𝟐𝟎𝟎𝟎𝐉
𝐊. 𝐄 =
𝟐
Assuming mass to 100 kg average kinetic energy
𝟏
K.E. = 𝟐 × 𝟏𝟎𝟎 × 𝟏𝟎 𝟐
= 𝟓𝟎𝟎𝟎𝐉
40. A2 kg block slides on ahorizontal floor with a speed 𝐨𝐟𝟒𝐦/𝐬. It strikes a uncompressed
spring, and compresses it till the block is motionless. The kinetic fiiiction force is 𝟏𝟓𝐍
and spring constant is 10,000 𝐍/𝐦. The spring compresses by [2007]
(a) 𝟖. 𝟓 cm (b) 𝟓. 𝟓 cm (c) 𝟐. 𝟓 cm (d) 𝟏𝟏. 𝟎 cm
SOLUTION : . (b)
Suppose the spring gets compressed by 𝒙 before stopping.
kinetic energy of the block = 𝐏. 𝐄. stored in the spring + work done against fiiction.
𝟏 𝟐
𝟏
⇒ ×𝟐× 𝟒 = × 𝟏𝟎, 𝟎𝟎𝟎 × 𝒙𝟐 + 𝟏𝟓 × 𝒙
𝟐 𝟐
⇒ 𝟏𝟎, 𝟎𝟎𝟎𝒙𝟐 + 𝟑𝟎𝒙 − 𝟑𝟐 = 𝟎
⇒ 𝟓𝟎𝟎𝟎𝒙𝟐 + 𝟏𝟓𝒙 − 𝟏𝟔 = 𝟎
−𝟏𝟓 ± 𝟏𝟓
− 𝟒 × 𝟓𝟎𝟎𝟎 −𝟏𝟔𝟐
𝒙=
𝟐 × 𝟓𝟎𝟎𝟎
= 𝟎. 𝟎𝟓𝟓𝐦 = 𝟓. 𝟓𝐜𝐦.
41. Aparticle 𝐢𝐬𝐩rojected a𝐭𝟔𝟎𝐨 𝐭𝐨 the horizontal with akinetic energy 𝑲. The kinetic energy
at the highest point is
(a) 𝑲 l2 (b) 𝑲 [2007] (c) Zero (d) 𝑲 l4
SOLUTION : (d)
Let 𝐮 be the velocitywith which the particle is thrown and 𝐦 be the mass ofthe particle.
Then
𝟏
𝑲 = 𝟐 𝒎𝒖𝟐 ... (1)
At the highest point the velocity is 𝐮 𝐜𝐨𝐬 𝟔𝟎∘ (only the horizontal component remains, the
vertical component being zero at the top‐most point). Therefore kinetic energy at the
highest point.
𝟏 𝟏 𝑲
𝑲‘ = 𝟐 𝒎 𝒖 𝐜𝐨𝐬 𝟔𝟎∘ 𝟐
= 𝟐 𝒎𝒖𝟐 𝐜𝐨𝐬𝟐 𝟔𝟎∘ = [From 1]
𝟒
42. A particle ofmass 𝟏𝟎𝟎𝐠 is thrown verticallyupwards with a speed of 5 𝐦/𝐬. The work
done by the force ofgravity during the time the particle goes up is [2006]
(a) −𝟎. 𝟓𝐉 (b) −𝟏. 𝟐𝟓𝐉 (c) 𝟏. 𝟐𝟓𝐉 (d) 𝟎. 𝟓𝐉
SOLUTION : . (b)
Given, Mass ofthe particle, 𝒎 = 𝟏𝟎𝟎𝐠 Initial speed ofthe particle, 𝝁 = 𝟓𝐦/𝐬
Final speed of the particle, 𝒗 = 𝟎
Work done by the force of gravity
= Loss in kinetic energy ofthe body.
𝟏 𝟏 𝟏𝟎𝟎 𝟐
= 𝒎 𝒗𝟐 − 𝒖 𝟐 = × 𝟎 − 𝟓𝟐
𝟐 𝟐 𝟏𝟎𝟎𝟎
= −𝟏. 𝟐𝟓𝐉
43. The potential energyofa 1 kg particle free to move along

(𝒙𝟒 𝒙𝟐)
the 𝐱‐axis is given by 𝑽 𝒙 = 𝟒 − 𝟐 𝑱. The total mechanical energy ofthe particle is
2 J. Then, the maximum speed (in 𝐦/𝐬) is [2006]
𝟑 𝟏
(a) (b) 𝟐 (c) (d) 2
𝟐 𝟐
SOLUTION : . (a)
Potential energy
𝒙𝟒 −𝒙𝟐
𝑽 𝒙 = joule
𝟒𝟐
Formaxima ofminima
𝒅𝑽
= 𝟎 ⇒ 𝒙𝟑 − 𝒙 = 𝟎 ⇒ 𝒙 = ±𝟏
𝒅𝒙
𝟏 𝟏 𝟏
⇒ 𝐌𝐢𝐧 . P.E. = 𝟒 − 𝟐 = − 𝟒 𝐉
𝐊. 𝐄 𝐦𝐚𝐱 . + 𝐏. 𝐄 𝐦𝐢𝐧 . = 𝟐 (Given)

𝟏 𝟗
𝐊. 𝐄 𝐦𝐚𝐱 . =𝟐+ =
𝟒 𝟒
𝟏
K.E. 𝐦𝐚𝐱 . = 𝟐 𝒎𝒗𝟐𝐦𝐚𝐱 .
𝟏 𝟗 𝟑
⇒ 𝟐 × 𝟏 × 𝒗𝟐𝐦𝐚𝐱 . = 𝟒 ⇒ 𝒗 𝐦𝐚𝐱 . = 𝟐
44. A mass 𝐨𝐟𝑴 kg is suspended by a weightless string. The horizontal force that is
required to displace it until the string makes an angle of 𝟒𝟓∘ with the initial vertical
direction is [2006]
𝑴𝒈
(a) 𝑴𝒈 𝟐+𝟏 (b) 𝑴𝒈 𝟐
(c) (d) 𝑴𝒈 𝟐−𝟏
𝟐
SOLUTION : (d)
Work done by tension + Work done by force (applied) + Work done by gravitational
force = change in kinetic energy
Work done by tension is zero
⇒ 𝟎 + 𝑭 × AB‐Mg × 𝑨𝑪 = 𝟎
𝟏
𝟏−
𝐀𝐂 𝟏 𝟐
⇒ 𝜞 = 𝐌𝐠 = 𝐌𝐠
𝐀𝐁 𝟐
𝓵
[⋅.⋅ 𝑨𝑩 = 𝓵 𝐬𝐢𝐧 𝟒𝟓∘ = and
𝟐
𝟏
𝑨𝑪 = 𝑶𝑪 − 𝑶𝑨 = 𝓵 − 𝓵 𝐜𝐨𝐬 𝟒𝟓∘ = 𝓵 𝟏 −
𝟐
where 𝓵 = length ofthe string.]
⇒ 𝑭 = 𝑴𝒈 𝟐−𝟏
45 . A spherical ball ofmass 20 kg is stationary at the top of a hill of height 100 𝐦. It rolls
down a smooth surface to the ground, then climbs up another hill ofheight 30 𝐦 and
finally rolls down to a horizontal base at a height of20 𝐦 above the ground. The velocity
attained by the ball is [2005]
𝟏𝟎 𝟑𝟎𝐦
(a) 20 𝐦/𝐬 (b) 40 𝐦/𝐬 (c) (d) 10 𝐦/𝐬
𝐬
SOLUTION : . (b)
𝟏
𝐦𝐠𝐇 𝐦𝐯 𝟐 + 𝐦𝐠𝐡
𝟐
Using conservation of energy,
Total energy at 100 𝐦 height
= Total energy at 𝟐𝟎𝐦 height
𝟏 𝟐
𝐦 𝟏𝟎 × 𝟏𝟎𝟎 = 𝒎 𝒗 + 𝟏𝟎 × 𝟐𝟎
𝟐
𝟏
or 𝟐 𝒗𝟐 = 𝟖𝟎𝟎 or 𝒗 = 𝟏𝟔𝟎𝟎 = 𝟒𝟎𝐦/𝐬
Note :
Loss in potential energy = gain in kinetic energy
𝟏
𝒎 × 𝒈 × 𝟖𝟎 = 𝒎𝒗𝟐
𝟐
𝟏 𝟐
𝟏𝟎 × 𝟖𝟎 = 𝒗
𝟐
𝒗𝟐 = 𝟏𝟔𝟎𝟎 or 𝒗 = 𝟒𝟎𝐦/𝐬
46. A particle moves in a straight line with retardation proportional to its displacement.
Its loss ofkinetic energy for any displacement 𝒙 is proportional to [2004]
(a) 𝒙 (b) 𝒆𝒙 (c) 𝑹𝒆𝒋𝒆𝒄𝒕 (d) 𝐥𝐨𝐠 𝒆 𝒙
SOLUTION : . (c)
Given: retardation ∝ displacement 𝒊. 𝒆., 𝒂 = −𝒌𝒙 [Here, 𝒌 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭]
𝒅𝒗
But 𝒂 = 𝒗 𝒅𝒙
𝒗𝒅𝒗 𝒗𝟐 𝒙
= −𝒉 ⇒ 𝒗𝟏
𝒗 𝒅𝒗 =− 𝟎
kxdx
𝒅𝒙
𝒗𝟐 𝒙
𝒗𝟐 𝒙𝟐
⇒ = −𝒌
𝟐 𝒗𝟏 𝟐 𝟎
𝒌𝒙𝟐
⇒ 𝒗𝟐𝟐 − 𝒗𝟐𝟏 = − 𝟐
𝟏 𝟏 −𝒌𝒙𝟐
⇒ 𝟐 𝒎 𝒗𝟐𝟐 − 𝒗𝟐𝟏 = 𝟐 𝒎 𝟐
Loss in kinetic energy, 𝜟𝑲 ∝ 𝒙𝟐

47. A particle is acted upon by a force of constant magnitude which is always
perpendicular to the velocity ofthe particle, the motion ofthe particles takes place in a
plane. It follows that [2004]
(a) its kinetic energy is constant (b) its acceleration is constant
(c) its velocity is constant (d) it moves in a straight line
SOLUTION : . (a)
Work done by such force is always zero since force is acting in a direction perpendicular
to velocity.
From work‐energy theorem = 𝜟𝐊 = 𝟎
𝐊 remains constant.
48. A wire suspended vertically from one of its ends is stretched by attaching a weight
𝐨𝐟𝟐𝟎𝟎𝐍 to the lower end. The weight stretches the wire by 1 mm. Then the elastic
energy stored in the wire is [2003]
(a) 𝟎. 𝟐𝐉 (b) 𝟏𝟎𝐉 (c) 𝟐𝟎𝐉 (d) 𝟎. 𝟏𝐉
SOLUTION : (d)
The elastic potential energy
𝟏
= 𝟐 ×Force × extension
𝟏
= ×𝑭×𝒙
𝟐
𝟏
=
× 𝟐𝟎𝟎 × 𝟎. 𝟎𝟎𝟏 = 𝟎. 𝟏𝐉
𝟐
49. Aball whose kinetic energy 𝐢𝐬𝑬, is projected at an angle of 𝟒𝟓∘ to the horizontal. The
kinetic energy ofthe ball at the highest point ofits flight will be [2002]
(a) 𝑬 (b) 𝑬𝒍 𝟐 (c) El2 (d) zero
SOLUTION : . (c)
Let 𝐮 be the speed with which the ball ofmass 𝐦 is projected. Then the kinetic energy (E) at the
point of projection is
𝒖
𝟐
𝟏
𝑬 = 𝟐 𝒎𝒖𝟐 (i)
𝒖
When the ball is at the highest point ofits flight, the speed of the ball is (Remember that the
𝟐
horizontal component of velocity does not change during a projectile motion).

The kinetic energy at the highest point
𝟏 𝒖 𝟐 𝟏 𝒎𝒖𝟐 𝑬
= 𝟐𝒎 =𝟐 = 𝟐 [From (i)]
𝟐 𝟐
POWER
50. A body ofmass 2 kg is driven by an engine delivering a constant power of 1 𝐉/𝐬. The
body starts from rest and moves in a straight line. After 9 seconds, the body has moved
a distance (in m) [5 Sep. 2020 (II)]
SOLUTION : . (18)
Given, Mass ofthe body, 𝒎 = 𝟐 kg Power delivered by engine, 𝑷 = 𝟏𝐉/𝐬
Time, 𝒕 = 𝟗 seconds
Power, 𝑷 = 𝑭𝒗
⇒ 𝑷 = 𝒎𝒂𝒗 ⋅.⋅ 𝑭 = 𝒎𝒂
𝒅𝒗 𝒅𝒗
⇒ 𝒎 𝒅𝒕 𝒗 = 𝑷 ⋅.⋅ 𝒂 = 𝒅𝒕
𝑷
⇒ 𝒗𝒅𝒗 =
𝒅𝒕
𝒎
Integrating both sides we get
𝒗
𝑷 𝒕
⇒ 𝒗 𝒅𝒗 = 𝒅𝒕
𝟎 𝒎 𝟎
𝟏/𝟐
𝒗𝟐 𝑷𝒕 𝟐𝑷𝒕
⇒ = ⇒𝒗=
𝟐 𝒎 𝒎
𝒅𝒙 𝟐𝑷 𝟏/𝟐 𝒅𝒙
⇒ = 𝒕 ⋅.⋅ 𝒗 =
𝒅𝒕 𝒎 𝒅𝒕
𝒙 𝒕
𝟐𝑷
⇒ 𝒅𝒙 = 𝒕𝟏/𝟐 𝒅𝒕
𝟎 𝒎 𝟎
𝟐𝑷 𝒕𝟑/𝟐 𝟐𝑷 𝟐
Distance, 𝒙 = = × 𝟑 𝒕𝟑/𝟐
𝒎 𝟑/𝟐 𝒎
𝟐×𝟏 𝟐 𝟐
⇒𝒙= × × 𝟗𝟑/𝟐 = × 𝟐𝟕 = 𝟏𝟖.
𝟐 𝟑 𝟑
51. Aparticle is moving unidirectionally on ahorizontal plane under the action of a

constant power supplying energy source. The displacement 𝒔 ‐ time(t) graph that
describes the motion ofthe particle is(graphs are drawn schematically and are not to
scale): [3 Sep. 2020 (II)]
SOLUTION : . (b)
We know that
Power, 𝑷 = 𝑭𝒗
𝒅𝒗
But 𝑭 = 𝒎𝒂𝒗 = 𝒎 𝒅𝒕 𝒗
𝒅𝒗
𝑷 = 𝒎𝒗 ⇒ 𝑷𝒅𝒕 = 𝒎𝒗𝒅𝒗
𝒅𝒕
𝒕 𝒗
Integrating both sides 𝟎
𝑷 𝒅𝒕 =𝒎 𝟎
𝒗 𝒅𝒗
𝟏 𝟐𝑷
P. 𝐭 = 𝟐 𝒎𝒗𝟐 ⇒ 𝒗 = 𝒕𝟏/𝟐
𝒎
𝒕 𝟐𝑷𝒕 𝟏/𝟐 𝟐𝑷 𝒕𝟑/𝟐

Distance, 𝒔 = 𝟎
𝒗 𝒅𝒕 = 𝒕 𝒅𝒕 = ⋅ 𝟑/𝟐
𝒎 𝟎 𝒎
𝟖𝑷 𝟑/𝟐
⇒𝒔= ⋅𝒕 ⇒ 𝐬 ∝ 𝐭 𝟑/𝟐
𝟗𝒎
So, graph (b) is correct.
52. A 60 HP electric motor lifts an elevator having a maximum total load capacity of200𝐤𝐠.
Ifthe fiictional force on the elevator is 4000 𝐍, the speed ofthe elevator at full load is
close to: (1 HP = 𝟕𝟒𝟔𝐖, 𝐠 = 𝟏𝟎 ms 2) [7 Jan. 2020 I]
(a) 𝟏. 𝟕 ms 1 (b) 𝟏. 𝟗 ms 1 (c) 𝟏. 𝟓 ms 1 (d) 𝟐. 𝟎 ms 1
SOLUTION : . (b)
Total force required to lift maximum load capacity against fiictional force = 𝟒𝟎𝟎𝑵
𝑭𝐭𝐨𝐭𝐚𝟏 = 𝑴𝒈 + friction
= 𝟐𝟎𝟎𝟎 × 10+4000
= 𝟐𝟎, 𝟎𝟎𝟎 + 𝟒𝟎𝟎𝟎 = 𝟐𝟒𝟎𝟎𝟎𝑵
Using power, 𝑷 = 𝑭 × 𝒗
𝟔𝟎 × 𝟕𝟒𝟔 = 𝟐𝟒𝟎𝟎𝟎 × 𝒗
⇒ 𝒗 = 𝟏. 𝟖𝟔𝐦/𝐬 ≈ 𝟏. 𝟗𝒎𝒍𝒔
Hence speed ofthe elevator at full load is close to 𝟏. 𝟗 ms 1
53. Aparticle ofmass Mis moving in a circle offixed radius 𝐑 in such a way that its
centripetal acceleration at time 𝐭 is given by 𝐧𝟐 𝐑𝐭 𝟐 where 𝐧 is a constant. The power
delivered to the particle by the force acting on it, is: [Online April 10, 2016]
𝟏
(a) 𝟐 𝐌𝐧𝟐 𝐑𝟐 𝐭 𝟐 (b) 𝐌𝐧𝟐 𝐑𝟐 𝐭 (c) 𝐌𝐧𝐑𝟐 𝐭 𝟐 (d) 𝐌𝐧𝐑𝟐 𝐭
SOLUTION : . (b)
Centripetel acceleration 𝐚𝐜 = 𝐧𝟐 𝐑𝐭 𝟐
𝐯𝟐
𝐚𝐜 = = 𝐧𝟐 𝐑𝐭 𝟐
𝐑
𝐯 𝟐 = 𝐧𝟐 𝐑𝟐 𝐭 𝟐
𝐯 = 𝐧𝐑𝐭
𝐝𝐯
𝐚𝐜 = = 𝐧𝐑
𝐝𝐭
Power = 𝐦𝐚𝐭 𝐯 = 𝐦 nRnRt = 𝐌𝐧𝟐 𝐑𝟐 𝐭.
54. A car ofweight 𝐖 is on an inclined road that rises by 100 𝐦 over a distance of 1 Km
𝐖
and applies a constant fiictional force on the car. While moving uphill on the road at
𝟐𝟎
𝐏
a speed of 10 𝐬 −𝟏 , the car needs power P. If it needs power while moving downhill at
𝟐
speed 𝒗 then value of 𝒗 is: [Online April 9, 2016]

(a) 20 𝐦𝐬−𝟏 𝐛 𝟓𝐦𝐬−𝟏 (c) 15 𝐦𝐬−𝟏 (d) 10 𝐦𝐬−𝟏
SOLUTION : . (c)
While moving downhill power
𝒘
𝑷 = 𝒘 𝐬𝐢𝐧 𝜽 + 𝟏𝟎
𝟐𝟎
𝐰 𝐰 𝟑𝐰
𝐏= + 𝟏𝟎 =
𝟏𝟎 𝟐𝟎 𝟐
𝐏 𝟑𝐰 𝐰 𝐰
= = − 𝐕
𝟐 𝟒 𝟏𝟎 𝟐𝟎
𝟑 𝒗
= ⇒ 𝒗 = 𝟏𝟓𝐦/𝐬
𝟒 𝟐𝟎
Speed ofcar while moving downhill 𝐯 = 𝟏𝟓𝐦/𝐬.
55. Awind‐poweredgeneratorconvertswindenergyintoelectrical energy. Assume that the

generator converts a fixed fiiaction of the wind energy intercepted by its blades into
electrical energy. For wind speed 𝐯, the electrical power output will be most likely
proportional to [Online April 25, 2013]
(a) 𝐯 𝟒 (b) 𝐯 𝟐 (c) 𝐯 (d) 𝐯
SOLUTION : . (d)
56. 𝐀𝟕𝟎𝐤𝐠𝐦𝐚𝐧 leaps vertically into the air from acrouching position. To take the leap the
man pushes the ground with a constant force 𝑭 to raise himself. The center ofgravity
rises by 𝟎. 𝟓𝐦 before he leaps. After the leap the 𝐜. 𝐠. rises by another 1 𝐦. The
maximum power delivered by the muscles is : (Take 𝐠 = 𝟏𝟎𝐦𝐬 −𝟐 )
(a) 𝟔. 𝟐𝟔 × 𝟏𝟎 Watts at the start
𝟑
(b) 𝟔. 𝟐𝟔 × 𝟏𝟎 Watts at take off
𝟑
(c) 𝟔. 𝟐𝟔 × 𝟏𝟎𝟒 Watts at the start (d) 𝟔. 𝟐𝟔 × 𝟏𝟎𝟒 Watts at take off
SOLUTION : . (b)
57. Abody ofmass 𝒎’, accelerates uniformlyfrom rest to 𝒗𝟏 ’ in time 𝒕𝟏 ’. The instantaneous
power delivered to the body as a 𝐡𝐧𝐂𝐭𝟓𝐨𝐧𝐨𝐟𝐭𝐢𝐦𝐞𝒕 is [2004]
𝒎𝒗𝟏 𝒕 𝒎𝒗𝟐𝟏 𝒕 𝒎𝒗𝟏 𝒕 𝒎𝒗𝟐𝟏 𝒕
(a) (b) (c) (d)
𝒕𝟏 𝒕𝟐𝟏 𝒕𝟏 𝒕𝟏
SOLUTION : (b)
Let a be the acceleration ofbody Using, 𝒗 = 𝒖 + 𝒂𝒕
𝒗𝟏
𝒗𝟏 = 𝟎 + 𝒂𝒕𝟏 ⇒ 𝒂 =
𝒕𝟏
Velocity ofthe body at instant 𝒕,
𝒗 = 𝒂𝒕
𝒗𝟏 𝒕
⇒𝒗=
𝒕𝟏
Instantaneous powr, 𝐏 = 𝑭 ⋅ 𝒗 = 𝒎𝒂 ⋅ 𝒗
𝒎𝒗𝟏 𝒗𝟏 𝒕 𝒗𝟏 𝟐
= =𝒎 𝒕
𝒕𝟏 𝒕𝟏 𝒕𝟏
58. A body is moved along a straight line by a machine delivering a constant power. The
distance moved by the body in time 𝒕’ is proportional to [2003]
(a) 𝒕𝟑/𝟒 (b) 𝒕𝟑/𝟐 (c) 𝒕𝟏/𝟒 (d) 𝒕𝟏/𝟐
SOLUTION : . (b)
Power, 𝑷 = 𝑭𝒗 = 𝒎𝒂. 𝒗
𝒎𝒅𝒗
⇒𝑷= 𝒗 = 𝒄 = contant
𝒅𝒕
𝒎𝒅𝒗
⋅.⋅ 𝑭 = 𝒎𝒂 =
𝒅𝒕
𝒎𝒗𝟎 𝒗 = 𝒄𝒅𝒕
Integrating both sides, we get
𝒗 𝒕
𝒎 𝒗 𝒅𝒗 = 𝒄 𝒅𝒕
𝟎 𝟎
𝟏
⇒ 𝒎𝒗𝟐 = 𝒄𝒕
𝟐
𝒗𝟐 𝒄. 𝒕
⇒ =
𝟐 𝒎
𝟐𝒄. 𝒕
⇒ 𝒗𝟐 =
𝒎
/
𝟐𝒄
⇒𝒗= × 𝒕𝟏𝟐
𝒎
𝒅𝒙 𝟐𝒄 𝒅𝒙
⇒ = × 𝒕𝟏𝟐 / where 𝒗 =
𝒅𝒕 𝒎 𝒅𝒕
𝒙 𝒕
𝟐𝒄
⇒ 𝒅𝒙 = × 𝒕 /𝟏𝟐 𝒅𝒕
𝒆 𝒎 𝟎
𝟐𝒄 𝟐𝒕𝟑𝟐 /
⇒𝒙= × ⇒ 𝐱 ∝ 𝐭 𝟑𝟐 /
𝒎 𝟑
COLLISIONS
59. Two bodies of the same mass are moving with the same speed, but in different
directions in a plane. They have a completely inelastic collision and move together
thereafter with a final speed which is halfoftheir initial speed. The angle between the
initial velocities of the two bodies (in degree) is . [NA 6 Sep. 2020 (I)]
SOLUTION : . (120)
Momentum conservation along 𝒙 direction,

𝒗𝟎 𝟏
𝟐𝒎𝒗𝟎 𝐜𝐨𝐬 𝜽 = 𝟐𝒎 ⇒ 𝐜𝐨𝐬 𝜽 = 𝟐 or 𝜽 = 𝟔𝟎∘
𝟐
Hence angle between the initial velocities ofthe two bodies

= 𝜽 + 𝜽 = 𝟔𝟎∘ + 𝟔𝟎∘ = 𝟏𝟐𝟎∘ .
60. ParticleAofmass m movingwith velocity 𝟑𝒊 + 𝒋 𝐦𝐬 −𝟏 collides with another particle 𝐁

ofmass 𝐦𝟐 which is at rest initially. Let 𝑽𝟏 and 𝑽𝟐 be the velocities ofparticles A and 𝐁
after collision respectively. 𝐈𝐟𝐦𝟏 = 𝟐𝐦𝟐 and after collision 𝑽𝟏 = 𝒊 + 𝟑𝒋 𝐦𝐬−𝟏 , the
angle between 𝑽𝟏 and 𝑽𝟐 is : [6 Sep. 2020 (II)]
(a) 𝟏𝟓𝐨 (b) 𝟔𝟎𝐨 (c) −𝟒𝟓 𝐨
(d) 𝟏𝟎𝟓𝐨
SOLUTION : (d)
Before collision,
Velocity ofparticle 𝑨, 𝒖𝟏 = 𝟑𝒊 + 𝒋 𝐦/𝐬
Velocity ofparticle 𝑩, 𝒖𝟐 = 𝟎
After collision,
Velocity ofparticle 𝑨, 𝒗𝟏 = 𝒊 + 𝟑𝒋
Velocity of particle 𝑩, 𝒗𝟐 = 𝟎
Using principal ofconservation of angular momentum
𝒎𝟏 𝒖𝟏 + 𝒎𝟐 𝒖𝟐 = 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐
⇒ 𝟐𝒎𝟐 𝟑𝒊 + 𝒋 + 𝒎𝟐 × 𝟎 = 𝟐𝒎𝟐 𝒊 + 𝟑𝒋 + 𝒎𝟐 × 𝒗𝟐
⇒ 𝟐 3î+2j = 𝟐𝒊 + 𝟐 𝟑𝒋 + 𝒗𝟐
⇒ 𝒗𝟐 = 𝟑 − 𝟏 𝒊 − 𝟑 − 𝟏 𝒋
⇒ 𝒗𝟏 = î+ 3j
For angle between 𝒗𝟏 and 𝒗𝟐 ,
𝒗𝟏 ⋅ 𝒗𝟐 𝟐 𝟑 − 𝟏 𝟏 − 𝟑 𝟏− 𝟑
𝐜𝐨𝐬 𝜽 = = =
𝒗𝟏 𝒗𝟐 𝟐×𝟐 𝟐 𝟑−𝟏 𝟐 𝟐
⇒ 𝜽 = 𝟏𝟎𝟓∘
Angle between 𝒗𝟏 and 𝒗𝟐 is 𝟏𝟎𝟓∘
61. Blocks ofmasses m, 𝟐𝒎, 𝟒𝒎 and 𝟖𝒎 are arranged in aline on a fiictionless floor.
Another block ofmass 𝒎, moving with speed 𝒗 along the same line (see figure) collides
with mass 𝒎 in perfectly inelastic manner. All the subsequent collisions are also
perfectly inelastic. By the time the last block of mass 𝟖𝒎 starts moving the total energy
loss is p% of the original energy. Value of 𝒑’ is close to: [4 Sep. 2020 (I)]
→𝒗
𝒎 𝒎 𝟐𝒎 𝟒𝒎 𝟖𝒎
(a) 77 (b) 94 (c) 37 (d) 87
SOLUTION : . (b)
According to the question, all collisions are perfectly inelastic, so after the final collision, all
blocks are moving together.
𝐦 𝟐𝒎 𝟒𝒎 𝟖𝒎
Let the final velocity be 𝒗′ , using momentum conservation
𝒗
𝒎𝒗 = 𝟏𝟔𝒎𝒗′ ⇒ 𝒗† =
𝟏𝟔
𝟏
Now initial energy 𝑬𝒊 = 𝟐 𝒎𝒗𝟐
𝟏 𝒗 𝟐 𝟏 𝒎𝒗𝟐
Final energy: 𝑬𝒇 = 𝟐 × 𝟏𝟔𝒎 × =𝟐
𝟏𝟔 𝟏𝟔
𝟏 𝟏 𝒗𝟐
Energy loss : 𝑬𝒊 − 𝑬𝒇 = 𝟐 𝒎𝒗𝟐 − 𝟐 𝒎 𝟏𝟔
𝟏 𝟏 𝟏 𝟏𝟓
⇒ 𝒎𝒗𝟐 𝟏 − ⇒ 𝒎𝒗𝟐
𝟐 𝟏𝟔 𝟐 𝟏𝟔
The total energy loss is 𝑷𝟎 /𝟎 of the original energy.
𝐄𝐧𝐞𝐫𝐠𝐲𝟏𝐨𝐬𝐬
%𝑷 = × 𝟏𝟎𝟎
𝐎𝐫𝐢𝐠𝐢𝐧𝐚𝟏𝐞𝐧𝐞𝐫𝐠𝐲
𝟏 𝟏𝟓
𝒎𝒗𝟐 𝟏𝟔
= 𝟐 × 𝟏𝟎𝟎 = 𝟗𝟑. 𝟕𝟓%
𝟏 𝟐
𝒎𝒗
𝟐
Hence, value 𝐨𝐟𝑷 is close to 94.
62. A block ofmass 𝟏. 𝟗 kg is at rest at the edge ofa table, of height 1 𝐦. Abullet ofmass 𝟎. 𝟏
kg collides with the block and sticks to it. If the velocity of the bullet is 20 𝐦/𝐬 in the
horizontal direction just before the collision then the kinetic energy just before the
combined system strikes the floor, is [Take 𝒈 = 𝟏𝟎𝐦/𝐬𝟐 . Assume there is no rotational
motion and losss of energy after the collision is negligiable.] [3 Sep. 2020 (II)]
(a) 20 𝐉 (b) 21 𝐉 (c) 19 𝐉 (d) 23 𝐉
SOLUTION : (b)
Given,
Mass ofblock, 𝒎𝟏 = 𝟏. 𝟗 kg
Mass ofbullet, 𝒎𝟐 = 𝟎. 𝟏 kg
Velocity of bullet, 𝒗𝟐 = 𝟐𝟎𝐦/𝐬
Let 𝒗 be the velocity of the combined system. It is an inelastic collision.
Using conservation oflinear momentum
𝒎 𝟏 × 𝟎 + 𝒎 𝟐 × 𝒗𝟐 = 𝒎 𝟏 + 𝒎 𝟐 𝒗
⇒ 𝟎. 𝟏 × 𝟐𝟎 = 𝟎. 𝟏 + 𝟏. 𝟗 × 𝒗
⇒ 𝒗 = 𝟏𝐦/𝐬
Using work energy theorem
Work done = Change in Kinetic energy
Let 𝑲 be the Kinetic energy of combined system.
𝐦𝟏 + 𝐦𝟐 𝐠𝐡
𝟏
=𝐊− 𝐦 + 𝐦𝟐 𝐕 𝟐
𝟐 𝟏
𝟏
⇒ 𝟐×𝒈×𝟏 =𝑲− × 𝟐 × 𝟏𝟐 ⇒ 𝑲 = 𝟐𝟏
𝟐
63. A particle ofmass 𝒎 with an initial velocity 𝒖 î collides perfectly elastically with a mass
3 𝐦 at rest. It moves with a velocity 𝒗𝒋 after collision, then, 𝒗 is given by:
[2 Sep. 2020 (I)]
𝟐 𝒖 𝒖 𝟏
(a) 𝒗 = 𝒖 (b) 𝒗 = (c) 𝒗 = (d) 𝒗 = 𝒖
𝟑 𝟑 𝟐 𝟔
SOLUTION : (c)
From conservation oflinear momentum
𝒎𝒖𝒊 + 𝟎 = 𝒎𝒗𝒋 + 𝟑𝒎𝒗′
−, 𝒖 ∧ 𝒗 ∧
𝒗 = −𝒊 − −𝒋𝟑𝟑
𝒎 − 𝒖 𝟑𝒎 𝒎𝟏𝒗 𝟑𝒎
∼ 𝐀 𝒗′
Before
After collision
collision From kinetic energy conservation,
𝟏 𝟏 𝟏 𝒖 𝟐 𝒗 𝟐
𝒎𝒖𝟐 = 𝒎𝒗𝟐 + 𝟑𝒎 +
𝟐 𝟐 𝟐 𝟑 𝟑
𝒎𝒖𝟐 𝒎𝒗𝟐
or, 𝒎𝒖𝟐 = 𝒎𝒗𝟐 + +
𝟑 𝟑
𝒖
𝒗=
𝟐
64. Aparticle ofmass 𝒎 is moving along the x‐axis with initial velocity 𝒖𝒊. It collides
elastically with a particle ofmass 10 𝐦 at rest and then moves with half its initial
kinetic energy (see figure). If 𝐬𝐢𝐧 𝜽𝟏 = 𝒏 𝐬𝐢𝐧 𝜽𝟐 , then value of 𝒏 is
[NA 2 Sep. 2020 (II)]
𝐦.
𝟏𝟎𝐦
SOLUTION : . (𝟏𝟎. 𝟎𝟎)
From momentum conservation in perpendicular direction of initial motion.
𝒎𝒖𝟏 𝐬𝐢𝐧 𝜽𝟏 = 𝟏𝟎𝒎𝒗𝟏 𝐬𝐢𝐧 𝜽𝟐 (i)
It is given that energy of 𝒎 reduced by half. If 𝒖𝟏 be velocity of 𝒎 after collision, then
𝟏 𝟏 𝟏
𝒎𝒖𝟐 = 𝒎𝒖𝟐𝟏
𝟐 𝟐 𝟐
𝒖
⇒ 𝒖𝟏 =
𝟐
If 𝒗𝟏 be the velocity of mass 10 𝐦 after collision, then
𝟏 𝟏𝒖𝟐 𝒖
× 𝟏𝟎𝒎 × 𝒗𝟐𝟏 = 𝒎 ⇒ 𝒗𝟏 =
𝟐 𝟐 𝟐 𝟐𝟎
From equation (i), we have
65. Two particles of equal mass 𝒎 have respective initial (î+j) velocities 𝒖𝒊 and 𝒖 𝟐 . They
collide completely inelastically. The energylost in the process is:
[9 Jan. 2020 I]
𝟏 𝟏 𝟑 𝟐
(a) 𝟑 𝐦𝐮𝟐 (b) 𝟖 𝐦𝐮𝟐 (c) 𝟒 𝐦𝐮𝟐 (d) 𝐦𝐮𝟐
𝟑
SOLUTION : [b]
𝒙‐direction
𝒎𝒖 𝟑𝒖
𝒎𝒖 + = 𝟐𝒎𝒗𝒙 ⇒ 𝑽𝒙 =
𝟐 𝟒
𝒎𝒖 𝒖 𝟏 𝟏 𝒖 𝟐 𝒖 𝟐
𝐲‐direction 𝟎 + = 𝟐𝒎𝒗𝒚 ⇒ 𝒗𝒚 = 𝟒 K.E. 𝒊 = 𝟐 𝒎𝒖𝟐 + 𝟐 𝒎 +
𝟐 𝟐 𝟐
𝟏 𝟐
𝒎𝒖𝟐 𝟑𝒎𝒖𝟐
= 𝒎𝒖 + =
𝟐 𝟒 𝟒
𝟏 𝟏 𝟐
K.E. 𝒇 = 𝟐 𝟐𝒎 𝒗𝒙 ′ 𝟐
+ 𝟐 𝟐𝒎 𝒗𝒚 ′
𝟏 𝟑𝒖 𝟐 𝒖 𝟐 𝟓
= 𝟐𝒎 + = 𝒎𝒖𝟐
𝟐 𝟒 𝟒 𝟖
Loss in 𝑲𝑬 = 𝑲𝑬𝒇 − 𝑲𝑬𝒋
𝟔 𝟓 𝒎𝒖𝟐
= 𝒎𝒖𝟐
− =
𝟖 𝟖 𝟖
66. 𝐀𝐛𝜶𝟏𝐲𝑨, ofmass 𝒎 = 𝟎. 𝟏 kg has an initialvelocityof3 î ms 1. It collides elastically with
another body, 𝑩 of the same mass which has an initial velocity of 5 𝒋 ms 1. After
collision, 𝑨 moves with a velocity 𝒗 = 𝟒 𝒊 + 𝒋 . The
𝒙
energy 𝐨𝐟𝑩 after collision is written as 𝟏𝟎 𝑱. The value of 𝒙 is . [NA 8 Jan. 2020 I]
SOLUTION : . (a)
For elastic collision 𝑲𝑬𝐢 = 𝕸𝒇
𝟏 𝟏 𝟏 𝟏
𝒎 × 𝟐𝟓 + × 𝒎 × 𝟗 = 𝒎 × 𝟑𝟐 + 𝒎𝒗𝟐𝑩
𝟐 𝟐 𝟐 𝟐
𝟑𝟒 = 𝟑𝟐 + 𝑽𝟐𝑩 ⇒ 𝑽𝑩 = 𝟐
𝟏 𝟏 𝟏
𝑲𝑬𝑩 = 𝒎𝒗𝟐𝑩 = × 𝟎. 𝟏 × 𝟐 = 𝟎. 𝟏𝑱 = 𝑱 𝒙=𝟏
𝟐 𝟐 𝟏𝟎
67. Aparticle ofmass 𝒎 is dropped 𝐟𝐢𝐢 𝐨𝐦 a height 𝒉 above the ground. At the same time
another particle of the same mass is thrown vertically upwards from the ground with a
speed of 𝟐𝒈𝒉. If they collide head‐on completely inelastically, the time taken for the
combined mass to reach
𝒉
the ground, in units of is: [8 Jan. 2020 II]
𝒈
𝟏 𝟑 𝟏 𝟑
(a) (b) (c) 𝟐 (d)
𝟐 𝟒 𝟐
SOLUTION : (d)
Let 𝐭 be the time taken by the particle dropped from height 𝐡 to collide with particle thrown
upward.
Using,
𝒉
𝑺𝟏 𝐓
𝟒
𝟑𝒉
𝒔𝟐 𝐢 𝐓
𝟒
𝒗𝟐 − 𝒖𝟐 = 𝟐𝒈𝒉
⇒ 𝒗𝟐 − 𝟎𝟐 = 𝟐𝒈𝒉
⇒ 𝒗 = 𝟐𝒈𝒉
Downward distance travelled
𝟏 𝟏 𝒉 𝒉
𝑺𝟏 = 𝒈𝒕𝟐 = 𝒈. =
𝟐 𝟐 𝟐𝒈 𝟒
Distance of collision point 𝐟𝐢𝐢 𝐨𝐦 ground
𝒉 𝟑𝒉
𝒔𝟐 = 𝒉 −
=
𝟒 𝟒
Speed of (A) just before collision
𝒈𝒉
𝒗𝟏 = 𝒈𝒕 =
𝟐
And speed of(B) just before collision
𝒈𝒉
𝒗𝟐 = 𝟐𝒈𝒉 −
𝟐
Using principle of conservation of linear momentum

𝒎𝒗𝟏 + 𝒎𝒗𝟐 = 𝟐𝒎𝒗𝒇
𝒈𝒉 𝒈𝒉
⇒ 𝒗𝒇 = 𝒎 𝟐𝒈𝒉 − −𝒎 =𝟎
𝟐 𝟐
𝟐𝒎
After collision, time taken 𝒕𝟏 for combined mass to reach the ground is
𝟑𝒉 𝟏 𝟑𝒉
⇒ = 𝟐 𝒈𝒕𝟐𝟏 ⇒ 𝒕𝟏 =
𝟒 𝟐𝒈
68. A man (mass = 𝟓𝟎𝐤𝐠) and his son (mass = 𝟐𝟎 kg) are standing on a frictionless surface
facing each other. The man pushes his son so that he starts moving at a speed of 𝟎. 𝟕𝟎
ms 1 with respect to the man. The speed ofthe man with respect to the surface is:
[12 April 2019 I]
(a) 𝟎. 𝟐𝟖𝐦𝐬 −𝟏 (b) 𝟎. 𝟐𝟎 ms 1 (c) 𝟎. 𝟒𝟕 ms 1 (d) 𝟎. 𝟏𝟒 ms 1
SOLUTION : (b)
𝐏𝒊 = 𝐏𝒇
or 𝟎 = 𝟐𝟎 𝟎. 𝟕 − 𝒗 = 𝟓𝟎𝒗 or 𝒗 = 𝟎. 𝟐𝐦/𝐬
69. Two particles, ofmasses 𝐌 and 𝟐𝐌, moving, as shown, with speeds of10 𝐦/𝐬 and 5 𝐦/𝐬,
collide elastically at the origin. After the collision, they move along the indicated
directions with speeds 𝐯𝟏 and 𝐯𝟐 , respectively. The values 𝐨𝐟𝐯𝟏 and 𝐯𝟐 are nearly:
[10 April 2019 I]
(a) 𝟔. 𝟓 𝐦/𝐬 and 𝟔. 𝟑 𝐦/𝐬 (b) 𝟑. 𝟐 𝐦/𝐬 and 𝟔. 𝟑 𝐦/𝐬

(c) 𝟔. 𝟓𝐦/𝐬𝐚𝐧𝐝𝟑. 𝟐𝐦/𝐬 (d) 𝟑. 𝟐𝐦/𝐬 and 𝟏𝟐. 𝟔𝐦/𝐬
SOLUTION : . (a)
Apply concervation oflinear momentum in X and 𝐘 direction for the system then
𝐌 𝟏𝟎 𝐜𝐨𝐬 𝟑𝟎∘ + 𝟐𝐌 𝟓 𝐜𝐨𝐬 𝟒𝟓∘ = 𝟐𝐌 𝐯𝟏 𝐜𝐨𝐬 𝟑𝟎∘
+𝐌 𝐯𝟐 𝐜𝐨𝐬 𝟒𝟓∘
𝐯𝟐
𝟓 + 𝟓 𝟑𝐯𝟏 + (1)
𝟐
Also
𝐯𝟐
𝟐𝐌 𝟓 𝐬𝐢𝐧 𝟒𝟓∘ − 𝐌 𝟏𝟎 𝐬𝐢𝐧 𝟑𝟎∘ = 𝟐𝐌𝐯𝟏 𝐬𝐢𝐧 𝟑𝟎∘ − 𝐌𝐯𝟐 𝐬𝐢𝐧 𝟒𝟓∘ 𝟓 𝟐 − 𝟓 = 𝐯𝟏 − (2)
𝟐
Solving equation (1) and (2)

𝟑 + 𝟏 𝐯𝟏 = 𝟓 𝟑 + 𝟏𝟎 𝟐 − 𝟓 ⇒ 𝐯𝟏 = 𝟔. 𝟓𝐦/𝐬
𝐯𝟐 = 𝟔. 𝟑𝐦/𝐬
70. A body of mass 2 kg makes an elastic collision with a second body at rest and continues
to move in the original direction but with one fourth ofits original speed. What
is the mass ofthe second body? [9 April 2019 I]
(a) 𝟏. 𝟎𝐤𝐠 (b) 𝟏. 𝟓𝐤𝐠 (c) 𝟏. 𝟖𝐤𝐠 (d) 𝟏. 𝟐𝐤𝐠
SOLUTION : . (b)
𝒖
𝟐𝒖 + 𝟎 = 𝟐 + 𝒎𝒗𝟐
𝟒
𝟏 𝟏 𝒖 𝟐 𝟏
and 𝟐 × 𝟐 × 𝒖𝟐 + 𝟎 = 𝟐 × 𝟐 × + 𝟐 𝒎𝒗𝟐𝟐 On solving, we get 𝐦 = 𝟏. 𝟓𝐤
𝟒
71. A particle of mass ‘m’ is moving with speed ‘2v’ and collides with a mass ‘2m’
moving with speed ‘v’ in the same direction. After collision, the first mass is stopped
completely while the second one splits into two particles each ofmass 𝒎’, which move at
angle 𝟒𝟓∘ with respect to the original direction. [9 April 2019 II]
The speed of each ofthe moving particle will be:
(a) 𝟐𝐯 (b) 𝟐 𝟐𝐯 (c) 𝒗𝒍 𝟐 𝟐 (d) 𝐯/ 𝟐
SOLUTION : . (b)
𝒎 𝟐𝒗 + 𝟐𝒎𝒗 = 𝟎 + 𝟐𝒎𝒗 ’ 𝐜𝐨𝐬 𝟒𝟓∘
or 𝒗’ = 𝟐 𝟐𝒗
72. A body ofmass 𝐦𝟏 moving with an unknown velocity of 𝐯𝟏 î , undergoes a collinear
collision with a body ofmass 𝒎𝟐 moving with a velocity 𝐯𝟐 î. After collision, 𝒎𝟏 and 𝒎𝟐
move with velocities of 𝐯𝟑 î and 𝐯𝟒 î , respectively. If 𝒎𝟐 = 𝟎. 𝟓𝐦𝟏 and 𝐯𝟑 = 𝟎. 𝟓𝐯𝟏 , then 𝐯𝟏
is: [8 April 2019 𝐈]
𝐯𝟐 𝐯𝟐
(a) 𝐯𝟒 − 𝐛 𝐯𝟒 − 𝐯𝟐 (c) 𝐯𝟒 − (d) 𝐯𝟒 + 𝐯𝟐
𝟐 𝟒
SOLUTION : . (b)
𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐 = 𝒎𝟏 𝒗𝟐 + 𝒎𝟐 𝒗𝟏
or 𝒎𝟏 𝒗𝟏 + 𝟎. 𝟓𝒎𝟏 𝒗𝟐 = 𝒎𝟏 𝟎. 𝟓𝒗𝟏 + 𝟎. 𝟓𝒎𝟏 𝒗𝟒
On solving, 𝒗𝟏 = 𝒗𝟒 − 𝒗𝟐
73. An alpha‐particle ofmass 𝐦 suffers 1‐dimensional elastic collision with a nucleus at
rest of unknown mass. It is scattered directly backwards losing, 64% of its initial
kinetic energy. The mass ofthe nucleus is: [12 Jan. 2019 II]
(a) 𝟐𝐦 (b) 𝟑. 𝟓𝐦 (c) 𝟏. 𝟓𝐦 (d) 𝟒𝐦
SOLUTION : . (d)
Using conservation ofmomentum,
𝐦𝐯𝟎 = 𝐦𝐯𝟐 − 𝐦𝐯𝟏
𝜶𝐕𝟎 ↔
After collision
𝟏 𝟏
𝐦𝐯𝟏𝟐 = 𝟎. 𝟑𝟔 × 𝐦𝐯𝟎𝟐
𝟐 𝟐
⇒ 𝐯𝟏 = 𝟎. 𝟔𝐯𝟎
The collision is elastic. So,
𝟏 𝟏
𝐌𝐕𝟐𝟐 = 𝟎. 𝟔𝟒 × 𝟐 𝐦𝐯𝟎𝟐 [.⋅. 𝐌 = mass ofnucleus]
𝟐
𝐦
⇒ 𝐕𝟐 = × 𝟎. 𝟖𝐕𝟎
𝐌
𝐦𝐕𝟎 = 𝐦𝐌 × 𝟎. 𝟖𝐕𝟎 − 𝐦 × 𝟎. 𝟔𝐕𝟎
⇒ 𝟏. 𝟔𝐦 = 𝟎. 𝟖 𝐦𝐌
⇒ 𝟒𝐦𝟐 = 𝐦𝐌 ; 𝐌 = 𝟒𝐦
74. A piece of wood of mass 𝟎. 𝟎𝟑 kg is dropped from the top of a 100 𝐦 height building. At
the same time, a bullet ofmass 𝟎. 𝟎𝟐 kg is fired vertically upward, with a velocity 100
𝐬 −𝟏 , from the ground. The bullet gets embedded in the wood. Then the maximum
height to which the combined system reaches above the top of the building before
falling below is: 𝐠 = 𝟏𝟎𝐦𝐬−𝟐 [𝟏𝟎 Jan. 2019 I]
(a) 𝟐𝟎𝐦 (b) 𝟑𝟎𝐦 (c) 𝟒𝟎𝐦 (d) 𝟏𝟎𝐦
SOLUTION : (c)
10
𝟎𝐦/𝐬
𝟎. 𝟎𝟐 kg
Time taken for the particles to collide,
𝐝 𝟏𝟎𝟎
𝐭= = = 𝟏 𝐬𝐞𝐜
𝐕𝐫𝐞𝟏 𝟏𝟎𝟎
Speed of woodjust before collision = 𝐠𝐭 = 𝟏𝟎𝐦/𝐬 and speed of bullet just before collision
= 𝐯 − 𝐠𝐭
= 𝟏𝟎𝟎 − 𝟏𝟎 = 𝟗𝟎𝐦/𝐬
𝟏
𝐒 = 𝟏𝟎𝟎 × 𝟏 − × 𝟏𝟎 × 𝟏 = 𝟗𝟓𝐦
𝟐
Now, using conservation oflinear momentumjust before and after the collision
‐(𝟎. 𝟎𝟑) (10) + 𝟎. 𝟎𝟐 𝟗𝟎 = 𝟎. 𝟎𝟓 𝐯
⇒ 𝟏𝟓𝟎 = 𝟓𝐯 𝐯 = 𝟑𝟎𝐦/𝐬
𝐌𝐚𝐱 . height reached by body
𝟑𝟎 × 𝟑𝟎
𝐡 == = 𝟒𝟓𝐦
𝟐 × 𝟏𝟎
Before After
𝟎. 𝟎𝟑𝐤𝐠𝑹𝒆𝒋𝒆𝒄𝒕 ↓ 𝟏𝟎𝐦/𝐬 𝑹𝒆𝒋𝒆𝒄𝒕𝐕
𝑹𝒆𝒋𝒆𝒄𝒕
↑ 𝟗𝟎𝐦/𝐬
𝟎. 𝟎𝟐 kg
𝟎. 𝟎𝟓 kg
Height above tower = 𝟒𝟎𝐦
5. There block 𝐀, 𝐁 and 𝐂 are lying on a smooth horizontal surface, as shown in the figure.
A and 𝐁 have equal masses, 𝐦 while 𝐂 has mass M. Block A is given an inital speed 𝐯
towards 𝐁 due to which it collides with 𝐁 perfectly inelastically. The combined mass
collides with
(a) 5 (b) 2 (c) 4 (d) 3

SOLUTION : (c)
Kinetic energy of block 𝐀
𝟏
𝐤𝟏 =𝐦𝐯𝟎𝟐
𝟐
From principle oflinear momentum conservation
𝐦𝐯𝟎
𝐦𝐯𝟎 = 𝟐𝐦 + 𝐌 𝐯𝐟 ⇒ 𝐯𝐟 =
𝟐𝐦 + 𝐌
𝟓
According to question, of 𝟔 th the initial kinetic energy is lost in whole process.
𝟏
𝐤𝐢 𝐦𝐯𝟎𝟐
=𝟔⇒ 𝟐 =𝟔
𝐤𝐟 𝟏 𝐦𝐯𝟎 𝟐
𝟐 𝟐𝐦 + 𝐌 𝟐𝐦 + 𝐌
𝟐𝐦 + 𝐌 𝐌
⇒ = 𝟔.⋅. = 𝟒
𝐦 𝐦
76 . In a collinear collision, aparticle with an initial speed 𝐯𝟎 strikes a stationary particle

ofthe same mass. Ifthe fmal total kinetic energy is 50% greater than the original
kinetic energy, the magnitude of the relative velocity between the two particles, after
collision, is: [2018]
𝐯𝟎 𝐯𝟎 𝐯𝟎
(a) (b) 𝟐𝐯𝟎 (c) (d)
𝟒 𝟐 𝟐
SOLUTION . (b)
Before Collision After Collision
∞𝐦𝐯𝟎𝐎𝐦 ⇒ ∞𝐦𝐯𝟏 ∝ 𝐦𝐯𝟐
Stationary
𝟏 𝟏 𝟑 𝟏
𝐦𝐯𝟏𝟐 + 𝐦𝐯𝟐𝟐 = 𝐦𝐯𝟎𝟐
𝟐 𝟐 𝟐 𝟐
𝟑
⇒ 𝐯𝟏𝟐 + 𝐯𝟐𝟐 = 𝟐 𝐯𝟎𝟐 (i)
From momentum conservation

𝐦𝐯𝟎 = 𝐦 𝐯𝟏 + 𝐯𝟐 (ii)
Squarring both sides,
⇒ 𝐯𝟏 + 𝐯𝟐𝟐 + 𝟐𝐯𝟏 𝐯𝟐 = 𝐯𝟎𝟐 𝐯𝟏 + 𝐯@ 𝟐 = 𝐯𝟎𝟐
𝐯𝟎𝟐
𝟐𝐯𝟏 𝐯𝟐 = −
𝟐
𝟑 𝟐 𝐯𝟎𝟐
𝐯𝟏 − 𝐯𝟐 = 𝟐
𝐯𝟏𝟐 𝐯𝟐𝟐
+ − 𝟐𝐯𝟏 𝐯𝟐 = 𝐯𝟎 +
𝟐 𝟐
Solving we get relative velocity between the two particles
𝐯𝟏 − 𝐯𝟐= 𝟐 𝐯𝟎
77. The mass ofa hydrogen molecule is 𝟑. 𝟑𝟐 × 𝟏𝟎−𝟐𝟕𝐤𝐠. 𝐈𝐟𝟏𝟎𝟐𝟑 hydrogen molecules strike,
per second, a fixed wall ofarea 2 𝐜𝐦𝟐 at an angle of 𝟒𝟓∘ to the normal, and rebound
elastically with a speed of103 𝐦/𝐬, then the pressure on the wall is nearly: [2018]
(a) 𝟐. 𝟑𝟓 × 𝟏𝟎𝟑 𝐍/𝐦𝟐 (b) 𝟒. 𝟕𝟎 × 𝟏𝟎𝟑 𝐍/𝐦𝟐
(c) 𝟐. 𝟑𝟓 × 𝟏𝟎𝟐 𝐍/𝐦𝟐 (d) 𝟒. 𝟕𝟎 × 𝟏𝟎𝟐 𝐍/𝐦𝟐
SOLUTION . (a]
𝐏 𝐏 𝐏 𝐏
𝜟𝐏 = 𝐉+ 𝐉+ î‐ î
𝟐 𝟐 𝟐 𝟐
𝟐𝐏
𝜟𝐏 = 𝐉 = 𝐈𝐇 molecule
𝟐
𝟐𝐏
⇒ 𝐈𝐰𝐚𝐥𝟏 = − 𝐉
𝟐
Pressure, 𝐏
𝜞 𝟐𝐩
=𝐀= 𝐧 (⋅.⋅ 𝐧 = 𝐧𝐨. of particles)
𝐀
𝟐 × 𝟑. 𝟑𝟐 × 𝟏𝟎−𝟐𝟕 × 𝟏𝟎𝟑 × 𝟏𝟎𝟐𝟑

= = 𝟐. 𝟑𝟓 × 𝟏𝟎𝟑 𝐍/𝐦𝟐
𝟐 × 𝟏𝟎−𝟒
78. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest,
fractional loss ofits energy is 𝐩𝐝 ; while for its similar collision with carbon nucleus at
rest, fractional loss of energy is 𝐏𝐜 . The values ofPdand 𝐏𝐜 are respectively: [2018]
(a) ⋅ 𝟖𝟗,⋅ 𝟐𝟖 𝐛 ⋅ 𝟐𝟖,⋅ 𝟖𝟗 𝐜 𝟎, 𝟎 (d) 𝟎, 𝟏
SOLUTION (a)
For collision ofneutron with deuterium:
𝐦 𝟐𝐦 𝐦 𝟐𝐦
𝐯 ∙ 𝐯𝟏 𝐯𝟐
Applying conservation ofmomentum:
𝐦𝐯 + 𝟎 = 𝐦𝐯𝟏 + 𝟐𝐦𝐯𝟐 (i)
𝐯𝟐 − 𝐯𝟏 = 𝐯 (ii)
Collision is elastic, 𝐞 = 𝟏
𝐯
From eqn (i) and eqn (ii) 𝐯𝟏 = − 𝟑
𝟏 𝟏
𝐦𝐯 𝟐 − 𝟐 𝐦𝐯𝟏𝟐 𝟖
𝐏𝐝 = 𝟐 = = 𝟎. 𝟖𝟗
𝟏 𝟐 𝟗
𝟐 𝐦𝐯
Now, For collision of neutron with carbon nucleus
𝐦 𝟏𝟐𝐦 𝐦 𝟏𝟐𝐦
𝐯 ∙ 𝐯𝟏 𝐯𝟐
Applying Conservation ofmomentum
𝒎𝒗 + 𝟎 = 𝒎𝐯𝟏 + 𝟏𝟐𝒎𝐯𝟐 (iii)
𝐯 = 𝐯𝟐 − 𝐯𝟏 (iv)
From eqn (iii) and eqn (iv)
𝟏𝟏
𝐯𝟏 = − 𝐯
𝟏𝟑
𝟐
𝟏 𝟏 𝟏𝟏
𝐦𝐯 𝟐 − 𝟐 𝐦 𝟏𝟑 𝐯 𝟒𝟖
𝐏𝐜 = 𝟐 = ≈ 𝟎. 𝟐𝟖
𝟏 𝟐 𝟏𝟔𝟗
𝟐 𝐦𝐯
79. A proton ofmass 𝐦 collides elastically with a particle of unknown mass at rest. After
the collision, the proton and the unknown particle are seen moving at an angle of 𝟗𝟎∘
with respect to each other. The mass ofunknown particle is: [Online April 15, 2018]
𝐦 𝐦
(a) (b) (c) 𝟐𝐧 (d) 𝐦
𝟑 𝟐
SOLUTION : . (d)
Applyprinciple ofconservation ofmomentum along 𝒙‐direction,
𝒎𝒖 = 𝒎𝐯𝟏 𝐜𝐨𝐬 𝟒𝟓∘ + 𝑴𝐯𝟐 𝐜𝐨𝐬 𝟒𝟓∘
𝟏
𝒎𝒖 = 𝒎𝒗𝐢 + 𝑴𝐯𝟐 ..... (i)
𝟐
Along 𝒚‐direction,
𝐨 = 𝒎𝐯𝟏 𝐬𝐢𝐧 𝟒𝟓∘ − 𝑴𝐯𝟐 𝐬𝐢𝐧 𝟒𝟓∘
𝟏
𝐨 = 𝒎𝐯𝟏 − 𝑴𝐯𝟐 ..... (ii)
𝟐
Protonm,𝒖𝟏 = 𝒖 Unknown massM,𝒖𝟐 = 𝟎
Before collision
After collision
𝐯𝟐 −𝐯𝟏 𝐜𝐨𝐬 𝟗𝟎
Coefficient of restution 𝐞 = 𝟏 = 𝒖 𝐜𝐨𝐬 𝟒𝟓
(Collision is elastic)
𝐯𝟐
⇒ =𝟏
𝒖
⇒ 𝒖 = 𝟐𝐯𝟐 (iii)
Solving eqs (i), (ii), & (iii), we get mass of unknown particle, 𝑴 = 𝒎
80. Two particles A and 𝐁 of equal mass 𝐌 are moving with the same speed 𝐯 as shown in
the figure. They collide completely inelastically and move as a single particle C. The
angle 𝜽 that the path of 𝐂 makes with the 𝐗‐axis is [Online April 9, 2017]
given by:
𝟑+ 𝟐 𝟑− 𝟐
(a) 𝐭𝐚𝐧 𝜽 = (b) 𝐭𝐚𝐧 𝜽 =
𝟏− 𝟐 𝟏− 𝟐
𝟏− 𝟐 𝟏− 𝟑
(c) 𝐭𝐚𝐧 𝜽 = (d) 𝐭𝐚𝐧 𝜽 = 𝟏+
𝟐 𝟏+ 𝟑 𝟐
SOLUTION : (a)
For particle 𝐂,
According to law of conservation of linear momentum, verticle component,
2 𝐦𝐯 ′ 𝐬𝐢𝐧 𝜽 = 𝐦𝐯 𝐬𝐢𝐧 𝟔𝟎∘ + 𝐦𝐯 𝐬𝐢𝐧 𝟒𝟓∘
𝐦𝐯 𝐦𝐯 𝟑
𝟐𝐦𝐯 ′ 𝐬𝐢𝐧 𝜽 = + ...... (i)
𝟐 𝟐
Horizontal component,
2 𝐦𝐯 𝐜𝐨𝐬 𝜽 = 𝐦𝐯 𝐬𝐢𝐧 𝟔𝟎∘ − 𝐦𝐯 𝐜𝐨𝐬 𝟒𝟓∘
′
𝐦𝐯 𝐦𝐯
𝟐𝐦𝐯 ↑ 𝐜𝐨𝐬 𝜽 = + ...... (ii)
𝟐 𝟐
For particle A For particle 𝐁
Dividing 𝐞𝐪𝐧 (i) by 𝐞𝐪𝐧 (ii),
𝟏 𝟑
+ 𝟐
𝟐 𝟏 𝟐+ 𝟑
𝐭𝐚𝐧 𝜽 = =
𝟏 𝟐 𝟏− 𝟐
𝟐−
81. A neutron moving with a speed 𝐯 ↑ makes a head on collision with a stationary hydrogen
atom in ground state. The minimum kinetic energy of the neutron for which inelastic
collision will take place is : [Online Apri110, 2016]
(a) 𝟐𝟎. 𝟒𝐞𝐕 ( 𝐛 𝟏𝟎. 𝟐𝐞𝐕 (c) 𝟏𝟐. 𝟏𝐞𝐕 (d) 𝟏𝟔. 𝟖𝐞𝐕
SOLUTION : . (a)
𝐦𝟏
For inelastic collision 𝐯 ′ = 𝐯
𝐦𝟏 +𝐦𝟐
𝟏 𝐯
= 𝐯=
𝟏+𝟏 𝟐
𝐧 → 𝐯 𝐇 Before
𝐯
(n) 𝐇 → 𝟐 After
𝟏 𝟏 𝐯 𝟐 𝟏
Loss in K.E. = 𝟐 𝐦𝐯 𝟐 − 𝟐 𝟐𝐦 = 𝟒 𝐦𝐯 𝟐
𝟐
K.E. lost is used tojump fiiom 1st orbit to 2nd orbit

𝜟𝐊. 𝐄. = 𝟏𝟎. 𝟐𝐞𝐯
Minimum K.E. ofneutron for inelastic collision
𝟏
𝐦𝐯 𝟐 = 𝟐 × 𝟏𝟎. 𝟐 = 𝟐𝟎. 𝟒𝐞𝐕
𝟐
82. Aparticle ofmass 𝐦 moving in thex direction with speed 𝟐𝒗 is hit by another particle
ofmass 𝟐𝒎 moving in the 𝒚 direction with speed 𝐯. Ifthe collision is perfectly inelastic,
the percentage loss in the energy during the collision is close to : [2015]
(a) 56% (b) 62% (c) 44% (d) 50%
SOLUTION :[a]
Initial momentum of the system
𝐩𝐢 = 𝐦 𝟐𝐕 𝟐 × 𝟐𝐦 𝟐𝐕 𝟐
= 𝟐𝐦 × 𝟐𝐕
Final momentum of the system = 𝟑𝐦𝐕
By the law ofconservation ofmomentum
𝟐 𝟐𝐯
𝟐 𝟐𝐦𝒗 = 𝟑𝐦𝐕 ⇒ = 𝐕𝐜𝐨𝐦𝐛𝐢𝐧𝐞𝐝
𝟑
Loss in energy
𝟏 𝟏 𝟏 𝟐
𝜟𝐄 = 𝐦𝟏 𝐕𝟏𝟐 + 𝐦𝟐 𝐕𝟐𝟐 − 𝐦𝟏 + 𝐦𝟐 𝐕𝐜𝐨𝐦𝐛𝐦𝐞𝐝
𝟐 𝟐 𝟐
𝟒 𝟓
DE = 𝟑𝐦𝐯 𝟐 − 𝟑 𝐦𝐯 𝟐 = 𝟑 𝐦𝐯 𝟐 = 𝟓𝟓. 𝟓𝟓%
Percentage loss in energyduring the collision = 𝟓𝟔%
83. A bullet of mass 𝟒𝐠 is fired horizontally with a speed of 300 𝐦/𝐬 into 𝟎. 𝟖 kg block
ofwood at rest on a table. Ifthe coefficient of fiiction between the block and the table is
𝟎. 𝟑, how far will the block slide approximately? [Online Apri112, 2014]
(a) 𝟎. 𝟏𝟗𝐦 (b) 𝟎. 𝟑𝟕𝟗𝐦 (c) 𝟎. 𝟓𝟔𝟗𝐦 (d) 𝟎. 𝟕𝟓𝟖𝐦
SOLUTION : (b)
Given, 𝐦𝟏 = 𝟒𝐠, 𝐮𝟏 = 𝟑𝟎𝟎𝐦/𝐬
𝐦𝟐 = 𝟎. 𝟖𝐤𝐠 = 𝟖𝟎𝟎𝐠, 𝐮𝟐 = 𝟎𝐦/𝐬
From law of conservation ofmomentum,
𝐦𝟏 𝐮𝟏 + 𝐦𝟐 𝐮𝟐 = 𝐦𝟏 𝐯𝟏 + 𝐦𝟐𝐕 𝟐
Let the velocity of combined system = 𝐯𝐦/𝐬 then,
𝟒 × 𝟑𝟎𝟎 + 𝟖𝟎𝟎 × 𝟎 = 𝟖𝟎𝟎 + 𝟒 × 𝐯
𝟏𝟐𝟎𝟎
𝐯= = 𝟏. 𝟒𝟗𝐦/𝐬
𝟖𝟎𝟒
Now, 𝝁 = 𝟎. 𝟑 (given)
𝐚 = 𝝁𝐠
𝐚 = 𝟎. 𝟑 × 𝟏𝟎 (take 𝐠 = 𝟏𝟎𝐦/𝐬𝟐 )
= 𝟑𝐦/𝐬𝟐
then, from 𝐯 𝟐 = 𝐮𝟐 + 𝟐𝐚𝐬
(1.49)2 = 𝟎 + 𝟐 × 𝟑 × 𝐬
𝟏. 𝟒𝟗 𝟐
𝐬=
𝟔
𝟐. 𝟐𝟐
𝐬= = 𝟎. 𝟑𝟕𝟗𝐦
𝟔
84. Three masses m, 2mand3mare moving in x‐y plane with speed 𝟑𝐮, 𝟐𝐮 and 𝐮
respectively as shown in figure. The three masses collide at the same point at 𝐏 and
stick together. The velocity ofresulting mass will be: [Online Apri112, 2014]
𝟐𝐮
)𝐮
𝐮 𝐮 𝐮 𝐮
(a) 𝟏𝟐(î+ 3j) (b) 𝟏𝟐(î‐ 3j) (c) 𝟏𝟐(‐î+ 3j) d) 𝟏𝟐(‐î‐ 3j)
SOLUTION : . (d)
From the law of conservation ofmomentum we know that,
𝐦𝟏 𝐮𝟏 + 𝐦𝟐 𝐮𝟐 + ⋯ . = 𝐦𝟏 𝐯𝟏 + 𝐦𝟐 𝐯𝟐 + ⋯.
Given 𝐦𝟏 = 𝐦, 𝐦𝟐 = 𝟐𝐦 and 𝐦𝟑 = 𝟑𝐦
and 𝐮𝟏 = 𝟑𝐮, 𝐮𝟐 = 𝟐𝐮 and 𝐮𝟑 = 𝐮
→
Let the velocity when they stick = 𝐯
Then, according to question,
𝐦 × 𝟑𝐮 (î) +𝟐𝐦 × 𝟐𝐮 (‐îcos 𝟔𝟎∘ − 𝐣 𝐬𝐢𝐧 𝟔𝟎∘ )

+𝟑𝐦 × 𝐮 (‐îcos60∘ +jsin60∘) = 𝐦 + 𝟐𝐦 + 𝟑𝐦 → 𝐯
𝐢 𝐢
⇒ 𝟑𝐦𝐮𝐢 − 𝟒𝐦𝐮 − 𝟒𝐦𝐮 − 𝟑𝐦𝐮
𝟐 𝟐
𝟑
+𝟑𝐦𝐮 𝐣 = 𝟔𝐦𝐯 →
𝟐
𝟑 𝟑
⇒ 𝐦𝐮𝐢 − 𝟐 𝐦𝐮𝐢 − 𝐦𝐮𝐣 = 𝟔𝐦𝐯 →
𝟐
𝟏 𝟑
⇒ − 𝟐 𝐦𝐮𝐢 − 𝐦𝐮𝐣 = 𝟔𝐦𝐯 →
𝟐
𝐮
⇒ → 𝐯 = 𝟏𝟐(‐î‐ 3j)
85. This question has statement I and statement II. Ofthe four choices given after the
statements, choose the one that best describes the two statements.
Statement‐ I: Apoint particle of mass 𝐦 moving with speed 𝐮 collides with stationary
𝟏
point particle of mass M. If the maximum energy loss possible is given as 𝒇 𝐦𝐯 𝟐
𝟐
𝐦
then = .
𝐌+𝐦
Statement‐ II: Maximum energy loss occurs when the particles get stuck together as a
result ofthe collision. [2013]
(a) Statement‐I is true, Statment‐ II is true, Statement ‐ II is the correct
explanation of Statement‐I.
(b) Statement‐I is true, Statment‐ II is true, StatementII is not the correct
explanation of Statement‐ II.
(c) Statement‐I is true, Statment‐ II is false.
(d) Statement‐I is false, Statment‐ II is true.
SOLUTION : . (d)
𝐏𝟐 𝐏𝟐
Maximum energy loss = 𝟐𝐦 − 𝟐 𝐦+𝐌
𝐏𝟐 𝟏
[⋅.⋅ K.E. = 𝟐𝐦 = 𝟐 𝐦𝐯 𝟐 ]
𝐏𝟐 𝐌 𝟏 𝐌
= = 𝐦𝐯 𝟐
𝟐𝐦 𝐦 + 𝐌 𝟐 𝐦+𝐌
Statement II is a case ofperfectly inelastic collision.
By comparing the equation given in statement I with above equation, we 𝐠𝐞𝐭\
𝐌 𝐦
𝐟= instead of
𝐦+𝐌 𝐌+𝐦
Hence statement I is wrong and statement II is correct.

86. A projectile of mass 𝑴 is fired so that the horizontal range is 4 km. At the highest point
the projectile explodes in two parts of masses Ml4 and 𝟑𝑴𝒍𝟒 respectively and the
heavier part starts falling down vertically with zero initial speed. The horizontal range
(distance from point of firing) of the lighter part is : [Online April 23, 2013]
(a) 16 km (b) 1 km (c) 10 km (d) 2 km
SOLUTION : . (c)
87. A moving particle of mass 𝒎, makes a head on elastic collision with another particle
ofmass 𝟐𝒎, which is initially at rest. The percentage loss in energy of the colliding
particle on collision, is close to [Online May 19, 2012]
(a) 33% (b) 67% (c) 𝕬)% (d) 10%
SOLUTION : . (c)
Fractional decrease in kinetic energy of mass 𝐦
𝟐
𝟐
𝟐−𝟏
=𝟏− =𝟏−
𝟐+𝟏
𝟏 𝟐 𝟏 𝟖
=𝟏− =𝟏− =
𝟑 𝟗 𝟗
Percentage loss in energy
𝟖
= × 𝟏𝟎𝟎 = 𝟗𝟎%
𝟗
88. Two bodies 𝑨 and 𝑩 of mass 𝒎 and 𝟐𝒎 respectively are placed on a smooth floor. They
are connected by a spring of negligible mass. 𝑨 third body 𝑪 of mass 𝒎 is placed on the
floor. The body 𝑪 moves with a velocity 𝒗𝟎 along the line joining 𝑨 and 𝑩 and collides
elastically with 𝑨. At a certain time after the collision it is found that the instantaneous
velocities of 𝑨 and 𝑩 are same and the compression of the spring is 𝒙𝟎 . The spring
constant 𝒌 will be [Online May 12, 2012]
𝒗𝟐 𝒗 𝒗 𝟐
(a) 𝒎 𝒙𝟐𝟎 (b) 𝒎 𝟐𝒙𝟎 (c) 𝟐𝒎 𝒙𝟎 (d) 𝟑 𝒎 𝟐
𝟎 𝟎 𝟎
SOLUTION : (d)
𝒗𝟎 →
Initial momentum of the system block (C) = 𝒎𝒗𝟎 . After striking with 𝑨, the block 𝑪 comes to
rest and now both block A and 𝐁 moves with velocity 𝐯 when compression in sprin 𝐠 is 𝒙𝟎 .
By the law ofconservation oflinear momentum
𝒗𝟎
𝒎𝒗𝟎 = 𝒎 + 𝟐𝒎 𝒗 ⇒ 𝒗 =
𝟑
By the law of conservation of energy
K.E. ofblock 𝑪 = 𝐊. 𝐄. of system +𝐏. 𝐄. of system
𝟏 𝟏 𝒗𝟎 𝟐 𝟏
𝒎𝒗𝟐𝟎 = 𝟑𝒎 + 𝒌𝒙𝟐𝟎
𝟐 𝟐 𝟑 𝟐
𝟏 𝟏 𝟏
⇒ 𝒎𝒗𝟐𝟎 = 𝒎𝒗𝟐𝟎 + 𝒌𝒙𝟐𝟎
𝟐 𝟔 𝟐
𝟏 𝟏 𝟏 𝒎𝒗𝟐𝟎
⇒ 𝟐 𝒌𝒙𝟐𝟎 = 𝟐 𝒎𝒗𝟐𝟎 − 𝟔 𝒎𝒗𝟐𝟎 = 𝟑
𝟐 𝟐
𝒌= 𝒎
𝟑
89. A projectile moving vertically upwards with a velocity of 200 𝐦𝐬−𝟏 breaks into two
equal parts at a height of490 𝐦. One part starts moving vertically upwards with a
velocity of400 𝐦𝐬−𝟏 . How much time it will take, after the break up with the other part
to hit the ground? [Online May 12, 2012]
(a) 𝟐 𝟏𝟎𝐬 (b) 𝟓𝐬 (c) 10 𝐬 (d) 𝟏𝟎𝐬
SOLUTION : . (c)
𝐧=𝒎
490
𝐦/𝐬 (vertically)
Momentum before explosion = Momentum after explosion
𝒎 𝒎
𝒎 × 𝟐𝟎𝟎𝒋 = × 𝟒𝟎𝟎𝒋 + 𝒗
𝟐 𝟐
𝒎
= 𝟒𝟎𝟎𝒋 + 𝒗
𝟐
⇒ 𝟒𝟎𝟎𝒋 − 𝟒𝟎𝟎𝒋 = 𝒗
𝒗=𝟎
i. e., the velocity ofthe other part ofthe mass, 𝒗 = 𝟎
Let time taken to reach the earth by this part be 𝒕
𝟏
Applying formula, 𝒉 = 𝒖𝒕 + 𝟐 𝒈𝒕𝟐
𝟏
𝟒𝟗𝟎 = 𝟎 + × 𝟗. 𝟖 × 𝒕𝟐
𝟐
𝟗𝟖𝟎
⇒ 𝒕𝟐 = = 𝟏𝟎𝟎
𝟗.𝟖
𝒕 = 𝟏𝟎𝟎 = 𝟏𝟎 𝐬𝐞𝐜
90 . Statement ‐𝟏: Two particles moving in the same direction do not lose all their energy
in a completely inelastic collision.
Statement −𝟐 : Principle of conservation of momentum holds true for all kinds of
collisions. [2010]
(a) Statement‐l is true, Statement ‐𝟐 is true; Statement −𝟐 is the correct explanation
of Statement ‐𝟏.
(b) Statement‐l is true, Statement ‐𝟐 is true; Statement ‐𝟐 is not the correct
explanation of Statement‐l
(c) Statement‐l is false, Statement ‐𝟐 is true.
(d) Statement‐l is true, Statement ‐𝟐 is false.
SOLUTION : . (a)
In completely inelastic collision, all initial kinetic energy is not lost but loss in kinetic energy
15 as large as it can be. Linear momentum remain conserved in all types of collision. Statement
‐𝟐 explains statement ‐𝟏 correctly because applying the principle of conservation
ofmomentum, we can get the common velocity and hence the kinetic energy of the combined
body.
91. A block ofmass 𝟎. 𝟓𝟎 kg is moving with a speed of 𝟐. 𝟎𝟎𝐦𝐬 −𝟏 on a smooth surface. It
strikes another mass of 𝟏. 𝟎𝟎 kg and then they move together as a single body. The
energy loss during the collision is [2008]
(a) 𝟎. 𝟏𝟔 𝐉 (b) 𝟏. 𝟎𝟎 𝐉 (c) 𝟎. 𝟔𝟕 𝐉 (d) 𝟎. 𝟑𝟒 𝐉
SOLUTION : (c)
Initial kinetic energy of the system
𝟏 𝟏
𝐊. 𝐄𝒋 = 𝒎𝒖𝟐 + 𝑴 𝟎 𝟐
𝟐 𝟐
𝟏
= × 𝟎. 𝟓 × 𝟐 × 𝟐 + 𝟎 = 𝟏𝐉
𝟐
Momentum before collision
= Momentum after collision
𝒎𝟏 𝒖 𝟏 + 𝒎𝟐 𝒖 𝟐 = 𝒎 + 𝑴 × 𝒗
𝟐
𝟎. 𝟓 × 𝟐 + 𝟏 × 𝟎 = 𝟎. 𝟓 + 𝟏 × 𝒗 ⇒ 𝒗 = 𝐦/𝐬
𝟑
Final kinetic energy of the system is
𝟏 𝟐
K.E 𝒇=𝟐 𝒎+𝑴 𝒗
𝟏 𝟐 𝟐 𝟏
= 𝟎. 𝟓 + 𝟏 × × = 𝐉
𝟐 𝟑 𝟑 𝟑
Energy loss during collision
𝟏−𝟏
= 𝐉 = 𝟎. 𝟔𝟕𝐉
𝟑
92. A bomb ofmass 𝟏𝟔𝐤𝐠 at rest explodes into two pieces of masses 4 kg and 12 𝐤𝐠. The
velolcity ofthe 12 kg mass is 4 𝐦𝐬−𝟏 . The kinetic energy ofthe other mass is [2006]
(a) 144 𝐉 (b) 288 𝐉 (c) 192 𝐉 (d) 96 𝐉
SOLUTION : . (b)
Let the velocity and mass of 4 kg piece be 𝒗𝟏 and 𝒎𝟏 and that of 12 kg piece be 𝒗𝟐 and 𝒎𝟐 .
Situation 1𝐎𝐥𝟔𝐤𝐠 Initial 𝐦𝐨𝐦𝐞𝐧𝐭𝐮𝐦 = 𝟎
𝟒𝐤𝐠 =𝐦 𝟏
𝐯𝟏 𝝍 ∗ 𝐯𝟐 𝐦𝟐 = 𝟏𝟐𝐤𝐠𝐅𝐢𝐧𝐚𝟏𝐦𝐨𝐦𝐞𝐧𝐭𝐮𝐦 = 𝐦𝟐 𝐯𝟐 − 𝐦𝟏 𝐯𝟏
𝐀𝐩𝐩𝐥𝐲𝐢𝐒𝐢𝐭𝐮𝐚𝐭𝐢𝐨𝐧𝟐 𝐧𝐠𝐜onservation oflinear momentum
𝟏𝟔 × 𝟎 = 𝟒 × 𝒗𝟏 + 𝟏𝟐 × 𝟒
𝟏𝟐×𝟒
⇒ 𝒗𝟏 = − = −𝟏𝟐𝒎𝒔−𝟏
𝟒
Kinetic energy of 4 kg mass

𝟏 𝟏
𝑲. 𝑬. = 𝒎𝟏 𝒗𝟐𝟏 = × 𝟒 × 𝟏𝟒𝟒 = 𝟐𝟖𝟖𝑱
𝟐 𝟐
93. The block ofmassMmoving on the fiiictionless horizontal surface collides with the
spring of spring constant 𝒌 and compresses it by length 𝑳. The maximum momentum of
𝒌𝑳𝟐 𝑴𝑳𝟐
(a) (b) 𝑴𝒌𝑳 (c) (d) zero
𝟐𝑴 𝒌
SOLUTION : . (b)
When the spring gets compressed by length L.
K.E. lost bymass 𝐌 = 𝐏. 𝐄. stored in the compressed spring.
Momentum ofthe block, = 𝑴 × 𝒗
𝒌
= 𝑴× ⋅ 𝑳 = 𝒌𝑴 ⋅ 𝑳
𝑴
94. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another
identical mass. After collision
𝒗
the 𝟏𝐬𝐭 mass moves with velocity in a direction perpendicular to the initial direction
𝟑
ofmotion. Find the speed of the 𝟐𝐧𝐝 mass after collision. [↔𝒎 ∙ 𝒎 ↓ 𝒗/ 𝟑 after
before collision [2005]
𝒗 𝟐
(a) 𝟑𝒗 (b) 𝒗 (c) (d) 𝒗
𝟑 𝟑
SOLUTION : . (d)
Considering conservation ofmomentum along x‐direction,
𝒎𝒗 = 𝒎𝒗𝟏 𝐜𝐨𝐬 𝜽 (1)
where 𝒗𝟏 is the velocity of second mass
In 𝒚‐direction,
𝒎𝒗
𝟎= − 𝒎𝒗𝟏 𝐬𝐢𝐧 𝜽
𝟑
𝒎𝒗
or 𝒎𝟏 𝒗𝟏 𝐬𝐢𝐧 𝜽 = (2)
𝟑
Squaring and adding eqns. (1) and(2) we get
𝟐
𝒗𝟐 𝟐
𝒗𝟏𝟐 = 𝒗 + ⇒ 𝒗𝟏 = 𝒗
𝟑 𝟑
95. Consider the following two statements : [2003] A Linear momentum of a system of
particles is zero B. Kinetic energy of a system of particles is zero.
Then
(a) A does not imply 𝑩 and 𝑩 does not imply 𝑨
(b) 𝑨 implies 𝑩 but 𝑩 does not imply 𝑨
(c) 𝑨 does not imply 𝑩 but 𝑩 implies 𝑨
(d) 𝑨 implies 𝑩 and 𝑩 implies 𝑨
SOLUTION : . (c)
Kinetic energy of a system of particle is zero only when the speed of each particles is zero. This
implies momentum ofeach particle is zero, thus linear momentum of the system of particle has
to be zero.
Also iflinear momentum of the system is zero it does not mean linear momentum ofeach
particle is zero. This is because linear momentum is a vector quantity. In this case the kinetic
energy of the system ofparticles will not be zero.
A does not imply 𝐁 but 𝐁 implies A.
Given, force, 𝑭 = 𝟐𝟎𝟎𝐍 extension ofwire, 𝒙 = 𝟏𝐦𝐦.
SYSTERM OF PARTICLES AND ROTATIONAL MOTION
CENTRE OF MASS
C.M is the point which behaves as if total mass of the body is supposed to be concentrated at
that point. This point may lie inside or outside the material of body, but always lies within the
space occupied by the body.
Mass may exist or may not exist at the location of centre of mass.
Centre of gravity: The point through which weight of the body acts is called centre of gravity.
Coordinates of C.M.
Coordinates of C.M. of a system of ‘n’ particles
m1x1  m2x2  ......+mnxn

Xcm =
m1  m2 ......  mn
m1 y1  m2y2  ......+mn yn
Ycm =
m1  m2 ......  mn
m1 z1  m2z2  ......+mn zn
Zcm =
m1  m2  ......  mn
Case-1: Position of C.M. of two particle system:
In case of two bodies, the ratio of distance of centre of mass from the bodies is in the
inverseratio of their masses. If m1 and m2 are masses of two bodies separated by a
distance ‘d’ then
d1 m 2
m1d1  m 2 d 2  or  
d 2 m1
Thus C.M. locates near to heavier body.

In figure, d  d1  d 2 .
On solving, m1d1  m2 d 2
m1d1  m2   d  d1 
m2 d m1 d
 d1 = m  m and d2 = m  m
1 2 1 2
d1 , d2 are the distances of CM from m1 , m2.

1: If the distance between the centres of the atoms of potassium and bromine in KBr
(potassium-bromide) molecule is 0.282  10 9 m , find the centre of mass of this two particle
system from potassium mass of bromine = 80 u, and of potassium = 39 u)
Sol: Let position co-ordinate of potassium, xk  0
Position co-ordinate of bromine,
xBr  0.282  10 9 m .
 Position co-ordinate of centre of mass.
mk xk  mBr xBr
xc 
mk  mBr
39  0  80  0.282  10 9
 xc 
39  80
 xc  0.189  109 m .
2: Two blocks of masses 10 kg and 30 kg are placed on x-axis. The first mass is moved
on the axis by a distance of 2 cm right. By .what distance should the second mass be
moved to keep the position of centre of mass unchanged.
Sol:
Mass of the first block, m1  10 kg .

Mass of the second block, m2  30 kg .
m1x1  m2 x2
xcm 
m1  m2
10  2  30x2
0 .
40
2
x2   . Therefore the second block should be moved left through a distance of
3
2
cm to keep the position of centre of mass unchanged.
3
3.When ‘n’ number of particles of masses m, 2m,3m,....nm are at distances

x1  1, x2  2, x3  3....xn  n units respectively from origin on the x-axis, then find the
distance of centre of mass of the system from origin.
Sol:
m 1  2m  2   3m  3  ...   nm  n
xcm 
m  2m  3m  ...  nm
m 12  22  32  ....  n 2 
xcm 
m 1  2  3  ....  n 
 n  n  1 2 n  1 
 
xcm   6   2n  1
 n  n  1  3 .
 
 2 
4 When‘n’ number of particles of masses m, 2m,3m,....nm are at
distances x1  1, x2  4, x3  9,...xn  n 2 units respectively from origin on the x-axis, then
find the distance of their centre of mass from origin.
Sol:
m 1  2m  4   3m  9   ....  nm  n 2 
xcm 
m  2m  3m  ....  nm
m 1  23  33  .....  n3 

m 1  2  3  ....  n 
 n  n  1 
2
 
  2   n  n  1
n  n  1 2 .
2
5.When ‘n’ number of particles each of mass ‘ m ’ a r e a t d i s t a n c e s
x1  a, x2  ar , x3  ar 2 ....xn  ar n units
from origin on the x-axis, then find the
distance of their centre of mass from origin.
ma  m  ar   m  ar 2   ....  m  ar n 
Sol: xcm 
m  m  m  ...  m  nterms 
m  a  ar  ar 2  ....  ar n 
xcm 
mn
If r  1 then
1  a  r  1  a  r  1
n n
xcm   
n  r  1  n  r  1
If r  1 then
1  a 1  r   a 1  r 
n n
xcm   
n  1  r  n 1  r 
6.A 1m long rod having a constant cross sectional area is made of four materials.
The first 0.2 m are made of iron, the next 0.3 m of lead, the following 0.2 m of
aluminium and the remaining part is made of copper. Find the centre of mass of
the rod. The densities of iron, lead, auminium and copper are
7.9  103 kg / m3 , 11.4  103 kg / m3 ,
2.7  103 kg / m3 and 8.9  103 kg / m3
respectively.
Sol:
mass  m   volume  v   density  d 

m  Area  A  length  l   density  d 
m  Ald .
Mass of iron part, m1  A  0.2  7.9  103  1.58  103 A .
Mass of lead part, m2  A  0.3  11.4  103  3.42  103 A .
Mass of aluminium part,
m3  A  0.2  2.7  103  0.54  103 A .
Mass of copper part,
m4  A  0.3  8.9  103  2.67  103 A .
Co-ordinate of iron part from end “O” of the rod x1  0.1 m .
Co-ordinate of lead part from end “O” of the rod x2  0.35 m .
Co-ordinate of aluminium part from end “O” of the rod, x3  0.6 m .
Co-ordinate of copper part from end “O” of the rod, x4  0.85 m .
 Centre of mass of the rod,
m1 x1  m2 x2  m3 x3  m4 x4
Xc 
m1  m2  m3  m4
 Xcm 
1.58 10  0.1  3.42 10  0.35  0.54 10  0.6  2.67 10  0.85 A
3 3 3 3
1.5810  3.42 10  0.5410  2.67 10  A

3 3 3 3
 X cm  0.481 m from the end “O” of the rod.

7: If the centre of mass of three particles of masses of 1kg, 2kg, 3kg is at (2,2,2), then
where should a fourth particle of mass 4kg be placed so that the combined
centre of mass may be at (0,0,0).
Sol. Let  x1 , y1 , z1  ,  x2 , y2 , z2  and  x3 , y3 , z3  be the positions of masses 1kg, 2kg, 3kg
and let the co-ordinates of centre of mass of the three particle system is  xcm , ycm , zcm 
respectively.
m1 x1  m2 x2  m3 x3
xcm 
m1  m2  m3
1 x1  1 x2  1 x3
2 ,
1 2  3
 or  x1  2 x2  3x3  12 .............(1)
Suppose the fourth particle of mass 4kg is placed at  x4 , y4 , z4  so that centre of mass
of new system shifts to (0,0,0).
For X co-ordinate of new centre of mass we have
1 x1  2  x2  3  x3  4  x4
0
1 2  3  4
 x1  2 x2  3 x3  4 x4  0 ..........(2)
from equations (1) and (2)
12  4 x4  0  x4  3
similarly, y4  3 and z4  3
Therefore 4kg should be placed at (-3,-3,-3).
Case-3: Center of mass of a system of particles in

(two dimensional) Plane:
Consider n-particles in x-y plane having masses m1 , m2 ,...., mn with co-ordinates

 x , y  ,  x , y  ,.....,  x , y  respectively.The distance of centre of mass from origin in a
1 1 2 2 n n
plane is d  2
xcm  ycm
2
n n
m x i i m y i i
where x  i 1 and y  i 1
cm cm
M M
8.Find position of centre of mass of four particle system, which are at the vertices
of a parallelogram, as shown in figure.
Sol: From figure
DC  b sin  , OD  b cos .
m1 x1  m2 x2  m3 x3  m4 x4
xcm 
m1  m2  m3  m4
 a  b cos  b sin
xcm  , similarly ycm 
2 2
 a  b cos  b sin  
  xcm , ycm  =  ,
2 
 2
.
Case-4: Centre of mass of a system of ‘n’ particles in (Three dimensional)
Space:
Consider n-particles in space having masses w i t h
cordinates.  x1 , y1 , z1  ,  x2 , y2 , z2  ......  xn , yn , zn  respectively, then distance of centre of mass
from origin in space is d  2

xcm  ycm
2
 zcm
2
n n n
 mi xi  mi y i  mi zi
Where xcm  i 1
, y cm  i 1 , z cm  i  1
M M M
Case-5: Position vector of Centre of mass:
   
Let r1 , r2 , r3 ,....., rn be the position vectors of n-particles having masses m1,m2,......,mn

respectively. If rc.m. is position vector of their C.M., then
n


  
m1r1  m2 r2  ......  mn rn m r i 1
rc.m.   i 1
m1  m2  ......  mn M
  
Where r1  x1iˆ  y1 ˆj  z1kˆ, r2  x2iˆ  y2 ˆj  z2 kˆ, ,....., rn  xn iˆ  yn ˆj  zn kˆ
 rcm  xcmiˆ  ycm ˆj  zcmkˆ

MOTION OF CENTRE OF MASS
(a) Velocity of centre of mass  v cm  :
If v1,v2,....,vn are velocities of particles of masses
m1, m 2 ,...m n respectively and vcm is velocity of their centre of mass then
dr cm 1 n  dr i 
vcm    mi  
dt M i1  dt 
m1 v1  m 2 v2  ...  m n v n
vcm 
m1  m 2  ...  m n
1 n
  m i vi
M i1
where M  m1  m2  ...  mn = Total mass of the system.
When two particles of masses m1 and m 2 aremoving from a point with velocities v1 and v2 at
an angle ‘  ’ with each other, then the velocity of their centre of mass is given by
m12 v12  m 22 v22  2  m1v1  m 2 v2  cos 
vcm 
 m1  m2 
m1v1  m 2 v 2
If they move in the same direction, then   0 and vcm 
m1  m 2
m12 v12  m 22 v2 2
f they move at right angles to each other, then   90 and vcm 
m1  m 2
m1v1  m 2 v2
f they move in opposite directinos, then   180 and vcm  m1  m 2
(b)Linear momentum of centre of mass  p cm  :
If p1, p2 ,...., pn are linear momenta of particles of masses m1, m 2 ,...m n respectively and pcm is
n
linear momentum of their centre of mass then p cm  p1  p 2  ...  p n   p i ,
i 1
n
pcm  Mvcm   mi vi  Psystem
i 1
M  m1  m2  ...  m n  Total mass of the system.
(c) Acceleration of centre of mass  a cm  :
If a1, a2 ,....., an are accelerations of particles of masses m1, m2 ....mn respectively and acm is
the acceleration of their centre of mass then
dvcm 1 n dvi
a cm    mi
dt M i1 dt
m1 a1  m 2 a 2  ...  m n a n 1 n
a cm    mi a i
m1  m 2  ...  m n M i1
Where M  m1  m2  ...  m n  Total mass of a system.
When two particles of masses m1 and m 2 are moving from a point with accelerations a1 and
a 2 at an angle  with each other, then the acceleration of their centre of mass is given
m12 a12  m 22a 22  2  m1a1  m 2a 2  cos 

by a cm 
m1  m 2
m1a1  m 2 a 2
f they move in the same direction, then   0 and a cm 
m1  m 2
m12a12  m 22a 22
If they move at right angles to each other, then   90 and a cm
0 
m1  m 2
m1a1  m 2a 2
If they move in opposite directions, then   180 and then a cm  m1  m 2
9.An object A is dropped from rest the top of a 30m high building and at the same
moment another object B is projected vertically upwards with an inital speed
of 15m/s from the base of the building. Mass of the object A is 2kg while mass of
the object B is 4kg. Find the maximum height above the ground level attained
by the centre of mass of the A and B system (take g = 10m/s2)
Sol. m1  4kg , m2  2kg
Initially 4kg is on the ground, therefore x1  0
2kg is on top of the building, therefore x2  30m
m1 x1  m2 x2 0  2  30
xc m    10m
m1  m2 42
Initital height of CM = 10m
m1u1  m2u2
Initial velocity of CM, ucm  m1  m2
4  15  0
ucm   10m / s upward
42
Acceleration of CM, ac m  g  10m / s 2 downwards
Maximum height attained by CM from initial
2
ucm 102
position, hc m    5m
2 g 20
Maximum height attained by CM of 4kg and 2kg from the ground = 10+5=15m
10.Find the acceleration of centre of mass of the blocks of masses

m1 and m2  m1  m2  in Atwood’s machine.
Sol. We know from Newton’s laws of motion magnitude of acceleration of each block
 m  m2  m1a  m2   a 
a 1 g ac m 
 m1  m2  m1  m2
2
 m  m2 
Acceleration of centre of mass ac m  1  g
 m1  m2 
1
Note: The magnitude of displacement of centre ofmass in time ‘t’ is Sc m  ac m t 2
2
Effect of external forces on C.M.
1
We know a cm   mi a i
M i
Therefore Ma cm   Fext   Fint

But the internalforces are in the form of
action - reaction pairs. Hence they cancel each other. Thus Fint 0
 Ma cm   Fext
Thus centre of mass is effected by only external force acting on the system. Internal
forces will have no effect on the motion of centre of mass.
W hen no external force acts on the system
then
a) acceleration of centre of mass is zero i.e.,
Fext  Ma cm  Ma cm  O  a cm  O .
b) Velocity of centre of mass is constant
vcm = constant
c) Linear momentum of the system is constant
pcm = constant. It is called the law of conservation of linear momentum.
Characteristics of centre of mass
1.Centre of mass of system of particles depend on mass of particles and their relative positions.
2. For continuous distribution of mass, centre of mass depends on mass distribution and
shape of the body.
Sum of moments of masses about centre of mass is zero i.e., m r

i
i i 0
3. Centre of mass is independent of frame of reference chosen to locate it.

4.Mass need not be present at centre of mass.
5.The motion of centre of mass is purely translational.
6.External forces only can effect the motion of C.M., but internal forces have no effect.
7.The motion of centre of mass is according to Newton’s 2 nd law.
Examples for the motion of centre of mass
(a) When a bomb at rest at origin of xyz-coordinate system explodes due to internal forces into
many ragments. These fragments fly off randomly with different velocities in different directions.
But C.M. is not effected and remains at rest at the origin.
  m i ri  0 , where ri is position vector of i th particle about origin.
(b) A bomb is projected on the ground to follow parabolic path. When it explodes during the
motiondue to internal forces into many fragments, they move randomly in different directions.
But the centre of mass follows the same parabolic path as unexploded bomb. So at any
moment the vector sum of the moments of mass of all the fragments about centre of mass is zero.
(c) When a wheel is rolling on a road, then the paths of various particles are complicated as they
are in combined motion (translational + rotational). But the motion of centre of mass is purely
translational and it follows straight line path.
Note: Gravitational force between two masses, electric force between two charges are the
examples of internal forces for the system, Which cannot produce acceleration in centre
of mass of the system.
Mutual forces between two bodies :

When two particles approach each other due to their mutual interaction, then they always
meet at their centre of mass.
To a system of particles m1(x1y1), m2(x2y2) another particle of mass m3 is added so that
(m1 x1  m2x2 )
centre of mass shifts to the origin then coordinates of third particle arex3 =  ;
m3
 m1 y1  m2 y2 
y3 
m3
In a system of two particles of mass m1 and m2 ,
when m1 is pushed towards m2 through a distance d then shift in m2 towards m1 without altering
m1
C.Mposition is m d .
2
A boy of mass m is at one end of a flat boat of mass M and length l which floats stationary
on water. If boy moves to the other end, the boat moves through a distance d in the opposite
ml
direction with respect to ground (or shore), such that d  M  m .
 
Ml
The displacement of boy with respect to ground is d1 
 M  m
A boy of mass m is standing on a flat boat floating stationary on the surface of water. If the boy
starts moving on the boat with velocity Vr with respect to boat, then
mVr
Velocity of the boat w.r.t. ground is V  ,
M m
‘  ’ indicates boat moves opposite to the velocity of the boy.
MVr
Velocity of boy w.r.t. ground is V 
1
M m
11: A 10kg boy standing in a 40kg boatfloating on water is 20m from the shore
of the river. If he moves 8 meters on the boat towards the shore, then how far is
he from the shore now?
Sol. Mass of the boy (m)=10kg
Mass of the boat (M)=40kg
Distance travelled by boy (l)=8m
Distance travelled by the boat in the
ml 10  8
opposite direction    1.6m
M  m 10  40
Distance of the boy from the shore is
(20-8+1.6) = 13.6m
Shift in centre of mass in different cases:

Shift is the distance of final location of centre of mass of the system from its initial location.
Shift in the centre of mass generally occurs due to
a) Addition of matter
b) Removal of matter
c) Change in shape
d) Change in mass distribution
a) Addition of mass : Due to addition of mass, the C.M of a system generally shifts
towards or into the region where mass is added. If C1 is C.M before addition and C2 is the C.M of
added mass and C1 C2  d .,then
 madded  d 
X shift   
 minitial  madded 
C.M shifts towards the side of added mass
b) Removal of mass : Due to removal of mass the C.M of a system shifts away from the
region where mass is removed. If C1 is C.M of the body before removal and C2 is the C.M of the
removed part and C1C2  d then
 mremoved  d 
X shift   
 M initial  mremoved 
‘  ’ indicates CM shifts opposite to the side of removed massOut of a uniform circular disc of
radius R, if a circular sheet of radius ‘r’ is removed, then the centre of massof remaining part shifts
r 2 d
by a distance
R2  r 2
where ‘d’ is the distance of the C.M. of the removed part from the centre of the original disc.
In this case the circular sheet is removed from the edge of disc, then the shift in centre of mass is
maximum. Here d  Rr.
r 2
Maximum shift =
Rr
2.Out of a uniform solid sphere of radius R, if a sphere of radius ‘r’, is removed, then the centre
r 3 d
 
of mass of the ramaining part shifts by R 3  r 3 ,
where ‘d’ isthe distance of the C.M. of removed part from thecentre of the original sphere. In this
case spherical cavity is made at the edge of large sphere,
r 3  R  r 
then shift in C.M. is maximum. It is given by
R 3
 r3  .
3.To a circular disc of radius R1 another of radius R2 and of the same material is added then shift
2
R2 (R 1  R 2 )
in the CM is x = 2 2
R1  R 2
4.If two spheres of same material and radii r1 and r2 are kept in contact, distance of centre of
mass from the centre of the first sphere is equal to
r23
 r1  r2  .
r13  r23
r13
Similarly distance of centre of mass from the centre of the second sphere is 3 3  r1  r2  .
r1  r2
The location of C.M. of system depends on the mass distribution within the system. Due to this
the location of C.M. changes whenever the shape of system changes and also the relative positions
of particles change.
Methods to locate C.M:
Locating the Centre of Mass can be done in four different ways. They are
1) Method of symmetry
2) Method of Decomposition
3) Method using theorems of Pappu’s
4) Method of integration
For continuous distribution of mass, the coordi inates of centre of mass are given by
xcm 
 xdm ; y 
 ydm ; z 
 zdm
.
 dm  dm  dm
cm cm
12. Distance of centre of mass of a uniform cone of height ‘h’ and base radius R,
3h
from the vertex on the line of symmetryis .
4
Sol. Consider a cone of height ‘h’ base radius ‘R’ and density  . To find centre of mass of the
cone imaging a small element of radius ‘r’ and thickness ‘dx’ at a distance x from ‘O’. Mass of
small element, dm   r 2  dx 
from figure
r x Rx
 r
R h h
 dm x    r  dx  x
h
2
xc m  0
 dm
h
  r dx 
2
0
 R2 x2 
h
0  h2  x dx 
h
x 3 dx
 2 2
 0
h
h R x
0 h2 dx 
0
x 2 dx
h
 x4  h4
 
4 3h
  h0  43 
 x3  h 4
  3
 3 0
3h
Therefore, centre of mass of cone is at a distance from vertex on its line of symmetry..
4
13: If the linear density of a rod of length L varies as   A  Bx , find the position of its
centre of mass.
Sol. Let the x-axis be along the length of the rod and origin at one of its ends. As rod is
along x-axis, for all points on it y and z coordinates are zero.
Centre of mass will be on the rod. Now consider an element of rod of length dx at a distance
x from the origin, then dm   dx   A  Bx  dx
L L
X cm 

0
x dm

 x  A  Bx  dx
0
or
L L
 0
dm   A  Bx  dx
0
AL2 BL3

3  3 AL  2 BL  L  3 A  2 BL 
2
X cm  2
BL2 3  2 A  BL  3  2 A  BL 
AL 
2
14: Identical blocks each of mass M and length L are placed one above the other
such that each extends out by maximum length as shown in figure. Find the
maximum extension of the nth block from the top. So that the blocks will not fall.
Sol. For a two block system, the centre of mass (C1) of upper block should be at the edge of
l
lower block i.e. at distance. But if center of mass of upper block is not resting on the lower
2
block then, the upper block falls down because of unbalaned torque created by gravitational
force.
If a third block (EF) is arranged below the two blocks then
The centre of mass (C2) of (AB) and (CD) block system must lie on the edge E of third
block. To find x2 consider C as origin. Then
l
M  0  M  
x2  2  l
2M 2
l l
x2  So, center of mass of upper two blocks is at distance from edge of lower block.
4 4
Also, ifanother block (GH) is placed below the three blocks in equilibrium, then
The center of mass (C3) of the upper three block must lie on the edge of the lower fourth
block i.e. at G. To find x3 consider E as origin.
l
2M 0   M  
x3  2  l  x3 
l
3M 6 6
l l
similarly x4  , x5  ,........
8 10
l
for nth block xn 
2n
Note-1: When the above blocks are arranged in such a manner, that each block projects out
by same distance, so that the blocks will not fall then the distance of projection of each block
l
form the edge of its bottom block is   .
n
Note-2: If the entire system is placed at the edge of a table, so that the blocks will not fall then
 l 
the equal distance of projection of each block from the edge of its bottom block is  
 n 1
POSITION OF CENTRE OF
S.NO SHAPE OF THE BODY FIGURE
MASS
1 Circular ring At the centre of the ring
2 Circular disc A the centre of the disc
Thin uniform straight

3 At the geometric centre
rod
4 Triangular plate At the centroid
At the point of intersection

5 Square plate
of the diagonals
At the point of intersection

6 Rectangular plate
of the diagonals
Hollow or solid
7 At the centre of the sphere
sphere
h
At a height of , from the
8 Hollow cone 3
base
Solid cone h
At a height of from the
9 or 4
Pyramid base
Solid (or) hollow At the mid-point of its own

10
cylinder axis
POSITION OF CENTRE OF
S.NO SHAPE OF THE BODY FIGURE
MASS

sin  
An arc of radius R 2R
At a distance of
subtending an angle  2
 at its centre from its centre of curvature
Of curvature on the axis of symmetry 2R  
OC  sin  
  2
11
2R
At a distance of from its
i) A semi-circle of 
radius ‘R’ centre on the axis of
symmetry
4R
At a distance of from
ii) A quadrant of a  2
circle of radius ‘R’ its centre ‘o’ on the axis of
symmetry
4R
12 Semi-circular disc 3
centre ‘o’ on the axis of
symmetry
3R
13 Solid hemi-sphere 8
centre ‘o’ on the axis of
symmetry
R
Hollow hemi-sphere At a distance of from its
14 (or) 2
Hemi-spherical shell centre ‘o’ on the axis of
symmetry
At its centre within the

15 Horse-shoe magnet
boundary limits
At a distance of
Semi-Circular 4  R12  R1 R2  R22 
16 OC 
annular plate 3  R1  R2 
from its centre of symmetry
16.Find position vectors of mass center of a system of three particle of masses 1 kg, 2
kg and 3 kg located at position vectors r1   4 iˆ  2 ˆj  3 kˆ  m, r2   iˆ  4 ˆj  2 kˆ  m and


r3  2 iˆ  2 ˆj  kˆ  m respectively..
Solution.
From eq. , we have

 m i ri
rc  ®
     
 1 4 î  2 ˆj  3 kˆ  2 î  4 ˆj  2 kˆ  3 2 î  2 ˆj  kˆ
rc 
2
 2 î  2 ˆj  kˆ
M 1 2 3 3
17.Find coordinates of mass center of a quarter ring of radius r placed in the first
quadrant of a Cartesian coordinate system, with centre at origin.
Solution.
Making use of the result obtained in the previous example, distance
y
r sin   / 4  2 2 r yc C
OC of the mass center form the center is OC  
/ 4 
/ 4
O xc x
 2r 2r 
Coordinates of the mass center (xc, yc) are  , 
 
18.Find coordinates of mass center of a semicircular ring of radius r placed symmetric

to the y-axis of a Cartesian coordinate system.
Solution.
The y-axis is the line of symmetry, therefore mass center of the ring
y
lies on it making x-coordinate zero.
C
Distance OC of mass center from center is given by yc
the result obtained in example 4. Making use of this / 2
O x
result, we have
r sin  r sin   / 2  2 r
OC  ® yc  
 / 2 
19.Find coordinates of mass center of a quarter sector of a uniform disk of radius r

placed in the first quadrant of a Cartesian coordinate system with centre at origin.
Solution.
Making use of the result obtained in the previous
example, distance OC of the mass center form the
center is
y
2 r sin   / 4  4 2 r
OC  
3 / 4 3 yc C
/ 4
O xc x
 4r 4r 
Coordinates of the mass center (xc, yc) are  , 
3 3 
20.Find coordinates of mass center of a uniform semicircular plate of radius r placed

symmetric to the y-axis of a Cartesian coordinate system, with centre at origin.
Solution.
The y-axis is the line of symmetry, therefore mass
center of the plate lies on it making x-coordinate
zero.
Distance OC of mass center from center is given by
the result obtained in example 7. Making use of this
result, we have
2 r sin  2 r sin   / 2  4 r
OC  ® yc  
3 3 / 2 3 y
C
yc
/ 2
O x
21. Find coordinates of mass center of a non-uniform rod of length L whose linear
mass density l varies as l=a+bx, where x is the distance from the lighter end.
Solution.
y
Assume the rod lies along the x-axis with its lighter end on the dm = dx 
O xc x
dx x=L
origin to make mass distribution equation consistent
with coordinate system.
L L
 xdm 
 x dx   x ax  b  dx
0 0

2 aL  3 b  L
Making use of eq. , we have x c  ® xc L L
3  aL  2 b 
M  dx  ax  b  dx
0 0
22.A jeep of mass 2400 kg is moving along a straight stretch of road at 80 km/h. It is
followed by a car of mass 1600 kg moving at 60 km/h.
(a) How fast is the center of mass of the two cars moving?
(b) Find velocities of both the vehicles in centroidal frame.
 
 m jeep v jeep  m car v car
Solution.(a) Velocity of the mass center vc 
m jeep  m car
Assuming direction of motion in the positive x-direction, we have

 
 m jeep v jeep  m car v car  2400  80  1 60 0  60
vc  ® vc   72 km/h
m jeep  m car 2400  1 60 0
(b) Velocity of the jeep in centroidal frame v jeep / c  8 0  72  8 km/h in positive x-direction.
Velocity of the car in centroidal frame v car / c  60  72  1 2 km/h 12 km/h negative x-

direction direction.
23.A 2.0 kg particle has a velocity of v1   2.0 iˆ  3.0 ˆj  m/s, and a 3.0 kg particle has a
velocity v2  1.0 iˆ  6.0 ˆj  m/s.
(a) How fast is the center of mass of the particle system moving?
(b) Find velocities of both the particles in centroidal frame.
Solution.
 
(a) Velocity of the mass center  m v  m 2v 2
vc  1 1
m1  m 2
 
 m v  m 2v 2 2  2 .0 iˆ  3.0 ˆj   3 1.0 î  6 .0 ˆj 
 1.4 iˆ  2 .4 ˆj  m/s
vc  1 1 
m1  m 2 ® vc 
2 3
(b) Velocity of the first particle in centroidal frame
v 1 / c   2.0 iˆ  3.0 ˆj   1 .4 iˆ  2.4 ˆj   0 .6  iˆ  ˆj  m/s
   
v1 / c  v1  v c ®
Velocity of the second particle in centroidal frame

v 2 / c  1.0 iˆ  6.0 ˆj   1.4 iˆ  2.4 ˆj     0 .4 iˆ  3.6 ˆj  m/s
   
v 2 / c  v1  v c ®
24.Two particles of masses 2 kg and 3 kg are moving under their mutual interaction in
free space. At an instant they were observed at points (-2 m, 1 m, 4 m) and (2 m, -
3 m, 6 m) with velocities  3 î  2 ˆj  kˆ  m/s and  î  ˆj  2kˆ  m/s respectively. If after 10
sec, the first particle passes the point (6 m, 8 m, -6 m), find coordinate of the
point where the second particle passes at this instant?
Solution.
System of these two particles is in free, therefore no external forces act on them. There total
linear momentum remains conserved and their mass center moves with constant velocity
relative to an inertial frame.
Velocity of the mass center
vc 

 
m i v i 2 3 iˆ  2 ˆj  kˆ  3 iˆ  ˆj  2 kˆ


3 iˆ  ˆj  4 kˆ 
m i 23 5
m/s

Location rco of the mass center at the instant t = 0 s

 m i ri
rc   
rco 
  
2 2 î  ˆj  4 kˆ  3 2 î  3 ˆj  6 kˆ


2 î  7 ˆj  26 kˆ
m i 2 3 5

New location rc of the mass center at the instant t = 10 s
    2 iˆ  7 ˆj  26 kˆ 3 iˆ  ˆj  4 kˆ 32 iˆ  17 ˆj  14 kˆ
rc  rco  v c t  rc    10 
5 5 5
New location (x, y, z) of the second particle.


 m iri
rc    
3 2 iˆ  1 7 ˆj  1 4 kˆ 2 6 iˆ  8 ˆj  6 kˆ  3 xiˆ  yjˆ  zkˆ

 
m i 5 2 3
Solving the above equation, we obtain the coordinates of the second particle (20/3, –11,
–2/3)
25.Find the x coordinate of the centre of mass of the bricks shown in figure :
y

 6 m
 4 m
2 m
m x
Solution
             
m   m    m     m    
2 2 2 2 4 2  2 4 6 2  25
X cm   
mmmm 24
26.If the distance between the centres of the atoms of potassium and bromine in KBr
(potassium -bromide) molecule is 0.282 ×10–9m, find the centre of mass of this two particle
system from potassium (mass of bromine = 80 u, and of potassium = 39 u).
Solution: Mass of bromine, mBr = 80 units
Mass of potassium, mK = 39 units
Position co-ordinate of potassium, xk = 0
Position co-ordinate of bromine, xBr = 0.282 × 10–9m
y-axis
cm Br
K
0.282 ×10–9m x-axis
 Position co-ordinate of centre of mass,
m x  m Br x Br 39  0  80  0.282  109
xc  k k  x 
m k  m Br c
39  80
 xc = 0.189 × 10–9 m
27.When ‘n’ number of particles of masses m, 2m, 3m,..... nm are at distances x1=1, x2 =2,
x3 = 3, ...... xn = n units respectively from origin on the x-axis, then find the distance of
centre of mass of the system from origin.
Solution:
m(1)  2m(2)  3m(3)  ........  (nm)n
x cm 
m  2m  3m  ........  nm
m(12  22  32  ......  n 2 )
x cm 
m(1  2  3  .......  n)
 n ( n  1)(2 n  1) 
  2n  1

6 
X cm
 n ( n  1)  3
 
 2 
28.When ‘n’ number of particles each of mass ‘m’ are at distances x1=1, x2=2, x3=3, ......xn=n
units from origin on the x-axis, then find the distance of their centre of mass from origin.
m(1)  m(2)  m(3)  .......  m(n)
Solution: x cm 
m  m  m  .......  m(n terms)
 n ( n  1) 
m 
m (1  2  3  .......  n )
  
2

nm nm
n 1
x cm 
2
29.When ‘n’ number of particles each of mass ‘m’ are at distances x1=a, x2 = ar, x3 = ar2.....
xn = arn units from origin on the x-axis, then find the distance of their centre of mass
from origin.
Solution:
ma  m(ar)  m(ar 2 )  ......  m(ar n )
x cm 
m  m  m  ......  m(n terms)
m(a  ar  ar 2  ......  ar n )
x cm 
mn
1  a  r  1  a(r n  1)
n
If r > 1 then xcm   

n  r  1  = n(r  1)
 
1  a(1  rn )  a(1  r n )
If r < 1 then xcm  
n  1  r  n(1  r)
=
30.Two blocks of masses 10 kg and 30 kg are placed on x - axis. The first mass is moved
on the axis by a distance of 2 cm. By what distance should the second mass be moved
to keep the position of centre of mass unchanged.
Solution: mass of the first block, m1 = 10kg
mass of the second block, m2 = 30 kg
Let x1 and x1 are positions of m1
x2 adn x1 are positions of m2
y-axis
(0,0) (2,0) cm x1 , 0 (x,0) x-axis
m 1 x1  m 2 x 2
In this case if ‘xcm’ is the position of centre of mass then x cm  m1  m 2
m 1x1  m 2 x 2
then the new position of CM when blocks are shifted x cm  m1  m 2
subtracting the above equations
m1 x1  x1   m 2 x 2  x 2 
x cm  x cm 
m1  m 2
m1x1  m 2 x 2
x cm 
m1  m 2
10  2  30x 2
0  x 2   2 3 Therefore the second block should be moved left
40
2
through a distance of cm to keep the position of centre of mass unchanged.
3
31.A 1 m long rod having a constant cross sectional area is made of four materials. The
first 0.2 m are made of iron, the next 0.3 m of lead, the following 0.2m of aluminium and
the remaining part is made of copper. Find the centre of mass of the rod. The densities
of iron, lead, aluminium and copper are 7.9 × 103kg/m3,11.4 × 103 kg/m3, 2.7 × 103 kg/m3
and 8.9 × 103 kg/m3 respectively.
1m
Solution: O Iron Lead Aluminium Copper
0.2m 0.3m 0.2m 0.3
0.35m
0.6m
0.85m
mass(m) = volume(v) x density (d)
m = Area (A) x length (l) x density (d)
m = Ald
Mass of iron part, m1 = A × 0.2 × 7.9 ×103 = 1.58 × 103 A
Mass of lead part, m2 = A × 0.3 × 11.4 × 103 =3.42 × 103 A
Mass of aluminium part, m3 = A × 0.2 × 2.7 × 103 =0.54 × 103 A
Mass of copper part, m4 = A ×0.3 × 8.9 × 103
= 2.67 × 103A
Co - ordinate of iron part from end “O” of the rod, x1 = 0.1m
Co - ordinate of lead part from end “O” of the rod, x2 = 0.35m
Co - ordinate of aluminium part from end “O” of the rod, x3 = 0.6m
Co - ordinate of copper part from end “O” of the rod, x4= 0.85m
 Centre of mass of the rod,
m1x1  m 2 x 2  m 3 x3  m 4 x 4
Xc 
m1  m 2  m 3  m 4
Xcm 
1.58103 0.13.42103 0.35 0.54103 0.6 2.67103 0.85A
1.58103 3.42103 0.54103 2.67103A
 X cm  0.481m from the end “O” of the rod.
32.Find the position of centre of mass of the system of 3 objects of masses 1 kg, 2kg
and 3 kg located at the corner of an equilateral triangle of side 1 m. Take 1 kg mass
object at the origin and 2 kg along x-axis.
Solution:
Y- axis
m3 = 3kg 
1 3
(x3, y3)   , 
2 2 
x-axis
m1 =1kg m2 = 2kg
(x1, y1) = (0,0) (x2, y2) = (1,0)
m1x1  m 2 x 2  m 3 x 3
x cm 
m1  m 2  m 3
1
1 0  2 1 3  7
 x cm  2  x cm  m
1 2  3 12
m y  m 2 y 2  m3 y3
Ycm  1 1
m1  m 2  m 3
3
1 0  2  0  3  3
 Ycm  2  Ycm  m
1 2  3 4
 7 3 
 Co- ordinates of centre of mass (xcm, ycm)   12 m, 4 m
 
33.Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A,B,
C and D of the square of side 1 m. Find the position of centre of mass of the particles.
Solution: Assuming A as the origin, AB as x-axis and AD as y-axis we have

Y
x3 , y3   1,1
x 4 , y 4   0,1 D C m =3
3
m4=4
B
m1=1 A m2=2 X
x1, y1   0,0 x 2 , y2   1, 0
Co-ordinates of their CM are

m 1 x1  m 2 x 2  m 3 x 3  m 4 x 4
xCM =
m1  m 2  m 3  m 4
(1)(0)  2(1)  3(1)  4(0)
= = 0.5m
1 2  3  4
m 1 y1  m 2 y 2  m 3 y 3  m 4 y 4
Similarly, yCM =
m1  m 2  m 3  m 4
1(0)  2(0)  3(1)  4(1)
 = 0.7m
1 2  3  4
\ Co-ordinates of centre of mass (x CM, y CM) = (0.5 m, 0.7 m)
34.A system consists of three particles located at the corners of a right triangle as shown in
the figure. Find the position
Y vector of centre of mass of the system.
3m
Solution :
h
2m
O d b m
Using the equation
2md mb d  3md b
Xc 
mx
M
i i

6m
d  2 b
3  
 mi yi 2m o  m o  3mh h
Yc   
M 6m 2
Zc = 0; because the particles are in X - Y plane we can express the position of centre of
mass from the origin using a position vector as
  2  h 
k , rc  d  3 bi  2 j

rc  X c i  Yc j  Z c 
35.The position vectors of three particles of mass m1 = 1 kg, m2 = 2 kg and m3 = 4 kg are


 
  
  
r1  i  4 j  k m, r2  i  j  k m, and r3  2i  j  2k m respectively. Find the position
vector of their centre of mass.
Solution :
The position
 vector
 of
 centre of mass of the three particles is given by
 m r  m2r2  m3r3
rc  1 1
m1  m2  m3
rc 
 
 
    
 
 1 i  4 j  k  2 i  j  k  4 2i  j  2 k
  

1 2  4
 
11i  2 j  5k

7
 1

 11i  2 j  5k m
7
36.Two 3 kg masses have velocities v1  2 i  3 j m / s and v2  4 i  6 j m / s . Find a) velocity of
centre of mass, b) the total momentum of the system, c) The velocity of centre of mass 5s
after application of a constant force F  24 i N , d) position of centre of mass after 5s if it is at
the origin at t = 0
Solution :
a) vc 
 
  
 m1v1  m2 v2  3 2i  3 j  3 4i  6 j
, vc 

m1  m2 6
 Velocity of centre of mass Vc  3i 1.5 j ms1 .
 
b) The momentum of the system  Mvc  6kg 3i 1.5 j ms  18i  9 j kgms
1 1

c) To find the velocity of centre of mass after 5 s of application of the force F  24iN we first find
the acceleration of the centre of mass. It is given by

 F 24i
ac    4i ms2
M 6

The velocity of centre of mass before the force is applied is vc .
and from the equation vc   vc  ac .t
     
vc   3i 1.5 j  4i 5  3i 1.5 j  20i

 

Vc1  23i 1.5 j ms1
37.Two 3 kg masses have velocities v1  2 i  3 j m / s and v2  4 i  6 j m / s . Find a) velocity of

centre of mass, b) the total momentum of the system, c) The velocity of centre of mass 5s
after application of a constant force F  24 i N , d) position of centre of mass after 5s if it is at
the origin at t = 0
Solution :
 m1v1  m2 v2  3 2i  3 j   34i  6 j 
 
a) vc  , vc 
m1  m2 6
 Velocity of centre of mass Vc  3i 1.5 j ms1 .
 
b) The momentum of the system  Mvc  6kg 3i 1.5 j ms  18i  9 j kgms
1 1

c) To find the velocity of centre of mass after 5 s of application of the force F  24iN we first find
the acceleration of the centre of mass. It is given by

 F 24i
ac    4i ms2
M 6

The velocity of centre of mass before the force is applied is vc .
and from the equation vc   vc  ac .t
    
vc   3i 1.5 j  4i 5  3i 1.5 j  20i


 

Vc1  23i 1.5 j ms1
   12  
d) From the equation of the position vector r  r0  v0t  at where r0  0 (origin at t = 0);
2
   
v0  vc ; a  ac and t = 5 s

  1
r  3i 1.5 j 5  4i 25
2
 


r  15i  7.5j  50i  r 
  65i  7.5 j m

The coordinates of the centre of mass after 5 s of application of the force F are (65 m, – 7.5 m)
38. Find the acceleration of center of mass of the blocks of masses m1 and m2 (m1 > m2)
in Atwood’s machine:
Solution: We know from Newton’s laws of motion magnitude of acceleration of each block
 m1  m2 

a =  m  m  g
 1 2
m1a  m2 a
acm 
m1  m2
a
a
 m  m2 
acm   1 a m2
 m  m 
1 2 m1
 m  m2 
 m1  m2  g
acm   1  m  m 
 m  m 
2
 m  m2 
1 2 1 2
 Accleration of centre of mass acm   1  g

 m1  m2 
   12  
d) From the equation of the position vector r  r0  v0t  at where r0  0 (origin at t = 0);
2
   
v0  vc ; a  ac and t = 5 s

  1
r  3i 1.5 j 5  4i 25
2
 


r  15i  7.5j  50i  r 
  65i  7.5 j m

The coordinates of the centre of mass after 5 s of application of the force F are (65 m, – 7.5 m)
39.Find the acceleration of center of mass of the blocks of masses m1 and m2 (m1 > m2) in
Atwood’s machine:
Solution: We know from Newton’s laws of motion magnitude of acceleration of each block
 m1  m2 

a =  m  m  g
 1 2
m1a  m2 a
acm 
m1  m2
a
a
 m  m2 
acm   1 a m2
 m  m  m1
1 2
 m  m2 
 m1  m2  g
acm   1  m  m 
 m  m 
2
 m  m2 
1 2 1 2
 Accleration of centre of mass acm   1  g

 m  m  1 2
40.An object A is dropped from rest the top of a 30m high building and at the same
moment another object B is projected vertically upwards with an initial speed of 15 m/s
from the base of the building. Mass of the object A is 2 kg while mass of the object B is
4 kg. Find the maximum height above the ground level attained by the centre of mass of
the A and B system (take g = 10 m/s2)
So lu tio n : m1 = 4kg, m2 = 2 kg
Initially 4 kg is on the ground  x1 = 0
2 kg is on top of the building  x2 = 30 m
2kg
m x  m2x2
 1 1
A
x cm
m1  m 2
ucm Initial position
of CM
0  2 30
= 10cm
42 B
4kg
= 10 m
 Initial height of CM = 10m.
m1u1  m 2 u 2
Initial velocity of CM, u cm  m1  m 2
4 15  0
ucm  = 10 m/s upward.
42
Acceleration of CM, acm = g = 10 m/s2 downwards
u2cm 102
 Maximum height attained by CM from initial position, h cm   =5m
2g 20
 Maximum height attained by CM of 4 kg and 2 kg from the ground = 10 + 5 = 15 m
.
41. Two masses m1 and m2 are connected by a spring of force constant k and are placed
on a frictionless horizontal surface. Initially the spring is stretched through a distance
x0, when the system is released from rest. Find the distance moved by two masses
before they again comes to rest.
Solution :
k 0
m2 m1
0  x0
m2 m1
0  x0
m2 m1
x 1 x 2
Blocks again come to rest when spring is compressed by x0. Since no external force is
acting on the system, so there is no change in the position of c.m. of the system. i.e. xcm  0 .
Let mass m1 displaces by x1 and m2 displaces by x2 , then
We have x1  x2  2 x0 .............. (i)
m1x1  m2 x2
and xcm  m1  m2
m1x1  m2 x2
As xcm  0  m1  m2
 0 .........(ii)
After solving equation (i) & (ii), we get

2m2 x0 2m1 x0
x1  , x2 
m1  m2 m1  m2
42.A projectile is fired at a speed of 100 m/s at an angle of 370 above the horizontal. At
the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the smaller
piece coming to rest. Find the distance from the launching point to the point where
the heavier piece lands.
Sol : Internal forces do not affect the motion of the centre of mass, the centre of mass hits the
ground at a position where the original projectile would have landed. The range of the original
projectile is
4m
m
( 0 ,0 )
x1 CM 3m
xcm
x2
3 4
2104  
2u 2 sin  cos  5 5 m  960m
x CM  
g 10
The centre of mass will hit the ground at this position. As the smaller block comes to rest
after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x = 480 m. If
the heavier block hits the ground at x2, then
m x  m2 x 2 m480  3m x 2 
x CM  1 1  960 
m1  m 2 m  3m
 x 2  1120m
43. A man of mass ‘m’ is standing on a boat of mass M which is at rest in still water. If the
man walks a distance L on the boat towards the shore the boat moves back through a
mL
distance, d 2 
mM
V
Proof :
VB cm L
d2
V
VB L cm

As the system is initially at rest and no external force acts on the system (horizontally)  vCM  0 .
 
mv1  Mv2  
(or)  0 (or) mv1  Mv2  0
mM
 
r1 r2  
or m M  0 (or) mr1  M r2  0
dt dt
 
or md1 – Md2 = 0 [ as d 2 is opposite to d1 ]
md1 = Md2
Now when man moves a distance L towards the shore relative to boat, the boat will shift a
distance d2 relative to shore opposite to the displacement of man. The displacement of man
relative to shore (towards shore) will be
d1 = L – d2 (i.e., d1 + d2 = drel = L)
so, md1 = Md2  m(L – d2) = Md2
hence d 2  mL
M m
44.If the linear density of a rod of length L varies as  A+Bx , find the position of its
centre of mass.
Solution: Let the x-axis be along the length of the rod and origin at one of its ends. As rod is
along x–axis, for all points on it y and z co-ordinates are zero.
y
O
x
dx
z L
Centre of mass will be on the rod. Now consider an element of rod of length dx at a distance
x from the origin, then dm =  dx  A + Bxdx
L L
 xdm  xA+ Bxdx

X CM  0
L  0
L
 dm  A + Bxdx
0 0
AL BL3 2

or X CM  2 3  L3A +2BL
AL+
BL2 32A +BL 
2
kx 2
45.If the linear density of a rod of length L varies as   where k is a constant and x
L
is the distance of any point from one end, then find the distance of centre of mass from
the end at x = 0.
Solution: Let the x-axis be along the length of the rod and origin at one of its ends. As rod is
along x–axis, for all points on it y and z co-ordinates are zero.
y
O
x
dx
z L
Centre of mass will be on the rod. Now consider an element of rod of length dx at a
Kx 2
distance x from the origin, then dm =  dx  dx
L
L L
kx 2
 xdm  x
L
dx
so, X CM  0
L  0
L
kx 2
 dm  L
dx
0 0
L
x L4
3
dx
4 3L
= 3 
0
= L
 x dx 3
2 L 4
PREVIOUS JEEMAINS QUESTIONS

0
46.The centre of mass of a solid hemisphere of radius 8 cm is x cm from the centre of the flat
surface. Then value of x is_______ [6SEP 2020 MAINS]
3R
Solution: Centre of mass of solid hemisphere of radius R lies at a distance above the
8
centre of flat side of hemisphere.
a a
47.A square shaped hole of side l  is carved out at a distance d  from the
2 2
centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of
a
the remaining portion from O is  , value of X (to the nearest integer) is_______
X
[2 SEP 2020 MAINS]
Solution: Let  be the mass density of circular disc.

Original mass of the disc, m0  a 2 
a2
Removed mass, m  
4
 2 a2  2  4  1 
Remaining, mass, m '   a     a  
 4   4 
New position of centre of mass
a2 a
m0 x0  mx a 2  0  
X cm   4 2
m0  m a2
a 2 
4
a 3 / 8 a a a
   
 1  2 2  4  1 8  2 23
 a
 2
 x  23
2
x
48.A rod of length L has non-uniform linear mass density given by   x   a  b   ,
L
where a and b are constants and 0  x  L . The value of x for the centre of mass of the
rod is at:
3 ab  3  2a  b 
(1)  L (2)  L [9JAN 2020 MAINS]
2  2a  b  4  3a  b 
3 ab  3  2a  b 
(3)  L (4)  L
4  2a  3b  2  3a  b 
2
x
solution Linear mass density,   x   a  b  
L
X CM 
 xdm
 dm
L
 dm     x  dx
0
 dm     x  dx
0
L 2
x bL
  a  b  L   dx  aL 
 3
0
L L
 bx3   aL2 bL2 
 xdm   ax  L2  dx   2  4 
0 0   
 aL2 bL2 
  
 2 4 
 X CM 
bL
aL 
3
3 L  2a  b 
 X CM   
4  3a  b 
49.The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plale)of
mass 4kg. (The coordinates of the same are shown in figure) are
[8 JAN 2020 MAINS]
(1) (1.25, 1.50 m) (2) (0.75 m, 1.75 m)

(3) (0.75 m, 0.75 m) (4) (1 m, 1.75 m)
(Solution: 4)
For given Lamina

x y
m1  1, C1  1.5, 2.5 
m2  3, C2   0.5,1.5 
m1 x1  m2 x2 1.5  1.5
X cm    0.75
m1  m2 4
m1 y1  m2 y2 2.5  4.5
Ycm    1.75
m1  m2 4
 Coordinate of centre of mass of flag shaped lamina
(0.75, 1.75)
50. As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a
uniform sphere of radius R (centred at (C), the centre of maa ss of remaining
(shaped) part of sphere in at G, i.e. on the surface of the cavity. R and be determined
by the equation [8 JAN 2020 MAINS]
 
(1) R  R  1  2  R   1
2
(2)  R  R  1  2  R   1
2
Solution (1) Mass of sphere = volume of sphere x density of sphere
4 3
 R 
3
4
 1 
3
Mass of cavity M cavity 
3
4 3 4
R    1 
3
M  Re maining  
3 3
Centre of mass of remaining part,
M1r1  M 2 r2
X COM 
M1  M 2
4 3  4 
 3 R  0   3  1      R  1
3
 R  1  2  R
  2  R   3
4 3 4
R    1   
3
 R 1
3 3

 R  1  2R
 R  1  R 2  R  1
 
 R2  R  1  2  R   1
(3)  R  R  1  2  R   1
2
(4)  R  R  1  2  R   1
2
51. Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at thre
corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the
figure. The center of mass of the system is at a point: [7 JAN 2020MAINS]
(1) 0.6 cm right and 2.0 cm above 1 kg mass
(4) 0.9 cm right and 2.0 cm above 1 kg mass [7 JAN 2020 MAINS]
Solution
m1 x1  m2 x2  m3 x3
X cm 
m1  m2  m3
1 0  1.5  3  2.5  0 1.5  3
X cm    0.9cm
1  1.5  2.5 5
m1 y1  m2 y2  m3 y3
Ycm 
m1  m2  m3
1 0  1.5  0  2.5  4 2.5  4
Ycm    2cm
1  1.5  2.5 5
Hence, centre of mass of system is at point (0.9, 2)
52. Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an
equilateral triangle of side 1m (as shown in the figure). The (x,y) coordinates of the
centre of mass will be
[APR 12 2019 MAINS]
 3 5  7 3 

(1)  4 m, m (2)  m, m 
 12   12 8 
7 3   3 7 

(3)  12 m, m 
(4)  8 m, m
 4   12 
50  0  100  1  150  0.5 7
Solution (3) xcm   m
50  100  150 12
3
50  0  100  0  150 
ycm  2  3m
50  100  150 4
53.Four particles A, B, C and D with masses mA  m, ma  2m, mc  3m and mD  4m are

at the comers of a square. They have accelerations of equal magnitud with directions
as shown. The acceleration of the centre of mass of the particles is
[APR8 2019 MAINS]
(1)
5
 
a  
i j (2) a
(3) Zero (4)

5
 
a  
i j
SolutionAcceleration of centre of mass  acm  is given by

 
 m1 a1  m2 a 2  ........
 a cm 
m1  m2  ........
 2m  a j  3m  aiˆ  ma  iˆ   4m  a   ˆj 

2m  3m  4m  m
2aiˆ  2ajˆ a ˆ ˆ

10
 ij
5
 
54 A uniform rectangular thin sheet ABCD of mass M has length a and breadth b,
as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the
centre of mass of the remaining portion will be [8APR 2019 MAINS]
 3a 3b   5a 5b 
(1)  ,  (2)  , 
 4 4  3 3
 2a 2b   5a 5b 
(3)  ,  (4)  , 
 3 3   12 12 
Solution. With respect to point , the CM of the cut-off portion
a b
 , 
4 4
MX  mx
Using, xCM 
M m
M a
M 0  
 4 4  a
M 12
M
4
b
and yCM  
12
So CM coordinates one
a a 5a
x0   
2 12 12
b b 5b
and y0   
2 12 12
55. The position vector of the centre of mass rcm of an asymmetric uniform bar of
negligible area of crosssection as shown in figure is: [12 JAN 2019 MAINS]
 13 5
(1) rcm  Lx  L y
8 8
 5 13
(2) rcm  Lx  L y
8 8
 3 11
(3) rcm  Lx  L y
8 8
 11 3
(4) rcm  Lx  L y
8 8
Solution. x-coordinate of centre of mass is
5mL
2mL  2mL 
X cm  2  13 L
4m 8
56. A force of 40 N acts on a point B at the end of an L-shaped object, as shown in
the figure. The angle that will produce maximum moment of the force about point A is
given by: [APR15 2018 MAINS]
1
(1) tan   (2) tan   2
4
1
(3) tan   (4) tan   4
2
Solution.
To produce maximum moment of force line of action of force must be perpendicular to line AB.
2 1
 tan   
4 2
57. In a physical balance working on the principle of moments, when 5 mg weight
is placed on the left pan, the beam becomes horizontal. Both the empty pans of the
balance are of equal mass. Which of the following statements is correct?
(1) Left arm is longer than the right arm [APR8 2017 MAINS]
(2) Both the arms are of same length
(3) Left arm is shorter than the right arm
(4) Every object that is weighed using this balance appears lighter than its actual weight
Solution.
According to principle of moments when a system is stable or balance, the anti-clockwise
moment is equal to clockwise moment.
i.e., load × load arm = effort × effort arm
When 5 mg weight is placed, load arm shifts to left side,
hence left arm becomes shorter than right arm.
58.In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically
BC
below point A, then is close to :
AB
[APR11 2016 MAINS]
(1) 1.85 (2) 1.5 (3) 1.37 (4) 3
 x  
x1 y /2
x   y
Solution.Centre of mass xcm  2
  2
2  x  y
1 y y2
  
2 x x2
Bc y 1  3
   1.37
A x 2
59. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the
radius of its base is R and its height is h then z0 is equal to
5h 3h 2 h2 3h
(1) (2) (3) (4) [APR11 2015 MAINS]
8 8R 4R 4
Solution.Let density of cone  .
 ydm
Centre of mass, ycm 
 dm
h
 yr
2
dy h 2
 0 r
 0
ydy
1 2 1 2 ...(i)
R h R h
3 3
For a cone, we know that

r y y
 r  R
R n n
h
h
 y4 

3
3 y dy 3  
0  4 0 3
ycm    h
h3 h3 4
bx
60 A uniform thin rod AB of length L has linear mass density   x   a  , where x is
L
7
measured from A. If the CM of the rod lies at a distance of   L from A, then a and b
 12 
are related as:
(1) a = 2b (2) 2a = b (3) a = b (4) 3a= 2b [APR 2015 MAINS]
Solution.Centre of mass of the rod is given by:
L
bx 2
 (ax  L
)dx
0
xcm  L
bx
 (a  L
)dx
0
aL2 bL2 L  a  b 
  
2 3   2 3
bL b
aL  a
2 2
a b

7L 2 3

Now 12 b
a
2
On solving we get, b  2a
61. A thin bar of length L has a mass per unit length , that increases linearly with
distance from one end. If its total mass is M and its mass per unit length at the lighter
end is O , then the distance of the centre of mass from the lighter end is
L  0 L2 L  0 L2
(1)  (2)  [APR 2014 MAINS]
2 4M 3 8M
L  0 L2 2 L  0 L2
(3)  (4) 
3 4M 3 6M
Solution.(3)
62.A boy of mass 20 kg is standing on a 80 kg free to move long cart. There is negligible
friction between cart and ground. Initially, the boy is standing 25 m from a wall. If
he walks 10 m on the cart towards the wall, then the final distance of the boy from
the wall will be
(1) 15 m (2) 12.5 m (3) 15.5 m (4) 17 m [APR 2013 MAINS]
Solution. (4)
63.A thin rod of length ‘L’ is lying along the x-axis with its ends at x = 0 and x=L. Its
n
x
linear density (mass/length) varies with x as k   , where n can be zero or any
L
positive number. If the position xCM of the centre of mass of the rod is plotted against
‘n’, which of the follwing graphs best aproximates the dependence of xCM on n?
(1) [MAINS 2008]
(2)
(3)
(4)
n
x
Solution.The linear mass density   k  
L
x
Here 1
L
With increase in the value of n, the centre of mass shift towards the end x = L This is satisfied
by only option (a).
L L L n
x
 xdm  x  dx   k  L  .xdx
0 0 0
xCM  L
 L
 n
L
x
 dm  dx  k  L  dx
0 0 0
L
 x n2 
k n
  n  2  L  0 L  n  1
 
 kx n 1 
L
n2
 n
  n  1 L  0
L
For n  0, xCM  ; n  1,
2
2L 3L
xCM  ; n  2, xCM  ;.....
3 4
For n  xcm  L
Moment of inertia of a square plate about an axis through its centre and perpendicular
to its plane is.
64. A circular disc of radius R is removed from a bigger circular disc of radius 2R
such that the circumferences of the discs coincide. The centre of mass of the new
disc is  / R from the centre of the bigger disc. The value of  is
(1) 1/4 (2) 1/3 (3) 1/2 (4) 1/6 [MAINS 2007]
Solution.Let  be the mass per unit area of the disc. Then the mass of the complete disc

    2R 
2

2
The mass of the removed disc   R  R
2
 
Let us consider the above situation to be a complete disc of radius 2R on which a disc
of radius R of negative mass is superimposed. Let O be the origin. Then the above figure can
be redrawn keeping in mind the concept of centre of mass as :
xcm 
 6  2R    0   6  R  R
2 2
4R 2  R 2
R 2  R
 xc.m 
3R 2
R 1
 xc.m    R   
3 3
65 Consider a two particle system with particles having masses m1 and m2 . If the
first particle is pushed towards the centre of mass through a distance d, so as to keep
the centre of mass at the same position?
m2 m1
(1) m d (2) m  m d [MAINS 2006]
1 1 2
m1
(3) m d (4) d
2
Solution.Initially,
m1   x1   m2 x2
0  m1 x1  m2 x2 .......(1)
m1  m2
Let the particles is displaced through distanced away from centre of mass
m1  d  x1   m2  x2  d '
0 
m1  m2
 0  m1d  m1 x1  m2 x2  m2 d '
m1
d' d
m2
21. A body A of mass M while falling vertically downwards under gravity breaks
1 2
into two parts; a body B of mass M and a body C of mass M . The centre of mass
3 3
of bodies B and C taken together shifts compared to that of body A towards
(1) does not shift
(2) depends on height of breaking [MAINS 2005]
(3) body B
(4) body C
Solution.(1) The centre of mass of bodies B and C taken together does not shift as no external
force acts. The centre of mass of the system continues its original path. It is only the internal
forces which comes into play while breaking.
22. A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth

floor. A force ‘ F ’ is applied at the point P parallel to AB, such that the object has only
the translational motion without rotation. Find the location of P with respect to C
[MAINS 2005]
3 2 4
(1)  (2)  (3)  (4) 
2 3 3
Solution.

To have translational motion without rotation, the force F has to be applied at centre of
mass. i.e. the point ‘P ’has to be at the centre of mass Taking point C at the origin position,
positions of y, and y2 are r1  2l , r2  l and ml  m and m2  2m
m1 y1  m2 y2 m  2  2m   4
y  
m1  m2 3m 3
ROTATIONAL DYNAMICS
Rigid body : If there is no relative motion between any two particles of the body along the line
joining them by the application of external force, then that body is called rigid body.
A No real body is truly rigid, since real bodies deform under the influence of external forces.
Types of motion of a rigid body
A Translational motion : All particles of the body move in parallel paths such that
displacements of all the particles are same as that of the body then its motion is said to be
translational.
A Rotational motion : A body is said to be in pure rotation if every particle of the body
moves in a circle and the centres of all the particles lie on a straight line called the axis of
rotation.
A Rolling motion : The combination of rotational and translational motion with regard to
certain constraints is called rolling motion.
Kinematics of rotational motion about a fixed axis :
A The kinetic equations for rotational motion with uniform angular acceleration
1 2
1)   0   t 2)   0t   t
2
 1
3)  2  02  2 4)  n  0    n  
 2
1: The motor of an engine is rotating about its axis with angular velocity of 120 rpm. It
comes to rest in 10s, after being switched off. Assuming constant deceleration,
calculate the number of revolutions made by it before coming to rest.
Sol. Here n=120rpm=2rps
0  2 n  4 rad s 1 ;   0 and t  10s
  0   t
0  4    10 or   0.4 rad s 2
Also, the angle covered by the motor,
1
  0t   t 2
2
1
  4  10    0.4   10 2  40  20  20 rad Hence, the number of revolutions completed,
2
 20
N   10
2 2
2: A wheel rotates with an angular acceleration given by   4at 3  3bt 2 , where t is the
time and a and b are constants. If the wheel has initial angular speed 0 , write the
equations for the (a) angular speed (b) angular displacement
d
Sol. (a) Since    d   dt
dt
Integrating both sides, we get
 t
 d    dt    4at  3bt  dt
3 2
0 0
t4 t3
  0  4  3b    0  at 4  bt 3
4 3
d
(b) Since    d   dt
dt
On integrating both sides, we get
 t t
 d   dt   
0 0 0
0  at 4  bt 3  dt
at 5 bt 4
    0t  
5 4
Calculation of Angular velocityof a Rigid Body :
A The angular velocity of the rigid body can be given as the relative angular velocity between
any two points of the rigid bodies,
 
  v
   AB   BA  AB
l
Where l = distance of separation between A and B and vAB = magnitude of relative velocity
between A and B. The vectorial representation of above equation can be given as
  
v AB   AB  r AB

Where, r AB = position vector of A relative to B.
 
3.The velocities at the ends of rod of length l are given as v and v ' , making angles 300
and 600 with rod. Find
600
V0
v B
300 l
A

(a) v ' (b) angular velocity of the rod
Sol. Since rod is rigid, equate the velocities of the points A and B along the rod to obtain,
v A cos 300  v 'cos 600
3 v'
or v  or v '  3v
2 2
B
VAB 
v AB
(b) the angular velocity of the rod is  
l
1 3
 2
v
2
v ' v  3 3v

 
l 2l
v
 2 (clock wise)
l
MOMENT OF FORCE (TORQUE):
A Torque is the turning effect of a force about a fixed point.
A Magnitude of the torque is given as the product of magnitude of force and perpendicular

distance of line of action of a force from the fixed point.   F  r sin      r  F .
S.I. Unit : Nm Dimensional formula:  ML2T 2 

z
 F
P 
r
O y
r sin 
x
Application: A force of given magnitude applied at right angles to the door at its outer
edge is most effective in producing rotation.
A The moment of a force vanishes if either the magnitude of the force is zero, or if the line
action of the force pass through the fixed point.
A If the direction of F is reversed, the direction of the moment of force is also reversed.
A If directions of both r and F are reversed, the direction of the moment of force remains the same.
Sign convention : Torque that produces anti clockwise rotation is taken as positive and
clockwise rotation taken as negative.
4.A particle is projected at time t=0 from a point ‘O’ with a speed ‘u’ at an angle ‘  ’ to
horizontal. Find the torque of a gravitational force on projectile about the origin at
time ‘t’. (x, y) plane is vertical plane)
  1 2
Sol. r   u cos   tiˆ    u sin   t  gt  ˆj
 2 
   
F    mg  ˆj;   r  F
y

u
P(x,y)
 x
O
  1  
   u cos  tiˆ    u sin   t  gt 2  ˆj   mg   ˆj 

  2  
   u cos  t  mg   iˆ   ˆj 


   mg  u cos  t kˆ 
MOMENT OF COUPLE:
A pair of equal and opposite forces with different lines of action is known as a couple. A couple
produces rotation without translation. If an object is not on pivot (unconstrained) a couple causes
the object to rotate about its centre of mass.
F
d
F
This couple can produce turning effect
(or) torque on the body. Moment of couple is the measure of turning effect   .
  moment of couple = magnitude of either force  perpendicular distance between the
forces
  Fd
Mechanical Equilibrium of a rigid body:
A A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular
momentum are not changing with time, or equivalently, body has neither linear acceleration
nor angular acceleration.
Condition for translational equilibrium
   n 
A The vector sum of the forces, on the rigid body is zero; 1 F  F 2  ....  Fn   F1  0 i 1
If the total force on the body is zero, then the total linear momentum of the body does not
change with time. P=constant
Condition for rotational equilibrium :
n
   
A The vector sum of the the torques on the rigid body is zero;  1   2  ....   n   1  0
i 1
If the total torque on the rigid body is zero, the total angular momentum of the body does not
change with time.
A The rotational equilibrium condition is independent of the location of the origin about which
the torques are taken.
Principle of moments :
A An ideal is lever essentially a light rod pivoted at a point along its length. This point is called the
fulcrum. Two forces F1 and F2, parallel to each other and usually perpendicular to the lever act on
the lever at distances d1 and d2 respectively from the fulcrum. N is directed opposite to the
forces F1 and F2. (N=Reaction at fulcrum) For translational equilibrium. N-F1-F2=0
N
d1 d2
F1 F2
A For rotational equilibrium take the moments about the fulcrum; the sum of moments must
be zero, d1F1=d2F2=0
N acts at the fulcrum itself and has zero moment about the fulcrum.
A In the case of the lever force F1 is usually some weight to be lifted. It is called the load and its
distance from the fulcrum d1 is called the load am. Force F2 is the effort applied to lift the
load; distance d2 of the effort from the fulcrum is the effort arm.
Principle of moments for a lever
Load arm x load = effort arm x effort
Mechanical advantage :
A The ratio F1/F2 is called the Mechanical Advantage
F d
M . A.  1  2
F2 d1
If the effort arm d2 is larger than the load arm, the mechanical advantages is greater than
one. It means that a small effort can be used to lift a large load.
5. PQR is a rigid equilateral triangle frame of a side length ‘L’. Forces F1,F2 and F3 are
acting along PQ, QR, PR. If the system is in rotational equilibrium find the relation
between the forces.
Sol. Perpendicular distance of any force from centroid ‘C’ of triangle is L / 2 3 . The forces F1, F2
produce anti-clockwise turning effect where as F3 clock wise turning effect about ‘C’.
F2
R
C
P Q
F1
F3
Since the system is in rotational equilibrium the total torque acting on the system about the
centroid is zero
L L L
F1   F2   F3  0
2 3 2 3 2 3
Hence F1  F2  F3  0;  F3  F1  F2
6.A meter stick is balanced on a knife edge at its centre. When two coins, each of mass
5g are put one on top of the other at the 12cm mark, the stick is found to be balanced
at 45cm. What is the mass of the metre stick?
Sol. R
10g
mg
Since the stick is in rotational equilibrium, the total torque of all the forces about the resultant
‘R’ is zero. Taking the turning effects about the point of action of the resultant R we have
10g x 33=mg x 5; m= 66g
7.A uniform ladder of mass 10 Kg leans against a smooth vertical wall making an angle
530 with it. The other end rests on a rough horizontal floor. Find the normal force
and the frictional force that the floor exerts on the ladder.
Sol. The ladder is in equilibrium.
N1 A
N2
B O
f C
 N1  f and N 2  W
Taking torque about ‘B’; N1(AO)=W(CB)
 AB 
N1  AB  cos 530  W   sin 53
0
 2 
2
N1  W and N 2  W  10  9.8  98 N .
3
2
The frictional force is f  N1  W  65 N
3
Toppling :
N N
B D F D F
B a
C b C b
f A E f A E
W=mg W=mg
N
B D F
a
C b
f A E
W=mg
Suppose a force F is applied at a height b above the base AE of the block. Further, suppose
the friction ‘f’ is sufficient to prevent sliding. In this case if the normal reaction N also passes
through C then despite the fact that the block is in translational equilibrium (F=f and N=mg)
an unbalanced torque (due to the couple of forces F and f) is there. This torque has tendency
to topple the block about point E. To cancel the effect of this unbalanced torque the normal
reaction N is shifted towards right a distance ‘a’ such that, net anti clock wise torque is
equal to the net clock wise torque.
Fb
Fb  mg  a   a 
mg
Now, as F(or) b (or) both are increased distance a also increases. But it can not go beyond
the right edge of the block. So in extreme case the normal reaction passes through E. Now
if F or b are further increased, the block will topple down. This is why the block having the
broader base has less chances of toppling in comparison to a block of smaller base.
8 A uniform cylinder of height h and radius r is placed with its circular face on a rough
inclined plane and the inclination of the plane to the horizontal gradually in creased.
If  is the coefficient of friction, then under what conditions the cylinder will
a) slide before toppling
b) topple before sliding
Sol.
a) The cylinder will slide if
Mg sin    Mg cos   tan   ....1

The cylinder will topper if
h 2r
 Mg sin     Mg cos   r  tan   .... 2 
2 h
N
f  N
Mg sin   Mg Mg cos 
2r
Thus, the condition of sliding is tan    and condition of toppling is tan   . Hence, the
h
2r
cylinder will slide before toppling if  
h
2r
b) The cylinder will topple before sliding if  
h
9.A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal
force F is applied normal to one of the face at a point directly above the centre of the
3a
face, at a height above the base. What is the minimum value of F for which the
4
cube begins to topple about an edge?
Sol. In the limiting case normal reaction will pass through O. The cube will topple about O if
torque of F exceeds the torque of mg.
N
a
2 F
3a
4
O
mg
 3a  a 2
 F    mg   ;  F  mg
 4  2 3
2
So, the minimum value of F is mg
3
10. A force F is applied on the top of a cube as shown in the figure. The coefficient of
friction between the cube and the ground is  . If F is gradually increased, find the
value of  for which the cube will topple before sliding.
F a
P
f
mg
Sol. Let m be the mass of the cube and ‘a’ be the side of the cube.
The cube will slide if F   mg ......(1)
and it will topple if torque of F about P is greater than torque of ‘mg’ about P i.e,
a 1
Fa    mg or F  mg .....  2 
2 2
1
From equations (1) and (2) we see that cube will topple before sliding if   .
2
MOMENT OF INERTIA [ROTATIONAL INERTIA]
A A body at rest cannot start rotating itself or a rotating body cannot stop rotating on its own.
Hence, a body has inertia of rotational motion.
A The quantity measuring the inertia of rotational motion is known as moment of inertia.
A Moment of inertia of a particle of mass m is
I  mr 2
Where r=perpendicular distance of particle from axis of rotation.
S.I Unit : kgm2; Its D.F-ML2
Dimensional formula :  ML2 
A Moment of inertia of a group or system of particles is I  m1r12  m2 r22  ......  mn rn2 I   mr 2
Where m1, m2,....,mn are masses of particles and r1, r2,.....,rn are their perpendicular distance
from axis of rotation.
A Motional or Inertia in rotational motion is analogous (similar) to mass in translatory motion.
A Moment of inertia of a rigid body depends on the following three factors.
a) mass of the body b) position of axis of rotation
c) Nature of distribution of mass.
Note-1 : Moment of inertia of a rotating rigid body is independent of its angular velocity.
Note-2 : Moment of inertia of a metallic body depends on its temperature
11.Four holes of radius R are cut from a thin square plate of the side 4R and mass M in
XY plane as shown. Then moment of inertia of the remaining portion about z-axis
is
Sol. M is the mass of the square plate before cutting the holes.
 M  2 
Mass of one hole m   2
R  M
16 R  16
R R
R R
moment of inertia of remaining portion

I  I square  4 I hole
M  mR 2 
I 16 R 2  16 R 2   4   m  2R 2 
12  2 
8  8 10 
 MR 2  10mR 2    MR
2
3  3 16 
Radius of Gyration(K) : Radius of gyration of a rigid body about an axis of rotation is
distance between the axis of rotation and a point at which the whole mass of the body can
be supposed to be concentrated so that its moment of inertia would be the same with the
actual distribution of mass.
A Moment of inertia of a rigid body of mass M is I=MK2
Where K=radius of gyration
r12  r22  ......  rnn
K
n
When n is total number of particles in the body and r1, r2,.....,rn are their perpendicular
distances from axis of rotation.
S.I. Unit : metre CGS Unit: cm
Dimensional formula : [M0LT0]
Note : K is not the distance of centre of mass of body from the axis considered.
A Radius of gyration of a rigid body depends on the following two factors
a) Position of axis of rotation
b) Nature of distribution of mass.
PERPENDICULAR AXES THEOREM
Statement : It states that the moment of inertia of a plane lamina about an axis perpendicular
to its plane is equal to the sum of its moments of inertia about two mutually perpendicular
axes concurrent with perpendicular axis and lying in the plane of the body.
z
Planar Body
O N
x
M
P(x,y)
y
Iz  Ix  I y
A This theorem is applicable to bodies which are planar.

A This theorem applies to flat bodies whose thickness is very small compared to their other
dimensions.
A K z  K x2  K y2
12. Two identical rods each of mass M and length L are joined in cross position as
shown in figure. The moment of inertia of a system about a bisector would be.
B1 B2
Sol. Moment of inertia of a system about an axis which is perpendicular to plane of rods and
passing through the common centre of rods
ML2 ML2 ML2
Iz   
12 12 6
Again from perpendicular axes theorem
I z  I B1  I B2  2 I B1  2 I B2  as I B1  I B2 
I z ML2
 I B1  I B2  
2 12
PARALLEL AXES THEOREM

Statement: The moment of inertia of a body about an axis is equal to the sum of the moment
of inertia of the body about a parallel axis passing through its centre of gravity and the
product of its mass and the square of the distance between the two parallel axes.
I IG
M
I  I G  Ma 2 a
G
A This theorem is applicable to a body of any shape.

A K  KG2  a 2
13. The moment of inertia of a rod of length l about an axis passing through its centre of
mass and perpendicular to rod is I. The moment of inertia of hexagonal shape
formed by six such rods, about an axis passing through its centre of mass and
perpendicular to its plane will be
Sol. M.I. of rod AB about its centre and
ml 2
perpendicular to length=  I  ml 2  12 I
12
A l B
Now moment of inertia of rod about the axis which is passing through O and perpendicular
to the pane of hexagon
ml 2
I rod   mx 2 [ from parallel axes theorem]
12
2
ml 2  3  5ml 2
  m l 
12  2  6
Now moment of inertia of system
I system  6  I rod  5ml 2  5  12 I  60 I
14.The radius of gyration of a body about an axis at a distance of 12cm from its centre of
mass is 13cm. Find its radius of gyration about a parallel axis through its centre of
mass.
Sol. By parallel axes theorem
M 13  I 0  M 12 
2 2
I 0  M 132  12 2   M  25 
I0
Its radius of gyration about a parallel axes through its centre of mass K   25  5cm
M
15. Find the moment of inertia of a thin uniform rod about an axis perpendicular to its
l
length and passing through a point which is at a distance of from one end. Also
3
find radius of gyration about that axis.
l
Sol. i) Let P be the point at a distance from one end.
3
l l l
It is a distance of     from the centre as shown in the figure.
 2 3 6
l I l Ic
3 6
P C
l
2
By parallel axes theorem I=IC+Mr2
2
Ml 2 l Ml 2
 M  
12 6 9
I Ml 2 l
ii) The radius of gyration, K   
M 9M 3
16..A uniform cylinder has radius R and length L. If the moment of inertia of this cylinder
about an axis passing through its centre and normal to its circular face is mg equal
to the moment of inertia of the same cylinder about an axis passing through its
centre and normal to its length, then
Sol. Moment of inertia of a cylinder about an axis passing through centre and normal to circular
MR 2
face =
2
Moment of inertia of a cylinder about an axis passing through centre and normal to its
length
 L2 R 2 
M  
12 4 
MR 2  L2 2
But M   
2 12 4
R 2 L2 R 2 R 2 L2
    ;
2 12 4 4 12
 L  3R
17. A metal piece of mass 120g is stretched to form a plane rectangular sheet of area of
cross section 0.54m2. If length and breadth of this sheet are in the ratio 1:6, find its
moment of inertia about an axis passing through its centre and perpendicular to
its plane.
Sol. Mass M=120g=120x10-3 kg
b  b
Area = lb  0.54m  b  0.54  l  
2
6  6
b 2  0.54  6  b  3.24  1.8m
M l 2  b2 
I  33.3  1023 kgm 2
12
18.: The moment of inertia of HCl molecule about an axis passing through its centre of
mass and perpendicular to the line joining the H+ and Cl- ions will be (if the inter
atomic distance is 1A0).
Sol. r  1A0  10 10 m; m1  1amu; m2  35.5amu
m1m2
Reduced mass   m  m  0.9726amu
1 2
 1.624  10 27 kg 1amu  1.67 x10 27 kg 

Moment of inertia about an axis passing through centre of mass of two particle system and
perpendicular to the line joining them is
1   r 2  1.624  1047 kgm 2
19.Four solid spheres each of diameter 2a and mass m are placed with their centers on
the four corners of a square of side b. Calculate the moment of inertia of the system
about any side of the square.
4 3
Sol.
1 2
a
b
2 2 2
I1  ma ; I 2  ma 2  mb 2
5 5
2 2 2
I3  ma  mb 2 ; I 4  ma 2
5 5
Moment of inertia of the system I  I1  I 2  I 3  I 4
2 2 2 2
 ma2  ma2  ma2  ma2  ma2
5 5 5 5
8
I  ma 2  2mb 2
5
20.A rod PQ of mass ‘m’ and length L is rotated about an axis through ‘P’ as shown in
figure. Find the moment of inertia of the rod about the axis of rotation.
Sol. Consider a small element ‘dx’ of the rod which is at a distance ‘x’ from the end ‘P’. If ‘  ’ is
the inclination of rod w.r.t the axis of rotation, the radius of the circle in which the element
r
rotates is given by sin    r  x sin 
x
Q
x sin 
 dx
P x
M.I. of the element about the axis of rotation is dI=dm.r2

m
where dm is the mass of element dm  dx
L
L
mL m 2 2 mL2
 x sin   . Total M.I. of the rod is given by   L
2
dI  I  dI  sin  x dx ,I  sin 2 
dx 0 3
21. Two uniform circular disc, each of mass 1kg and radius 20cm, are kept in contact
about the tangent passing through the point of contact. Find the moment of inertia
of the system about the tangent passing through the point of contact.
A
Sol.
B
Mass m+1kg, r=20x10-2m
MR 2 5MR 2
I1   MR 2 
4 4
5MR 2
Similarly I 2  , I  I1  I 2
4
10 MR 2 10  1   20  10 
2 2
I    0.1kgm 2
4 4
22. Two solid sphere (A and B) are made of metals of different densities  A and  B
respectively. If their masses are equal, the ratio of their moments of inertia (IA/IB)
about their respective diameter is [E-2007]
Sol. As two solid spheres are equal in masses, so
1
4 4 R   3
mA  mB   RA3  A   RB3  B  A   B 
3 3 RB   A 
The moment of inertia of sphere about diameter
2 2
2 I R  I   3
I  mR 2  A   A   A   B 
5 I B  RB  IB   A 
23. The moment of inertia of a then circular disc about an axis passing through its center
and perpendicular to its plane is I. Then, the moment of inertia of the disc about an
axis parallel to its diameter and touching the edge of the rim is [E-2008]
MR 2
Sol. I   MR 2  2 I
2
M.I. of the disc about tangent in a plane
5 5
 MR 2  I
4 2
24. The moment of inertia of a disc, of mass M and radius R, about an axis which is a tangent
and parallel to its diameter is [E-2010]
Sol. About the tangent parallel to the diameter
MR 2 5
I  I g  MR 2   MR 2  MR 2
4 4
25.0Two solid spheres A and B each of radius R are made of materials of densities  A
and  B respectively. Their moments of inertia about a diameter are IA and IB
respectively. The value of IA/IB is [E-2012]
4 3
IA 3
 R A 
  A
Sol. I B 4 3
 R 3  B
3
26.From a complete ring of mass M and radius R, an are making 300 at centre is removed.
What is the moment of inertia of the incomplete ring about an axis passing through
the centre of the ring and perpendicular to the plane of the ring.
M  11M
Sol. Mass of incomplete ring= M   
2 6 12
300
R
11M  11
M.I. of incomplete ring I    R 2  MR 2
 12  12
Note : If a sector of mass m, rotates about its natural axis then its M.I. is mR2
27. A thin wire of length I having density  is bent into a circular loop with C as its centre,
as shown in figure. The moment of inertia of the loop about the line AB is [E-2014]
2
3 3  l  3 l 3
Sol. I  MR 2
  l     
2 2  2  8 2
28.For the given uniform square lamina ABCD, whose centre is O. Its moment of inertia
about an axis AD is equal to how many times its moment of inertia about an axis
EF? [AIEEE-2007]
D F C
A E B
1) 2 I AC  I EF 2) I AD  3I EF
3) I AC  4 I EF 4) I AC  2 I EF
Sol. I EF  I GH  due to symmetry 
I AC  I BD  due to symmetry 
I AC  I BD  I 0
 2 I AC  I 0 ....1 and I EF  I GH  I 0
 2 I EF  I 0 ....... 2 
From Eqs (1) and (2), we get
I AC  I EF
md 2 md 2 md 2
 I AD  I EF   
4 12 4
md 2
I AD   4 I EF
3
29.Consider a uniform square plate of side a and mass m. The moment of inertia of this
plate about an axis perpendicular to its plane and passing through one of its corners
is [AIEEE-2008]
5 2 1 7 2 2
1) ma 2) ma 2 3) ma 2 4) ma
6 12 12 3
Sol. Using parallel axes theorem,
Ml 2 Ml 2 7 Ml 2
I  I G  Mr 2   
12 2 12
30.A disc of moment of inertia 4kgm2 is spinning freely at 3rads-1. A second disc of moment
of inertia 2kgm2 slides down the spindle and they rotate together. a) What is the
angular velocity of the combination?
b) What is the change in kinetic energy of the system?
Sol. a) Since there are no external torques acting, we may apply the conservation of angular
momentum.
I f  f  I ii   6 kgm 2   f   4 kgm 2  3rad s 1 
Thus  f  2rads 2
b) The kinetic energies before and after the collision are
1 1
Ki  I ii2  18 J ; K f  I f  2f  12 J
2 2
The change is K  K f  K i  6 J
In order for the two discs to spin together at the same rate, there had to be friction between
them. The loss in kinetic energy is converted into thermal energy.
Angular momentum of a particle
Definition :The moment of linear momentum of a body w.r.t. an axis of rotation is known as
angular momentum.

A The angular momentum L of the particle with respect to the origin O is represented as
  
L  r  p  mr  v  .
A The magnitude of the angular momentum vector is L  r p sin  , where p is the magnitude of
 
p and  is the angle between r and p .
A It is always directed perpendicular to the plane of rotation and along the axis of rotation.
Angular momentum of rigid body:
When a rigid body is rotating, then the vector sum of angular momenta of all the particles of
body about the axis of rotation is called angular momentum of rigid body. It is equal to the
product of moment of inertia and angular velocity.
 L    ri  m i vi   I
i
S.I. Unit : kgm2/sec

Dimensional formula : ML2T-1
When a body is rolling its total angular momentum is the vector sum of its angular momentum
about centre of mass and the angular momentum of centre of mass about a fixed point on
the ground.
Moment of inertia of some regular rigid bodies
Moment Radius of
Rigid Body Axis Rotation of Inertia Gyration
(I) (K)
MR2 R
1) r to the plane of ring and passing through its
centre
2MR2 2R
2) r to the plane of ring and passing through its rim
1) Circular
ring of (or) passing through any tangent r to the plane of ring
mass M and 3) In the plane of the ring and passing through its MR2/2 R/ 2
radius R. centre
(or) passing through any diameter of ring
4) In the plane of the ring and passing through its edge 3MR2/2 3 / 2R
(or) passing through any tangent of ring in its plane.
1) r to the plane of plate and passing through its

centre MR2/2 R/ 2
2) Thin 2) r to the plane of plate and passing through its edge 32R
circular
(or) passing through any tangent r to its plane.
plane of
MR2/4 R/2
mass M and 3) In the plane of plate and passing through its centre
radius R (or) passing through any diameter of plate
4) In the plane of the plate and passing through its edge 5MR2/4 5R / 2
(or) passing through any tangent of plate in its plane.
3) Thin
hollow 1) Passing through its centre or any diameter 2MR2/3 2R / 3
sphere of
mass M and 2) Passing through any tangent 5MR2/3 5R / 3
radius R
4) Solid
1) Passing through its centre or any diameter 2MR2/5 2R / 5
sphere of
mass M and
2) Passing through any tangent 7MR2/5 7R / 5
radius R
5) Thin 1) r to the length of rod and passing through its ML2/12
L/2 3
uniform rod centre
of mass M
and L ML2/3 L/ 3
2) r to the length of rod and passing through its end
Moment of inertia of some regular rigid bodies
Radius of
Moment of
Rigid Body Axis Rotation Gyration
Inertia (I)
(K)
4)  r to the axis of cylinder and passing through  L3 R 2  L2 R2
M   
one end 3 2  3 2
MR2/2
1) About geometrical or natural axis R/ 2
2) Parallel to the length of cylinder and touching

9) Solid its surface (or) passing through line of contact of 3R / 2
3MR2/2
cylinder of cylinder with floor when it is rolling.
Mass M
 L2 R 2  L2 R2
radius R and 3)  r to the axis of cylinder and passing through its M   
length L  12 4  12 4
centre
4) r to the axis of cylinder and passing through one  L2 R 2  L2 R2

M  
end  3 4
 3 4 
Law of conservation of angular momentum

If there is no external torque acting on the rotating body (or system of particles), then its
angular momentum is conserved.
dL  dL 
If  ext  0 then  0    ext 
dt  dt 
 L  I   constant  I11  I 22
31. A ballet dancer spins about a vertical axis at 60 rpm with arms outstretched. When
her arms are folded the angular frequency increases to 90 rms. Find the change in
her moment of inertia
Sol. By the principle of conservation of angular momentum I  60  I 2  90
2I
Final moment of inertia, I 2 
3
2I I
Change in moment of inertia = I  
3 3
32.A horizontal disc is freely rotating about a vertical axis passing through its centre at
the rate of 100 rpm. A bob of wax of mass 20g falls on the disc and sticks to it a
distance of 5 cm from the axis. If the moment of inertia of the disc about the given
axis is 2x10-4 kgm2, find new frequency of rotation of the disc.
Sol. I1=Moment of inertia of disc = 2 x 10-4 kgm2
I2=moment of inertia of the disc + moment of inertia of the bob of wax on the disc
 2  10 4  mr 2  2  10 4  20  103  0.05 
2
 2  104  0.5  10 4  2.5  104 kgm2

By the principle of conservation of angular momentum
I1n1  I 2 n2  2  10 4  100  2.5  104 n2
100  2
n2   80rpm
2.5
33.A circular platform is mounted on a vertical frictionless axle. Its radius is r=2m and its
moment of inertia is I=200kg-m2. It is initially at rest. A 70kg man stands on the edge
of the platform and begins to walk along the edge at speed. V0=1.0 m/s relative to the
ground. Find the angular velocity of the platform.
Sol. Angular momentum of man=angular momentum of platform in opposite direction.
mv0 r  I     0.7 rad / s
EX-52: A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two
point masses m and 2m moving in the same horizontal plane with speeds 2v and v
respectively, strike the bar (as shown in fig) and stick to the bar after collisions.
Calculate (a) velocity of the centre of mass (b) angular velocity about centre of mass
and (c) total kinetic energy, just after collision.
2m
v
8m
2a a 2a a
2v
m
Sol. (a) As Fext=0 linear momentum of the system is conserved, i.e.,
2m  v  m  2v  0   2m  m  8m   V
or V=0 i.e. velocity of centre of mass is zero.
(b) As  ext  0 angular momentum of the system is conserved, i.e.
m1v1r1  m2 v2 r2   I1  I 2  I 3  
2mva  m  2v  2a    2m  a   m  2a   8m   6a  / 12 
2 2 2
 
 v 
i.e. 6mva  30ma    
2
 5a 
(c) From (a) and (b) it is clear that, the system has no translatory motion but only rotatory
motion.
1 2 1  v 2 3 2
E I    30ma 2     mv
2 2  5a  5
34.A hoop of radius r and mass m rotating with an angular velocity ‘ 0 ’ is placed on a
rough horizontal surface. The initial velocity of the centre of the hoop is zero. What
will be the velocity of the centre of the hoop when it ceases to slip. (JEE-2013)
v 0 r
Sol. mr 0  mvr  mr   v 
2 2
r 2
Rotational dynamics
Relation between Torque and angular momentum of a rigid body:
The vector sum of torques acting on various particles of rigid body gives the net torque
acting on body.
dL
   i and   , is total angular momentum of body..
dt L
The time rate of change of the angular momentum of a particle is equal to the torque acting
on it.
Relation between torque and angular acceleration:
dL d
  But L  I   I    I
dt dt
This equation is called equation of rotatory motion and analogous to Newton’s 2nd law in
dynamics.
35.A uniform rod of mass ‘m’ and length ‘l’ is suspended by means of two light
inextensible strings at the ends of a rod. Tension in one string immediately after
the other string is cut is
Sol. mg  T  ma.....1
T T
mg

mg
 2  3g ...... 2 
  2
I ml 2l
3
1
a   ...... 3
2
mg
solving eq (1), (2) and (3) we get, T 
4
36. The pulley of Atwoods machine has a moment of inertia ‘I’ about its axis and its
radius is ‘R’. Find the magnitude of acceleration of the two blocks assuming the
string is light and does not slip on the pulley.
T2
Sol. T1
m
M
Suppose the block of mass ‘M’ goes down with an acceleration ‘a’. The angular acceleration
a
of the pulley is,  
R
Mg  T1  Ma;T2  mg  ma
a
and T1 R  T2 R  I  I
R
 M  m  gR 2
Solving the equation, a 
I   M  m R2
ROTATIONAL KINETIC ENERGY

The sum of the kinetic energies of various particles of rotating body is called rotational
kinetic energy.
L2 1 2 1
KErot   I   L
2I 2 2
37. The angular momentum of rotating body is increased by 20%. What will be the
increase in its rotational kinetic energy?
L2
Sol. Kinetic energy KE =  E  L2
2I
2
E  120  E
   or   0.44
E  100  E
E
 100  44%
E
38.A uniform rod of length ‘l’ is held vertically on a horizontal floor fixing its lower end, the
rod is allowed to fall onto the ground. Find (i) its angular velocity at that instant of
reaching the ground (ii) The linear velocity with which the tip of rod hits the floor.
Sol. The rod rotates about an axis through one end. From the principle of conservation of
mechanical energy. Loss of P.E. of the rod is equal to its gain in rotational K.E.
l l
2
l 1 2 l 1 ml 2 2
 mg  I   mg  . 
2 2 2 2 3
3g
on solving  
l
iii) V  r or V  l  l 3g / l  3gl
39. A rigid body is made of three identical thin rods, each of length ‘L’ fastened together
in the form of the letter ‘H’. The body is free to rotate about horizontal axis that runs
along the length of one of the legs of ‘H’. The body is allowed to fall from rest from
a position in which the plane of ‘H’ is horizontal. What is the angular speed of the
body when the plane of ‘H’ is vertical?
mL2 4
Sol. The moment of inertia of the system about one rod as axis I   mL2 ; I  mL2
3 3
Potential energy decreases for B and C
B
Y
C
mgL 3
 mgL  mgL
2 2
By conservation of mechanical energy, the loss in PE of body is equal to the gain in
rotational KE
3 14 
 mgL   mL2   2 on solving   3 g
2 2 3  2 L
40.A uniform rod AB of mass ‘m’ length ‘2a’ is allowed to fall under gravity with AB in
horizontal. When the speed of the rod is ‘v’ suddenly the end ‘A’ is fixed. Find the
angular velocity with which it begins to rotate.
PREVIOUS MAINS QUESTIONS
Topic-2: Angular Displacement, Velocity and Aceleration
42. A bead of mass m stays at point P  a, b  on a wire bent in the shape of a

parabola y  4Cx 2 and rotating with angular speed  (see figure). The value of  is
(neglect friction [sep2 2020 MAINS ]
2gC 2g
(1) 2 2gC (2) 2 gC (3) (4)
ab C
2 dy
solution:. y  4Cx   tan   8Cx
dx
At P, tan   8Ca
For steady circular motion
m2 a cos   mg sin 
g tan 

a
g  8aC
   2 2 gC
a
43. A cylindrical vessel containing a liquid is rotated about its axis so that the the
liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the
an g u lar s p eed o f ro tatio n is  rad s–1. The difference in the height, h (in cm)of liquid at
the centre of vessel and at the side will be [sep2 2020 MAINS ]
22 52 252 22

(1) (2) (3) (4)
25g 2g 2g 5g
solution:. (3) Here, dr 2 r  gdh
R h
 2  rdr  g  dh
0 0
2 R 2
  gh (Given R = 5 cm)
2
2 R 2 252
h  
2g 2g
44 A spring mass system (mass m, spring constant k and natural length l) rests in
equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the
disc. If thedisc together with spring mass system, rotates about it’s axis with an
 2

angular velocity  , k  m the relative change in the lenght of the spring is best
given by the option : [9 JAN2020 MAINS ]
2  m2  2m2
(1)   (2)
3  k  k
m2 m2
(3) (4)
k 3k
solution:.
25. (3) Free body diagram in the frame of disc
 m2   0  x   kx
m 02
x
k  m2
For k  m2
x m2
 
0 k
45.A particle of mass m is fixed to one end of a light spring having force constant k
and unstretched length l. The other end is fixed. The system is given an angular
speed  about the fixed end of the spring such that it rotates in a circle in gravity free
space. Then the stretch in the spring is [8JAN2020 MAINS ]
ml 2 ml2 ml2 ml 2
(1) (2) (3) (4)
k  m k  m2 k  m2 k  m
solution:.26. (2) At elongated position (x),
mv 2
Fradial   mr 2
r
 kx  m    x  2
 r    x here 
kx  m2  mx2
m2
x 
k  m2
46. A uniform rod of length l is being rotated in a horizontal plane with a constant
angular speed about an axis passing through one of its ends. If the tension generated
in the rod due to rotation is T(x) at a distance x from the axis, then which of the
following graphs depicts it most closely? [120APR 2019 MAINS ]
(1) (2)
(3) (4)
T x
  dT     dm   x
2
solution:. (4)
0 l
x
m 
T    dx  2 x
l 
l
m2 2
Or T 
l
l  x2 
It is a parabola between T and x.
47 A smooth wire of length 2r is bent into a circle and kept in a vertical plane. A
bead can slide smoothly on the wire. When the circle is rotating with angular speed 
about the vertical diameter AB, as shown in figure, the bead is is at rest with respect
to the circular ring at position P as shown. Then the value of 2 is equal to:
[12 APR 2019 MAINS ]
(1)
3g
2r

(2) 2 g / r 3 

(3) r 3 / r  (4) 2 g / r
solution:(2) N sin   m2  r / 2  .....(i)

r/2 1
sin       300
r 2
and N cos   mg ......(ii)
2 r
or tan  
2g
0 2 r
or tan 30 
2g
1 2 r
or 
3 2g
2g
2 
r 3
48. A long cylindrical vessel is half filled is half filled with a liquid. When the vessel
is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of
vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in
the heigthts between the centre and the sides, in cm, will be
(1) 2.0 (2) 0.1 (3) 0.4 (4) 1.2 [12 JAN 2019 MAINS ]
solution:((1) Using v 2  u 2  2 gy [ u  0 at (0, 0)]
v 2  2 gy [ v  x ]
2 x 2  2  2    0.05 
2 2
y   2cm
2g 20
49.. A particle is moving with a uniform speed in a circular orbit of radius R in a
central force inversely proportional to the nth power of R. If the period of rotation of
the particle is T, then [2018 MAINS ]
(1) T  R 3/2 for any n (2) T  R n /21
(3) T  R n1/2 (4) T  R n /2
2 1 mv 2
solution:.(3) m R  Force  (Force  )
Rn R
1 1
 2  n 1
   n 1
R
R 2
2
Time period T 

n 1
Time period,
T R 2
50.The machine as shown has 2 rods of length 1m connected by a pivot at the top.
The end of one rod is connected to the floor by a stationary pivot and the end of the
other rod has a roller that rolls along the floor in a slot. As the roller goes back and
forth, a 2kg weight moves up and down. If the roller is moving towards right at a
constant speed, the weight moves up with a:
[2017 MAINS ]
(1) constant speed (2) decreasingspeed (3) increasing speed
3
(4) speed which is th of that of the roller when the weight is 0.4 m above the ground
4
solution:.
50. (2) decreasing speed

51.A slender uniform rod of mass M and length  is pivoted at one end so that it can
rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free
end is held vertically above the pivot and then released. The angualr acceleration of
the rod when it makes an angle with the vetical is
[2017MAINS ]
3g 2g 3g 2g
(1) cos  (2) cos  (3) sin  (4) sin 
2 3 2 2
solution:.(3) Torques at angle 
l l
  Mg sin . Also   l   l   Mg sin 
2 2
Ml 2 l  Ml 2 
.  Mg sin   I rod  
3 2  3 
l sin  3g sin 
 g  
3 2 2l
52.Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical
drum. If the drum rotates too faste, the ingredients remain stuck to the wall of the
drum and proper mixing of ingredients does not take place. The maximum rotational
speed of te drum in revolutions per minute (rpm) to ensure proper mixing is close
to: (Take the radius of the drum to be 1.25 m and its axle to be horizontal):
(1) 27.0 (2) 0.4 (3) 1.4 (4) 8.0 [2016MAINS ]
solution:.(1) For just complete rotation

v  Rg at top point
The rotational speed of the drum
v g 10
  
R R 1.25
The maximum rotational speed of the drum in revolutions per minute
60 10
  rpm    27
2 1.25
53.A cubical block of side 30 cm is moving with velocity 2 ms–1 on a smooth horizontal
surface. The surface has a bump at a point O as shown in figure. The angular
velocity (in rad/s) of the block immediately after it hits the bump, is
[APR9 2016 MAINS]
(1) 13.3 (2) 5.0 (3) 9.4 (4) 6.7
solution:.(2) Angular momentum, mvr  I 
Moment of Inertia (I) of cubical block is given by
R
m.2
 R2  R 2    2
I  m    R  R 2 
2
 6  2   m   
 
 6  2  
12 3 10
    5 rad / s
8R 2  0.3 2
54.Two point masses of mass m1  fM and m2  1  f  M  f  1 are in outer space (far

from gravitational influence of other objects) at a distance R from each other. They
move in circular orbits about their centre of mass with angular velocities 1 for m1
and 2 for m2 . In the case [MAY9 2016 MAINS]
(1) 1  f  1  f 
(2) 1  2 and independent of f
(3) f 1  1  f  2
(4) 1  2 and depend on f

solution:.(2) Angular velocity is the angular displacement per unit time i.e.,  
t
Here 1  2 and independent of f .
Topic-3: Torque, Couple and Angular Momentum

55.Four point masses, each of mass m, are fixed at the corners of a square of side l.
The square is rotating with angular frequency  , aqbout an ax is passing through
one of the corners of the square and parallel to its diagonal, as shown in the figure.
The angular moemtnum of the square about this axis is: [SEP 6 2020 MAINS]
(1) ml 2  (2) 4ml 2  (3) 3ml 2  (4) 2ml 2 
solution:.(3) Angular momentum, L  I 
2
 l 
 2l 
2
I  m  0
2
 m  2 m
 2
2ml 2
  2ml 2  3ml 2 Angular momentum L  I   3ml 2
2
56.A thin rod of mass 0.9 kg and length 1m is suspended, at rest, from one end so that
it can freely oscillate in the vertical plane. Aparticle of move 0.1 kg moving in a straight
line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see
figure). The angular speed (in rad/s) of the rod immediately after the collision will be
_________ [SEP 6 2020 MAINS]
solution:.(20)
Using principal of conservation of angular momentum we have
 
Li  L f  mvL  I 
 ML2 
 mvL   mL2  
 3 
 0.9  12 
 0.1 80  1    0.1 12  
 3 
 3 1 4
 8     8  
 10 10  10
   20 rad/sec
57.A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg
rotating about its axis at 5 revolutions per minute (rpm). The person now starts
moving towards the centre of the plastform. What will be the rotational speed (in rpm)
of the platform when the person reaches its centre_________[SEP3 2020 MAINS]
solution:.Here M 0  200kg , m  80 kg
Using conservation of angular momentum, Li  L f

 M R2 
I11  I 22 I1   I M  I m    0  mR 2 
 2 
1
I2  M 0 R 2 and 1  5 rpm
2
 M R2  5
2   0  mR 2  
 2  M R2
  0
2
5R 2  80  100 
   9 rpm
R2 100
58..A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal

surface and collides with a uniform vertical rod and sticks to it as shown. The rod is
pivoted about O and swings as a result of the collision making angle  q before
momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value
of q is approximately: (take g = 10 m/s2) [SEP3 2020 MAINS]
(1) 630 (2) 550 (3) 690 (4) 490
solution:.(1) Using conservation of angular momentum

Now using energy conservation, after collision
2
1 2 l 15  9v
I   2mg 1  cos    mgl 1  cos     ml 2   2mgl 1  cos  
2 2 23  25l 2
3
 mv 2  2mgl 1  cos  
5 2
3 36 27
  1  cos   1   cos 
10 2 10 50
23
Or, cos     630
50
 2ml 2  5 2 3v 3  6 18
mvl   ml 2     mvl  ml     or    rad/s
 3  3 5l 5 1 5
59. A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical
shaft of negligible radius. When the shaft rotates at angular speed  the rod makes
an angle  with it (see figure). To find  equate the rate of change of angular
ml 2 2
momentum (direction going into the paper)  sin  cos  about the centre of mass
12
(CM) to the torque provided by the horizontal and vertical forces FH and FV about the
CM. The value of q is then such that : [SEP3 2020 MAINS]
2g g
(1) cos   (2) cos  
3l2 2l2
g 3g
(3) cos   (4) cos  
l2 2l2
solution:.(4) Vertical force = mg

2 l
Horizontal force = Centripetal force  m sin 
2
l
Torque due to vertical force  mg sin 
2
2 l l
Torque due to horizontal force  m sin  cos 
2 2
l 2 l l ml 2 2
Net Torque = Angular momentum mg sin   m sin  cos    sin  cos 
2 2 2 12
3 g
 cos  
2 2l
60.
Shown in the figure is rigid and uniform one meter long rod AB held in
horizontal position by two strings tied to its ends and attached to the ceiling. The rod
is of mass ‘m’ and has another weight of mass 2 m hung at a distance of 75 cm from
A. The tension in the string at A is : [SEP2 2020 MAINS]
(1) 0.5 mg (2) 2 mg
(3) 0.75 mg (4) 1 mg

solution:.(4) Net torque, net about B is zero at equilibrium
TA  100  mg  50  2mg  25  0
 TA  100  100mg
 TA  1mg (Tension in the string at A)
61.A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a
< R) by applying a force F at its centre ‘O’ perpendicular to the plane through the axes
of the cylinder on the edge of the step (see figure). The minimum value of F required
is :
[SEP 2 2020 MAINS]
2 2
 Ra  R 
(1) Mg 1    (2) Mg   1
 R   Ra
a a2
(3) Mg (4) Mg 1 
R R2
solution:.(1)
For step up, F  R  Mg  x
x  R 2   R  a  from figure
2
2
Mg  Ra
 R 2   R  a   Mg 1  
2
Fmin  
R  R 
62.Consider a uniform rod of mass M = 4m and length l pivoted about its centre. A
mass m moving with velocity v making angle to the rod’s long axis collides with one

end of the rod and sticks to it. The angular speed of the rod-mass   system just
4
after the collision is: [8 JAN 2020 MAINS]
3 v 3v 3 2v 4v
(1) (2) (3) (4)
7 2 l 7l 7 l 7l
solution:.(3)
About point O angular momentum
Linitial  L final
mV 1  4mL2 mL2 
    
2 2  12 4 
6V 3 2V
  
7 2L 7L
63.A particle of mass m is moving along a trajectory given by
x  x0  a cos 1t y  y0  b cos 2 t [10 APR 2019 MAINS]
The torque, acting on the particle about the origin, at t = 0 is:
(1) m   x0 b  ya  12 k (2)  my0 a12 k (3) zero  

(4)  m x0 b22  y0 a12 k
solution:.(2) Given that, x  x0  a cos 1t
y  y0  b sin 2t
dx
 Vx
dt
dy
 vx   a1 sin  1t  , and  v y  b2 cos  2t 
dt
dvx dv y
 ax  a12 cos  1t  ,  a y   a22 sin  2t 
dt dt
At t  0, x  x0  a, y  y0
ax   a12 , a y  0
    
Now,   r  F  m r  a  
 
  x0  a  i  y0 j   m  a12 i   my0 a12 k
 
64.The time dependence of the position of a particle of mass m  2 is given by


r  t   2ti  3t 2 j . Its angular momentum, with respect to the origin, at time t = 2 is
 
(1) 48 i  j (2) 36k  
(3) 34 k  i (4) 48k
[10 APR 2019 MAINS]

 
solution:. We have given r  2ti  3t 2 j r  at t  2   4i  12 j

 dr
Velocity, v   2i  6t j
dt

v  at t  2   2i  12 j
   
  
L  mvr sin n  m r  v   2 4i  12 j  2i  12 j  48k
65. metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible
mass as shown in the figure The system is initially at rest. The constant torque, that
will make the system rotate about AB at 25 rotations per second in 5s, is close to :
[10 APR 2019 MAINS]
(1) 4.0  10 6 Nm (2) 1.6  10 5 Nm
(3) 7.9  10 6 Nm (4) 2.0  10 5 Nm
solution:.(4) Angular acceleration,
  0 25  2  0
   10 rad / s 2
t 5
  I
5
4


5

    mR 2      5  103 104 10
4
 
 2.0 105 Nm
66.A particle of mass 20 g is released with an initial velocity 5 m/s along the curve
from the point A, as shown in the figure. The point A is at height h from point B. The
particle slides along the frictionless surface. When the particle reaches point B, its
angular momentum about O will be : (Take g = 10 m/s2)[12 JAN 2019 MAINS]
(1) 2 kg-m2/s (2) 8 kg-m2/s (3) 6 kg-m2/s (4) 3 kg-m2/s
solution:.(3) According to work-energy theorem
1 2 1 2
mgh  mvB  mvA
2 2
2 gh  vB2  v A2
2 10 10  vB2  52  vB  15 m / s
Angular momentum about O,
L0  mvr
L0  6kg .m2 / s
 
67.A slab is subjected to two forces F 1 and F 2 of same magnitude F as shown in the
 

 
figure. Force F 2 is in XY-plane while force F1 acts along z-axis at the point 2i  3 j .
The moment of these forces about point O will be : [11 JAN 2019 MAINS]

(1) 3i  2 j  3k F   
(2) 3i  2 j  3k F

(3) 3i  2 j  3k F   
(4) 3i  2 j  3k F

solution:.(1) Given, F1 
2
 
F  F 3 
i 
2
j  

r1  0i  6 j
Torque due to F1 force

   F  F 3  
 F1  r1  F 1  6 j  
2
i   
2
j     3F  k 
Torque due to F2 force

 
 F2  2i  3 j  F k  3F i  2 F  j 
  
 
net  F1   F2  3Fi  2 F  j  3F k 

 3i  2 j  3k F
68.The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the
force acting on it is 1 N, and the distance of the particle from the origin is 5m, the
angle between the force and the position vector is (in radians):
   
(1) (2) (3) (4) [11 JAN 2019 MAINS]
6 3 8 4
  
solution:.(1) Torque about the origin    r  F
 rF sin   2.5  1 5sin 
1
sin   0.5 
2


6
69.To mop-clean a floor, a cleaning machine presses a circular mop of radius R

vertically down with a total force F and rotates it with a constant angular speed about
its axis. If the force F is distributed uniformly over the mop and if coefficient of friction
between the mop and the floor is m, the torque, applied by the machine on the mop
is: [10JAN 2019 MAINS]
(1) FR / 3 (2) FR / 6
2
(3) FR / 2 (4)  FR
3
solution:.(4) Consider a strip of radius x and thickness dx, Torque due to friction on this strip
Net torque =  Torque on ring
R
F .2xdx 2F R 3
 d   R 2
 2 .
R 3
0
2FR

3
70.A rigid massless rod of length 3l has two masses attached at each end as shown
in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When
released from initial horizontal position, its instantaneous angular acceleration will
be: [10 JAN 2019 MAINS]
g g g 7g
(1) (2) (3) (4)
13l 3l 2l 3l
solution:.(1) Applying torque equation about point P.
  I    2 M 0  2   5 M 0  2  
2
 
 5 M 0 g   4 M 0 g    2 M 0  2   5 M 0  2  
2
 

 M 0 g   13M 0 g  2  
71.An L-shaped object, made of thin rods of uniform mass density, is suspended with
a string as shown in figure. If AB = BC, and the angle made by AB with downward
vertical is  , then: [9 JAN 2019 MAINS]
1 1
(1) tan   (2) tan  
2 3 2
2 1
(3) tan   (4) tan  
3 3
solution:.(4) Given that, the rod is of uniform mass density and AB = BC
Let mass of one rod is m
Balancing torque about hinge point
mg  C1P   mg  C2 N 
L  L 
mg  sin    mg  cos   L sin  
2  2 
 mgL
 mgL sin   cos 
2 2
sin  1 1
  or, tan  
cos  3 3
72.
A uniform rod AB is suspended from a point X, at a variable distance from x
from A, as shown. To make the rod horizontal, a mass m is suspended from its end A.
A set of (m, x) values is recorded. The appropriate variable that give a straight line,
when plotted, are:
1 1
(1) m, (2) m, (3) m, x (4) m, x 2 [15 APR 2018 MAINS]
x x2
solution:.(1) Balancing torque w.r.t. point of suspension
 
mgx  Mg   x 
2 

 mx  M  Mx
2
 1
m  M  M
 2 x
1
y   C Straight line equation
x
73.A particle of mass m is moving along the side of a square of side ‘a’, with a uniform
speed v in the x-y plane as shown in the figure : [11 JAN 2016 MAINS]

Which of the following statements is false for the angular momentum L about
the origin?
  R 
(1) L  mv   a  k when the particle is moving from B to C.
 2 
 mv
(2) L  Rk when the particle is moving from D to A.
2
 mv 
(3) L   Rk when the particle is moving from A to B.
2
  R 
(4) L  mv   a  k when the particle is moving from C to D.
 2 
solution:.(1) We know that L  mvr
 R 
In none of the cases, the perpendicular distance r is   a
 2 
74.A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path
of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed
of the particle is 12 rad s–1, the magnitude of its angular momentum about a point on
the ground right under the centre of the circle is : [11 APR 2015 MAINS]
(1) 14.4 kg m2s-1 (2) 8.64 kg m2s-1
(3) 20.16 kg m2s-1 (4) 11.52 kg m2s-1
solution:.(1) Angular momentum, L0  mvr sin 900
 2  0.6  12  1 1
[As V  r ,sin 900  1 ]

So, L0  14.4 kgm 2 / s
75.A bob of mass m attached to an inextensible string of length l is suspended from a

vertical support. The bob rotates in a horizontal circle with an angular speed  rad/s
about the vertical. About the point of suspension: [11 JAN 2014 MAINS]
(1) angular momentum is conserved
(2) angular momentum changes in magnitude but not in direction
(3) angular momentum changes in direction but not in magnitude
(4) angular momentum changes both in direction and magnitude.
solution:.(c) Torque working on the bob of mass m is,   mg   sin  . (Direction parallel to plane
of rotation of particle)

As  is perpendicular to L, direction of L changes but magnitude remains same
76.A ball of mass 160 g is thrown up at an angle of 60° to the horizontal at a speed of
10 ms–1. The angular momentum of the ball at the highest point of the trajectory with
respect to the point from which the ball is thrown is nearly (g = 10 ms–2)
(1) 1.73 kg m2/s (2) 3.0 kg m2/s [19 APR 2014 MAINS]
(3) 3.46 kg m2/s (4) 6.0 kg m2/s
solution:..(3) Given : m  0.160 kg
  600
v  10 m / s
 
Angular momentum L  r  mv
 H mv cos 
v 2 sin 2   v 2 sin 2  
 cos   H  
2g  2g 
102  sin 2 600  cos 600


2  10
 3.46 kg m2 / s
77.A particle of mass 2 kg is moving such that at time t, its position, in meter, is given

by r  t   5i  2t 2 j . The angular momentum of the particle at t = 2s about the origin in kg
m–2 s–1 is : [23 APR 2013 MAINS]
(1) 80k 
(2) 10i  16 j 
(3) 40k (4) 40k
solution:.(1) Angular momentum L  m  v  r 
 dr 
   
 2kg   r   2kg 4tJ  5i  2t 2 j  2kg 20tk  2kg  20  2m 2 s 1 k
 dt   80k
78.A thin horizontal circular disc is rotating about a vertical axis passing through its
centre. An insect is at rest at a point near the rim of the disc. The insect now moves
along a diameter of the disc to reach its other end. During the journey of the insect,
the angular speed of the disc. [9APR 2013 MAINS]
(1) continuously decreases (2) continuously increases
(3) first increases and then decreases (4) remains unchanged
solution:.(3) Angular momentum, L  I   L  mr 2
As insect moves along a diameter, the effective mass and hence moment of inertia (I)
first decreases then increases so from principle of conservation of angular momentum, angular
speed  first increases then decreases
79.A small particle of mass m is projected at an angle  with the x-axis with an initial
v0 sin 
velocity n0 in the x-y plane as shown in the figure. At a time t  , the angular
g
momentum of the particle is [23 APR 2010MAINS]
1 1
(3)  mgv0 t cos k (4) mgv0 t cos k
2 2
(1)  mgv0 t 2 cos  j (2) mgv0 t cos k
2 2
Where i, j and k are unit vectors along x, y and z-axis respectively
  

solution:.(3) L  m r  v 
   1  
L  m v0 cos ti   v0 sin t  gt 2  j 
  2  
 v0 cos i   v0 sin   gt  j 

 
 1  1
 mv0 cos t   gt  k   mgv0t 2 cos k
 2  2
80.Angular montum of of particle rotating with a central force is constant due to
(1) constant torque [2007 MAINS]
(2) constant force
(3) constant linear momentum
(4) zero torque


 d Lc
solution:.(4) We know Torque c 
dt

Where L c  Angular momentum about the center of mass of the body. Central forces
act along the center of mass. Therfore torque about center of mass is zero
dL 
  0  L c  constt
dt
81.A thin circular ring of mass m and radius R is rotating about its axis with a constant
angular velocity  . Two objects each of mass M are attached gently to the opposite
ends of a diameter of the ring. The ring now rotates with an angular velocity  ' 
 m  2M  m  2M  m m
(1) (2) (3)  m  M  (4)  m  2 M  [2006 MAINS]
m  m  2M 
solution:.(4) Applying conservation of angular momentum I '  '  I 
 mR 2

 2 MR 2  '  mR 2
  m  2m  R 2  '  mR 2 
 m 
 '   
 m  2M 
82.A force of  Fk acts on O, the origin of the coordinate system. The torque about the
point (1, –1) is [2006 MAINS]
 
(1) F i  j  
(2)  F i  j
 
(3) F i  j  
(4)  F i  j
  
  
solution:.(3) Torque   r  F  i  j   F k 
     
 F  i  k  j  k   F j  i  F i  j
[Since k  i  j and j  k  i ]
83. A solid sphere is rotating in free space. If the radius of the sphere is increased
keeping mass same, which one of the following will not be affected ?[2004 MAINS]
(1) Angular velocity (2) Angular momentum
(3) Moment of inertia(4) Rotational kinetic energy
solution:.(2) Angular momentum will remain the same since no external torque act in free
space.
  
84.Let F be the force acting on a particle having position vector r , and T be the
torque of this force about the origin. Then [2003 MAINS]
   
(1) r.T  0 and F.T  0
   
(2) r.T  0 and F.T  0
   
(3) r.T  0 and F.T  0
   
(4) r.T  0 and F.T  0
 
solution:.(4) We know that   r  F
 
Vector  is perpendicular to both r and F . We also know that the dot product of two
vectors which have an angle of 900 between them is zero.
   
 r.T  0 and F .T  0
85.A particle of mass m moves along line PC with velocity v as shown. What is the
angular momentum of the particle about P? [2002 MAINS]
(1) mvL (2) mvl (3) mvr (4) zero

solution:.(4) Angular momentum (L)
= (linear momentum) x (perpendicular distance of the line of motion passes through

 L  mv  r
 mv  0
0
Topic-4: Moment of Inertia and Rotational K.E.
86.hown in the figure is a hollow icecream cone (it is open at the top). If its mass is M,
radius of its top, R and height, H, then its moment of inertia about its axis is :
[SEP 6 2020 MAINS]
(1)
MR 2
(2)

M R2  H 2 
2 4
MH 2 MR 2
(3) (4)
3 3
solution:.(4) Hollow ice-cream cone can be assume as several parts of discs having different
radius, so
 
I   dI   dm r 2 .....(i)
From diagram,
r R R
 tan   or r  h
h H H
2
 
Mass of element, dm   r dh ....(iii)
From eq. (i), (ii) and (iii),
From eq. (i), (ii) and (iii),
dh
Area of element, dA  2rdl  2r
cos 
2Mh tan dh
Mass of element, dm 
R R 2  H 2 cos 
(here, r  h tan  )
H H 2
I   dI     
R 
 dm r    r dh  H .h 
2 2
0 0
H   R 4 
      .h   dh
 H  
0   
MR 2
Solving we get, I 
2
87.The linear mass density of a thin rod AB of length L varies from A to B as

 x
  x    0 1   , where x is the distance from A. If M is the mass of the rod then its
 L
moment of inertia about an axis passing through A and perpendicular to the rod is :
5 7
(1) ML2 (2) ML2 [SEP 6 2020 MAINS]
12 18
2 3
(3) ML2 (4) ML2
5 7
solution:.(2)
Mass of the small element of the rod
dm  .dx
Moment of inertia of small element
 x
dI  dm.x 2   0 1   .x 2 dx
 L
Moment of inertia of the complete rod can be obtained by integration
L
 x3 
I   0   x 2   dx
0
L
L
x3 x 4  L3 L3 
 0   0   
3 4L 0 3 4
7 0 L3
I ......(i)
12
Mass of the thin rod,
L L
 x 3 L
M   dx    0 1   dx  0
0 0  L 2
2M
 0 
3L
7  2M  3 7 2
I    L  I  ML
12  3L  18
88.A wheel is rotating freely with an angular speed w on a shaft. The moment of inertia
of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of
moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The
resultant fractional loss in the kinetic energy of the system is :
5 1 3
(1) (2) (3) 0 (4) [SEP 6 2020 MAINS]
6 4 4
solution:.(4) By angular momentum conservataion, Lc  L f


I  3I  0  4 I  '   ' 
4
1 2
 KE i  I
2
1
 KE  f   3 I  I   '2
2
2
1   I 2
   4I     
2 4 8
1 2 3 2
 KE  I  I
2 8
3 2
I
 KE 8 3
  
 Fractional loss in K.E. KEli 1
I 2 4
2
89.ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of
AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an
axis passing through G and perpendicular to the plane ABC is I 0 . If part ADE is
NI 0
removed, the moment of inertia of the remaining part about the same axis is
16
where N is an integer. Value of N is _______.
[SEP 4 2020 MAINS]
solution:.(4) By angular momentum conservataion, Lc  L f


I  3I  0  4 I  '   ' 
4
1 2
 KE i  I
2
1
 KE  f   3 I  I   '2
2
2
1   I 2
   4I     
2 4 8
1 2 3 2
 KE  I  I
2 8
3 2
I
 KE 8 3
  
 Fractional loss in K.E. KEli 1
I 2 4
2
90.A circular disc of mass M and radius R is rotating about its axis with angular
R
speed 1 . If another stationary disc having radius and same mass M is dropped co-
2
axially on to the rotating disc. Gradually both discs attain constant angular speed 2 .
The energy lost in the process is p% of the initial energy. Value of p is ___________.
solution: [SEP4 2020 MAINS]
Using angular momentum conservation
I11  I 22   I1  I 2    f
MR 2  MR 2 MR 2  4
   0      f   f  
2  2 8  5
1 2 1  MR 2  2 MR 22
Initial K.E., i 2 I   2  2   
K 
4
 
1 2 1  MR 2  2 MR 22
Final K.E., f 2 I   2  2   
K 
4
 
Percentage loss in kinetic energy % loss
MR 2 2 MR 22

 4 5  100  20%  P %
2 2
MR 
4
Hence, value of P = 20
91.Consider two uniform discs of the same thickness and different radii R1 = R and R2
=  R made of the same material. If the ratio of their moments of inertia I1 and I2,
respectively, about their axes is I1:I2 = 1 : 16 then the value of a is :
(1) 2 2 (2) 2 (3) 2 (4) 4 [SEP 4 2020 MAINS]
solution:.(3) Let p be the density of the discs and t is the thickness of discs
Moment of inertia of disc is given by
I 
 
2  2
MR 2  R t  R 
2 2
I  R4 (As  and t are same)
4
I 2  R2  16
      4  2
I1  R1  1
92.
For a uniform rectangular sheet shown in the figure, the ratio of moments of
inertia about the axes perpendicular to the sheet and passing through O (the centre
of mass) and O’ (corner point) is : [SEP4 2020 MAINS]
(1) 2/3 (2) 1/4 (3) 1/8 (4) 1/2
solution:(b) Moment of inertia of rectangular sheet about an axis passing through O,
I0 
M 2
12

a  b2 
M

12 
 80    60  
2 2

From the parallel axis theorem, moment of inertia about O’,
I O '  I O  M  50 
2
IO
M
2

802  602
1 
 

I O ' M 802  602  M 50 2 4
12
  
93.Moment of inertia of a cylinder of mass M, length L and radius R about an axis
passing through its centre and perpendicular to the axis of the cylinder is
 R 2 L2 
I  M    . If such a cylinder is to be made for a given mass of a material, the
 4 12 
ratio L/R for it to have minimum possible I is : [SEP 3 2020 MAINS]
2 3 3 2
(1) (2) (3) (4)
3 2 2 3
solution:(b) Moment of inertia of rectangular sheet about an axis passing through O,
I0 
M 2
12

a  b2 
M
12


 80 2   60 2 
From the parallel axis theorem, moment of inertia about O’,
I O '  I O  M  50 
2
IO
M
2

802  602
1 
 

I O ' M 802  602  M 50 2 4
12
  
.94. A n m as s les s eq u ilater al tr ian g le EFG of side ‘a’ (As shown in figure) has three
particles of mass m situated at its vertices. The moment of inertia of the system about
N
the line EX perpendicular to EG in the plane of EFG is ma 2 where N is an integer..
20
The value of N is _________. [SEP 3 2020 MAINS]
solution(25)
Moment of inertia of the system about axis XE

I  I E  I F  IG
2
a 5 25 2
 I  m  0  m    ma 2  ma 2 
2
ma
2 4 20
 N  25
95.Two uniform circular discs are rotating independently in the same direction around
their common axis passing through their centres. The moment of inertia and angular
velocity of the first disc are 0.1 kg-m2 and 10 rad s–1 respectively while those for the
second one are 0.2 kg-m2 and 5 rad s–1 respectively. At some instant they get stuck
together and start rotating as a single system about their common axis with some
angular speed. The kinetic energy of the combined system is :
10 20 5 2
(1) J (2) J (3) J (4) J [SEP2 2020 MAINS]
3 3 3 3
solution:.(2) Initial angular momentum  I11  I 2 2
Let  be angular speed of the combined system
Final angular momentum  I1  I 2
According to conservation of angular momentum  I1  I 2    I11  I 2 2
I11  I 22 0.1 10  0.2  5 20

  
I2  I2 0.1  0.2 3
Final rotational kinetic energy

2
1 1 1  20 
K f  I12  I 22   0.1  0.2    
2 2 2  3 
20
 Kf  J
3
96.Three solid spheres each of mass m and diameter d are stuck together such that
the lines connecting the centres form an equilateral triangle of side of length d. The
I0
ratio I of moment of inertia I0 of the system about an axis passing the centroid and
A
about center of any of the spheres IA and perpendicular to the plane of the triangle is:
[9 JAN 2020 MAINS]
13 15 23 13
(1) (2) (3) (4)
23 13 13 15
solution:.[1)]Moment of inertia
2
2 d 
I1  m    m  AO 
2
5 2
d
And AO 
3
Moment of inertia about ‘O’

 2  d 2  d  
2
I 0  3I1  3  m    m   
 5  2   3  
13
 I0  Md 2
10
 2  d 2 2
 2  d 2
And I A  2  5 M  2   Md   5 M  2 
     
23
 IA  Md 2
10
13
Md 2
IO 10 13
  
I A 23 Md 2 23
10
97.One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is

released from rest when it makes an angle 30° from the horizontal (see figure). Its
1
angular speed when it hits the table is given as ns , where n is an integer. The
value of n is ________ [9 JAN 2020 MAINS]
solution:.(15) Here, length of bar, 1 = 1m
Angle,   300
1 2
PE  KE or mgh  2 I 
l 1  ml 2  2
  mg  sin 300   
2 2  3 
l 1 1  ml 2  2
 mg    
2 2 2  3 
   15 rad / s
98.A uniformly thick wheel with moment of inertia I and radius R is free to rotate about
its centre of mass (see fig). A massless string is wrapped over its rim and two blocks
of masses m1 and m2 (m1 > m2) are attached to the ends of the string. The system is
released from rest. The angular speed of the wheel when m1 descents by a distance h
is: [9 JAN 2020 MAINS]
1/2
 2  m1  m2  gh 
(1)  
  m1  m2  R  1 
2
1/2
 2  m1  m2  gh 
(2)  
  m1  m2  R  1 
2
1/2
  m1  m2  
(3)   gh
  m1  m2  R  1 
2
1/2
  m1  m2  
(4)   gh
  m1  m2  R  1 
2
solution:.(1) Using principal of conservation of energy
1 1 2
 m1  m2  gh   m1  m2  v 2  I
2 2
1 1
  m1  m2  gh   m1  m2  R 2  I 2
2 2
 v  R 
2 
  m1  m2  gh   m1  m2  R 2  I 
2 
2  m1  m2  gh

 m1  m2  R 2  I
99.As shown in the figure, a bob of mass m is tied by a massless string whose other
end portion is wound on a fly wheel (disc) of radius r and mass m. When released
from rest the bob starts falling vertically. When it has covered a distance of h, the
angular speed of the wheel will be:
[7JAN 2020 MAINS]
1 4 gh 3
(1) (2) r
r 3 2 gh
1 2 gh 3
(3) (4) r
r 3 4 gh
solution:.(1) When the bob covered a distance ‘h’
1 2 1 2
Using mgh  mv  I 
2 2
1 1 mr 2
m  r     2  v  r no slipping 
2

2 2 2
3
 mg  m2 r 2
4
4 gh 1 4 gh
 
3r 2 r 3
100.The radius of gyration of a uniform rod of length l, about an axis passing through
l
a point away from the centre of the rod, and perpendicular to it, is:
4
1 1 7 3
(1) l (2) l (3) l (4) l [ 7 JAN 2020 MAINS]
4 8 48 8

solution:.(3) Moment inertia of the rod passing through a point away from the centre of the rod
4
I  Ig  m 2
MI 2  I 2  7 MI 2
I  M    
12  16  48
2 7 MI 2
Using I  MK  (K = radius of gyration)
48
7
K I
48
101.Mass per unit area of a circular disc of radius a depends on the distance r from its
centre as  (r) = A + Br. The moment of inertia of the disc about the axis,
perpendicular to the plane and passing through its centre is:[7 JAN 2020 MAINS]
4  A aB  4  aA B
(1) 2a    (2) 2a   
4 5   4 5
4  A aB  4 A B
(3) a    (4) 2a   
4 5  4 5
solution:.(1) Given,
Mass per unit area of circular disc,   A  Br
Area of the ring  2rdr
Mass of the ring, dm  2rdr

Moment of inertia,
I   dmr 2   2rdr.r 2
a
 Aa 4 Ba 5 
 I  2   A  Br  r dr  2 
3
 
0  4 5 
 A Ba 
 I  2a 4  
 4 5 
102.A circular disc of radius b has a hole of radius a at its centre (see figure). If the
 0 
mass per unit area of the disc varies as   , then the radius of gyration of the disc
 r 
about its axis passing through the centre is : :[7 JAN 2020 MAINS]
a 2  b 2  ab ab
(1) (2)
2 2
a 2  b 2  ab ab
(3) (4)
3 3
  dm  r
2
solution:.(3) I 
a
b
  20 3 b
   0  2r dr  r 2  r
a  r  3 a
20 3 3

3

b a 
Mass of the disc,
b
0
m  2r dr  20  b  a 
a
r
Radius of gyration,
I
k
m
 20 / 3  b3  a3  a 2  b2  ab
 
20  b  a  3
I1
103.Two coaxial discs, having moments of inertia I1 and , are rotating with
2
1
respective angular velocities 1 and , about their common axis. They are brought
2
in contact with each other and thereafter they rotate with a common angular velocity.
If Ef and Ei are the final and initial total energies, then (Ef – Ei) is ::[10 APR 2019MAINS]
I112 I112 3 2 I112

(1)  (2) (3) I11 (4) 
12 6 8 24
solution:.(d) As no external torque is acting so angular momentum should be conserved

 I1  I 2    I11  I 22 [ c  common angular velocity of the system, when discs are in contact]
I11
I11 
c  4  5  2 
I   1
I1  1  4 3 
2
51
c 
6
1 1 1
E f  Ei   I1  I 2  c2  I112  I122
2 2 2
51
Put I1 / 2 and c  51 / 6
6
We get:
I112
E f  Ei  
24
104.A thin disc of mass M and radius R has mass per unit area   r   kr 2 where r is
the distance from its centre. Its moment of inertia about an axis going through its
centre of mass and perpendicular to its plane is :: [10 APR 2019 MAINS]
MR 2 2 MR 2 MR 2 MR 2
(1) (2) (3) (4)
3 3 6 2
solution:.(2) As from the question density     kr 2
R
R 4 kR 4
Mass of disc M   kr 2
2 rdr 
 2 k
4

2
0
2M
k ....(i)
R 4
 Moment of inertia about the axis of the disc
1   d1    dm  r 2  dAr 2
   2rdr  r
  kr 2 2
 2M 
R   4   R 6
6
kR  R  2
  2k r 5 dr    MR 2
0
3 3 3
[putting value of k from equ .....(i)]
105.A solid sphere of mass M and radius R is divided into two unequal parts. The first
7M
part has a mass of and is converted into a uniform disc of radius 2R. The second
8
part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the
new sphere about its axis. The ratio I1/I2 is given by : [10 APR 2019 MAINS]
(1) 185 (2) 140 (3) 285 (4) 65
solution:.2)
 7M  2 1 7  14
  2 R     4  MR  mR
2 2
I1  
 8  2  16  8
2 M  2
I2   r
5 8 
 R  MR 2
2
2 M
 I2      
5 8  4  80
4 3 14 3 
 3 r   8 3 R   
 
 r  R / 2 
I1 14  80
  140
I2 8
106.A stationary horizontal disc is free to rotate about its axis. When a torque is
applied on it, its kinetic energy as a function of  , where  is the angle by which it has
rotated, is given as k 2 . If its moment of inertia is I then the angular acceleration of the
disc is: [9 APR 2019 MAINS]
k k k 2k
(1)  (2)  (3)  (4) 
4I I 2I I
1 2
solution:.92. (4) I   kQ 2
2
 2k 
Or    I  Q
 
d 2 K  dQ   2k 
Or      
 I  dt   I 
 2k   2k  2k 
     
 I  I  I
107.Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at
rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration
of 20 rad/s2 must be applied about the axis for a duration of:
(1) 2.5 g (2) 2s (3) 5s (4) 3s [9APR 2019 MAINS]
solution:.(2)   t  20t
1 2
Given, I  1200
2
1
 1.5   20t   1200
2
Or
2
Or t  2 s
108.A thin smooth rod of length L and mass M is rotating freely with angular speed 0
about an axis perpendicular to the rod and passing through its center. Two beads of
mass m and negligible size are at the center of the rod initially. The beads are free to
slide along the rod. The angular speed of the system, when the beads reach the
opposite ends of the rod, will be: [9 APR 2019 MAINS]
M 0 M 0 M 0 M 0
(1) (2) (3) (4)
M m M  3m M  6m M  2m
solution:.(c) I i i  I f  f
 ML2   ML2  L 
2
  
Or  12  0  12  2 m     f
   2 
 M 0 
 f   
 M  6m 
109.A thin circular plate of mass M and radius R has its density varying as   r   0 r
with 0 as constant and r is the distance from its center. The moment of Inertia of the
circular plate about an axis perpendicular to the plate and passing through its edge is
I = a MR2. The value of the coefficient a is:
(1) 1 2 (2) 3 5 (3) 8 5 (4) 3 2 [8APR 2019 MAINS]
solution:.(3) Taking a circular ring of radius r and thickness dr as a mass element, so total
mass,
R
20 R 3
M   0 r  2rdr 
0
3
R
20 R5
I C   0 r  2rdr  r 2 
0
5
Using parallel axis theorem
5
 1 1  160 R
 I  I C  MR 2  20 R5    
3 5 15
8 2  8
  0 R 3  R 2  MR 2
5 3  5
110. The moment of inertia of a solid sphere, about an axis parallel to its diameter
and at a distance of x from it, is ‘I(x)’. Which one of the graphs represents the variation
of I(x) with x correctly? [12 JAN 2019 MAINS]
(1) (2)
(3) (4)
solution:.(4) According to parallel axes theorem
2
I mR 2  mx 2
5
Hence graph (4) correctly depicts I vs x.

111.An equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E
and F are the mid-points of its sides as shown and G is the centre of the triangle. The
moment of inertia of the triangle about an axis passing through G and perpendicular
to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the
moment of inertia of the remaining figure about the same axis is I. Then :
[11 JAN 2019 MAINS]
15 3 9 I
(1) I  I 0 (2) I  I 0 (3) I  I 0 (4) I  0
16 4 16 4
solution:.(1) Let mass of the larger triangle = M
Side of larger triangle = 
Moment of inertia of larger triangle = ma2
M
Mass of smaller triangle 
4

Length of smaller traiangle 
2
2
M a
Mometn of removed triangle   
4 2
2
M a
I removed  
2
  4 .  2
I original M a
I0
I removed 
16
I 0 15I 0
So, I  I 0  
16 16
112.a string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the
string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without
slipping on a horizontal surface (see figure), then the angular acceleration of the
cylinder will be (Neglect the mass and thickness of the string)
[11 JAN 2019 MAINS]
(1) 20 rad/s2 (2) 16 rad/s2
(3) 12 rad/s2 (4) 10 rad/s2
solution:.(2)
From newton’s second law
40 +f = m  R  .....(i)
Taking torque about 0 we get
40  R  f  R  I 
40  R  f  R  mR 2
40  f  mR .....(ii)
Solving equation (i) and (ii)
40
  16 rad / s 2
mR
113. A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of
the same mass M and radius R attached rigidly at its opposite ends (see figure). The
moment of inertia of the system about the axis OO’, passing through the centre of D1,
as shown in the figure, will be : [11 JAN 2019 MAINS]
4 2
(1) MR 2 (2) 3MR 2 (3) MR 2 (4) MR 2
5 3
MR 2
solution:. (2) Moment of inertia of disc D1 about OO '  I1 
2
1  MR 2  2 MR 2
 I 2     MR   MR 2
2  2  4
M.O.I of D3 about OO’
1  MR 2  2 MR 2
 I 3     MR   MR 2
2 2  4
So, resultant M.O.I about OO’ is I  I1  I 2  I 3
MR 2  MR 2 
I  2   MR 2 
2  4 
MR 2 MR 2
   2 MR 2  3MR 2
2 2
114.Two identical spherical balls of mass M and radius R each are stuck on two ends
of a rod of length 2R and mass M (see figure). The moment of inertia of the system
about the axis passing perpendicularly through the centre of the rod is:
137 17 209 152
(1) MR 2 (2) MR 2 (3) MR 2 (4) MR 2 [10 JAN 2019 MAINS]
15 15 15 15
solution:.(1) For Ball
Using parallel axes theorem, for ball moment of inertia,
2 22
MR 2  M  2 R  
2
I ball  MR 2
5 5
22
For two balls I ball  2  MR 2
5
M  2R 
2
MR 2
Irod  
12 3
I system  I balls  I rod
44 2 MR 2 137
 MR   MR 2
5 3 15
m
115.Two masses m and are connected at the two ends of a massless rigid rod of
2
length l. The rod is suspended by a thin wire of torsional constant k at the centre of
mass of the rod-mass system (see figure). Because of torsional constant k, the
restoring toruque is   k  for angular displacement  . If the rod is rotated by 0 and
released, the tension in it when it passes through its mean position will be:
[ 9JAN 2019 MAINS]
3k 02 2k 02 k 02 k 02

(1) (2) (3) (4)
l l l 2l
k
solution:.(3) As we know,  
I
3k  1 2
  I rod  3 m 
m 2
Tension when it passes through the mean position,
 3k 2  k 02
 m2 02  m 2 0 
3 m 3 
116.A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle
of 30° from the horizontal as shown and released from rest. Its angular speed when it
passes through the horizontal (in rads–1) will be (g = 10 ms–2)
[9 JAN 2019 MAINS]
30 20 30
(1) (2) 30 (3) (4)
7 3 2
solution:.(4)
By the low of conservation of energy,
P.E. of rod = Rotational K.E.

 1
mg sin   I 2
2 2
 1 m 2 2  1 1 m 2 2
 mg sin 300    mg   
2 2 3 2 2 2 3
For complete length of rod,
30
  3g / 2  2   rod s 1
2
117.Seven identical circular planar disks, each of mass M and radius R are welded
symmetrically as shown. The moment of inertia of the arrangement about the axis
normal to the plane and passing through the point P is: [2018 MAINS ]
19 55 73 181
(1) MR 2 (2) MR 2 (3) MR 2 (4) MR 2
2 2 2 2
solution:.4) Using parallel axes theorem, moment of inertia about ‘O’ I 0  I cm  md 2

7 MR 2
2

 6 M   2R  
2
2

55MR 2
Again, moment of inertia about point P, I p  I 0  md 2

55MR 2 181
 7 M  3R  
2
 MR 2
2 2
R
118.From a uniform circular disc of radius R and mass 9 M, a small disc of radius is
3
removed as shown in the figure.The moment of inertia of the remaining disc about an
axis perpendicular to the plane of the disc and passing,through centre of disc is :
[2018 MAINS ]
40 37
(1) 4 MR2 (2) MR 2 (3) 10MR 2 (4) MR 2
9 9
solution:.1) Let  be the mass per unit area.
The total mass of the disc   R 2  9M
Let us consider the above system as a complete disc of mass 9M and a negative mass
M super imposed on it.
1
Moment of inertia  I1  of the complete disc = 9MR 2 about an axis passing through O
2
2
1 R
and perpendicular to the plane of the disc   M   
2 3
 M .I .  I 2  of the cut out portion about an axis passing through O and perpendicular to
the plane of disc
1 R
2
 2R  
2
   M    M   
 2 3  3  
[Using perpendicular axis theorem]
 The total M.I. of the system about an axis passing through O and perpendicular to the
plane of the disc is I  I1  I 2
1 2
1 R
2
 2R  
2
 9MR    M     M    
2  2 3  3  
9MR 2 9MR 2  9  1 MR
2
    4 MR 2
2 18 2
119.A thin circular disk is in the xy plane as shown in the figure. The ratio of its
moment of inertia about z and z˘ axes will be
[2018 MAINS ]
(1) 1 : 2 (2) 1 : 4 (3) 1 : 3 (4) 1 : 5
solution:.(3) As we know, moment of inertia of a disc about an axis passing through C.G. and
perpendicular to its plane,
mR 2
Iz 
2
Moment of inertia of a disc about a tangential axis perpendicular to its own plane,
3
I z'  mR 2
2
mR 2 3mR 2
 Iz / I z'  /  1/ 3
2 2
120.A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held
horizontal. When the end M is released the speed of this end, when the rod makes an
angle a with the horizontal, will be proportional to: (see figure)
[2018 MAINS ]
(1) cos  (2) cos  (3) sin  (4) sin 
l
solution:.(1) When the rod makes an angle  Displacement of ventre of mass  cos 
2
l l
mg cos   I 2
2 2
l ml 2 2
mg cos    ( M.I. of thin uniform rod about an axis passing through its centre of
2 6
ml 2
mass and perpendicular to the rod I  )
12
3g cos 

l
Speed of end   l  3 g cos l
i.e., speed of end,
121.The moment of inertia of a uniform cylinder of length  and radius R about its
perpendicular bisector is I. What is the ratio  /R such that the moment of inertia is
minimum ?
3 3 3
(1) 1 (2) (3) (4) [ 2017 MAINS ]
2 2 2
solution:.(3) As we know, moment of inertia of a solid cylinder about an axis which is perpendicular
bisector
mR 2 ml 2
I 
4 12
m  2 l2 
I  R  
4 3
m  V l2  dl m  V 2l 
       2  0
4  l 3  dl 4  l 3
V 2l 2l 3
  V 
l 2 3 3
2l 3 l2 3 l 3
R 2l   2  or, 
3 R 2 R 2
122.Moment of inertia of an equilateral triangular lamina ABC, about the axis passing
th r o u g h its c en tre O an d p erp en d ic u lar to its p lan e is I o as shown in the figure. A
cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides.
Moment of inertia of the remaining part of lamina about the same axis is :
[ APR 8 2017 MAINS ]
7 15 3I o 31I o
(1) Io (2) Io (3) (4)
8 16 4 32
solution:.(2) According to theorem of perpendicular axes, moment of inertia of triangle (ABC)
I 0  km / 2 .....(i)
BC=1
Moment of inertia of a cavity DEF
2
m l 
I DEF  K  
4 2
k
 ml 2
16
From equation (i),
I0
I DEF 
16
Moment of inertia of remaining part
I 0 15I 0
I remain  I 0  
16 16
R
123.A circular hole of radius is made in a thin uniform disc having mass M and
4
radius R, as shown in figure. The moment of inertia of the remaining portion of the
disc about an axis passing through the point O and perpendicular to the plane of the
disc is :
[ APR92017 MAINS ]
219 MR 2 237 MR 2 19 MR 2 197 MR 2

(1) (2) (3) (4)
256 512 512 256
solution:.(2) Moment of Inertia of complete disc about ‘O’ point
MR 2
I total 
2
Radius of removed disc  R / 4

 Mass of removed disc  M / 16
 As M  R 2 
 
M.I. of removed disc about its own axis (O’)
2
1 M R MR 2
   
2 16  4  512
M.I. of removed disc about O
I removed disc  I cm  mx 2
2
MR 2 M  3R  19 MR 2
    
512 16  16  512
M.I. of remaining disc
MR 2 19 237
I remaining   MR 2  MR 2
2 512 512
124.From a solid sphere of mass M and radius R a cube of maximum possible volume
is cut. Moment of inertia of cube about an axis passing through its center and
perpendicular to one of its faces is : [ APR 8 2015 MAINS ]
4MR 2 4MR 2 MR 2 MR 2
(1) (2) (3) (4)
9 3 3 3 32 2 16 2
2
solution:.(1) Here a  R
3
4 3
R
M
Now, 3 3
M' a
4 3
R
3 3
 3
 .
 2  2
 R
 3 
2M
M '
3
Moment of inertia of the cube about the given axis,
M ' a2
I
6
2
2M  2 
 R
3  3  4MR 2

6 9 3
125.Consider a thin uniform square sheet made of a rigid material. If its side is ‘a’
mass m and moment of inertia I about one of its diagonals, then:[ APR 2015 MAINS ]
ma 2 ma 2 ma 2 ma 2 ma 2
(1) I  (2) I (3) I  (4) I 
12 12 12 24 12
solution:.d) For a thin uniform square sheet
ma 2
I1  I  2  I 3
12
126. A ring of mass M and radius R is rotating about its axis with angular velocity w.
Two identical bodies each of mass m are now gently attached at the two ends of
a diameter of the ring. Because of this, the kinetic energy loss will be:
[ APR 8 2013MAINS ]
m  M  2m  2 2 Mm 2 2 Mm 2 2  M  m  M 2 R 2
(1)  R (2) M  m  R (3)  M  2m   R (4)
M    M  2m 
1 2
solution:.(3) Kinetic energy (rotational) K R  I
2
1
Kinetic energy(translational) K T  Mv 2 (v  R)
2
M .I .(initial ) I ring  MR 2 ; initial  
M
M .I .( new) I '( system)  MR 2  2 MR 2  '( system) 
M  2m
Solving we get loss in K.E.
Mm
 2 R 2
( M  2m )
127. This question has Statement 1and Statement 2. Of the four choices given after the
Statements, choose the one that best describes the two Statements.
Statement 1: When moment of inertia I of a body rotating about an axis with angular
speed  increases, its angular momentum L is unchanged but the kinetic energy K increases
if there is no torque applied on it. [ 2012 MAINS ]
1 2
Statement 2: L = Iw, kinetic energy of rotation  I
2
(1) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation
of Statement 1.
(2) Statement 1 is false, Statement 2 is true.
(3) Statement 1 is true, Statement 2 is true, Statement 2 is correct explanation of the

Statement 1.
(4) Statement 1 is true, Statement 2 is false
solution:.(2) As I  so L increases with invrease in  .
depends on an angular velocity and moment of inertia of the body

SIMPLE HARMONIC MOTION (S.H.M.)
PERIODIC MOTION
(i) Any motion which repeats itself after regular interval of time is called periodic motion or
harmonic motion.
(ii) The constant interval of time after which the motion is repeated is called time period.
Examples : (i) Motion of planets around the sun.
(ii) Motion of the pendulum of wall clock.
OSCILLATORY MOTION
(i) The motion of body is said to be oscillatory or vibratory motion if it moves back and forth
(to and fro) about a fixed point after regular interval of time.
(ii) The fixed point about which the body oscillates is called mean position or equilibrium
position.
Examples : (i) Vibration of the wire of 'Sitar'.
(ii) Oscillation of the mass suspended from spring.
Note : Every oscillatory motion is periodic but every periodic motion is not oscillatory.
SIMPLE HARMONIC MOTION (S.H.M.)
Simple harmonic motion is the simplest form of vibratory or oscillatory motion.
(i) S.H.M. are of two types
(a) Linear S.H.M.
When a particle moves to and fro about a fixed point (called equilibrium position)
along a straight line then its motion is called linear simple harmonic motion.
Example : Motion of a mass connected to spring.
(b) Angular S.H.M.

When a system oscillates angularly with respect to a fixed axis then its motion is
called angular simple harmonic motion.
Example :- Motion of a bob of simple pendulum
(ii) Necessary Condition to execute S.H.M.
(a) Motion of particle should be oscillatory.
(b) Total mechanical energy of particle should be conserved (Kinetic energy + Potential
energy = constant)
(c) extreme position should be well defined.
(d) In linear S.H.M.
The restoring force (or acceleration) acting on the particle should always be
proportional to the displacement of the particle and directed towards the equilibrium
position
F  –y or a  –y
Negative sign shows that direction of force and acceleration is towards equilibrium
position and y is displacement of particle from equilibrium position.
(e) In angular S.H.M.
The restoring torque (or angular acceleration) acting on the particle should always
be proportional to the angular displacement of the particle and directed towards the
equilibrium position
  – or a –
(iii) Comparision between linear and angular S.H.M.
Linear S.H.M. Angular S.H.M.

F  –x   –
F= –kx  = – C
Where k is the restoring force constant Where C is the restoring torque
constant
k C
a x  
m I
d2 x k d2  C
 x0  0
dt 2 m dt 2 I
It is known as differential equation of It is known as differential equation of

linear S.H.M. angular S.H.M.
x = A sinwt  =  0 sin wt
 = – w2x  = – w2 
where w is the angular frequency
k C
2  2 
m I
k 2 C 2
   2 n    2 n
m T I T
where T is time period and n is frequency
m I
T  2 T  2
k C
1 k 1 C
n n
2 m 2 I
This concept is valid for all This concept is valid for

types of linear all types of angular
S.H.M. S.H.M.
SOME BASIC TERMS

Mean Position
The point at which the restoring force on the particle is zero and potential energy is
minimum, is known as its mean position.
Restoring Force
(a) The force acting on the particle which tends to bring the particle towards its mean
position, is known as restoring force.
(b) This force is always directed towards the mean position.
(c) Restoring force always acts in a direction opposite to that of displacement.
Displacement is measured from the mean position.
(d) It is given by F = –kx and has dimension MLT–2.
Amplitude
The maximum displacement of particle from mean position is define as amplitude.
Time period (T)
(a) The time after which the particle keeps on repeating its motion is known as time
period.
(b) The smallest time taken to complete one oscillation or vibration is also define as
time period.
(c) It is given by T  2  , T  1 where w is angular frequency and n is frequency..
 n
Oscillation or Vibration
When a particle goes on one side from mean position and returns back and then it goes
to other side and again returns back to mean position, then this process is known as one
oscillation.
one one
oscillation oscillation
displacement (x)
time (t)
T T
Frequency (n or f)
(a) The number of oscillations per second is define as frequency.
1 
(b) It is given by n  ,n
T 2
(c) SI UNIT : hertz (Hz)
1 hertz = 1 cycle per second (cycle is a number not a dimensional quantity).
–1
(d) Dimension : M0 L0 T .
Phase
Y

N t
y A
t=0
t

X' O x X
Y'
(a) Phase of a vibrating particle at any instant is the state of the vibrating particle
regarding its displacement and direction of vibration at that particular instant.
(b) Projection of particle's position on Y-axis.
y = Asin(wt +  ) or y = Acos(wt +  ')
The quantity (wt +  ) represents the phase angle at that instant.
(c) The phase angle at time t = 0 is known as initial phase or epoch.
(d) The difference of total phase angles of two particles executing S.H.M. with respect
to the mean position is known as phase difference.
(e) If the phase angles of two particles executing S.H.M. are (wt +  1) and (wt +  2)
respectively, then the phase difference between two particles is given by
  = (wt +  2 ) – (wt +  1) or  = 2 – 1
(f) Two vibrating particles are said to be in same phase if the phase difference between
them is an even multiple of  , i.e.,   = 2n  Same phase.
(g) Two vibrating particle are said to be in opposite phase if the phase difference between
them is an odd multiple of  i.e.,   = (2n + 1)  opposite phase.
Angular frequency (w)
(a) The rate of change of phase angle of a particle with respect to time is define as its
angular frequency.
–1
(b) SI UNIT : radian/second Dimension : M0 L0 T
k
(c) 
m
Instantaneous displacement
(a) The displacement of the particle from mean position in a particular direction at any
instant of time is known as instantaneous displacement.
(b) At time t the instantaneous displacement
x = A sin (wt +  ),
where  is initial phase and A is amplitude.
 Every peroidic motion can be resolved into a number of simple harmonic motions.
 Oscillatory motion can be treated as simple harmonic motion only in the limit of small
amplitudes because in this limit the restoring force (or torque) becomes linear.
 Harmonic oscillations is that oscillations which can be expressed in terms of single
harmonic function.(i.e. sine functions or cosine function)
 The motion of the molecules of a solid, the vibration of the air columns and the vibration
of string of music instruements are either simple harmonic or superposition of simple
harmonic motions.
DISPLACEMENT, VELOCITY AND ACCELERATION IN S.H.M.
DISPLACEMENT IN S.H.M.
(i) The displacement of a particle executing linear S.H.M. at any instant is defined as
the distance of the particle from the mean position at that instant.
(ii) It can be given by relation y = Asinwt or x = Acoswt.
The first relation is valid when the time is measured from the mean position and the
second relation is valid when the time is measured from the extreme position of the
particle executing S.H.M. along a straight line path.
Note : (i) The direction of displacement is always away from the mean position whether the
particle is moving from or coming towards the mean position.
(ii) In linear S.H.M. the length of S.H.M. path = 2A
(iii) In. S.H.M. total work done in one complete oscillation is zero but total covered length
is 4A
VELOCITY IN S.H.M.
(i) It is define as the time rate of change of the displacement of the particle at the given
instant.
dx d
(ii) Velocity in S.H.M. is given by v  (A sin t)  v = Awcoswt
dt dt
x2
v   A 1  sin t 2  v   A 1   x  A sin t 
A2
v   (A 2  x 2 )
v2
Squaring both the sides v2 = w2 (A2 – x2)   A 2  x2
2
v2 x2 x2 v2
1  2   2 2 1
A2 2
A A 2
A 
This is equation of ellipse. So curve between displacement
and velocity of particle executing S.H.M. is ellipse.
A
(iii) The graph between velocity and displacement is shown in figure.

If particle oscillates with unit angular frequency (w = 1)
then curve between V and x will be circular.
Note: (i) The direction of velocity of a particle in S.H.M. is either towards or away from
the position
(ii) At mean position (x = 0), velocity is maximum (=Aw) and at extreme position (x =
 A), the velocity of particle executing S.H.M. is zero
ACCELERATION IN S.H.M.
(i) It is define as the time rate of change of the velocity of
the particle at given instant.
dv d
(ii) Acceleration in S.H.M. is given by a   (A cos t)
dt dt
a = –w2 A sin wt  a = – w2x
(iii) The graph between acceleration and displacement
as shown in figure
a
2A
+A
x
–A
– A
2
Note :
(i) The acceleration of a particle executing S.H.M. is always directed towards the mean
position.
(ii) The acceleration of the particle executing S.H.M. is maximum at extreme position
(= w2A) and minimum at mean position
GRAPHICAL REPRESENTATION
Graphical study of displacement, velocity, acceleration and force in S.H.M.
S. No. Graph In form of t In from of x Maximum

value
1. Displacement x = Asinwt x=x x= A
displacement (x)
A T
2 T
t
2. Velocity v = Awcoswt v   A2  x2 v  A

velocity (v)
T A
2
T t
3. Acceleration a = –w2Asinwt a = –w2x a = + w2A
a
2 A
acceleration (a)
+A
T 2A x
–A
2
T t
– A
2
4. Force (F = ma) F = – mw2Asinwt F = – mw2x F = + mw2A
F=ma
m A
2
T
force (F)
2
T t
 In linear S.H.M., the length of S.H.M. path = 2A

 In S.H.M., the total work done and displacement in one complete oscillation is zero but
total traversed length is 4A.
 In S.H.M., the velocity and acceleration varies simple harmonically with the same
frequency as displacement.

 Velocity is always ahead of displacement by phase angle radian
2
 Acceleration is ahead of displacement by phase angle  radian i.e., opposite to

displacement.

 Acceleration leads the velocity by phase angle radian.
2
 The velocity of a particle in S.H.M. at position x1 and x2 are v1 and v2 respectively.

Determine value of time period and amplitude.
v   A 2  x2  v2 = w2 (A2 – x2)
At position x1 velocity v12 = w2 (A2 – x12 ) ... (i)
At position x2 velocity v22 = w2 (A2 – x22 ) ... (ii)
Subtracting (ii) from (i) v12  v22  2 (x22  x12 )
v12  v22

x 22  x12
Time period 2
T

x 22  x12
T  2
v12  v22
v12 A 2  x12
Dividing (i) by (ii) 
v22 A 2  x22
v12 A 2  v12 x22  v22 A2  v22 x12

So A2 (v12  v22 )  v12 x22  v22 x12
v12 x 22  v22 x12

A
v12  v22
PROBLEMS
 
1. A particle executes SHM represented bythe equation, y=0.02sin  3.14t   metre.
2
Find (i) amplitude (ii) time period (iii) frequency (iv) epoch (v) maximum
velocity and (vi) maximum acceleration.
 
SOLUTION : Compare the equation y=0.02sin 3.14t   with the general form of the
2
equation, y  A sin t  0 
i) Amplitude : A = 0.02m
ii) Time period (T) is given by
2 2
T or T=  2s
 3.14
1 1
iii) Frequency v   Hz  0.5 Hz
T 2
 3.14
iv) Epoch  0    1.57 rad
2 2
v) Maximum velocity vmax  A  0.023.14 = 0.68 ms–1
vi) Maximum acceleration
amax=  A 2  0.023.14 = 0.1972ms–2

2

2. The equation of a simple harmonic wave is given by :y = 3 sin 2
(50 t – x),where x and y
are in metres and t is in seconds. The ratio of maximum particle velocity to the wave velocity
is :-
2 3
(1) 3  (2)  (3) 2  (4) 
3 2
SOLUTION :..

y = 3sin (50 t – x)
2
y 
Particle velocity = = 3    50  cos (50t–x)
t 2  2
Maximum particle velocity = 75  m/s
 50
Wave velocity v = = = 50 m/s
k 1
75 3
Required ratio = = 
50 2
3. A particle executing simple harmonic motionof amplitude 5 cm has maximum speed of 31.4
cm/s. The frequency of oscillation is :
(1) 1 Hz (2) 3 Hz (3) 2 Hz (4) 4 Hz
SOLUTION :
Vmax = Aw
Vmax = A(2  n)
31.4 = 5 (2 × 3.14) n
3 1.4
n=  1Hz
3 1.4
4.. The phase difference between the instantaneous velocity and acceleration of a particle
executing simple harmonic motion is :-
(1) Zero (2) 0.5  (3)  (4) 0.707 
SOLUTION :
v

 0.5  
2 2
a x
5.. A particle executes simpe harmonic oscillation with an amplitude a. The period of
oscillation is T. The minimum time taken by the particle to travel half of the amplitude
from the equilibrium position is :-
(1) T/2 (2) T/4 (3) T/8 (4) T/12
SOLUTION :
. .DISPLACEMENT x = A sin wt from mean position
A 2 t A
 A sin x
2 T 2
 2 t T
 t= sec.
6 T 12
6. A point performs simple harmonic oscillation of period T and the equation of motion is
given by x = a sin(wt +  /6). After the elapse of what fraction of the time period the velocity
of the point will be equal to half of tis maximum velocity ?
(1) T/3 (2) T/12 (3) T/8 (4) T/6
SOLUTION :.
  
x = a sin  t  6 
 
 T
t  t 
6 12
7.. The period of oscillation of a mass M suspended from a spring of negligible mass is T.
If along with it another mass M is also suspended, the period of oscillation will now be
:-
T
(1) 2 T (2) T (3) (4) 2T
2
m
SOLUTION :.Time period T = 2 
K
2m  m
T ' = 2 = 2 2
   2T
K K
8. The instantaneous displacement of a simple pendulum oscillator is given by
 
x  A cos  t   . Its speed will be maximum at time
 4
   2
(a) 4
(b) 2
(c) 
(d) 
SOLUTION :
  dx   
x  A cos  t   and v    A  sin   t  
 4 dt  4 
 
For maximum speed, sin   t    1
 4
   
t  or t  
4 2 2 4

t
4
9. A particle in S.H.M. is described by the displacement function x (t )  a cos(t   ) . If the

initial (t  0 ) position of the particle is 1 cm and its initial velocity is  cm/s . The angular
frequency of the particle is  rad / s , then it’s amplitude is
(a) 1 cm (b) 2 cm (c) 2 cm (d) 2.5 cm
SOLUTION : ….(i)
x  a cos( t   )
dx
and v    a  sin(  t   ) ….(ii)
dt
Given at t  0 , x  1 cm and v  and  
1 1
Putting these values in equation (i) and (ii) we will get sin   and cos  
a a
2 2
 1 1
sin 2   cos 2        
 a a
a  2 cm
10. A particle executes a simple harmonic motion of time period T. Find the time taken by the
particle to go directly from its mean position to half the amplitude
(a) T / 2 (b) T / 4 (c) T/8 (d) T / 12
A sin 2
SOLUTION : y  A sin t 
T
t
A 2t
 A sin
2 T
T
t
12
.
11. A particle executing S.H.M. of amplitude 4 cm and T = 4 sec. The time taken by it to move
from positive extreme position to half the amplitude is
(a) 1 sec (b) 1/3 sec (c) 2/3 sec (d) 3/2 sec
SOLUTION :
Equation of motion y  a cos t
a 1 
 a cos t  cos t   t 
2 2 3

T
2t  4 2
  t 3   sec
T 3 2 32 3
12. Two simple harmonic motions are represented by the equations

 
y1  0 . 1 sin  100  t   and y 2  0 . 1 cos  t. The phase difference of the velocity of
 3
particle 1 with respect to the velocity of particle 2 is

   
(a) 3
(b) 6
(c) 6
(d) 3
SOLUTION :
dy 1  
v1   0 .1  100  cos  100 t  
dt  3
dy 2  
v2   0 .1 sin t  0 .1 cos  t  
dt  2
Phase difference of velocity of first particle with respect to the velocity of 2nd particle at t = 0 is
  
  1   2    .
3 2 6
 
13. A particle has simple harmonic motion. The equation of its motion is x  5 sin 4 t   ,
 6
where x is its displacement. If the displacement of the particle is 3 units, then it

velocity is
2 5
(a) 3
(b) 6
(c) 20 (d) 16
SOLUTION : From the given equation, a = 5 and w = 4
\ v   a 2  y 2  4 (5 )2  (3)2  16
14. The maximum velocity and the maximum acceleration of a body moving in a simple
harmonic oscillator are 2 m/s and 4 m /s 2 . Then angular velocity will be
(a) 3 rad/sec (b) 0.5 rad/sec

(c) 1 rad/sec (d) 2 rad/sec
SOLUTION :
Amax 4
v max  a and Amax  a 2 Ţ   v   2 rad / sec
2
max
15. A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean
position the velocity of the particle is 10 cm/s. The distance of the particle from the mean
position when its speed becomes 5 cm/s is
(a) 3 cm (b) 5 cm (c) 2( 3 ) cm (d) 2( 5 ) cm
v max 10
SOLUTION : v max  a Ţ  
a 4
Now, v   a2  y 2 Ţ v2   2(a2  y 2)
v2
y 2  a2 
2
v2 52
y  a2   42  cm
2 (10 / 4 ) 2  2 3
16. Two particles P and Q start from origin and execute Simple Harmonic Motion along X-
axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The
ratio of the velocities of P and Q when they meet is
(a)1 : 2 (b)2 : 1( c)2 : 3 (d)3 : 2

SOLUTION :SOL..The particles will meet at the mean position when P completes one
oscillation and Q completes half an oscillation
vP a P TQ 6 2
So    
vQ a Q TP 3 1
17. The displacement of a particle moving in S.H.M. at any instant is given by y  a sin t .
T
The acceleration after time t
4
is (where T is the time period)
(a) a (b) a (c) a 2 (d)  a 2
SOLUTION :.  a 2 when it is at one extreme point.
18. The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The
maximum value of the acceleration of the particle is
144
(a) 144  2m /sec 2 (b) 144 m /sec 2 (c) m /sec 2 (d) 288  2m /sec 2
2
SOLUTION : (a) Maximum acceleration  a 2  a  4  2 n 2
19. A small body of mass 0.10 kg is executing S.H.M. of amplitude 1.0 m and period 0.20
sec. The maximum force acting on it is
(a) 98.596 N (b) 985.96 N (c) 100.2 N (d) 76.23 N
SOLUTION : Maximum acceleration
a  4 2 1  4  (3 .14 ) 2
A max  a 
2

T 2
0 .2  0 .2
0 .1  4  (3 .14 ) 2
Fmax  m  A max   98 .596 N
0 .2  0 .2
20.. A particle of mass 10 grams is executing simple harmonic motion with an amplitude
of 0.5 m and periodic time of ( / 5 ) seconds. The maximum value of the force
acting on the particle is
(a) 25 N (b) 5 N (c) 2.5 N (d) 0.5 N

 4 2 
SOLUTION :Maximum force  m (a 2 )  ma  2 

 T 
 4 2 
 0 . 5  2   0 . 01  0 . 5 N

  / 25 
21. A body is executing simple harmonic motion with an angular frequency 2rad / s . The
velocity of the body at 20 mm displacement, when the amplitude of motion is 60
mm, is .
(a) 40 mm /s (b) 60 mm / s (c) 113 mm / s (d) 120 mm / s
SOLUTION :. v   (a 2  y 2 )  2 60 2  20 2  113 mm / s .
22. A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum
velocity is 100 cm/sec. Its velocity will be 50 cm/sec at a distance [CPMT 1976]
(a) 5 (b) 5 2 (c) 5 3 (d) 10 2
SOLUTION : It is given v max  100 cm / sec , a = 10 cm.

v max  a
100
  10 rad / sec
10
Hence v   a2  y 2 Ţ 50  10 (10 )2  y 2
y  5 3 cm
23. A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The
magnitude of the velocity in m sec 1 at the centre of oscillation is
(a) 20 (b) 100 (c) 40p (d) 100 
2 0 . 2  2
SOLUTION : At centre v max  a  a.   40 
T 0 . 01
24. A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The
magnitude of the velocity in m sec 1 at the centre of oscillation is
(a) 20 (b) 100 (c) 40p (d) 100 
2 0 . 2  2
SOLUTION : At centre v max  a  a.   40 
T 0 . 01
25. A body executes SHM, such that its velocity at the mean position is 1 ms-1 and
acceleration at extreme position is 1.57 ms-2. Calculate the amplitude and the time
period of oscillation.
amax A 2 1.57
SOLUTION :  
vmax A 1
2
or   1.57   1.57 rad.
T
2 23.14
 Time period T    4s.
1.57 1.57
1
but A  1 i.e., A(1.57) = 1 or A 
1.57
 Amplitude A = 0.637 m.
26. Figure given below shows the displacement versus time graph for two particles
A and B executing simple harmonic motions. Find the ratio of their maximum
velocities,
Y
10
A
y(cm) B
5
24 26
0
2 6 10 14 18 22 t(s)
r
-5
-10
SOLUTION :For A, time peroid TA = 16s [Distance between two adjacent crets]
for B, time period TB = (26–2) = 24s
[length between the crest and trough shown =20s –8s = 12s]
Also, amplitudes aA = 10cm; aB = 5cm
Ratio of maximum velocities
VA a A  A a A TB 10  24 3
  
VB a B B a B TA = 516 4
27. Acceleration displacement graph of a particle executing S.H.M is as shown in
given figure. Find the time period of its oscillation (in sec)
SOLUTION :
Acceleration= -  2 x , i.e.,  2 =tan 450 =1
2
or  1 or T  2s

28. .Two particles execute SHM of same amplitude and frequency on parallel lines.
They cross each another when moving in opposite directions each time their
displacement is half their amplitude. What is the phase difference between them
?
SOLUTION :
If we assume that the particles are initially at the mean position, their equation for
displacement.
A
x  A sin t But x 
2
A 1
  A sin t (or) sin t 
2 2
Phase = t  300 ,1500
 sin 180 0
  
   sin  ;sin 1800  300  sin 30 0 
One of the particles has phase of 300 and the other has phase of 1500
2
Phase difference between them = 1200 = radian
3
29. The displacement of SHO at which its velocity is half of maximum velocity is...
y2 vmax y2
SOLUTION : v  vmax 1    vmax 1 
A2 2 A2
1 y2 3
 1 2 y A
4 A 2
30. What is average speed and average velocity of SHO in one oscil-
lation
SOLUTION :
In one oscillation SHO travels a distance of 4A
4A 4A A 2
 average speed in one oscillation =   2.  vmax
T 2  

Average velocity = 0 (as displacement is 0)
31. Figure shows the graph of velocity versus displacement of a particle executing
simple harmonic motion. Find the period of oscillation of the particle
SOLUTION :
0.6 v
(ms -1 )
-10 O (cm ) x
-0.6
xmax = A=10 cm and vmax = A  0.6ms 1

vmax 6  10 1
   2
 6rad s1
x max 10  10
2 2 
T   sec
 6 3
32. A particle is performing simple harmonic motion along x-axis with amplitude 4 cm and
time period 1.2 sec. The minimum time taken by the particle to move from x =2 cm to x =
+ 4 cm and back again is given by
(a) 0.6 sec (b) 0.4 sec (c) 0.3 sec (d) 0.2 sec
SOLUTION :.
Time taken by particle to move from x=0 (mean position) to x = 4 (extreme
T 1.2
position)  4

4
 0 .3 s
Let t be the time taken by the particle to move from x=0 to x=2 cm
2 1 2
y  a sin t  2  4 sin t   sin t
T 2 1 .2
 2
  t  t  0 .1 s .
6 1 .2
Hence time to move from x = 2 to x = 4 will be equal to 0.3 - 0.1 = 0.2 s

Hence total time to move from x = 2 to x = 4 and back again  2  0 . 2  0 . 4 sec
33. A large horizontal surface moves up and down in SHM with an amplitude of 1 cm. If a
mass of 10 kg (which is placed on the surface) is to remain continually in contact with it,
the maximum frequency of S.H.M. will be
(a) 0.5 Hz (b) 1.5 Hz (c) 5 Hz (d) 10 Hz

SOLUTION : For body to remain in contact a max  g
  2 A  g  4 2 n 2 A  g
g 10
 n2    25  n  5 Hz
4 A
2
4 (3 .14 )2 0 .01
34. Due to some force F1 a body oscillates with period 4/5 sec and due to other force F2
oscillates with period 3/5 sec. If both forces act simultaneously, the new period will
be
(a) 0.72 sec (b) 0.64 sec (c) 0.48 sec (d) 0.36 sec
SOLUTION :
Under the influence of one force F1  m 12 y and under the action of another force,
F2  m  22 y .
Under the action of both the forces F  F1  F2
 m  2 y  m  12 y  m  2 y
2 2 2
 2   2   2 
  T    T  
 T


   1   2 
2 2
 4 3
   
 T
T12 T22
T12  T22
 5 5
2
 4  3
   
2
= 0.48S
 5 5
35. An object performs S.H.M. of amplitude 5 cm and time period 4 s. If timing is started
when the object is at the centre of the oscillation i.e., x = 0 then calculate.
(i) Frequency of oscillation
(ii) The displacement at 0.5 sec.
(iii) The maximum acceleration of the object.
(iv) The velocity at a displacement of 3 cm.
SOLUTION :
1 1
(i) Frequency f    0.25 Hz
T 4
(ii) The displacement equation of object x = Asinwt
 5
so at t = 0.5 s x = 5sin(2p × 0.25 × 0.5) = 5 sin 4  cm
2
(iii) Maximum acceleration amax = w2A = (0.5 p )2 × 5 = 12.3 cm/s2

(iv) Velocity at x = 3 cm is v   A2  x2  0.5  5 2  3 2  6.28 cm/s
36. A particle executes S.H.M. from extreme position and covers a distance equal to half of
its amplitude in 1 s. Determine the time period of motion.
SOLUTION :
For particle starts S.H.M. from extreme position y = Acoswt
A
 A cos( 1)
2

cos   cos
3


3
2 2 3
T  6s
 
37. A particle executing S.H.M. having amplitude 0.01 m and frequency 60 Hz. Determine
maximum acceleration of particle.
SOLUTION : Maximum acceleration amax. = w2A = 4p2n2A
= 4p2(60)2 × (0.01)
= 144 p2 m/s2
38. A particle is executing S.H.M. of frequency

300 Hz and with amplitude 0.1 cm. Its maximum velocity will be :
(1) 60  cm/s (2) 0.6  cm/s (3) 0.50  cm/s (4) 0.05  cm/s
SOLUTION :.Vmax = AW= 2 nA = 2 (300)0.1  60 cm/s
39. Which of the following equation does not represent a simple harmonic motion :
(1) y = asinwt (2) y = bcoswt
(3) y = asinwt + bcoswt (4) y = atanwt

SOLUTION :
.y = atanwt is not a periodic function
 t 
40. A particle moves according to the equation x  a cos   . The distance covered by
 2
it in the time interval between t  0 to t  3 s is
1) 2a 2) 3a 3) 4a 4) a
SOLUTION :

Here   , T  4sec ;
2
Amplitude = a
Between t  0 to t  3 s It Covers a distance 3a

ENERGY OF PARTICLE IN S.H.M.POTENTIAL ENERGY (U OR P.E.)
In terms of displacement
The potential energy is related to force by the relation
dU
F
dx
  dU    Fdx
for S.H.M. F = – kx
1
 dU    (kx)dx   kx dx  U=
2
kx2 + C
1
At x = 0, U = U0 r C = U0 So U= kx2 + U0
2
the potenital energy at equilibrium position = U0

1
When U0 = 0 then U= kx2
2
U U
Umax=
1 k A2
2
U0
–A O +A
-A O +A displacement(x)
 displacement 
In terms of time
Since x = Asin(wt +  )
Umax 1 k A2
Umax =
2
potential energy
0 T T 3T T time (t)
4 2 4
One cycle
1
P. E. U= kA2sin2(wt +  )
2
1
If initial phase (  ) is zero U = kA2sin2wt
2
1
 U=
2
mw2A2sin2wt
Note :
(i) In S.H.M. the potential energy is a parabolic function of displacement, the potential
energy is minimum at the mean position (x = 0) and maximum at extreme position (x
=  A)
(ii) The potential energy is the periodic function of time.
T 3T T 3T 5T
It is minimum at t = 0, , T, ... and maximum at t = , , ...
2 2 4 4 4
KINETIC ENERGY (K. E.)

In terms of displacement
If mass of the particle executing S.H.M. is m and
Its velocity is v then kinetic energy at any instant.
1 1
K.E. = mv2 = mw2 (A2 – x2)
2 2
1
K.E. = k(A2 – x2)
2
KE 1 m 2 A 2
KEmax=
2
–A +A
In terms of time
KEmax 1 m 2 A2
KEmax=
2
Kinetic Energy
0 T T 3T T time (t)
4 2 4
One cycle
v = Awcos(wt +  )
1
K. E. = mw2A2 cos2 (wt +  )
2
If initial phase  is zero
1
K. E. = mw2A2 cos2wt
2
Note :
(i) In S.H.M. the kinetic energy is a inverted parabolic function of displacement.
1
The kinetic energy is maximum ( kA2) at mean position (x = 0) and minimum
2
(zero) at extreme position (x =  A)
(ii) The kinetic energy is the periodic function of time. It is maximum at
T T 3T 5T
t = 0, , T, ..............and minimum at t = , , ...
2 4 4 4
TOTAL ENERGY (E)

Total energy in S.H.M. is given by ; E = P. E. + K. E.
(i) w.r.t. position
1 1 1
E=
2
kx2 +
2
k (A2 – x2)  E=
2
kA2
(ii) w.r.t. time

1 1
E= mw2A2 sin2 wt + mw2A2 cos2 wt
2 2
1 1
E=
2
mw2A2  E=
2
kA2
1 k A2 1 k A2
TE TE 2
2
displacement time
Note
(i) Total energy of a particle in S.H.M. is same at all instant and at all displacement.
(ii) Total energy depends upon mass, amplitude and frequency of vibration of the particle
executing S.H.M.
AVERAGE ENERGY IN S.H.M.

(i) The time average of P.E. and K.E. over one cycle is
1 1 1
(a) <K. E.>t = kA2 (b) <P. E.>t = kA2 + U0 (c) <T. E.>t = kA2 + U0
4 4 2
 Both K. E. and P. E. varies periodically but the variation is not simple harmonic.
 The frequency of oscillation of P. E. and K. E. is twice as that of displacement or velocity
or acceleration of a particle executing S.H.M.
 Frequency of total energy is zero
PROBLEMS
1. A linear harmonic oscillator of force constant 2  106 Nm 1 and amplitude 0.01 m
has a total mechanical energy of 160J. Then find maximum and
minimum values of P.E and K.E?
(1) 60 J ,60 J (2) 0 J ,60 J (3) 60J , 0 J (4) 0 J , 0 J
1 1
KA2   2  106   0.01  100 J
2
SOLUTION :
2 2
Since total energy is 160J. Maximum P.E. is 160 J.
From this it is understood that at the mean position potential energy of the
simple harmonic oscillator is minimum which need not be zero.
PEmin  TE  KEmax  160  100  60 J
KEmin  0
2. A particle of mass 1 kg is executing SHM with an amplitude of 1m and time
period  s. Calculate kinetic energy of the particle at the moment when the
displacement is 0.8m from mean position
(1) 0.72J (2) 72 J (3) 7.2 J (4) 0 J
SOLUTION :
2 2
   2s
T 
We have, v   A2  x 2
v  2 1   0.8  2  0.6  1.2m / s

2 2
1 2 1
mv   1 1.2   0.72 J
2
Kinetic energy =
2 2
3. A particle of mass 10 g executes a linear SHM of amplitude 5 cm with a period

1
of 2s. Find the P.E. and K.E. s after it has passed through the
6
mean position.
(1) 9.25 105 J , 3.085 106 J (2) 3.085 106 J , 9.25 105 J
(3) 9.25 106 J , 3.085 105 J (4) 8.5 105 J , 9 106 J
SOLUTION :
Mass of particle m 10g 102 kg
2 2
Time period T =2s,      rad / s
T 2
Amplitude A=5cm = 5  102 m
1 1
K.E. = mA2 2 cos 2 t when t  s
2 6
1 
1102   5 102   2  cos2
2
K.E.=
2 6
2
25  106  3
=   2    = 3.085 106 J
2  2 
1
PE  mA2 2 sin 2 t
2
1 
1 102   5 10 2   2 sin 2
2
=
2 6
2
25  10 6 1
=   2    = 9.25 105 J
2 2
4. The potential energy of a harmonic oscillator of mass 2 kg at its mean position is
5J. If its total energy is 9J and its amplitude is 0.01m, find its time period
  100
(1) s (2) s (3)  s (4) s
  
SOLUTION :
1 2
kA = (9 –5 ) =4J,
2
8
k  8  10 4 N/m
0.01
2
m  100
T  2  2  s
k  

5. A particle is describing SHM with amplitude 'a'. When the potential energy of particle
is one fourth of the maximum energy during oscillation, then its displacement from mean
position will be:
a a a 2a
(1) (2) (3) (4)
4 3 2 3
SOLUTION :
2 E0
SOL.. P.E(U )  E0 sin wt 
4
1
whereE0  KA2
2
1
sin wt 
2
a
x  a sin wt 
2
6.. Displacement between max. P.E. position and max. K.E. position for a particle excuting
simple harmonic motion is :
a
(1)  (2) + a (3)  a (4)
2
–1
SOLUTION :
–a O +a
(P.E.)max (K.E.)max (P.E.)max
a
the displacement between max P.E and max K.E is 
2
7. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis
25
with a frequency Hz. At position x = 0.04m, the object has kinetic energy 0.5J

and (potential energy is zero at mean position) .Find its amplitude of vibration.
(1) 6 cm (2) 9 cm (3) 3 cm (4) 0.6 cm
SOLUTION :
1
U m 2 x 2  0.4 J
2
k
  2 f   k= 2 f  m
2
Total energy of oscillation is (0.5+0.4) = 0.9 J
1 2
 0.9 = kA or A  1.8  1.8
2 2 f  m)
2
k
1 1.8 1 1.8 3
= 
2 f 0.2  25  0.2 = 50 m=6 cm
2  
  
8. The displacement of two identical particles executing SHM are represented by
 
equations x1 = 4 sin 10t   and x2 =5 cos  t For what value of  energy of both

the particles is same
(1) 6 unit (2) 9 unit (3) 3 unit (4) 8 unit
SOLUTION :
1
E= mA2  2
2
i.e., E   A 
2
 A11    A2  
2 2
A1  A2
410  5
   unit
9. A particle of mass ‘m’ is executing oscillation about the origin on the x -axis. Its
P.E.varies with position as U  x   K x , here K is constant. The amplitude
3
of oscillation is ‘a’, then how does its time period T vary with amplitude.
1
(4) a
2
(1) a (2) a (3) a
SOLUTION :
U  x  K x ,
3
1
but U  mA2 2 sin 2 t
2
ma 2 2  Ka 3
2  a
1
T 
a
Alternate method:
U
K   ML1T 2
x3
Now, time period may depend on
 mass   amplitude   K 
x y z
T
z
 M 0 L0T 1   M x Ly  ML1T 2 
on solving x= 1/2, y = z = -1/2
Hence T   amplitude 
1/2
1
T 
a
10. An object of mass 4 kg is moving along X-axis and its potential energy as a
functionof x varies as U  x   4 1  cos 2 x  J then time period for small
oscillation is :
1
(1)  seconds (2) seconds

(3) 2 seconds (4)  2 seconds
SOLUTION : -
dU d
F     4  4 cos 2 x 
dx dx
F  4  2   sin 2x   8sin 2x
here ‘x’ is small
sin 2 x  2 x
So F  16x  m2  16  42  16
  2
T  2     seconds
11. The displacement of SHO is , y  6sin  t   / 3 find 1) Instants at which
PE is min. (or) KE max (or) velocity is max.2) Instants at which PE is max
(or) KE is zero (or) velocity is zero
SOLUTION : PE is min (or) KE is max (or) velocity is max

when SHO is at mean position. i.e., y = 0
y  0  6sin  t   / 3
 1
   t    n here n = 1,2,3 ........
 3
1
t n
3
PE is max (or) KE is min (or) velocity is zero
12.. In case of simple harmonic motion –

(a) What fraction of total energy is kinetic and what fraction is potential when
displacement is one half of the amplitude.
(b) At what displacement the kinetic and potential energies are equal.
SOLUTION :
1 1 1
In S.H.M. K. E. = k(A2 - x2) P. E. = kx2 T.E. = kA2
2 2 2
K.E. A 2  x 2 P.E. x 2
(a) fK.E. =  fP. E. = 
T.E. A2 T.E A2
A A 2  A2 4 3 A2 4 1
at x = fK.E. =  and fP. E. = 
2 A2 4 A2 4
1 1 A
(b) K. E. = P. E. r k (A2 -x2) = kx2 r 2x2 = A2 r x= 
2 2 2
13. The total energy of a particle executing S.H.M. is proportional to

(a) Displacement from equilibrium position
(b) Frequency of oscillation
(c) Velocity in equilibrium position
(d) Square of amplitude of motion
SOLUTION :
1
E m 2a2 Ţ E  a2
2
14. The angular velocity and the amplitude of a simple pendulum is  and a
respectively. At a displacement X from the mean position if its kinetic energy is T
and potential energy is V, then the ratio of T to V is
(a) X  /(a  X  )
2 2 2 2 2 (b) X 2 /(a 2  X 2 )
(c) (a 2  X 2 2 ) / X 2 2 (d) (a 2  X 2 ) / X 2
SOLUTION : .
1 1 T a2  x 2
E m 2a2 Ţ E  a 2 and potential energy,, V m2x 2  
2 2 V x2
15.. When the potential energy of a particle executing simple harmonic motion is one-fourth
of its maximum value during the oscillation, the displacement of the particle from the
equilibrium position in terms of its amplitude a is
(a) a/4 (b) a/3 (c) a/2 (d) 2a / 3
SOLUTION :
1
m  2y 2
U
 2
Umax 1
m  2a 2
2
1 y2

4 a2
a
y
2
16. A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec
and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position
is
(a) 37 .5 2 ergs (b) 3 .75 2 ergs (c) 375  2 ergs (d) 0 .375  2 ergs
SOLUTION :
1
Kinetic energy K m  2 (a 2  y 2 )
2
2
1  2  2 2
  10    [10  5 ]  375  2 ergs
2  2 
17. When the displacement is half the amplitude, the ratio of potential energy to the total
energy is
1 1 1
(a) 2
(b) 4
(c) 1 (d) 8
SOLUTION :
2
1 a
m  2y 2 2  
U y 2 1
 2  2   
E 1 a 4
m  2a2 a
2
18. For any S.H.M., amplitude is 6 cm. If instantaneous potential energy is half the total energy
then distance of particle from its mean position is
(a) 3 cm (b) 4.2 cm (c) 5.8 cm (d) 6 cm

1
SOLUTION : If at any instant displacement is y then it is given that U
2
E Ţ
1 1 1 
m  2 y 2    m  2a2 
2 2 2 
a 6
y   4 .2 cm
2 2
19. A body of mass 1 kg is executing simple harmonic motion. Its displacement y(cm ) at t
seconds is given by y  6 sin(100 t   /4 ) . Its maximum kinetic energy is
(a) 6 J (b) 18 J (c) 24 J (d) 36
SOLUTION :
So a  6 cm ,   100 rad / sec
1 1
K max  m  2 a 2   1  (100 )2  (6  10  2 )2  18 J
2 2
20. A particle is executing simple harmonic motion with frequency f. The frequency at which
its kinetic energy change into potential energy is
(a) f/2 (b) f (c) 2f (d) 4f
SOLUTION :
n S.H.M., frequency of K.E. and P.E. = 2 ´ (Frequency of oscillating particle)
21. There is a body having mass m and performing S.H.M. with amplitude a. There is a
restoring force F   Kx , where x is the displacement. The total energy of body
depends upon
(a) K, x (b) K, a (c) K, a, x (d) K, a, v
SOLUTION : .
1
Total energy U Ka 2
2
22. The total energy of a particle executing S.H.M. is 80 J. What is the potential energy when
the particle is at a distance of 3/4 of amplitude from the mean position
(a) 60 J (b) 10 J (c) 40 J (d) 45 J
SOLUTION :
1
m  2y 2
U y2
 2  2
E 1
m  2a2 a
2
2
3 
 a
U 4  9
 
80 a2 16
U  45 J
23. When a mass M is attached to the spring of force constant k, then the spring stretches by
l. If the mass oscillates with amplitude l, what will be maximum potential energy stored in
kl 1
the spring (a) 2
(b) 2kl (c) 2
Mgl
(d) Mgl
SOLUTION :
Mg  Kl
1 2 1
U max  Kl  mgl
2 2
24. A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what
displacement from the equilibrium position, is its energy half potential and half kinetic
(a) 1 cm (b) 2 cm (c) 3 cm (d) 2 2 cm
SOLUTION :
Let x be the point where K.E. = P.E.
1 1
Hence 2
m  2 (a 2  x 2 )  m  2 x 2
2
2x 2  a2
a 4
x    2 2 cm
2 2
25. For a particle executing simple harmonic motion, the kinetic energy K is given by
K  K o cos 2 t . The maximum value of potential energy is

K0
(a) K0 (b) Zero (c) 2
(d) Not obtainable
SOLUTION :
Since maximum value of cos 2 t is 1.
 K max  K o cos 2 t  K o
Also K max  PE max  K o
26. The potential energy of a particle with displacement X is U(X). The motion is simple
harmonic, when (K is a positive constant)
KX 2
(a) U (b) U  KX 2 (c) UK (d) U  KX
2
SOLUTION :
F   kx
dW  Fdx   kxdx
W x
So  0
dW   0
 kx dx
1 2
W U  kx
2
27. The kinetic energy and potential energy of a particle executing simple harmonic motion
will be equal, when displacement (amplitude = a) is
a a a 2
(a) 2
(b) a 2 (c) 2
(d)
3
SOLUTION :
Suppose at displacement y from mean position potential energy = kinetic energy
1 1
m (a 2  y 2 ) 2  m  2 y 2
2 2
Ţ a2  2y 2
a
y
2
28. The total energy of the body executing S.H.M. is E. Then the kinetic energy when the
displacement is half of the amplitude, is
E E 3E 3
(a) 2
(b) 4
(c) 4
(d) E
4
SOLUTION :
1
Total energy in SHM E
2
m  2a2 ; (where a = amplitude)
1 1 a
Potential energy U m  2 (a 2  y 2 )  E  m  2 y 2 When y
2 2 2
1  a2 
U  E m  2    E  E  3E
2 
 4  4 4
29. The potential energy of a particle executing S.H.M. is 2.5 J, when its displacement is
half of amplitude. The total energy of the particle be
(a) 18 J (b) 10 J (c) 12 J (d) 2.5 J
SOLUTION : .
2
a
 
2 .5  2 

E a2
E  10 J
30. A particle starts oscillating simple harmonically from its equilibrium position with time
T
period T. Determine ratio of K.E. and P.E. of the particle at time t = .
12
(a) 1:3 (b) 3:1 (c) 9:1 (d) 16:9

SOLUTION :
T
at t =
12
2 T  A
x = A sin   A sin 
T 12 6 2
1 3 1
so K.E. = k (A2 – x2) =  kA 2 and
2 4 2
1 1 1
P.E. = kx2 =  kA 2
2 4 2
K.E. 3
\ 
P.E. 1
31.. The potential energy of a particle executing S.H.M. is 2.5 J, when its displacement is half
of the amplitude, then determine total energy of particle.
(a) 8 J (b) 16 J (c) 12 J (d) 10 J
SOLUTION : .
1
P.E. = kx2
2
1 A2
k = 2.5
2 4
1
total energy kA2 = 2.5 × 4 = 10 J
2
32. A harmonic oscillator of force consant 4 × 106 Nm and amplitude 0.01 m has total energy
240 J. What is maximum kinetic energy and minimum potential energy ?
(a) 18 J (b) 26 J (c) 40 J (d) 20 J
SOLUTION :
k = 4 × 106 N/m,
a = 0.01 m, T.E. = 240 J,
k
As 2 
m
1 1 2 1
Maximum kinetic energy = m 2 a 2 = ka   4  10 6  (0.0 1)2 = 200 J
2 2 2
Minimum potential energy = T.E. – maximum kinetic energy = 40 J
33. When a spring is stretch by 4 cm then its potential energy is U. If the spring is stretched
b y
20 cm then determine its potential energy.
(a) 18 U (b) 25U (c) 40U (d) 2U
SOLUTION :
1
U= kx2
2
U ' x '2 (20)2
 2  = 25
U x (4 )2
U' = 25U
SPRING SYSTEM AND SPRING PENDULUM

SPRING SYSTEM
halical spring spiral spring
(i) When spring is given small displacement by stretching or compressing it, then restoring
elastic force is daveloped in it because it obeys Hook's law.
F  –x  F = – kx Here k is spring constant
(ii) Spring is assumed massless, so restoring elastic force in spring is assumed same
everywhere.
(iii) Spring constant (k) depends on length, radius and material of wire used in spring.
spring constant (k)
k 1

hyperbola
length of spring ()
(iv) When spring is stretched or compressed then work done on it is stored as elastic potential
energy.
F F F F
compressing a spring spring without deformation stretching a spring
1
W =  Fdx   kx dx and U=W= kx2
2
When spring is stretched from l1 to l2 then

1
Work done W= k(X22 –X12) U
2
parabola
O x
(v) If there are two springs of force constant k1 and k2 with k1 > k2 then work done :
(a) When they are stretched by same amount (x1 = x2)
1
k1 x12
W1 2 k
  1 1  W 1 > W2
W2 1 k
k 2 x 22 2
2
(b) When they are stretched by same force

F2 W1 k 2

W= as (F1 = F2) W2 k1 <1  W 1 < W2
2K
SPRING PENDULUM
(i) When a small mass is suspended from a mass less spring then this arrangement
is known as spring pendulum.

For small linear displacement the motion of spring pendulum is simple harmonic.
(ii) For a spring pendulum
d2 x d2 x
F = – kx  m
dt 2
= – kx [ F = ma = m
dt 2
]
k
d2 x k d2 x k
or =– x or = – w2x with w2 =
dt 2 m dt 2 m
This is standard equation of linear S.H.M.
2 m 1 k
Time period T=  T = 2 Frequency n=
 k 2 m
(iii) Time period of a spring pendulum is independent of acceleration due to gravity. This is
why a clock based on oscillation of spring pendulum will keep proper time everywhere on
a hill or moon or in a satellite or different places of earth.
(iv) By increasing the mass, time period of spring pendulum increases (T  m ) , but by
 1 
increasing the force constant of spring (k). Its time period decreases  T   whereas
 k
frequency increases (n  k )
(v) If the spring has mass M and mass m is suspended from it then effective mass is given
by
M m eff mM 3
meff. = m + and T = 2  = 2
3 k k
M
If spring oscillates by its own weight then T = 2 
3k
(vi) If two masses m1 and m2 are connected by a spring and made to oscillate then time
period

T = 2
k k
m1 m2
m1 m 2
Here,  = reduced mass
m1  m 2
(vii) If a spring pendulum oscillates in a vertical plane is made to oscillate on a
horizontal surface or on an inclined plane then time period will remain unchanged.
m1
m
(viii) When body of mass m attached to spring then its elongation is y0 ky0 = mg i.e.,
m y0
 then it is further streched by y
k g
If the stretch in a vertically loaded spring is y then for equilibrium of mass m.
Net force = mg  K  y  y0  = ma
K
a=  y
m
m
So, time period T = 2 
k
k
ky0
y0 m
m mg
y
T = 2 g
0
but remember time period of spring pendulum is independent of acceleration due to

gravity.
(ix) If two particles are attached with spring in which only one is oscillating
m1
mass of oscillating particle
Time period = 2  force constant
k
m1
T= 2
k
m2
VARIOUS SPRING ARRANGEMENTS
Series combination of springs
Total displacement x = x1 + x2
m
k1 k2
Force acting on both springs
F = –k1x1 = – k2x2
F F F F
x1 = – k and x2   k x = – k  k  ....(i)
1 2  1 2 
If equivalent force constant is ks then F = – ksx
F F F 1 1 1
so by equation (i)      
ks k1 k 2 k s k1 k 2
k1
k1 k 2
 ks 
k1  k 2
k2
m
m m m(k1  k 2 )
Time period T= 2 = 2  2
ks ks k1 k 2
1 ks ks
Frequency n= , Angular frequency  
2 m m
 In series combination same force exerts in all springs but extension will be different.
 In series combination extension of spring will be reciprocal of its spring constant.
 Spring constant of spring is reciprocal of its length
1
k  k1l1 = k2l2 = k3l3

Parallel Combination of springs
k1
m
k1 k2
k1 k2 m
m k2
In this arrangement displacement on each spring is same but restoring force is different.
Force acting on the system F = F1 + F2  F = – k1x – k2x ...(i)
If equivalent force constant is kP then, F = – kPx
so by equation (i) – kPx = – k1x – k2x  kP = k1 + k2
m m
Time period T = 2 k = 2 k  k
P 1 2
1 kP k1  k 2
Frequency n= Angular frequency  
2 m m
 In parallel combination, different forces exerts in all springs but extension will be same.
 In parallel combination, forces on spring will be proportional of its spring constant.
1
 If the length of the spring is made n times then effective force constant becomes
n
times and the time period becomes n times.
 If a spring of spring constant k is divided into n equal parts, the spring constant of each
1
part becomes nk and time period becomes times.
n
 In case of a loaded spring the time period comes out to be the same in both horizontal
and vertical arrangement of spring system
 The force constant k of a stiffer spring is higher than that of a soft spring. So the time
period of a stiffer spring is less than that of a soft spring.
 n
A spring of force constant k is cut into two unequal parts l1 and l2. Where   n then
1 1

2 2
determine force constant of each part.

Initial length of the spring = l and cut into two pieces of length l1 and l2.
1 1 C
since force constant  length of spring so k  k (C = constant)
 
C C C C , k 1, k1 2, k2
k1  , k 2   and k      
1 2 1 2
k1  1   2    k2 1   2   
k

1  k1  k 1  2 
1  and k

2  k 2  k 1  1 
2 
 
1 n 1  n   n  n2 

2 n 2 So, k1  k 1  2 
n1   k1  k  1 
  n1 
 n   n  n2 
and k 2  k 1  n   k2  k  1
1

 2   n2 
PROBLEMS
-1
1. A spring of force constant 1200 Nm is mounted on a horizontal table as shown in
Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways
to a distance 2.0 cm and released. Determine
(a) the frequency of oscillation of the mass.
(b) the maximum acceleration of the mass.
(c) the maximum speed of the mass.
(a) 3.2 Hz ,8 m s-2 , 0.4 ms-1 (b) 2 Hz ,8.5 m s-2 , 4 ms-1
(c) 3 Hz 1,8 m s-2 , 0.4 ms-1 (d) 32 Hz ,80 m s-2 , 4 ms-1
SOLUTION :
Here, k  1200 Nm 1 ; m  3.0 kg,
A  2.0 cm  0.02 m
(a) Frequency,
1 K 1 1200
f    3.2 Hz
2 m 6.28 3
k
(b) Acceleration a   2 y  y
m
Acceleration will be maximum when y is maximum i.e. y = A  Max.acceleration,
kA 1200  0.02
amax    8ms 2
m 3
(c) Max. speed of the mass will be when it is passing through the mean position, given by
k 1200
Vmax  A  A  0.02   0.4ms 1
m 3
2. A body of mass m attached to a spring which is oscillating with time period 4 seconds. If
the mass of the body is increased by 4 kg, its timer period increases by 2 sec. Determine
value of initial mass m.
1) 6.4Kg 2) 3.2Kg 3) 1.6Kg 4) 32Kg
SOLUTION :
m
T = 2
k
m
1st case 4 = 2 ...(i) and
k
m4
2nd case 6 = 2 ...(ii)
k
4 m
divide (i) by (ii) 
6 m 4
16 m

36 m  4
m = 3.2 kg
3. One body is suspended from a spring of length l, spring constant k and has time period
T. Now if spring is divided in two equal parts which are joined in parallel and the same
body is suspended from this arrangement then determine new time period.
T T T T
1) 2) 3) 4)
2 2 4 6
SOLUTION :
Spring constant in parallel combination
k' = 2k + 2k = 4k
m m
T' = 2  = 2
k' 4k
m 1 T T
and T' = 2   = 
2
k 4 4
4. A block is on a horizontal slab which is moving horizontally and executing S.H.M. The
coefficient of static friction between block and slab is m. If block is not separated from
slab then determine angular frequency of oscillation.
g   
1) 2) gA
3) g
4)
A A
SOLUTION :
If block is not separated from slab then restoring force due to S.H.M. should be less than
frictional force between slab and block.
F restoring = Ffriction
m amax. = mmg
amax. = mg
w2A =mg
g
w=
A
5. A light vertical spring is stretched by 0.2 cm when a weight of 10 g is attached to

its free end. The weight is further down by 1.0 cm and released. Compute the frequency
and maximum velocity of load.
1 1
1) ms 1 2) ms 1 3) 7000ms 1 4) 2000ms 1
700 7000
SOLUTION :
i) Force constant of the spring.
Restoring Force mg
k= =
Increase in length Increase in length
102 x9.8
  49 Nm 1.
2.0x103
1 k 1 49 70
Frequency f    Hz
2 m 2 10 2
2
ii) amplitude of motion = distance through which the weight is further pulled down = 1.0
cm
i.e., A = 1.0 cm.  Maximum velocity
1 1
 A  10 2 m x rads 1  ms 1
70 7000
6. A mass m = 8kg is attached to a spring passing over a pulley whose other end is
fixed to ground and held in position so that the spring remains unstretched.
The spring constant is 200 N/ m. The mass m is then released and begins to
undergo small oscillations. Find the maximum velocity of the mass (g
=10 m/s2)
1) 0.5 m 2) 0.6 m 3) 0.4 m 4) 4
m
SOLUTION :
Mean position will be at kx =mg
mg 810 2
or x     0.4m
k 200 5
This is also the amplitude of oscillation A= 0.4m

7. The spring constant of two springs are K1 and K2 respectively springs are stretch up to
that limit when potential energyof both becomes equal. The ratio of applied force (F1
and F2) on them will be :
(1) K1 : K2 (2) K2 : K1 (3) K1 : K 2 (4) K 2 : K1
SOLUTION :.
U1 = U2
1 1
K1 x12  K2 x 22
2 2
x1 K2

x2 K1
F1 K1 K2
 
F2 K2 K1
8. Force constant of a spring is K. one fourth part is detach then force constant of remaining
spring will be :
3 4
(1) K (2) K (3) K (4) 4K
4 3
SOLUTION :
1 3l 4k
k  aslengthbecomes springcons tan tbecomes
l 4 3
k 200
vmax  A  A  0.4  2 m/s
m 8
9. The spring constant of a spring is K. When it is divided into n equal parts, then what
is the spring constant of one piece :
nK (n  1)K
(1) nK (2) K/n (3) (n  1) (4)
n
SOLUTION :
1
K


' 
n
 K '  nK
10. The time period of a mass suspended from a spring is T. If the spring is cut into four
equal parts and the same mass is suspended from one of the parts, then the new time
period will be
T T
(1) (2) T (3) (4) 2T
4 2
SOLUTION :.
m
we know timeperiod T  2 
K
when the spring made in to 4 equal parts springconstantof each part becomes 4k
m 2 m T
T '  2  
4K 2 K 2
11. A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides
with a nearly weightless spring of force constant k = 50 N/m. The maximum compression
of the spring would be :
(1) 0.12 m (2) 1.5 m (3) 0.5 m (4) 0.15 m

SOLUTION :.
Kinetic energy of mass particle will convert into potential energy of spring.
1 1
mv2  kx2
2 2
mv2 m 0.5 1 .5
x v  1.5  = 0.15 m
k k 50 10 0
12. Frequency of a particle executing SHM is 10 Hz. The particle is suspended

from a vertical spring. At the highest point of its oscillation the spring is
unstretched. Find the maximum speed of the particle : (g = 10 m/s2)
1
1)
2
2) 20 3) 2 4) 
SOLUTION :
mg
Mean position of the particle is
k
distance below unstretched position of spring. therefore, amplitude of oscillation is
mg
A=
k
k
  2 f  20
m
m 1
 
k 400
Therefore, the maximum speed of particle will be
 g  1
vmax  A   
  20  m/s
 400 

2
13. A15gm ball is shot from a spring gun whose spring has a force constant 600 N/
m .The spring is compressed by 5cm. The greatest possible horizontal range of the
ball for this compression is (g = 10 m/sec2)
1) 20 m 2) 10 m 3) 1m 4)100 m
SOLUTION :
U2
Rmax = –– (1)
g
But K.E acquired by ball = P.E of spring gun
1 1
 m U2 = kx2
2 2
kx 2
 U2 = –– (2)
m
From equations (1) and (2)
Rmax =
kx 2
=

600  5  10 2
2

= 10m
mg 15x10 3  10
SIMPLE PENDULUM
If a heavy point mass is suspended by a weightless, inextensible and perfectly flexible

string from a rigid su pport, then this arrangement is called a simple pendulum
Expression for time period
For small angular displacement, sin    , so that
F = –mg sin 

= –mg  Effective  T
length
y
mg

si n
co
CM mg
s
mg
 
mg
= –  y
 
= –ky
(because y =   ), Thus, the time period of the simple pendulum is
m
T =2
k

or T = 2 
g
Time period is independent of mass of the pendulum.

 If angular amplitude (  0) is large (  0 > 15°) then time period is given by
 20 
T  2 1   here 0 is in radian.
g  16 
 If a simple pendulum of density ds is made to oscillate in a liquid of density dL then

its time period will increase as compare to that of air and is given by

T  2
 dL 
1   g
 dS 
 If the bob of simple pendulum has positive charge q and pendulum
is placed in uniform electric field which is in downward
direction then time period decreases

T  2
qE
g E m
+q
m
mg + qE
 If the bob of simple pendulum has positive charge q and is

made is oscillate in uniform electric field acting
in upward direction then time period increases

T  2 E m
+q
qE
g mg – qE
m
 If simple pendulum is taken above or below the surface of earth then value of
gravitational acceleration decreases and time period increases.
1
T
g
m
h
m
d
R
–d
R 0
earh
M R
mass M
2h
At height h gh = gs (1 – )
R
d
At depth d gd = gs (1 – )
R
SECONDS PENDULUM
A simple pendulum whose time period of oscillation is equal to two seconds, is called
“seconds pendulum”.
i.e., T = 2s
 g
or 2  2 or   2
g 
This is the length of a seconds pendulum and it changes from place to place.
If “g” is taken as 9.8ms - 2, then the length of the second’s pendulum is equal to
9.8
  0.994m   2  9.86
3.14
2
   (or) 1m
GRAPHS RELATED TO SIMPLE PENDULUM:
 2 
we have T  2 or T  4 g
2
g
l
and 2 = constant at a given place.
T
If a graph is drawn taking length “l” on
2
X - axis and time period “T ” on Y - axis it will be a straight line passing through origin as
shown.
T2

 T2
Slope of the graph “m” = tan   .


from T  2
g
T 2 4 2
slope =  . From this the value of ‘g’ at a given place can be found out.
 g
If a graph is drawn taking length “l” on X - axis and time period T on Y - axis it will be a
parabola as shown in figure.

When l – T & l– T2 graphs are plotted on the same graph paper they intersect at a point as
shown in the diagram. At the point of intersection,
T=T 2
 T = 1s.
The length of the pendulum is given by
9.8 1
2
gT 2
 2  (or)   25cm .
4 4 3.14
2
 T 2
T=1s
T
 T
(or)
T2
  25cm

Therefore for T = 1 s and   25cm,   T and   T 2 graphs intersect.
PERIODIC TIME OF SIMPLE PENDULUM IN REFERENCE SYSTEM

T  2
g eff
where, geff. = effective gravity acceleration in reference system

or total downward acceleration.
(a) If reference system is lift
(i) If velocity of lift v = constant

acceleration a = 0 and geff. = g T  2
g
(ii) If lift is moving upwards with acceleration a

geff. = g + a

T  2 T decreases
g a
(iii) If lift is moving downwards with acceleration a

geff. = g – a

T  2 T increases
g a
(iv) If lift falls downwards freely

geff. = g – g = 0
T = 0  simple pendulum will not oscillate
(b) A simple pendulum is mounted on a moving truck
(i) If truck is moving with constant velocity, no pseudo force acts on the pendulum and
time period remains same

T  2
g
(ii) If truck accelerates forward with acceleration a then a pseudo force acts in opposite
direction.
So effective acceleration, geff. = g 2  a 2 and

T' = 2 
g eff.

Time period T '  2 T' decreases
g  a2
2
vi) Consider a simple pendulum carrying a bob of mass ‘m’ and of length “  ” which is very
large and comparable to the radius “R” of earth. Let it be oscillating with small angular
amplitude.
The time period of oscillation of the pendulum is given by
1
T  2
1 1 
g  
  R 
a) If the pendulum is infinitely long, then using l  in the above expression we get,
R
T  2 Taking g = 9.8m/s2 and radius of earth as 6400km, the time period in this
g
case is nearly equal to 84.6 minutes
b) If the length of the pendulum is equal to radius of earth, then l = R and we get,
R
T  2
2g
and this value is nearly equal to 60 minutes.

c) If   R, T  2
g
vii) Consider a simple pendulum of length “l” suspended inside a trolley which is coming
down on an inclined plane of inclination “  ”.
a = acceleration of the point of suspension = g sin  (down the plane)
 geff  g2  g sin   2 gg sin  cos 90  

2
 g2  g2 sin 2   2g2 sin 2 
[ cos 90     sin  ]
 
geff = g 1 sin  = g cos 
2 2

But time period T = 2
geff
mg sin 
90  
mg


T  2
g cos 
viii) Consider two simple pendulums of different lengths. Let, lL and lS are the lengths of longer
and shorter pendulums and TL and TS are their time periods of oscillation respectively.
They are made to oscillate at the same instant starting from the same phase. By the time
they are again at the same phase, if “n” is the number of oscillations made by the
longer pendulum, the shorter pendulum completes (n+1) oscillations.
If “t” is the minimum time interval after which the pendulums are again in the same phase,
then
t = n TL = (n+1)TS ............(1)
 Simple pendulum performs angular S.H.M. but due to small angular displacement, it is
considered as simple S.H.M.
 If time period of clock based on simple pendulum increases then clock will be slow if
time period decreases then clock will be fast.
 If l is change in length and g is the change in acceleration then for small variation
(up to 5%) change in time period  T  will be
T  1  1 g 
 10 0      10 0
T 2  2 g 
 Due to change in shape of earth (not shperical but ellipitical) gravitational acceleration
is different at different places. So time period of simple pendulum varies with variation
of g.
 The time period of simple pendulum is independent of mass of bob.
1. If the length of the pendulum is increased by 2%. What is the percentage

change in time period of pendulum.
T 1 l
SOLUTION : 100  100  1%
T 2 l
2. A simple pendulum is suspended from the roof of a trolley which moves in a

horizontal direction with an acceleration a, then the time period is given by
l
T  2 , where g is equal to
g
(a) g (b) ga (c) ga (d) g2  a2
SOLUTION : d) g  g 2  a 2
a
g g
3. A second’s pendulum is placed in a space laboratory orbiting around the earth at a height
3R, where R is the radius of the earth. The time period of the pendulum is
(a) Zero (b) 2 3 sec (c) 4 sec (d) Infinite
SOLUTION :
In the given case effective acceleration geff. = 0
T 
4. The bob of a simple pendulum of mass m and total energy E will have maximum linear
momentum equal to
2E
(a) m
(b) 2mE (c) 2mE (d) mE 2
SOLUTION
p max  2m Emax
5.. The length of the second pendulum on the surface of earth is 1 m. The length of
seconds pendulum on the surface of moon, where g is 1/6th value of g on
the surface of earth, is
(a) 1 / 6 m (b) 6 m (c) 1 / 36 m (d) 36 m
SOLUTION
l l
T  2 Ţ = constant
g g
 l  g;
lm 1 g 1
  lm  m
1 6 g 6
6. If the length of second’s pendulum is decreased by 2%, how many seconds it will lose per
day
(a) 3927 sec (b) 3727 sec (c) 3427 sec (d) 864 sec
SOLUTION
T l Ţ
T 1 l 0 . 02
   0 . 01
T 2 l 2
T  0 . 01 T
Loss of time per day  0 . 01  24  60  60  864 sec
7. The period of simple pendulum is measured as T in a stationary lift. If the lift moves
upwards with an acceleration of 5g, the period will be
(a) The same (b) Increased by 3/5
(c) Decreased by 2/3 times (d) None of the above
SOLUTION
T g g 1
  
T g 'a g  5g 6
T
T 
6
8. The length of a simple pendulum is increased by 1%. Its time period will
(a) Increase by 1% (b) Increase by 0.5%
(c) Decrease by 0.5% (d) Increase by 2%
SOLUTION
T l
T 1 l 1
   1 %  0 .5 %
T 2 l 2
9.. A pendulum has time period T. If it is taken on to another planet having acceleration
due to gravity half and mass 9 times that of the earth then its time period on the other
planet will be
(a) T (b) T (c) T1/3 (d) 2T
SOLUTION
If initial length l1  100 then l 2  121
l T l1
By using T  2  1 
g T2 l2
T1 100
Hence,   T2  1 .1 T1
T2 121
T2  T1
% increase =  100  10 %
T1
10. A simple pendulum is executing simple harmonic motion with a time period T. If the length
of the pendulum is increased by 21%, the percentage increase in the time period of the
pendulum of increased length is
(a) 10% (b) 21% (c) 30% (d) 50%
SOLUTION
T1 l1 T 100
  1  (If l1  100 then l 2  400 )
T2 l2 T2 400
 T2  2T1
T2  T1
Hence % increase   100  100 %
T1
11. The mass and diameter of a planet are twice those of earth. The period of oscillation of
pendulum on this planet will be (If it is a second’s pendulum on earth)
1 1
(a) 2
sec (b) 2 2 sec (c) 2 sec (d) 2
sec
SOLUTION
GM
As we know g
R2
2
g earth Me R g 2 1 Te gp 2 1
 g   2
 e  Also T    
planet M p R e g p 1 g Tp ge Tp 2
12. A simple pendulum is set up in a trolley which moves to the right with an acceleration
a on a horizontal plane. Then the thread of the pendulum in the mean position makes
an angle  with the vertical
a
(a) tan 1 in the forward direction (b) in the backward direction
g
(c) in the backward direction (d) in the forward direction

SOLUTION
In accelerated frame of reference, a fictitious force (pseudo force) ma acts on the bob
of pendulum as shown in figure
ma
.
 mg
ma a a
Hence, tan       tan  1   in the backward direction.
mg g g
13. The time period of a second’s pendulum is 2 sec. The spherical bob which is empty from
inside has a mass of 50 gm. This is now replaced by another solid bob of same radius but
having different mass of 100 gm. The new time period will be
(a) 4 sec (b) 1 sec (c) 2 sec (d) 8 sec
SOLUTION
l
T  2 (Independent of mass)
g
14. If the length of pendulum is increased by 44% find the percentage change in
time period of simple pendulum
1) 24 % 2)36 % 3)14 % 4)20 %
SOLUTION
l2  l1
Percentage change in time period = 100
l1
1.44l1  l1
= 100 = 20%
l1
15. Two pendulums of lengths 1.69 m and 1.44 m start swinging together. After how
many vibrations will they again start swinging together?
1) they swing together after the shorter pendulum completes 13 oscillations
3) they swing together after the shorter pendulum completes 10oscillations

SOLUTION -
n l
 L
n 1 lS
n 1.69 1.3 13
  =
n 1 1.44 1.2 12
n  13
So they swing together after the shorter pendulum completes 13 oscillations or longer
pendulum completes 12 oscillations.
16. Weight of the bob of a simple pendulum is W. Length of the pendulum if l

and it is vibrating with an amplitude A. Maximum tension in the string dur-
ing oscillation will be:
 A2   A  l2   A2 
1)  W 1  2  2) W 1   3) W 1  2  4) W  2 
 l   l   A  l 
SOLUTION
mv 2
T  mg 
l
g
here v  A v A
l
mA 2 g
T  mg  l. l
 A2 
T  W 1 2  here W  mg
 l 
17. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest, its time
period is T. With what acceleration should lift be accelerated upwards in order to reduce
T
its time period to
2
SOLUTION

In stationary lift T  2  g ...(i)
T 
In accelerated lift 2  T '  2  g  a ...(ii)
ga
Divide (i) by (ii) 2 = g
g + a = 4g
a = 3g
18. The length of a second's pendulum at the surface of earth is 1m. Determine the length of
second's pendulum at the surface of moon.
e e e
1) m  2) m  e 3) m  4)  m 
5 2 6
SOLUTION
For second's pendulum at the surface of earth 2  2   e ...(i)

ge
m
For second's pendulum at the surface of moon 2  2  ...(ii)
gm
 
From (i) and (ii) g  g
e m
e m
g 
m   m  e
 ge 
e  ge 
m   gm 
6  6 
19.. If length of a simple pendulum is increased by 4%. Then determine percentage change
in time period.
1) 20 % 2) 2% 3) 0.2% 4) 4%
SOLUTION
T 1 
Percentage change in time period  100% =  100 [Q Dg = 0]
T 2 

According to question  100 = 4%

T 1
 100% = × 4% = 2%
T 2
20. A simple pedulum has time period 'T1' . The point of suppension is now moved
upwards according to the relation y = kt2, (k = 1m/sec2) where y is the vertical
T22
displacement . The time period now becomes 'T2' then find the ratio of 2
T1
1) 3:5 2) 6:7 3) 6: 5 4) 5 : 6
SOLUTION
y = kt2 = 1/2 at2
1
 a  k 1
2
 a  2m / sec 2 ( acceleration }
 
T1  2 and T2  2
g ga
T12 g  a 10  2 6
  
T22 g 10 5
21. A child swinging on a swing in sitting position, stands up, then the period of the swing
will be:
(1) Increase (2) Decrease
(3) Remain same (4) Increase if child is long and decrease if child is short
SOLUTION .
When the child stands up the length of pendulum increases.then timeperiod also increases
22. A simple pendulum 4 m long swing with an amplitude of 0.2 m. What is its acceleratin
at the ends of its path?
1) 0.5m/s2 2) 0.6m/s2 3) 5m/s2 4) 2m/s2
SOLUTION
T = mg cos 
 Fnet  mg sin  and
acceleration = g sin 
 0.2  0.5
= 10 m/s2
4
23. --A man measures the period of a simple pendulum inside a stationary lift and finds it
to be T sec. If the lift accelerates upwards with an acceleration g / 4 , then the
period of the pendulum will be
Pendulum
Lift Lift
T 2T
(a) T (b) 4
(c) 5
(d) 2T 5
l
SOLUTION In stationary lift T  2
g
l
In upward moving lift T   2
(g  a)
( a  Acceleration of lift)
T g g 4
   2T
 T ga  g 5  T 
g   5
 4
SOME SPECIAL CASES
 Consider a tunnel drilled along the diameter of the earth of radius “R”. Let “  ” be the mean
density of earth and “g” is the acceleration due to gravity on its surface. Let a body of mass
‘m’ be released at one end of the tunnel.
m M GM  mx 
F  GM x 2 but
M x  3 x3  F   2  
x R R  R 
g g m
F   mx  K   m  but T  2  
R R k
Hence, time period of oscillation of the body is given by,
R
 T  2
g
Taking R = 6400Km = 64 x 105m and g = 9.8 ms–2

64105
T  2 or T  84.6 min.
9.8
Thus, the body will be in SHM with a time period of nearly 84.6min. It can be recalled that
this is equal to time period of oscillation of a simple pendulum of infinite length.
Note :
1) The above expression for time period is valid even if the tunnel is dug along any
chord of the earth.
2) The above mentioned expression is not valid if the body is dropped into the tunnel
from certain height above the ground. In this case, the body executes oscillatory motion,
but it is not simple harmonic.
 Consider a body of mass “m” and cross sectional area “A” floating in a liquid of density “  ”
with a height “h” submerged in the liquid.
Under equilibrium,
Weight of the body = Buoyancy
mg = Ah  g
(or) m = Ah  .... (1)
When the body is slightly pushed down, through a small distance “y” and released, the
restoring force on it due to extra thrust is
F = – A  gy..
Which is in the form F = – Ky with K = A  g  The body executes SHM with time
period
m m m
T  2  2  T  2
K Ag A g
for this floating body mass m = Ah  .
h
 T  2
g
 Consider a liquid of density ‘  ’ taken into a ‘U’–tube such that the total length of the liquid
column is ‘L’ and height of the liquid column in each limb is ‘h’ (L=2h). The liquid column in
one of the limbs is depressed down by a small distance ‘y’ and released. Now the liquid
in the ‘U’–tube starts oscillating simple harmonically.
FR
A B y D
A y 2y B
C 2y
h
mg
In the new position, the difference in the
liquid levels in the two limbs is ‘2y’.
The restoring force on the liquid is given by
F = Pressure x area = 2y g A
F   A 2 y   g ........... (1)
The negative sign indicates that the direction of restoring force is opposite to the
displacement.
F   A 2 y   g  K  2Ag
The time period of oscillation is
m 2Ah h
T  2  2  2
K 2Ag g
h
T  2
g
 Consider a small sphere of radius ‘r’ released from the position shown on a bowl of radius
of curvature ‘R’ The motion of the sphere is analogus to oscillations of a simple
pendulum of length (R–r). At one instant of time let the imaginary pendulum is making
a small angle ‘  ’ with the vertical. At that instant the restoring force acting on the sphere
is given by,
F  mg sin  ............ (1)
The negative sign indicates that the restoring force acts in a direction opposite to the
displacement.
F = mg  ( sin    for small values of  )
y
= – mg
  r
R
mg
This is in the form F = –Ky with K 
Rr
R  r 
T  2
 The time period of oscillation of the sphere is given by, (or) g
 Consider a uniform cylinder of area of cross–section ‘A’ arranged horizontally, filled with a
gas at a pressure ‘P’ and enclosed by means of a piston of mass ‘M’ as shown in the figure.
Let, ‘h’ be the length of the gas column.
When the piston is pushed inwards through a small distance ‘x’, due to increase in pressure,
there acts a restoring force on the piston.
h m
M
Gas F P.A
P
x
from ideal gas equation PV = constant
 V 
PV  V P  0 (or) P    P
V 
If F is the restoring force acting on the piston of area of cross section ‘A’, then
F  Ax   AP 
    P (or) F     x ............ (1)
A  Ah   h 
AP M Mh
 K  T  2  2
h K AP
 The time period of oscillation of the piston can be obtained from
Mh
 T  2
AP
 Consider a massless elastic wire of length ‘L’, area of cross–section ‘A’ and Young’s
modulus ‘Y’ suspended from a rigid support. A small mass M is attatched at free
end. When the string is pulled down through a small distance ‘ L ’ and released
the string oscillates up and down. In equilibrium, the restoring force developed
in it is given by
YA  
F   L ............ (1) Q Y  F l 
L 
 Al  
The negative sign shows that, the restoring force acts in a direction opposite to the
elongation.
YA M ML
K  T  2  2
L K YA
 The time period of oscillation can be obtained from
ML
 T  2
YA
Expression for time period compound pendulum
Ą Any rigid body which is free to oscillate in a vertical plane about a horizontal axis passing
through a point, is define compound pendulum
Torque acting on a body  = – mglsin 
if angle is very small sin   
 = – mgl  ...(i) and  = Is a ...(ii)
m = mass of the body
l = distance between point of suspension and centre of mass
Is = moment of inertia about horizontal axis passes through point of suspension
from equation (i) and (ii) Is a = – mgl 
d2 
Is + mgl  = 0
dt 2
d2  mg
dt 2
+ Is  =0 ...(iii)
d2 
 + w2  = 0 ...(iv)
dt 2
mg
compare equation (iii) and (iv) w2 = I
s
mg 2 mg
 w= Is  T
= Is
Is
Time period of compound pendulum T = 2  mg
Applying parallel axis theorm Is = ICM + ml2  Is = mk2 + ml2
Is mk 2  m 2
T = 2 mg
= 2 
mg
k2

Time period T = 2  
g
S = point of suspension ; O = point of oscillation ; k = radius of gyaration

l = distance between point of suspension and centre of mass
k2
L= + l = equivalent length of simple pendulum = distance between point of suspension

and point of oscillation
k2

Time Period T = 2  
g
For maximum time period  = 0 maximum time period Tmax = 

dT
For minimum time period =0 then k= 
d
k2
k
T = 2 k
g
2k
Minimum time period of compound pendulum Tmin = 2 
g
k2
= distance from centre of mass to point of oscillation (standardresult)

Bar pendulum
It is a steel bar of 1 metre length with hole at regular interval for suspension. Time period
is measured by taking different value of l.
CM
 There are maximum four points for which time period of compound pendulum is same.
 Minimum time period is obtained at two points
 The point of suspension and point of oscillation are mutually interchangeable.
 Maximum time period will obtain at centre of gravity, which is infinite means compound
pendulum will not oscillate at this point.
 Compound pendulum executes angular S.H.M. about its mean position. Here restoring
torque is provided by gravitational force.
PROBLEMS
1.. A ring is oscillating about a horizontal axis passes through its rim. Determine time period
of oscillation.
= R
ring horizontal
axis
G
2R R
(1) T  2 (2) T  2 
g g
R 2R
(3) T  2  (4) T  
2g g
SOLUTION :
For ring I = Mk2 = MR2
So, k = R
k2 R2
L=  + =R+ = 2R
 R
2R
T  2
g
2. A sphere is made to oscillate about a horizontal tangential axis. Determine time period.
7R 5R
(1) T  2 5 g (2) T  2 
7g
R 2R
(3) T  2  (4) T  
g g
SOLUTION :
2
For sphere I = Mk2 = MR2
5
2
k2 = R2
5
2 2
k2 R 7
L=  + =R+ 5 = R
 R 5
7R
T  2
5g
3. A disc is made to oscillate about a horizontal axis passing through mid point of its radius.
Determine time period.
3R 5R
(1) T  2  2 g (2) T  2 
7g
R 2R
(3) T  2  (4) T  
g g
SOLUTION :
MR 2
For disc I = Mk2 =
2
R
 k=
2
k2 R R2 3R
L=  + = + =
 2 R  2
2 
2 
3R
 T  2
2g
4. A uniform rod of mass ‘m’ and length‘ l ’ is hinged at one end ‘A’. It can rotate
freely about a horizontal axis passing through A. If it is given a slight an-
gular displacement and left to itself then it oscillate. Find the time period.
l 3l
(1) T = 2 (2) T = 2
g 2g
2l 3l
(3) T  2 (4) T = 2
3g g
SOLUTION :
2
here I  ml , dl.
3 2
I ml 2
T  2  2 2l
mgd l  2
3.mg 3g
2
5. A uniform ring of radius ‘R’ is suspended from a horizontal nail ‘A’ as shown.
Find time period of its small oscillations.
3R 5R
(1) T  2  2 g (2) T  2 
7g
R 2R
(3) T  2  (4) T  
g g
SOLUTION :
here I  2mR 2 , d  R
I
, T  2
mgd
2mR 2 2R
T  2  2
mgR g
** 6.17. Damped Oscillation :

So far we have dealt oscillation of a body which are simple harmonic, in which total energy
of oscillation is constant.
But when the oscillation are in a medium which offer some resistance force, in which the
energy gradually decreases with time, and finally the oscillation will be stopped, such oscillations
are called “Damped oscillations”.
The damping force depends on velocity as f = –bv;
Hear, b = damping co efficient
Hence the total restoring force is
FR = – (kx + bv)  ma = –(kx + bv)
d 2x b k d 2 x b dx k
  v  x  0    x0
dt 2 m m dt 2 m dt m
for which the angular frequency is
k  b 
2
  
m  2m 
 Composition of two SHM’s of equal frequency in mutually perpendicular directons.
Let the two SHM’s be
x
(i) x  a sin t   sin t
a
y
(ii) y  b sin t      sin t   
b
y
sin t cos   cos t sin  
b
2
x x y
cos   1 2 sin  
a a b
Squaring on both sides
x2 y x 
1 sin 2     cos2 
a 2 
b a 
 x 2  2 2
1  sin   y  x cos   2 xy cos 
 a 2  b2 a2 ab
 x 2 y 2 2 xy 
   2  2  cos   sin 2 
 a b ab 
This is the equation representing resultant SHM. The path traversed by the particle,
depends on the values of a, b and  as shown in the table
Condition Resultant SHM Shape
a = b,   00 ,1800 Straight line y   x
  900 Circle x2 + y2 = a2
b
a  b,   900 , Straight line y   x
a
x2 y2
1800, Q = 900 Ellipse  1
a 2 b2
 Vector method of adding two linear SHM’s in a line :

Suppose two SHM’s of same frequencies are represented by Y1 = a1sin  t and Y2 = a2sin(  t
+ ) when these are super imposed on a particle, to get the resultant SHM consider a1
and a2 as vectors and the phase difference  as the angle between them. Now applying
parallelogram law to get the resultant amplitude and epoch
The resultant SHM equation is
Y  a sin  t   
The resultant amplitude
a  a12  a 22  2a1a 2 cos 

and epoch 
a 2 sin 
tan  
a1  a 2 cos 
EXERCISE PROBLEMS
1
1. When an oscillator completes 100 oscillation its amplitude reduced to of initial value.
3
What will be its amplitude, when it completes 200 oscillation
1 2 1 1
(1) (2) (3) (4)
8 3 6 9
SOLUTION :
A
\ After 100 oscillation amplitude(vk;ke) become 3
1 A A
Q After 200 oscillation it become 3
of 3

9
2. A body of mass 1.0kg is suspended from a weightless spring having force constant
600 N/m . Another body of mass 0.5 kg moving vertially upwards hits the
suspended body with velocity of 3.0 m/sec and gets embededin it. Find
the amplitude of oscillation ( neglect gravity )
(1) 10 cm (2) 5cm (3) 3 cm (4) 0.2 cm

SOLUTION :
By conservation of linear momentum in the collision
mv = ( m+ M) V
mv 0.5  3
V   1m / sec
m  M (1  0.5)
Now just after collision the system will have
1
KE = (m+M) V2 at equislibrium position .
2
So after collision by conservation of mechanical energy KEmax = PEmax
1 1
(m  M ) v 2  KA 2
2 2
mM 1.5 1
AV   1  m  5cm
 k  600 20
3. A block is kept on a rough horizontal plank. The coefficient of friction between

block and the plank is 1/2. plank is undegoing SHM of angular frequency 10 rad/s.Find
the maximum amplitude of plank in which the block does not slip over the
plank( g = 10 m/s2).
(1) 0.5m (2) 5cm (3) 5 m (4) 0.2 cm
SOLUTION :
Maximum acceleration in SHM is amax    A this will be provided to the block by friction .
Hence, amax   g or   A   g
 1 
 10
g  2 
or a   = 0.05m = 5 cm
 10
2
4. In the diagram shown find the time period of pendulum for small oscillations
l l
(1) T   (2) T  2
g sin  g tan 
l l
(3) T  2 (4) T  2
g sin  g cos 
SOLUTION :
For smaller values of  , sin   
y
f  mg sin  .  f   mg sin 
l
g sin 
a .y
l
g sin 

l
T l
 T   1 2
o
5. The trolly car having simple pendulum decelerated by friction. In consequence,

the pen dulum has time period T. If T o is time period of the simple pendulum
T
in the absence of any acceleration of the trolly car, the value of T is --
o
T l T l
(1) T  (2) 
o  1 2
To 
T l T l
(3)  (4) 
To  1 To  12
SOLUTION :
l
T  2 ; a  g
a2  g 2
l
T  2
g 2 1
1 T l
T  To  
 1
2 To 2 1
6. A cylindrical piston of mass M slides smoothly inside a long cylinder closed at one
end, enclosing a certain mass of gas. The cylinder is kept with its axis
horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple
harmonically. The period of oscillation will be
M
Gas
P A
 Mh   MA 
(a) T  2   (b) T  2  
 PA   Ph 
 M 
(c) T  2   (d) T  2 MPhA
 PAh 
SOLUTION :
Let the piston be displaced through distance x towards left, then volume decreases,
pressure increases. If P is increase in pressure and V is decrease in volume,
then considering the process to take place gradually (i.e. isothermal)
P1 V1  P2 V2  PV  (P  P)(V  V )
 PV  PV  PV  PV  PV
 P . V  P . V  0 (neglecting P.V )
P. x
P( Ah)  P( Ax )  P 
h
This excess pressure is responsible for providing the restoring force (F) to the piston of
mass M.
PAx
Hence F  P. A 
h
PA
Comparing it with | F |  kx  k  M  2 
h
PA Mh
   T  2
Mh PA
7.. Two particles executes S.H.M. of same amplitude and frequency along the same
straight line. They pass one another when going in opposite directions, and each time
their displacement is half of their amplitude. The phase difference between them is
(a) 30° (b) 60°

(c) 90° (d) 120°
ANSWER : (d)
SOLUTION :
. According to the question of SHM y  a sin(t  0 )
a a  5
Given y
2
 2
 a sin( t  0 )  ( t  0 )   
6
or 6
a/2

if 
6
 A O P B  Particle is at point P and it is going towards B
a
a/2
5
if 
6
 A O P B  Particle is at point Pand it is going towards O
a
5  2
So phase difference  
6
 
6 3
 120 
8.. The function sin 2 (t) represents
(a) A simple harmonic motion with a period 2 / 

(b) A simple harmonic motion with a period  /
(c) A periodic but not simple harmonic motion with a period 2 / 
(d) A periodic but not simple harmonic, motion with a period  /
ANSWER : (d)
1  cos 2t
SOLUTION : Given y  sin 2  t  y
2
1 cos 2t
 y 
2 2
Angular velocity  2
2 
Period, T  
2 
The given function is not satisfying the standard differential equation of S.H.M.
d 2y
   2y .
dx 2
Hence it represents periodic motion but not S.H.M.

9. Which of the following function represents a simple harmonic oscillation
(a) sin t  cos t (b) sin 2 t

(c) sin t  sin 2t (d) sin t  sin 2t
ANSWER : (a)
SOLUTION :
The function sin  t  cos  t. only satisfies the standard differential equation of
S.H.M.
d 2y
   2y .
dx 2
Hence it sin  t  cos  t. represents S.H.M.
10. The displacement of a particle varies with time as x  12 sin  t  16 sin 3  t (in cm). If its
motion is S.H.M., then its maximum acceleration is
(a) 12  2 (b) 36  2 (c) 144  2 (d) 192  2
ANSWER : (b )
SOLUTION :
Displacement x  12 sin  t  16 sin 3  t  4 [3 sin  t  4 sin 3  t]
By using sin 3  3 sin   4 sin 3  )
Displacement x  4[sin 3 t]
equation for maximum acceleration | Amax |   2 a
 maximum acceleration A max  (3 )2  4  36  2
11. A particle of mass m is executing oscillations about the origin on the x-axis. Its
potential energy is U(x )  k [x ]3 , where k is a positive constant. If the
amplitude of oscillation is a, then its time period T is
1
(a) Proportional to a
(b) Independent of a
(c) Proportional to a (d) Proportional to a3/2
ANSWER : (a )
SOLUTION :
dU
For concervSOLUTION :atieve force F  
dx
Given Potential Energy U  k | x |3
dk | x |3
Then F  
dx
F  3k | x |2 ...(i)
Also, for SHM F   m 2 x ...(ii)
3kx
From equation (i) & (ii) 
m
2
Time period T 

m
T  2
3kx
m
T  2
3k (a sin  t )
1
T .
a
12. For a particle executing S.H.M. the displacement x is given by x  A cos t . Identify the
graph which represents the variation of potential energy (P.E.) as a function of
time t and displacement x
PE PE
I II
III
IV
t x
(a) I, III (b) II, IV (c) II, III (d) I, IV

ANSWER : (a )
SOLUTION :
Here displacement x is given by x  A cos  t .
So Potential energy is
1) minimum (in this case zero) at mean position (x = 0)
2) maximum at extreme position (x   A).
From
At time t = 0, x = A,
hence potential should be maximum. Therefore graph I is correct. Further in graph
III. Potential energy is minimum at x = 0, hence this is also correct
 
13. A point mass oscillates along the x-axis according to the law x  x0 cos   t   . If the
 4
acceleration of the particle is written as a  A cos t    , then:-

3 
(1) A  x0 ,   (2) A  x0 ,   
2
4 4
 
(3) A  x0 ,   (4) A  x0 ,   
2 2
4 4
ANSWER : (1 )
d2x
SOLUTION : Acceleration of SHM particle a
dt 2
 
But displacement x  x0 cos  t  
 4
   
d 2  x0 cos  t   
 4 
a  2
dt
 3 
On solving we get a   2 x0 cos  t   ………(1)
 4 
But given acceleration ………(2)

a  A cos t   
3
From (1) and (2) A  x0 2 and  
4
14. On a smooth inclined plane, a body of mass M is attached between two springs. The
other ends of the springs are fixed to firm supports. If each spring has force constant K,
the period of oscillation of the body (assuming the springs as massless) is

1/2 1/ 2 1/2
 m   2M  Mg sin   2 Mg 
(a) 2   (b) 2   (c) 2 (d) 2  
 2K   K  2K  K 
ANSWER : (a )
SOLUTION :
Time period doesnot depends on inclination ….
1/ 2
 M 
Hence time period of arrangement is T  2  
 K eff 
Here two springs are arranged in parellel combination
Hence K eff  2 K
1/ 2
 m 
Time period becomes T  2  
 2K 
15. The displacement of a particle executing S.H.M. is given by

x  0.01 sin 100 (t  0.05) . The time period is
(1) 0.01 s (2) 0.02 s (3) 0.1 s (4) 0.2 s
ANSWER : (2 )
SOLUTION :
x  0.01 sin 100 (t  0.05)
Here w = 100  s
2 2 1
Time period T   T  T  T= 0.02 s
 100 50
PREVIOUS MAINS QUESTIONS
1. The position co-ordinates of a particle moving in a 3-D coordinate system is
given by [9 Jan
2019, II]
x  a cos t y  a sin t and z  at
The speed of the particle is:
(1) 2a (2) a
(3) 3a (4) 2a
SOLUTION :(1) Here, vx  a sin t , v y  a cos t and vz  a  v  vx2  v 2y  vz2
v   a sin t 2   a cos t 2   a2
2. Two simple harmonic motions, as shown, are at right angles. They are
combined to form Lissajous figures.
x  t   A sin  at    y  t   B sin  bt 
Identify the correct match below [Online April 15, 2018]
SOLUTION
(3) From the two mutually perpendicular S.H.M.’s, the general equation of Lissajous figure,
x 2 y 2 2 xy
2
 2 cos   sin 2 
A B AB
x  A sin  at   
y  B sin  bt  r 
Clearly A B hence ellipse.

(1) Parameters: A  B, a  2b;   ; Curve: Circle
2

(2) Parameters: A  B, a  b;   ; Curve: line
2

(3) Parameters: A  B, a  b;   ; Curve: Ellipse
2
(4) Parameters: A  B, a  b;   0; Curve: Parabola
3. The ratio of maximum acceleration to maximum velocity in a simple harmonic
motion is 10s 1 . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The

initial phase is [Online April 8, 2017]
4
(1) 500 m/s2 (2) 500 2m / s 2 (3) 750m / s 2 (4) 750 2m / s 2

SOLUTION
(2) Maximum velocity in SHM, vmax  a
Maximum acceleration in SHM, Amax  a2

where a and  are maximum amplitude and angular frequency..
Amax
Given that,  10
vmax
i.e.,   10s 1
Displacement is given by
x  a sin  t   / 4 
At t  0, x  5
5  a sin  / 4
Maximum acceleration Amax  a2  500 2m / s 2
4. A particle performs simple harmonic mition with amplitude A. Its speed is
2A
trebled at the instant that it is at a distance from equilibrium position. The new
3
amplitude of the motion is : [2016]
7A A
(1) A 3 (2) (3) 41 (4) 3A
3 3
SOLUTION (2) We know that V   A2  x 2
2
 2A 
2
Initially V   A   
 3 
2
 2A 
Finally 3V   A '2   
 3 
2A
Where A’ = final amplitude (Given at x , velocity to trebled)
3
On dividing we get
2
 2A 
A '2   
3  3 

1  2A 
2
A2   
 3 
 4 A2  2 4A
2
7A
9  A2    A '   A' 
 9  9 3
5. Two particles are performing simple harmonic motion in a straight line about
the same equilibrium point. The amplitude and time period for both particles are same
and equal to A and T, respectively. At time t = 0 one particle has displacement A while
A
the other one has displacement and they are moving towards each other. If they
2
cross each other at time t, then t is: [Online April 9, 2016]
5T T T T
(1) (2) (3) (4)
6 3 4 6
SOLUTION (at time t = 0)

Angel covered to meet   600  rad.
3
If they cross each other at time t then
  T
t  T
2  3  2 6
6. A simple harmonic oscillator of angular frequency 2 rad s–1 is acted upon by an

external force F = sin t N. If the oscillator is at rest in its equilibrium position at t = 0, its
position at later times is proportional to : [Online April 10, 2015]
1 1
(1) sin t  cos 2t (2) cos t  sin 2t
2 2
1 1
(3) sin t  sin 2t (4) sin t  sin 2t
2 2
SOLUTION (3) As we know,
F  ma  a  F
Or, a  sin t
dv
  sin t
dt
0 t
  dv   sin tdt
0 0
V  cos t  1
x t
 dx     cot t  1 dt
0 0
7. x and y displacements of a particle are given as  (t) = a sin  t and y (t) = a sin
2  t. Its trajectory will look like : [Online April 10, 2015]
(1) (2)
(3) (4)
SOLUTION (3) At t  0, x  t   0; y  t   0
x t  is a sinusoidal funciton

At t ; x t   a and y t   0
2
Hence trajectory of particle will look like as (3)
8. A body is in simple harmonic motion with time period half second (T = 0.5 s)
and amplitude one cm (A = 1 cm). Find the average velocity in the interval in which it
moves form equilibrium position to half of its amplitude.
(1) 4 cm/s (2) 6 cm/s
(3) 12 cm/s (4) 16 cm/s
SOLUTION
8. (3) Given: Time period, T = 0.5 sec
Amplitude, A = 1 cm
Average velocity in the interval in which body moves from equilibrium to half of its amplitude, v = ?
T
Time taken to a displacement A/2 where A is the amplitude of oscillation from the mean position ‘O’ is
12
A 1
Therefore, s  cm
2 2
A 1
v  2  2  12 cm / s
 Average velocity, t 0.5
12
9.
9.Which of the following expressions corresponds to simple harmonic motion along a
straight line, where x is the displacement and a, b, c are positive constants? [Online
April 12, 2014]
(1) a  bx  cx 2 (2) bx 2
(3) a  bx  cx 2 (4) bx

SOLUTION (4) In linear S.H.H., the restoring force acting on particle should always be proportional to the
displacement of the particle and directed towards the equilibrium position.
i.e., Fx
or F = - bx where b is a positive constant.
10. A particle which is simultaneously subjected to two perpendicular simple
harmonic motions represented by; x = a1 cos  t and y = a2 cos 2  t traces a curve
given by: [Online April 9, 2014]
(1) (2) (3) (4)
SOLUTION (2) Two perpendicular S.H.Ms are x  a1 cos t …..(1)
And y  a2 2 cos t …..(2)

Form eqation (1)
x
 cos wt
a1
And form equation (2)
y a
 2 cos t  y  2 2 x
a2 a1
2
11. The displacement y  t   A sin  t    of a pendulum for   is correctly
3
represented by [Online May 19, 2012]
(1) (2)
(3) (4)
SOLUTION (a) Displacement y  t   A sin  wt    Given
2
For 
3
2
At t  0; y  A sin   A sin
3
 A sin1200  0.87 A  sin1200  0.866 
Graph (1) depicts y  0.87 A at t0

12. Two particles are executing simple harmonic motion of the same amplitude A
and frequency w along the x-axis. Their mean position is separated by distance X0(X0
> A). If the maximum separation between them is (X0 + A), the phase difference
between their motion is: [2011]
   
(1) (2) (3) (4)
3 4 6 2
SOLUTION 1) Let, x1  A sin t and x2  A sin  t   
  
x2  x1  2 A cos  t   sin
 2 2

The above equation is SHM with amplitude 2 A sin
2

 2 A sin A
2
 1 
 sin  
2 2 3
13. A mass M, attached to a horizontal spring, executes S.H.M. with amplitude A1.
When the mass M passes through its mean position then a smaller mass m is placed
 A1 
over it and both of them move together with amplitude A2. The ratio of  A  is
 2
1 1
M m  M 2  M  m 2 M
(1) (2)   (3)   (4)
M M m  M  M m
SOLUTION (3) At mean position, F net = 0
Therefore, by principal of conseruation of linear momentum
 Mv1   M  m  v2
Mw, a,   M  m  w2 a2
k k
MA1   M  m  A2
M mM
 k 
  V  A 
 M 
 A1 M  A2 M  m
A1 M m
 
A2 M
14. A point mass oscillates along the x-axis according to the law x = x0 cos(  t - p /
4) . If the acceleration of the particle is written as a = A cos(  t +  ) ,then
[2007]
(1) A  x0 2 ,   3 / 4 (2) A  x0 ,    / 4
(3) A  x0 2 ,    / 4 (4) A  x0 2 ,    / 4
SOLUTION
(1) Given,
Displacement, x  x0 cos  t   / 4 
dx  
 Velocity, v  dt   x0 sin  t  4 
 
Acceleration,
dv  
a   x02 cos  t  
dt  4
   
 x0 2 cos     t   
  4 
 
 x02 cos  t   ….(1)
 4
Acceleration, a  A cos  t    …(2)

Comparing the two equations, we get
3
A  x0 2 and 
4
15. A coin is placed on a horizontal platform which undergoes vertical simple

harmonic motion of angular frequency w. The amplitude of oscillation is gradually
increased. The coin will leave contact with the platform for the first time (a) at the
mean position of the platform [2006]
(1) at the mean position of the platform
g
(2) for an amplitude of
2
g2
(3) for an amplitude fo 2

(4) at the highest position of the plastform
SOLUTION (2) For block A to move in SHM
mg  N  m2 x
Where x is the distance from mean position
For block to leave contact N 0
g
 mg  m2 x  x 
2
16. The maximum velocity of a particle, executing simple harmonic motion with an
amplitude 7 mm, is 4.4 m/s. The period of oscillation is [2006]
(1) 0.01 s (2) 10s (3) 0.1 s (4) 100 s
SOLUTION (1) Maximum velocity,
vmax  a
Here, a = amplitude of SHM
 = angular velocity of SHM
2  2 
vmax  a      
T  T 
2a 2  3.14  7 103

T    0.01s
vmax 4.4
17. The function sin2 (  t) represents [2005]

(1) a periodic, but not simple harmonic motion with a period

2
(2) a periodic, but not simple harmonic motion with a period


(3) a simple harmonic motion with a period

2
(4) a simple harmonic motion with a period


SOLUTION (1) Clearly sin 2t is a periodic function with period

d2y
For SHM  y
dt 2
1  cos 2t
y  sin 2 t 
2
1 1
  cos 2t
2 2
dy 1
v   2 sin 2t  2 sin t cos t
dt 2
1 1
  cos 2t
2 2
dy 1
v   2 sin 2t  2 sin t cos t
dt 2
  sin 2t
d2y
Aceleration, a 2
 22 cos 2t which is not proportional to –y. Hence, it is not in SHM.
dt
18. Two simple harmonic motions are represented by the equations y1 = 0.1 sin
 
100t   and y2  0.1cos t . The phase difference of the velocity of particle 1 with
 3
respect to the velocity of particle 2 is [2005]
   
(1) (2) (3) (4)
3 6 6 3
SOLUTION (2) Velocity of particle 1,
dy1  
v1   0.1100 cos  100t  
dt  3
Velocity of particle 2,
dy2  
v2   0.1 sin t  0.1 cos  t  
dt  2
 Phase difference of velocity of particle 1 with respect to the velocity of particle 2 is
  2  3 
 1  2    
3 2 6 6
19. Two particles A and B of equal masses are suspended from two massless
s p rin g s o f s p rin g c o n s tan ts k1 and k2, respectively. If the maximum velocities, during
oscillation, are equal, the ratio of amplitude of A and B is [2003]
k1 k2 k2 k1
(1) k2
(2) k (3) k1
(4) k
1 2
SOLUTION (3) Maximum velocity during SHM, vmax  A
But k  m2
k
 
m
k
 Maximum velocity ofh te body is SHM  A
m
As maximum velocities are equal
k1 k
 A1  A2 2
m m
A1 k
 A1 k1  A2 k2   2
A2 k1
20. The displacement of a particle varies according to the relation x = 4(cospt + sin
pt). The amplitude of the particle is [2003]
(1) – 4 (2) 4 (3) 4 2 (4) 8
SOLUTION (3) Diplacement, x  4  cos t  sin t 
 sin t cos t 
 2  4  
 2 2 

 4 2 sin t cos 450  cos t sin 450 
x  4 2 sin t  450 
On comparing it with standard equation x  A sin  t   
We get A4 2
Topic-2: Energy in Simple Harmonic Motion
21. The displacement time graph of a particle executing S.H.M. is given in figure :
(sketch is schematic and not to scale)
Which of the following statements is/are true for this motion? [Sep. 02, 2020
(II)]
3T
(1) The force is zero at t 
4
(2) The acceleration is maximum at t = T
T
(3) The speed is maximum at t 
4
T
(4) The P.E. is equal to K.E. of the oscillation at t 
2
(1) (1), (2) and (4) (2) (2), (3) and (4)
(3) (1), (2) and (3) (4) (1) and (4)
SOLUTION(3) From graph equation of SHM
X  A cos t
T
(1) At particle is at mean position.
4
 Acceleration = 0, Force = 0
(2) At T particle again at extreme position so acceleration ismaximum.
T
(3) At t , particle again at extreme position so velocity is maximum .
4
Acceleration = 0
(4) When KE = PE

1
2
 
1
k A2  x 2  kx 2
2
Here, A = amplitude of SHM
x = displacement from mean postion
A
 A2  2 x 2  x 
2
A T
  A cos t  t 
2 2
x  A which is not possible
x  A and 3 are correct.
22. A particle undergoing simple harmonic motion has time dependent
t
displacement given by x  t   A sin . The t = 210s will be: [11 Jan 2019, I]
90
1 1
(1) (2) 1 (3) 2 (4)
9 3
1
SOLUTION (4) Kinetic energy, k m2 A2 cos 2 t
2
1
Potential energy, U m2 A2 sin 2 t
2
k  1
 cot 2 t  cot 2  cot 2  210  
U 90 3
23. A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1.
If the length of the pendulum is doubled and it performs simple harmonic motion with the same
amplitude as in the first case, its maximum kinetic energy is K2.
K1
(1) K 2  2 K1 (2) K 2 
2
K1
(3) K 2  (4) K 2  K1
4
SOLUTION
1
(1) K m2 x 2
2
1
 K max  m2 A2
2
A  L
g

L
1
K
g
2m. .L2 2
L
1
 mgL2
2
K1 L 1
    K 2  2 K1
K2 2L 2
K1 L 1
    K 2  2 K1
K2 2L 2
24. A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis,
about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the
particle will be:
[9 Jan 2019, II]
A A A
(1) (2) (3) (4) A
2 2 2 2
1 2
SOLUTION (3) Potential energy U   kx
2
1 2 1 2
Kinetic energy K   kA  kx
2 2
1 1 1
 kx 2  kA2  kx 2
2 2 2
A
x A 2 2 or, x
2
25. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at
its position of equilibrium. The kinetic energy-time graph of the particle will look like:
[2017]
(1) (2)
(3) (4)
SOLUTION (2) For a particle executing SHM
At mean postion; t  0, t  0, y  0, V  Vmax  a
1
 K .E.  KEmax  m2 a 2
2
T 
At extreme postion : t , t  , y  A, V  Vmin  0
4 2
 K .E.  KEmin  0
Kinetic energy in SHM, KE 

1
2

m2 a 2  y 2 
1
 m2 a 2 cos 2 t
2
Hence graph (2) correctly depicts kinetic energy time graph.
26. A block of mass 0.1 kg is connected to an elastic spring of spring constant 640
Nm–1 and oscillates in a medium of constant 10-2 kg s-2. The system dissipates its
energy gradually. The time taken for its mechanical energy of vibration to drop to half
of its initial value, is closest to : [Online April 9, 2017]
(1) 2s (2) 3.5 s (3) 5 s (4) 7s
SOLUTION(2) Since system dissipates its energy gradually, and hence amplitude will also decreases with time
according to
a  a0e bt / m ….(i)
 Energy of virbration drop to half of its initial value

 E0  , as E  a2  a  E
a0 bt 102 t t
a   
2 m 0.1 10
From equation (i),
a0
 a0e t /10
2
1
 et /10 or
t
2 2  e10
t
ln 2  t  3.5 seconds
10
27. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and
potential energy (PE) against its displacement d. Which one of the following
r ep r es en ts th es e c o r r ec tly? ( graphs are schematic and not drawn to scale)
[2015]
(1) (2)
(3) (4)
SOLUTION (4) K .E 
1
2

k A2  d 2 
1 2
And P.E.  kd
2
At mean position d = 0. At extreme positions d = A
28. A pendulum with time period of 1s is losing energy. At certain time its energy is
45 J. If after completing 15 oscillations, its energy has become 15 J, its damping
constant (in s–1) is :
1 1 1
(1) (2) ln 3 (3) 2 (4) ln 3
2 30 15
bt

SOLUTION (4) As we know, E  E0 e m
b15
15  45e m
[As no. of oscillations = 15 so t = 15 sec]
b15
1 
e m
3
Taking log on both sides
b 1
  n3
m 15
29. This question has Statement 1 and Statement 2. Of the four choices given after
the Statements, choose the one that best describes the two Statements. If two
springs S1 and S2 of force constants k1 and k2 respectively, are stretched by the same
force, it is found that more work is done on spring S1 than on spring S2.
Statement 1 : If stretched by the same amount work done on S1
Statement 2 : k1 < k2 [2012]
(1) Statement 1 is false, Statement 2 is true.
(2) Statement 1 is true, Statement 2 is false.
(3) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for
Statement 1
(4) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation
for Statement 1
1 2
SOLUTION (2) Work done, w kx
2
1 2
Work done by spring S1 , w1  k1 x
2
1
Work done by spring S 2 , w2  k2 x 2
2
Since w1  w2 Thus  k1  k2 
30. A particle of mass m executes simple harmonic motion with amplitude a and
frequency n. The average kinetic energy during its motion from the position of
equilibrium to the end is [2007]
(1) 22 ma 2v 2 (2) 2 ma 2v 2
1 2 2
(3) ma v (4) 42 ma 2v 2
4
SOLUTION (2) The kinetic energy of a particle executing S.H.M. at any instant t is given by
1 2 2 2
K ma  sin t
2
Where, m = mass of particle
A = amplitude
 = angular frequency
T = time
1
The average value of sin 2 t 2 over a cycle is
2
1 1  1
 KE  m2 a 2    sin 2   
2 2  2
1 1
m2 a 2  ma 2  2v 
2

4 4
   2v 
or,  K  2 ma 2 v 2
31. Starting from the origin a body oscillates simple harmonically with a period of 2
s. After what time will its kinetic energy be 75% of the total energy? [2006]
1 1 1 1
(1) s (2) s (3) s (4) s
6 4 3 12
1 2 2
SOLUTION (1) K.E. of a body undergoing SHM is given by K .E.  ma  cos 2 t
2
Here, a = amplitude of SHM
 Angular velocity of SHM
1 2 2
Total energy in S.H.M  ma 
2
Given K.E. = 75% T.E.
1 2 2 75 1 2 2
ma  cos 2 t   ma 
2 100 2

 0.75  cos 2 t  t 
6
  2 1
t  t  t  s
6 6  2 6
32. The total energy of a particle, executing simple harmonic motion is [2004]
(1) independent of x (2)  x 2
(3)  x (4)  x1/2

1 2
SOLUTION (1) At any instant the total energy in SHM is kA0  constant,
2
Where A0 = amplitude
k = spring constant
hence total energy is independent of x.
33. A body executes simple harmoni

c motion. The potential energy (P.E), the kinetic energy (K.E) and total energy (T.E) are
measured as a function of displacement x. Which of the following statements is true?
[2003]
(1) K.E. is maximum when x = 0
(2) T.E is zero when x = 0
(3) K.E is maximum when x is maximum
(4) P.E is maximum when x = 0
SOLUTION (1) K.E. of simple harmonic motion 

1
2

m2 a 2  x 2 
34. In a simple harmonic oscillator, at the mean position [2002]
(1) kinetic energy is minimum, potential energy is maximum
(2) both kinetic and potential energies are maximum
(3) kinetic energy is maximum, potential energy is minimum
(4) both kinetic and potential energies are minimum
SOLUTION (3) The kinetic energy (K.E.) of particle in SHM is given by, K .E 
1
2
 
k A2  x 2 ;
1 2
Potential energy of particle in SHM is U kx
2
Where A = amplitude and k  m2

x = displacement from the mean postion
At the mean position x0
1 2
 K .E.  kA  Maximum
2
And U = 0
Topic-3: Time Period, Frequency, Simple Pendulum and Spring
Pendulum
35. An object of mass m is suspended at the end of a massless wire of length L
and area of cross-section, A. Young modulus of the material of the wire is Y. If the
mass is pulled down slightly its frequency of oscillation along the vertical direction is:
[Sep. 06, 2020 (I)]
1 mL 1 YA
(1) f  (2) f 
2 YA 2 mL
1 mA 1 YL
(3) f  (4) f 
2 YL 2 mA
SOLUTION (2) An elastic wqire can be treated as a spring and its spring constant.
YA  F l 
k  Y  
L  A l0 
Frequency of oscillation,
1 k 1 YA
f  
2 m 2 mL
36. When a particle of mass m is attached to a vertical spring of spring constant k
and released, its motion is described by y (t) = y0 sin2wt, where ‘y’ is measured from
the lower end of unstretched spring. Then w is : [Sep. 06, 2020 (II)]
1 g g g 2g
(1) 2 y0
(2) y0
(3) 2 y0
(4) y0
SOLUTION (3) y  y0 sin 2 t
y0  1  cos 2t 
y 1  cos 2t   sin 2 t  
2  2 
y0  y0
 y  cos 2t
2 2
y0  y0
y  cos 2t
2 2
 y  A cos 2t
y0
 Amplitude 
2
Angular velocity  2
ky0 k 2g
For equilibrium of mass,  mg  
2 m y0
k  m  2 
2
Also, spring constant
k 2g 1 2g g
 2    
m y0 2 y0 2 y0
37. A block of mass m attached to a massless spring is performing oscillatory

motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the
block breaks off when it is passing through its equilibrium point, the amplitude of
oscillation for the remaining system become fA. The value of f is :
[Sep. 03, 2020 (II)]
1 1
(1) (2) 1 (3) (4) 2
2 2
1 2
SOLUTION(1) Potential energy of spring  kx
2
Here, x = distance of block from mena postion,
k = spring constant
1 2
At mean postion, potential energy  kA
2
At equilibrium position, half of the mass of block breaks off, so its potential energy becomes half
11  1
Remaining energy   kA2   kA '2
22  2
A
Here, A’= New distance of block from mena postion  A' 
2
39. A simple pendulum oscillating in air has period T. The bob of the pendulum is
1
completely immersed in a non-viscous liquid. The density of the liquid is th of the
16
material of the bob. If the bob is inside liquid all the time, its period of oscillation in this
liquid is : [9 April 2019 I]
1 1
(1) 2T (2) 2T
10 14
1 1
(3) 4T (4) 4T
15 14
l
SOLUTION (3) T  2
g
When immersed non viscous liquid
 g  15 g
amt   g   
 16  16
l l 4
T '  2  2  T
Now 0net 15 g 15
16
40. Two light identical springs of spring constant k are attached horizontally at the
tw o en d s o f a u n ifo rm h o r izo n tal ro d A B o f len g th l and mass m. The rod is
pivoted at its centre ‘O’ and can rotate frreely in horizontal plane. The other ends
of two springs are fixed to rigid supports as shown in figure. The rod is gently
pushed through a small angle and released. The frequency of resulting
oscillation is:
[12 Jan 2019, I]
1 3k 1 2k
(1) (2)
2 m 2 m
1 6k 1 k
(3) (4)
2 m 2 m
SOLUTION (3) Net torque due to spring force:

  2 Kx cos 
2
 K 2   K 2 
       C let C  
 2   2 
 So, frequency of resulting oscillations
K 2
1 C 1 2  1 6K
f  
2 I 2 M  2 2 M
12
41. A simple pendulum, made of a string of length l and a bob of mass m, is

released from a small angle 1 . It strikes a block of mass M, kept on a horizontal
surface at its lowest point of oscillations, elastically. It bounces back and goes up to
an angle 1 . The M is given by : [12 Jan 2019, I]
m  0  1   0  1 
(1) 2      (2) m     
 0 1  0 1
 0  1  m  0  1 
(3) m      (4) 2     
 0 1  0 1
SOLUTION (2)
Velocity before collision v  2 g  1  cos 0 
Velocity after colistion
v1  2 g  1  cos 1 
Using momentum conservation
mv  MVm  mV1
m 2 g  1  cos 0   MVm  m 2 g  1  cos  
 m 2g  1  cos 0  1  cos 1  MVm 

Vm  2 g  1  cos 1 
And e 1
2 g  1  cos 0 
2g  1  cos 0  1  cos 1  Vm  …(i)
m 2g  1  cos 0  1  cos 1  MVM  …(ii)
Divinding (ii) by (i) we get
 1  cos 0  1  cos 1 M

 1  cos 0  1  cos   m
1
By compounendo and dividend rule
 
sin  1 
mM 1  cos 1
2
 
mM 1  cos 0 sin  0 
 2 
 
M 0  1  
  M m 0 1
m 0  1 0  1
42. 
A simple harmonic motion is represented by : y  5 sin 3t  3 cos 3t cm The
amplitude and time period of the motion are :
[12 Jan 2019, II]
2 3 3 2
(1) 10cm, s (2) 10cm, s (3) 5cm, s (4) 5cm, s
3 2 2 3
SOLUTION(1) Given: y  5 sin  3t   3 cos  3t  
 
 y  10sin  3t  
 3
 Amplitude = 10 cm
2 2 2
Time period, T   s
 3 3
43. A simple pendulum of length 1m is oscillating with an angular frequency 10 rad/
s. The support of the pendulum starts oscillating up and down with a small angular
frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular
frequency of the pendulum is best given by: [11 Jan 2019, II]
(1) 103 rad / s (2) 1 rad / s
(3) 101 rad / s (4) 105 rad / s
g
SOLUTION (1) Angular frequency of pendulum 

 Relative change in angular frequency
 1 g
 [ as length rmains constant]
 2 g
g  2 A2s [ s  angular frequency of support and , A = amplitude ]
 1 2 A2s
 
 2 g
1 2 12 102
    103 rad/sec
2 10
44. The mass and the diameter of a planet are three times the respective values for
the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The
period of oscillation of the same pendulum on the planet would be:[11 Jan 2019, II]
3 2 3
(1) s (2) s (3) s (4) 2 3s
2 3 2
GM
SOLUTION4) Acceleration due to gravity g
R2
gp M p  Re  1
2
1
    3   
ge M e  R p   3 3
1 Tp ge
Also
T    3
g Te gp
 Tp  2 3 s
45. A particle executes simple harmonic motion with an amplitude of 5 cm. When
the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units
is equal to that of its acceleration. Then, its periodic time in seconds is:
[10 Jan 2019, II]
4 3 8 7
(1) (2)  (3) (4) 
3 8 3 3
SOLUTION (3) Velocity, v   A2  x 2 …(i)
Acceleration, a  2 x …(ii)

And according to question,
v a
  A2  x 2  2 x
 A2  x 2  2 x 2
 52  42  2 42  
3
3   4   
4
2  8
 T  2 /   
3/ 4 3
47. A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits
torsional oscillations; If two masses each of ‘m’ are attached at distance ‘L/2’ from its
centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio
m/M is close to :
[9 Jan 2019, II]
(1) 0.77 (2) 0.57 (3) 0.37 (4) 0.17
SOLUTION (3)
1 C
f1  …(i)
2 1
1 3C

2 ML2
1 C
f2 
2 M M 
L2    …(ii)
 3 2 
As frequncey reduces by 80%
f2
 f 2  0.8 f1   0.8 …(iii)
f1
Solving equations (i), (ii) & (iii)
m
Ratio  0.37
M
As frequency reduces by 80%
f2
 f 2  0.8 f1   0.8 …(iii)
f1
Solving equation (i), (ii) & (iii)
m
Ratio  0.37
M
48. A silver atom in a solid oscillates in simple harmonic motion in some direction
with a frequency of 1012/sec. What is the force constant of the bonds connecting one
atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 ×1023 gm
mole–1)
(1) 6.4 N/m (2) 7.1 N/m
(3) 2.2 N/m (4) 5.5 N/m
SOLUTION(2) As we know, frequency in SHM
1 k
f   1012
2 m
Where m = mass of one atom
108
 103 kg
Mass of one atom of silver,
 6.02 10 
23
1 k
3
 6.02  1023  1012
2 108  10
Solving we get, spring constant,
K = 7.1 N/m
49. A particle executes simple harmonic motion and is located at x = a, b and c at

times t0, 2t0 and 3t0 respectively. The frequency of the oscillation is
1 1  a  b  1 1  a  b 
(1) 2t cos  2c  (2) 2t cos  3c 
0   0  
1 1  2a  3c  1 1  a  c 
(3) 2t cos  b  (4) 2t cos  2b 
0   0  
SOLUTION (4) Using y  A sin t
a  A sin 2t0
c  A sin 3t0
a  c  A sin t0  A sin 3t0 
 2 A sin 2t0 cos t0

ac
 2 cos t0
b
1 ac 1 ac
 cos 1   f  cos 1  
t0  2b  2t0  2b 
50. In an experiment to determine the period of a simple pendulum of length 1 m, it
is attached to different spherical bobs of radii r1 and r2. The two spherical bobs have
uniform mass distribution. If the relative difference in the periods, is found to be 5 ×
10 –4 s, the difference in radii, r1  r2 is best given by:

(1) 1 cm (2) 0.1 cm (3) 0.5 cm (4) 0.01 cm
SOLUTION (2) As we know, Time-perid of simple pendulum, T l
T 1 l
Differentiating both side, 
T 2 l
 Change in length l  r1  r2
1 r1  r2
5  104 
2 1
r1  r2  10  104
103 m  101 cm  0.1 cm

51. A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a
frictionless horizontal table. Two springs identical to the original spring are attached
in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of
the 8 kg block is : [Online April 8, 2017]
1 1 1
(1) Hz (2) Hz (3) Hz (4) 2 Hz
4 2 2 2
1 k
SOLUTION (3) Frequency of spring f   1Hz
2 m
k
 42 
m
If block of mass m = 1 kg is attached then, k  42
Now, identical springs are attached in parallel with mass m = 8 kg Hence,
keq  2k
1 k2 1
F  Hz
2 g 2
52. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a
day if the temperature is 20° C. The temperature at which the clock will show correct
time, and the co-efficient of linear expansion (  ) of the metal of the pendulum shaft
are respectively : [2016]
(1) 30°C;  = 1.85 × 10–3/°C
(2) 55°C; = 1.85 × 10–2/°C
(3) 25°C; = 1.85 × 10–5/°C
(4) 60°C; = 1.85 × 10–4/°C
1
SOLUTION 3) Time lost/gained per day    86400 second
2
1
12    40     86400 ….(i)
2
1
4     20   86400 ….(ii)
2
40  
On dividing we get, 3
  20
3  60  40  
4  100    250 C
53. In an engine the piston undergoes vertical simple harmonic motion with
amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor
speed is slowly increased. The frequency of the piston at which the washer no longer
stays in contact with the piston, is close to :
(1) 0.7 Hz (2) 1.9 Hz (3) 1.2 Hz (4) 0.1 Hz
SOLUTION (2) Washer contact with piston N 0
Given Amplitude A = 7 cm = 0.07 m.
amax  g  2 A
The frequency of piston
 g 1 1000 1
f     1.9 Hz
2 A 2 7 2
55. A particle moves with simple harmonic motion in a straight line. In first ts, after
starting from rest it travels a distance a, and in next t s it travels 2a, in same direction,
then: [2014]
(1) amplitude of motion is 3a
(2) time period of oscillations is 8
(3) amplitude of motion is 4a
(4) time period of oscillations is 6
SOLUTION
55. (d) In simple harmonic motion, starting from rest,
At t = 0 , x =A
x = A cos t …(i)
When t  , x  A  a
When t  2, x  A  3a
From equation (i)
A  a  A cos  ….(ii)
A  3a  A cos 2 ….(iii)
As cos 2  2 cos 2   1 …(iv)

From equation (ii), (iii), and (iv)
2
A  3a  Aa 
 2  1
A  A 
A  3a 2 A2  2a 2  4 Aa  A2
 
A A2
 A2  3aA  A2  2a 2  4 Aa
 2a 2  aA
 A  2a
a 1
 
A 2
Now, A  a  A cos 
A a
 cos  
A
1 2 
 cos   or 
2 T 3
 T  6
56. In an experiment for determining the gravitational acceleration g of a place with the help
of a simple pendulum, the measured time period square is plotted against the string length of
the pendulum in the figure. [Online April 19, 2014]
What is the value of g at the place?

(1) 9.81 m/s2 (2) 9.87 m/s2
(3) 9.91 m/s2 (4) 10.0 m/s2
SOLUTION (2) From graph it is clear that when
L  1m, T 2  4 s 2
As we know,
L
T  2
g
4 2 L
g
T2
2 2
 22  1  22 
 4     
 7  4  7 
484
g   9.87 m / s 2
49
57. The amplitude of a simple pendulum, oscillating in air with a small spherical
bob, decreases from 10 cm to 8 cm in 40 seconds. Assuming that Stokes law is valid,
and ratio of the coefficient of viscosity of air to that of carbon dioxide is 1.3. The time
in which amplitude of this pendulum will reduce from 10 cm to 5 cm in carbon dioxide
will be close to (In 5 = 1.601, In 2 = 0.693). [Online April 9, 2014]
(1) 231 s (2) 208 s (3) 161 s (4) 142 s
SOLUTION (4) As we know,
x  x0 e bt /2 m
From question,
40b
 ….(i)
8  10e 2m
bt
Similarly,  …(ii)
5  10e 2m
Solving equation (i) and (ii) we get

t  142 s
58. Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as
shown in the figure. The smaller mass executes simple harmonic motion of angular
frequency 25 rad/s, and amplitude 1.6 cm while the bigger mass remains stationary
on the ground. The maximum force exerted by the system on the floor is (take g = 10
ms–2) [Online April 9, 2014]
(1) 20 N (2) 10 N (3) 60 N (4) 40 N

SOLUTION (c) Mass of bigger body M  4kg
Mass of smaller body m = 1 kg
Smaller mass (m = 1kg) executes S.H.M of angular frequency   25 rad/s
Amplitude x = 1.6 cm=1.6 x 10 -2
As we know,
m
T  2
K
2 m
or,  2
 K
1 1
or,  [ m  1kg ;   25 rad/s]
25 K
or K =625 Nm-1
The maximum force exerted by the system on the floor  Mg  Kx  mg
 4 10  625  1.6  102  1 10

 40  10  10
= 60 N
Topic-4: Damped, Forced Oscillations and Resonance
76. A damped harmonic oscillator has a frequency of 5 oscillations per second.
The amplitude drops to half its value for every 10 oscillations. The time it will take to
drop to of the original amplitude is close to : [8 April 2019, II]
(1) 50 s (2) 100s (3) 20 s (4) 10 s
SOLUTION:(3) Time of half the amplitude is = 2s
Using, A  A0 e kt
A0
 Ae e  kx 2
2
A0
And  Ae e  kx 2
40
t  20 s
Divining (i) and (ii) and soving, we get
77. The displacement of a damped harmonic oscillator is given by . Here t is in

seconds. The time taken for its amplitude of vibration to drop to half of its initial value
is close to: [9 Jan 2019, II]
(1) 4s (2) 7s (3) 13s (4) 27s
SOLUTION:(2) Amplitude of vibration at time t = 0 is given by
A  Ae e 0.10  1 A0  a0
A0
Also at A  t  t, if A
0
1
  e 0.lt
2
t  10 ln 2  7s
79. The angular frequency of the damped oscillator is given by, where k is the
spring constant, m is the mass of the oscillator and r is the damping constant. If the
ratio is 8%, the change in time period compared to the undamped oscillator is
approximately as follows:
(1) increases by 1% (2) increases by 8%
(3) decreases by 1% (4) decreases by 8%
SOLUTION:(2) The change in time period compared to the undamped oscillator increases by 8%.
80. The amplitude of a damped oscillator decreases to 0.9 times its original
magnitude in 5s. In another 10s it will decrease to a times its original magnitude,
where a equals [2013]
(1) 0.7 (2) 0.81 (3) 0.729 (4) 0.6
SOLUTION:
bt
(3)  A  A0 e m
2
(where, A0 = maximum amplitude)

According to the quations, after 5 second,
b 5 
0.9 A0  Ae 2m ….(i)
After 10 more second,
b15 

A  A0e 2m …(ii)
From eqation (i) and (ii)
A  0.729 A0
  0.729
82. Bob of a simple pendulum of length l is made of iron. The pendulum is

oscillating over a horizontal coil carrying direct current. If the time period of the
pendulum is T then :
l
(1) T  2 and damping is smaller than in air alone.
g
l
(2) T  2 and damping is larger than in air alone.
g
l
(3) T  2 and damping is smaller than in air alone.
g
l
(4) T  2 and damping is larger than in air alone.
g
SOLUTION:4) When the pendulum is oscillating over a current carrying coil, and when the direction of oscillating
pendulum bob is opposite to the direction of current. Its instantaneous acceleration increases.

Hence time period T  2
g
and damping is larger than in air alone due energy dissipatio

83. In forced oscillation of a particle the amplitude is maximum for a frequency of
the force while the energy is maximum for a frequency of the force; then
[2004]
(1) 1  2 when damping is small and when damping is large 1  2
(2) 1  2 (3) 1  2 (4) 1  2

SOLUTION:3) As energy  (Amplitude)2, the maximum for both of them occurs at the same frequency and this is only
possible in case of resonance
In resonance state 1  2
84. A particle of mass m is attached to a spring (of spring constant k) and has a
natural angular frequency . An external force F(t) proportional to coswt is applied to
the oscillator. The displacement of the oscillator will be proportional to
[2004]
1 1 m m
(1) m  02  2
 (2) m  02  2
 (3) 2  2
0
(4)  02  2 
SOLUTION: :(2) Equation of displacement in forced oscillation is given by
F0
y
 
2
m 02  2
F0


m 02  2 
Here damping effect is considered to be zero
Gravitation
Basic Forces in Nature :
Basic forces are classified into four categories
a) Gravitational Force
b) Electromagnetic Force
c) Strong nuclear Force
d) Weak nuclear Force
Gravitational Force:
This force between any two massive particles.
It is always attractive force.
It is a conservative force.
It is independent of medium present between the masses.
It can provide radial acceleration.
It is communicated through a particle called as Gravitation.
Electro Magnetic Force :
This force exists between any two charged particles.
This force is either attractive or repulsive.
It is communicated through Photons.
Strong Nuclear Force :
This force may act between a pair of nucleons in the nucleus.
It is charge independent.
It is spin dependent.
It is communicated through  mesons.
Weak Nuclear Force :
They are responsible for radioactive decay like  -decay..
They acts between all leptons, positrons,  -mesons, neutrinos and Hadrons etc.
It is communicated through weak bosons.
Relative strengths of basic forces between protons :
R e la tiv e
B a sic f o r c e R a n g e
s tr e n g th
L o n g r a n g e
G r a v it a ti o n a l 1
( u p t o in f i n i ty )
S h o r t r a n g e 3 1
W e a k n u c le a r 1 0
( < < 1 fm )
L o n g r a n g e 3 6
E le c tr o m a g n e tic 1 0
( u p t o in f i n i ty )
S h o r t r a n g e 3 8
S tr o n g n u c le a r 1 0
(1 f m )
Newton’s law of Gravitation
Newton’s law of gravitation states that every body in this universe attracts every other body with a force,
which is directly proportional to the product of their masses and inversely proportional to the square of the
distance between their centres. The direction of the force is along the line joining the particles.
A  
B
F F
12 21
m1 m2
Thus the magnitude of the gravitational force F that two particles of masses m 1 and m 2 are
separated by a distance r exert on each other is given by
m1 m 2
F
r2
m 1m 2
or FG
r2
Vector form :
According to Newton’s law of gravitation
 Gm 1m 2  Gm 1m 2 
 Gm 1m 2  r21  r21
F 12  r̂21 r 3 
r2 | r21 | 3
Here negative sign indicates that the direction of F12 is opposite to that of r̂21 .
 Gm 1m 2   Gm m 
  Gm 1m 2  r12   1 2 r12 Gm 1m 2
Similarly F 21  r̂12 r 3
| r12 | 3
 r̂21 [rˆ12  rˆ21 ]
r2 r2
r̂12 = unit vector from A to B

r̂21 = unit vector from B to A,

F12 = gravitational force exerted on body A by body B

F 21 = gravitational force exerted on body B by body A
 
 It is clear that F12 = – F 21 . Which is Newton’s third law of motion.
Universal gravitational constant:
m 1m 2
FG
r2
Here G is constant of proportionality which is called ‘Universal gravitational constant’.
If m 1  m 2
r 1
then G  F
i.e. universal gravitational constant is equal to the force of attraction between two bodies each of
‘unit mass whose centres are placed unit distance apart.
 The value of G in the laboratory was first determined by Cavendish using the torsional balance.
 The value of G is 6.67×10–11 N–m2 kg–2
 Dimensional formula [M 1 L3 T 2 ] .
M
m l/2 
A B
r
M
From Cavendish experiment the value of universal gravitational constant (G) can be calculatedby
k r 2
G
Mml
M-Mass of heavier sphere
m-Mass of lighter sphere
k-Torsion constant ;  -Angle of twist
 The value of G does not depend upon the nature and size of the bodies.
 It also does not depend upon the nature of the medium between the two bodies.
 As G is very small, hence gravitational forces are very small, unless one (or both) of the mass is
huge.
Properties of Gravitational Force
 It is always attractive in nature while electric and magnetic force can be attractive or repulsive.
 It is independent of the medium between the particles while electric and magnetic force depend on
the nature of the medium between the particles.
 It holds good over a wide range of distances. It is found true for interplanetary to inter atomic
distances.
 It is a central force i.e. acts along the line joining the centres of two interacting bodies.
 It is a two-body interaction i.e. gravitational force between two particles is independent of the
presence or absence of other particles; so the principle of superposition is valid i.e. force on
a particle due to number of particles is the resultant of forces due to individual particles
   
i.e. F  F 1  F 2  F 3  ........
While nuclear force is many body interaction
 It is the weakest force in nature : As Fnuclear > F electromagnetic > F gravitational .
 The ratio of gravitational force to electrostatic force between two electrons is of the order of 10 43 .
 It is a conservative force i.e. work done by it is path independent or work done in moving a
particle round a closed path under the action of gravitational force is zero.
 It is an action reaction pair i.e. the force with which one body (say earth) attracts the second body (say
moon) is equal to the force with which moon attracts the earth. This is in accordance with
Newton’s third law of motion.
NOTE :
The law of gravitation is stated for two point masses, therefore for any two arbitrary finite size
bodies, as shown in the figure, It can not be applied as there is not unique value for the separation.
But if the two bodies are uniform spheres then the separation r may be taken as the distance between their
centres because a sphere of uniform mass behave as a point mass for any point lying outside it.
m1 m2 m1 m2
r
r=?
PROBLEMS
1. Two identical spheres are placed in contact with each other. The force of gravitation between the spheres
will be proportional to (R = radius of each sphere)
(a) R (b) R2 (c) R4 (d) None of these
SOLUTION :
2
4 
G   R 3   6
From law of gravitation MM  3   F  R  R4
FG 
R2 (2 R )2 R2
2. The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the
earth exerts on the moon, the gravitational force the moon exerts on earth
(a) Is the same (b) Is smaller (c)Is greater (d)Varies with its phase
SOLUTION :
Gm m m e
Force between earth and moon F 
r2
This amount of force, both earth and moon will exert on each other
i.e. they exert same force on each other.
3. If the distance between two masses is doubled, the gravitational attraction between them
(a) Is doubled (b)Becomes four times
(c) Is reduced to half (d)Is reduced to a quarter
SOLUTION :
1
F .
r2
F
If r becomes double then F reduces to
4
4. The gravitational force between two stones of mass 1 kg each separated by a distance of 1 metre in
vacuum is
(a) Zero (b) 6 .675  10 5 newton (c) 6 .675  10 11 newton (d) 6 .675  10 8 newton
SOLUTION :
m 1m 2 11
FG 2
 6 . 675   10 11  6 . 675  10 11 N
r 12
5. The distance of the centres of moon and earth is D. The mass of earth is 81 times the mass of the moon.
At what distance from the centre of the earth, the gravitational force will be zero
D 2D 4D 9D
(a) 2
(b) 3
(c) 3
(d) 10
SOLUTION :
d
P
m1
x m2
Force will be zero at the point of zero intensity
m1
x d
m1  m 2
81 M 9
 D D.
81 M  M 10
6. Mass M is divided into two parts xM and (1  x) M . For a given separation, the value of x for which
the gravitational attraction between the two pieces becomes maximum is
1 3
(a) 2 (b) 5 (c)1 (d)2
SOLUTION :
F  xm  (1  x )m  xm 2 (1  x )
dF
For maximum force 0
dx
dF
  m 2  2 xm 2  0  x  1/2
dx
7. If two particles each of mass ‘m’ are placed at the two vertices of an equilateral triangle of side
‘a’, then the resultant gravitational force on mass m placed at the third vertex is
SOLUTION :
C
300
FB
30 0
FA
FR
600 600
A B
FR  FA2  FB2  2 FA FB cos 600
 3F  FA  FB  F 
 Gm 2 
FR  3  2 
 a 
8. Three spherical balls of masses 1kg, 2kg and 3kg are placed at the corners of an equilateral
triangle of side 1m. Find the magnitude of the gravitational force exerted by 2 kg and 3 kg
masses on 1 kg mass.
SOLUTION :
If F1 is the force of attraction between 1kg, 2kg masses,
1 2
F1  G   F1  2G
then,
1
2
A1kg
F1 600 F2
B C
2kg 3kg
1 3
If F2 is the force of attraction between 1kg, 3kg masses, then, F2  G   F2  3G
1
2
The angle between the forces F1 and F2 is 600. If ‘FR’ is the resultant of these two forces then
FR  F12  F22  2 F1F2 cos
 2G    3G   2  2G  3G  cos 600
2 2
 FR 
 FR  19G
9. An infinite number of particle each of mass m are placed on the positive X-axis at
1m,2m,4m,8m,.... from the origin. Find the magnitude of the resultant gravitational force on
mass ‘m’ kept at the origin.
SOLUTION :
y
m m m m m
O 1 x
2 4 8
The resultant gravitational force

Gm 2 Gm 2 Gm 2
F    ....
1 4 16
 1 1 
 Gm 2  1    ... 
 4 16 
 
 1  4  a 
 Gm 2    Gm 2  S  
1 1 r 
1  3 
 4
10. If four identical particles each of mass m, are kept at the four vertices of a square of side
length a, the gravitational force of attraction on any one of the particle is
SOLUTION :
m a m
FR
a a
F1
F
m a F m
2Gm 2 Gm 2
FR  2 F  F '  
a2 2a 2
Gm2  1
2 
2  FR 
a  2
along the diagonal towards the opposite corner.
11. A particle of mass m is situated at a distance d from one end of a rod of mass M and length L
as shown in fig. Find the magnitude of the gravitational force between them.
x dm
m
d L
SOLUTION :
Consider an element of mass ‘dm’ and length ‘dx’ at a distance ‘x’ from the point mass.
M
Mass of the element dm  dx .
L
Gravitational force on ‘m’ due to this element is
M  M 
Gm  dx   d  L  Gm   dx
 L  ; F  L 
dF 
x 2 
d
x2
 d  L  d  L
GmM GmM  x 1 

2
F x dx   
L d L  1 
d  L
GmM  1  GmM 1 1 
F    
L  x  a L  d  d  L  
GmM d  L  d  GmM
F  
L   d  L  d  d  d  L 
12. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational
attraction. The speed of each particle is[CBSE PMT 1995; RPMT 2003]
1 1 Gm 1 Gm 4 Gm
(a) v  2R Gm
(b) v  2R
(c) v  2 R
(d) v  R
SOLUTION :
m
R
O R
m
Centripetal force provided by the gravitational force of attraction between two particles
mv 2 Gm  m 1 Gm
i.e.  v 
R (2 R )2 2 R
13. Four particles, each of mass M and equidistant from each other, move along a circle of radius
R under the action of their mutual gravitational attraction. The speed of each particle is
(2014A)
SOLUTION :
Let a be the distance between two particles.
M a M
a a
FR
F
M a F M
The resultant gravitational force on any one of the particle is given by

GM 2  1
FR  2 
2  .
a  2
Which provides necessary centripetal force for motion of mass M in circle, so
 1   GM 2  MV 2
 2    a
 2   a2 
2
 2 2  1  GM  2 2  1  GM
V2  
 2 2  a  2 2  2 R
   
V 
1 GM
2 R

1 2 2 
14. Two particles of masses 1Kg and 2Kg are placed at a distance of 50cm. Find the initial
acceleration of the first particle due to gravitational force.
SOLUTION :
Gravitational force between two particles is
Gm1m2 6.67  1011  1  2
F   5.3  1010 N
 
2
r2 0.5
The acceleration of 1Kg particle is
F 5.3  1010
a1    5.3  10 10 ms 2 towards the 2Kg mass
m1 1
15. If four different masses m1,m2,m3 and m4 are placed at the four corners of a square of side
‘a’ the resultant gravitational force on mass m kept at the centre is
SOLUTION :
2Gm
The force on m due to m 1 and m3 is F1   m1  m3 
a2
along the diagonal towards m1 if m12  m3 
m3 a m2
F2
a FR a
m 
F1
m4 a m1
2Gm
The force on m due to m2 and m4 is F2   m2  m4 
a2
along the diagonal towards m2 if m2  m4  .
The resultant force is F12  F22  F

2Gm
 m1  m3    m2  m4 
2 2
F 2
a
and the resultant force makes an angle  with F1
1  F2 
where,   tan   .
 F1 
16. A thin rod of mass M and length L is bent into semicircle as shown in figure. What is
gravitational force on a particle with mass m at the centre of curvature?
SOLUTION :
Consider an element of rod of length dl as shown in figure
treat it as a small particle of mass (M/L) dl
situated at a distance R from P.
Then gravitational force due to the element on the particle will be
Gm  M / L  Rd 
dF  along OP
R2
[as dl=Rd  ]
Y
dl dF
P
d
m 
X
So the components of this force along x and y axes will be
GmM cos d
dFx  dF cos  
LR
GmM sin  d
dFy  dF sin  
LR
So that
GmM  GmM
sin  0  0

Fx 
LR 
0
cos  d 
LR
GmM  GmM
  cos 0

Fy 
LR  0
sin  d 
LR
2 GmM  L
  asR   
L2  
2 GmM
F  Fx2  Fy2  Fy  (as Fx=zero)
L2
17. Mass M is split into two parts m and (M-m), which are then separated by a certain distance.
What is the ratio of (m/M) which maximises the gravitational force between the parts?
SOLUTION :
If r is the distance between m and (M-m), the gravitational force between them will be
m M  m
G
2 
F G 2
mM  m 2 

r r
For F to be maximum dF/dm=0 as M & r are constants.
d G 
2 
  mM  m 2    0;  M  2m  0
dm  r 
m 1
  .
M 2
So the force will be maximum when the parts are equal.
18. Imagine a light planet revolving around a very massive star in a circular orbit of radius r with
a period of revolution T. On what power of ‘r’ will the square of time period depend on the
gravitational force of attraction between the planet and the star is proportional to r-5/2?
SOLUTION :
mV 2 K K
The gravitational force provides necessary centripetal force  5/ 2  V 2  3/ 2
r r mr
2 r mr 3/ 2
But T   2 r ; T 2  r 7 / 2
V K
19. In a double star system, two stars of masses m1 and m2 separated by a distance ‘x’ rotates
about their centre of mass. Find the common angular velocity and Time period of revolution.
SOLUTION :
x
m1 r1 c.m r2 m2
The gravitational force between the masses provides the necessary centripetal force.
Gm1m2
i.e. 2
 m1r1 2 .....1
x
The distance of centre of mass from m1 is
m2 x
r1  ....... 2 
m1  m2
Gm1m2 mm x
From (1) and (2)  1 2 2
x 2
m1  m2
G  m1  m2  G  m1  m2 
2  2
 
x x3
2 x2
T ; T  2
 G  m1  m2 
20. In Cavendish’s experiment, let each small mass be 20 kg and each large mass be 5 kg. The
rod connecting the small masses is 50 cm long, while the small and the large spheres are
separated by 10.0 cm. The torsion constant is 4.8x10 -8 kgm2s-2 and the resulting angular
deflection is 0.40. Calculate the value of universal gravitational constant G from this data.
SOLUTION :
Here, m=20kg=0.02kg, M=5 kg
r=10cm=0.1m, l=50cm=0.5m
  0.40   0.40  2 / 360 0   0.007 rad ,
k=4.8x10-8kgm2s-2
k r 2
Thus, from G 
Mml
G
 4.8  10   0.007  0.1
8 2
5  0.02  0.5
=6.72x10-11 Nm2kg-2.
21. The mean orbital radius of the Earth around the Sun is 1.5x108 km. Estimate the mass of the
Sun.
SOLUTION :
As the centripetal force is provided by the gravitational pull of the Sun on the Earth
GM s M e 4 2 4 2 r 3
 M e r 2
 M e r  or  M S 
r2 T2 GT 2
given, r=1.5x108 km=1.5x1011m;
T=365 days=365x24x60x60s
4   22 / 7   1.5  1011 
2 3
MS   2  1030 kg
 6.67  10    365  24  60  60 
11 2
22. The gravitational force acting on a particle, due to a solid sphere of uniform density and radius
R, at a distance of 3R from the centre of the sphere is F1. A spherical hole of radius (R/2) is
now made in the sphere as shown in the figure. The sphere with hole now exerts a force F2 on
the same particle. Ratio F1 to F2 is (2013E)
R
2 m
R
3R
SOLUTION :
Let mass of the removed sphere = M.
Then mass of the original sphere = 8M (since mass  R 3 )
8GMm
F1 
9R2
8GMm GMm
F2   2
9R2  5R 
 
 2 
F1 50
Therefore, F  41  on simplifying 
2
Acceleration Due to Gravity
The acceleration produced in the motion of a body under the effect of gravity is called acceleration due
to gravity, it is denoted by g.
The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of
gravitational pull must accelerate.
mg R
Earth
Fig. 8.4
Consider a body of mass m is lying on the surface of earth then gravitational force on the body is
given by
GMm
F  …(i)
R2
Where M = mass of the earth
R = radius of the earth.
If g is the acceleration due to gravity,
then the force on the body due to earth is given by Force = mass  acceleration
F = mg …(ii)
GMm
From (i) and (ii) we have mg 
R2
GM
 g …(iii)
R2
G 4 
 g  R 3  
R2  3 
4
[As mass (M) = volume ( 3 R 3 ) × density ()]
4
 g
3
GR …(iv)
GM 4
 From the expression g  2  GR
3
it is clear that its value depends upon the mass radius and
R
density of planet and it is independent of mass, shape and density of the body placed on the
surface of the planet. i.e. a given planet (reference body) produces same acceleration in a
light as well as heavy body.
 The greater the value of (M / R 2 ) or R, greater will be value of g for that planet.
 Acceleration due to gravity is a vector quantity and its direction is always towards the centre of
the planet.
Dimension [g] = [LT–2]
 it’s average value is taken to be 9.8 m/s2 or 981 cm/sec2 or 32 feet/sec2, on the surface of the
earth at mean sea level.
 The value of acceleration due to gravity vary due to the following factors :
(a) Shape of the earth,
(b) Height above the earth surface,
(c) Depth below the earth surface
(d) Axial rotation of the earth.
Variation in g Due to Shape of Earth:
gp
Rp
ge
Re
Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial
GM
radius is about 21 km longer than polar radius, from g
R2
GM
At equator ge  …(i)
R e2
GM
At poles gp  …(ii)
R p2
ge R p2
From (i) and (ii)  2
gp Re
Since R equator  R pole
 g pole  g equator and g p  g e  0 .018 ms 2
Therefore the weight of body increases as it is taken from equator to the pole.
Variation in g With Height
g
h
r g
R
O
GM
Acceleration due to gravity at the surface of the earth g …(i)
R2
GM
Acceleration due to gravity at height h from the surface of the earth g'  …(ii)
(R  h)2
2
 R 
From (i) and (ii) g'  g  …(iii)
Rh
R2
=g …(iv)
r2
[As r = R + h]
1
 As we go above the surface of the earth, the value of g decreases because g  .
r2
 If r   then g  0 , i.e., at infinite distance from the earth, the value of g becomes zero.
) If h  R i.e., height is negligible in comparison to the radius then from equation (iii) we get
2
 R 
g  g 
R h
2
 h
 g1  
 R 
 2h 
 g 1  [As ]
 R  h  R
 If h  R then decrease in the value of g with height :

2 hg
Absolute decrease g  g  g  
R
g g  g  2 h
Fractional decrease g

g

R
g 2h
Percentage decrease g
 100 % 
R
 100 %
Variation in g With Depth:
g
d
P g
R
r
O
GM 4
Acceleration due to gravity at the surface of the earth g   GR …(i)
R2 3
4
Acceleration due to gravity at depth d from the surface of the earth g  G(R  d ) …(ii)
3
 d
From (i) and (ii) g   g 1  
 R 
 The value of g decreases on going below the surface of the earth.

From equation (ii) we get g  (R  d ) .
So it is clear that if d increase, the value of g decreases.
 At the centre of earth d  R
 g  0 ,
i.e., the acceleration due to gravity at the centre of earth becomes zero.
 Decrease in the value of g with depth
dg
Absolute decrease g  g  g  
R
g g  g  d
Fractional decrease g

g

R
g d
Percentage decrease g
 100 %   100 %
R
 The rate of decrease of gravity outside the earth ( if h  R ) is double to that of inside the earth.
Graphical representation of variation of ‘g’ with height and depth:
The variation of g with the distance r from the centre of the earth is shown below
i) Above the earth :
gR 2 gR 2 1
gh  : gh   R  h  r   gh 
 R  h
2
r2 r2
 g h versus r graph is a curve a shown.
g=9.8ms-2
due to gravity
acceleration
1
gr g
r2
inside
the earth
above the surface
x
i) Inside the earth :

g
gd  R  d 
R
g
gd   r  R  d  r    g d  r 
R
 g d versus r graph is a straight line passing through the origin as shown in fig.
NOTE:Lines joining the places on the earth having same values of g are called isograms.
Variation of ‘g’ with latitude :
Consider an object of mass m at latitude  of the earth due to rotation of a earth, the vale
of acceleration due to gravity g  at a given place is given by
g  g  r 2 cos 
where r 2 cos  is the component of centrifugal acceleration along the radius of the Earth.
pole FC cos 
r P
FC  mr 2
R a

Equator O
As the earth rotates, a body placed on its surface moves along the circular path and hence
experiences centrifugal force, due to it, the apparent weight of the body decreases.
Since the magnitude of centrifugal force varies with the latitude of the place, therefore the apparent
weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body.
If the body of mass m lying at point P, whose latitude is l, then due to rotation of earth its apparent
weight can be given by

m g   mg  Fc
m g   (mg )2  (Fc )2  2mg Fc cos(180   )
m g   (mg )2  (m  2 R cos  )2  2mg m  2 R cos  ( cos  )
[As Fc  m  2 r  m  2 R cos  ]
By solving we get g   g   2 R cos 2 
where r is the radius of the circle in which the object is revolving.

Here r  R cos 
 g   g   2 R cos 2 
where  is the angular velocity. R is radius of the earth and  is latitude of the place
 The latitude at a point on the surface of the earth is defined as the angle, which the line joining that
point to the centre of earth makes with equatorial plane. It is denoted by  .
 For the poles   90 o
for equator   0 o
Special cases :
At the poles   900
 g
  90 0  g   2 R  0
2
 cos90 0
 0
 g  g    maximum 
  90 0
At the equator   00
 g
 
 00
 g   2 R 1
2
 cos 0 0
 1
 g  g   2 R..... minimum 
0 
0
From equation (i) and (ii) g pole  g equator  R  2  0 .034 m / s 2
Here , R 2  0.034 ms 2 for the Earth.

The value of ‘g’ at poles does not depend on the speed of rotation of the earth, but at the equator ‘g’
decreases with the increase of speed of rotation of earth.
If earth suddenly stops its rotation, then the acceleration due to gravity at poles remains constant, and
acceleration due to gravity at equator increases by  2 R
Weightlessness:
Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases
due to rotation of earth. If  is the angular velocity of rotation of earth for which a body at the equator
will become weightless
g   g   2 R cos 2 
 0  g   2 R cos 2 0 o [As   0o for equator]
 g  2R  0
g
 
R
2 R
or time period of rotation of earth T  2
 g
Substituting the value of R  6400  10 3 m and g  10 m / s 2 we get

1 rad
  1 . 25  10  3
800 sec
T  5026 . 5 sec  1 . 40 hr.
1
 This time is about 17
times the present time period of earth. Therefore if earth starts rotating 17
times faster then all objects on equator will become weightless.
 If earth stops rotation about its own axis then at the equator the value of g increases by  2 R and
consequently the weight of body lying there increases by m  2 R .
 After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at
the poles and equator are related as
g p  g e  0 . 034  0 . 018 m / s 2
 g p  g e  0 . 052 m / s 2
Mass and Density of Earth
Newton’s law of gravitation can be used to estimate the mass and density of the earth.
GM
As we know g ,
R2
gR 2
so we have M
G
9 . 8  (6 . 4  10 6 )2
 M   5 . 98  10 24 kg  10 25 kg
6 . 67  10 11
4
and as we know g GR ,
3
3g
so we have 
4 GR
3  9 .8
   5478 . 4 kg / m 3
4  3 . 14  6 . 67  10 11  6 . 4  10 6
PROBLEMS
1. R is the radius of the earth and  is its angular velocity and gp is the value of g at the poles. The
‘effective value of g at the latitude   60  will be equal to
1 3 1
(a) gp  R 2 (b) gp  R 2 (c) g p  R 2 (d) gp  R 2
4 4 4
SOLUTION :
g  g p  R  2 cos 2 
= g p   2 R cos 2 60 
1
= gp  R 2
4
2. What is the time period of rotation of the earth around its axis so that the objects at the
equator becomes weightless?(g=9.8m/s2, Radius of earth = 6400km)
SOLUTION :’
When earth is rotating the apparent weight of a body at the equator is given by
Wapp  mg  mR 2
If bodies are weightless at the equator
0  mg  mR 2  g  R 2
g
 
R
2 R
Time period, T   2
 g
6.4 x106
T  2  5078s  82 minute 32s
9.8
1
3. The depth d at which the value of acceleration due to gravity becomes n
times the value at the
surface, is [R = radius of the earth]
R  n 1  R  n 
(a) (b) R  (c) (d) R 
n  n  n2  n 1 
SOLUTION :
 d g  d
g   g 1     g 1  
 R  n  R 
 n 1 
 d  R
 n 
4. The height at which the acceleration due to gravity becomes g/9 (where g is the acceleration
due to gravity on the surface of the earth) in terms of the radius of the earth (R) is(2009 A)
SOLUTION :
2
g  R  R 1
Given  g    
9  Rh Rh 3
3R  R  h  2 R  h
5. Find the percentage decrease in the weight of the body when taken for a depth of 32Km below
the surface of earth.
SOLUTION :
Weight of the body at depth d is
 d
mg '  mg  1  
 R
mg  mg '
% decrease in weight  mg
 100
d 32

 100   100  0.5%
R 6400
6. A man can jump 1.5 m on the Earth. Calculate the approximate height he might be able to jump
on a planet whose density is one-quarter that of the Earth and whose radius is one-third that
of the Earth.
SOLUTION :
We know that, it case of Earth,
g
GM
 G 
 4 / 3 R3    4  G  R 
 
R2 R2 3 
R 
Similarly, for the other planet whose radius and density is,
3 4
 4 G  R   
g'   
 3  3  4 
1  4 G  1 g
g'   R   g   12
12  3  12 g'
u2 1
h max   hmax  (here u is constant)
3g g
h' g
  12  h '  12h  12  1.5  18m
h g'
7. If the change in the value of ‘g’ at a height h above the surface of the earth is the same as at a depth x
below it, then (both x and h being much smaller than the radius of the earth)
h
(a) x h (b) x  2h (c) x  2
(d) x  h 2
SOLUTION :
 2h 
The value of g at the height h from the surface of earth g   g  1  R 
 
 x
The value of g at depth x below the surface of earth g   g  1  R 
 
 2h   x
These two are given equal, hence  1  R    1  R 
   
On solving, we get x  2 h
8. The time period of a simple pendulum on a freely moving artificial satellite is

(a) Zero (b) 2 sec (c) 3 sec (d) Infinite
SOLUTION :
l
Time period of simple pendulum T  2
g'
In artificial satellite g '  0
 T = infinite.
9. A body weighs 700 gm wt on the surface of the earth. How much will it weigh on the surface of a
1
planet whose mass is 7
and radius is half that of the earth
(a) 200 gm wt (b) 400 gm wt (c) 50 gm wt (d) 300 gm wt
SOLUTION :
GM
We know that g 
R2
GM / 7 4g 4
On the planet g p    g
R2 /4 7 7
4
Hence weight on the planet  700   400 gm wt
7
10. A spherical planet far out in space has a mass M 0 and diameter D0 . A particle of mass m falling
freely near the surface of this planet will experience an acceleration due to gravity which is equal to
(a) GM 0 / D02 (b) 4 mGM 0 / D02 (c) 4 GM 0 / D02 (d) GmM 0 / D02
SOLUTION :
GM GM 0 4 GM 0
g 2
 2

R (D0 /2) D 02
11. The mass and diameter of a planet have twice the value of the corresponding parameters of earth.
Acceleration due to gravity on the surface of the planet is
(a) 9 . 8 m / sec 2 (b) 4 . 9 m / sec 2 (c) 980 m / sec 2 (d) 19 . 6 m / sec 2
SOLUTION :
2 2
g' M '  R   2M   R  1
      
g M  R'   M  2 R  2
g 9.8
 g    4 . 9 m /s 2
2 2
12. If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density
of the earth is
(a) 4G / 3 gR (b) 3R / 4 gG (c) 3 g / 4RG (d) RG / 12 G
SOLUTION :
GM
g
R2
4
M  R 3  
3
4 R 3  G  3 g.
g  
3 R2 4 RG
13. The centres of a ring of mass m and a sphere of mass M of equal radius R, are at a distance
8 R apart as shown in fig. The force of attraction between the ring and the sphere is
m
R
8R
2 2 GmM GmM GmM 2 GmM

1) 2) 3) 4)
27 R 2 8R 2 9R2 9 9R2
SOLUTION :
GMdm
S dF  ; F   dF cos
3R 2
14. The radii of two planets are respectively R1 and R2 and their densities are respectively 1 and 2 .
The ratio of the accelerations due to gravity at their surfaces is
1 2
(a) g1 : g2  : (b) g1 : g 2  R1 R 2 : 1  2 (c) g1 : g 2  R1  2 : R2 1 (d) g1 : g 2  R1 1 : R2  2
R12 R 22
SOLUTION :
4
g GR
3
g1 R1  1
 g  R  ’
2 2 2
15. The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon.
The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth is
[MP PMT 1994]
(a) 0.15 (b) 0.04 (c) 1 (d) 6
SOLUTION :
GM
g (Given M e  81 M m , R e  3 . 5 R m )
R2
Substituting the above values,
gm
 0 . 15
ge
16. The magnitudes of the gravitational field at distance r1 and r2 from the centre of a uniform
sphere of radius R and mass M and E1 and E2 respectively. Then :
E1 r1 E1 r22
1) E  r if r1<R and r2<R 2)  if r1<R and r2<R
2 2 E2 r12
E1 r13 E1 r12
3)  if r1<R and r2<R 4)  if r1<R and r2<R
E2 r32 E2 r22
SOLUTION :
If r  R ,
GM
then E   r   E r
R3
E1 r1
 
E2 r2 if r1  R and r2  R
If r  R , then
GM 1
E 2
 E 2
r r
E1 r22
  if r1  R and r2  R
E2 r12
17. The value of g on the earth’s surface is 980 cm / sec 2 . Its value at a height of 64 km from the earth’ss
surface is(Radius of the earth R = 6400 kilometers)
(a) 960 . 40 cm / sec 2 (b) 984 . 90 cm / sec 2 (c) 982 . 45 cm / sec 2 (d) 977 . 55 cm / sec 2
SOLUTION :
2 2
g'  R   6400 
     g '  960 . 40 cm /s 2
g R h  6400  64 
18. The moon’s radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the
acceleration due to gravity on the surface of the earth, that on the surface of the moon is
(a) g/4 (b) g/5 (c) g/6 (d) g/8
SOLUTION :
GM
Using g  we get gm  g / 5
R2
19. How much above the surface of earth does the acceleration due to gravity reduce by 36% of its
value on the surface of earth.
SOLUTION :
Since g reduces by 36%, the value of g there is 100-36=64%.
64
It means, g '  g.
100
If h is the height of location above the surface of earth, then,
R2 64 R2
g'  g  gg
 R  h  R  h
2 2
100
8 R R 6.4  106
  h   1.6  106 m
10 R  h 4 4
20. Weightof 1kg becomes1/6on moon. If radius of moon is 1 .768  10 6 m ,then the mass of moon will be
(a) 1 . 99  10 30 kg (b) 7 . 56  10 22 kg (c) 5 . 98  10 24 kg (d) 7 . 65  10 22 kg
SOLUTION :
GM m
gm 
R m2
ge 9 .8
gm   m /s 2
6 6
 1 . 63 m /s 2
Substituting R m  1 .768  10 6 m ,
g m  1 . 63 m /s 2
and G  6 .67  10 11 N -m 2 /kg 2

We get M m  7 . 65  10 22 kg
21. Let g be the acceleration due to gravity at earth’s surface and K be the rotational kinetic energy of the
earth. Suppose the earth’s radius decreases by 2% keeping all other quantities same, then
(a) g decreases by 2% and K decreases by 4% (b)g decreases by 4% and K increases by 2%
(c) g increases by 4% and K increases by 4% (d)g decreases by 4% and K increases by 4%
SOLUTION :
GM L2
g 2 and K
R 2I
1
If mass of the earth and its angular momentum remains constant then g 
R2
1
K
R2
i.e. if radius of earth decreases by 2% then g and K both increases by 4%.
22. What should be the velocity of earth due to rotation about its own axis so that the weight at equator
become 3/5 of initial value. Radius of earth on equator is 6400 km
(a) 7 .4  10 4 rad / sec (b) 6 .7  10 4 rad / sec (c) 7 .8  10 4 rad / sec (d) 8 .7  10 4 rad / sec
SOLUTION :
3
Weight of the body at equator = of initial weight
5
3
 g'  g (because mass remains constant)
5
g '  g   2 R cos 2 
3
 g  g   2 R cos 2 (0 )
5
2g
  
2
5R
2g 2  10
  
5R 5  6400  10 3
4 rad
= 7 .8  10
sec
23. Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration
due to gravity on the surface of the earth. If R e is the maximum range of a projectile on the earth’ss
surface, what is the maximum range on the surface of the moon for the same velocity of projection
(a) 0 . 2 R e (b) 2 Re (c) 0 .5 R e (d) 5 Re
SOLUTION :
u 2 sin 2
Range of projectile R 
g
1
if u and  are constant then R  g
Rm g R 1 R
 e  m   R m  e  R  5 R
Re gm Re 0 .2 0 .2 m e
24. At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on
the surface (R = radius of earth)
(a) 2 R (b) R (c) 1.414 R (d) 0.414 R
SOLUTION :
2
 R  1 R
g'  g    
R h 2 Rh
 Rh 2R
 h  ( 2  1)R  0 .414 R
25. A star 2.5 times the mass of the sun is reduced to a size of 12km and rotates with a speed of 1.5
rps. Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of
the sun=2x1030 kg).
SOLUTION :
GM
Acceleration due to gravity, g 
R2
6.67  1011  2.5  2  1030
  2.3  1012 ms 2
12000 
2
Centrifugal acceleration  r 2  2  2 f 2  12000  2  1.5 2  1.1  10 6 ms 2

Since, g  r 2 , the body will remain stuck with the surface of star..
26. A man can jump to a height of 1.5 m on a planet A. What is the height he may be able to jump on another
planet whose density and radius are, respectively, one-quarter and one-third that of planet A
(a) 1.5 m (b) 15 m (c) 18 m (d) 28 m
SOLUTION :
u2 1
H  H 
2g g
B HA g
 H g
A B
gA
Now g B  as g  R
12
H
B A g
 H  g =12
A B
 H B  12  H A  12  1 . 5  18 m
27. Weight of a body of mass m decreases by 1% when it is raised to height h above the earth’s surface. If
the body is taken to a depth h in a mine, change in its weight is
(a) 2% decrease (b) 0.5% decrease (c) 1% increase (d) 0.5% increase
SOLUTION :
g 2h
For height  100 %   1 %;
g R
g d h 1
For depth  100 %    %  0 .5 %
g R R 2
28. At what depth below the surface of the earth, acceleration due to gravity g will be half its value 1600 km
above the surface of the earth
(a) 4 . 2  10 6 m (b) 3 . 19  10 6 m (c) 1 . 59  10 6 m (d) None of these
SOLUTION :
Radius of earth R = 6400 km
R
 h
4
Acceleration due to gravity at a height h
2
 
 
 R   g R
2
 16
g h  g   R   25 g
 R h R 
 4 
At depth ‘d’ value of acceleration due to gravity
1
gd  g h (According to problem)
2
1  16 
 g d  2  25  g
 
 d  1  16 
 g1     g
 R  2  25 
By solving we get d  4 .3  10 6 m
29. If earth is supposed to be a sphere of radius R, if g30 is value of acceleration due to gravity at
latitude of 30o and g at the equator, the value of g  g 30 o is
1 2 3 2 1 2
(a)  R (b)  R (c)  2 R (d)  R
4 4 2
SOLUTION :
Acceleration due to gravity at latitude  is given by
g   g  R  2 cos 2 
3
At 30 , g 30   g  R  cos 30  g  R 2
o 2 2 o
4
‘  g  g 30 
3 2
 R.
4
30. A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height
R above the earth’s surface, where R is the radius of the earth. The value of T2 / T1 is
(a) 1 (b) 2 (c) 4 (d) 2
SOLUTION :
If acceleration due to gravity is g at the surface of earth then at height R it value becomes
2
 R  g
g'  g  
 R  h 4
l l
T1  2 and T2  2
g g/4
T2
2
T1
31. Consider earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes
high up in a balloon. The value of g measured by
(a) A goes on decreasing and that by B goes on increasing
(b) B goes on decreasing and that by A goes on increasing
(c) Each decreases at the same rate
(d) Each decreases at different rates
SOLUTION :
 d
For scientist A which goes down in a mine g'  g 1  R 
 2h 
For scientist B, which goes up in a air g'  g 1  
 R 
So it is clear that value of g measured by each will decreases at different rates.

32. A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run
at the same rate on earth. On a planet having the same density as earth but twice the radius
(a) S will run faster than P (b)P will run faster than S
(c) They will both run at the same rate as on the earth (d)None of these
SOLUTION :
4
g GR .
3
If density is same then g  R
According to problem R p  2 R e
 g p  2ge
l
For clock P (based on pendulum motion) T  2
g
Time period decreases on planet so it will run faster because g p  g e
m
For clock S (based on oscillation of spring) T  2
k
So it does not change.
Gravitational Field:
It is the region or space around a massive particle in which its gravitational influence is left.
Gravitational field strength (or) Intensity of Gravitational Field:

Gravitational field strength at any point in a gravitational field is defined as the gravitational
force experienced by a unit mass placed

 at that point.
 F
 Gravitational field strength, E g  m
0
Units : Nkg or ms
1 2
Dimensional formula: LT 2
It is a vector quantity.
It is always directed radially towards the centre of mass of the body producing the field.
Note:
 
 F m g 
In the earth's gravitational field, E g    g.
0
m0 m0
Hence in the earth’s gravitational field, the intensity of gravitational field is equal to acceleration
due to gravity 'g'.
GM
The intensity of gravitational field at a distance r from a point mass ‘M’ is given by E g 
r2
The direction of the force F and hence of E is from P to O as shown in fig.
 GM 
‘ In vector form the above formula is E g  3 (r) .
r
r
Theoretically gravitational field due to a particle extends upto infinite distance around it
The value of Eg is zero at r   .
If the system has a number of masses, then resultant gravitational field intensity can be found out by using the
principle of super-position.
   
i.e. E g  Eg1  Eg2  Eg3 +.............
Null Point :
It is the point in a gravitational field at which resultant field intensity is zero.
If two particles of mass m1 & m2 are separated by a distance r, the distance of null point
from m1 is given by
Gm1 Gm2 r
E1  E2  0   ;x 
r  x
2 2
x m2
m1  1
m1 m2
E1 E2
x
r
Force and Field due to Circular Uniform Ring:

 A point mass m is at a distance x from the centre of the ring of mass M and radius R on its axis.
Gravitational force between the two is

GMmx
F
R  x2 
2 3/2
dm
R 2  x2
R
x m
GMm
 If x>>>R  F  , then for a distant point, ring behaves as point mass.
x2
GMm
 If x<<<R  F  , then force varies linearly as distance ‘x’
R3
R 2GMm
 Force is maximum, at x   maximum force Fmax 
2 3 3R 2
 Gravitational field intensity due to a uniform circular ring of mass M at any point at a distance ‘x’ (from
the centre of the ring) on its axis is

GMx
Eg  
 X 2  R2  along PO
3/ 2
dE
dE sin 
R
dE cos 

O dE cos  
dE dE sin 
x
 Gravitational field intensity is directed towards the centre of the circular ring.
 At the centre of the circular ring, E g  0
R
 Eg is maximum , at x  and
2
2GM
E max 
3 3R 2
Field due to Circular Disc:
Gravitational field intensity due to a circular disc of mass M at any point on the axial line
O
R 2  x2
R
Eg

p
O
2GM
or E g  1  Cos  (in terms of ‘  ’)
R2
Field due to Hollow sphere (or)Spherical Shell (E or I):
 Gravitational field intensity due to a uniform spherical shell
E
GM 1
E
R2 r2
r
R
 At a point inside the spherical shell,  Eg inside  0,  Eg centre  zero
GM
 At a point on surface of the spherical shell,  Eg  surface  (here r=R)
R2
GM
 At a point outside the spherical shell,  Eg outside  (here r>R)
r2
Field due to Solid Sphere (uniform mass density):
Gravitational field intensity due to a solid sphere
E
GM 1
E
r2
R2 E  r
r
R
 E g  0 ( at the centre of solid sphere )

GMr
 E 
g inside 
R3
 for r  R 
GM
 At a point on the surface of the solid sphere,  Eg    for r  R 
surface R2
GM
 At a point outside the solid sphere,  E g outside   for r  R 
r2
 E g  0 (at infinite distance)
Field due to Straight Rod:
A rod of length 2l, density  ,placed along x- axis, such that mid point of rod is coincides
with origin. The gravitational field intensity at a point P(a,o) is
2G l 2G  1
Eg  
a l a
2 2 a a2
1
l2
2G   a2 a2 
Eg  1  2  high powers of 2       
a  2l l 
EXAMPLE:
Calculate the gravitational field due to a uniform rod AB at a point P at perpendicular
distance a from the rod as shown in figure. Assume that the rod has a linear mass density  .
a 
N P

B
Solution :
Here it is very improtant to note that the unsymmetrical placement of the point P and hence
we must calculate the components of the gravitational fields Ex and Ey separately. For this, let us consider
an infinitesimal element of length dx at a distance x as shown
a 
N P

The net gravitation field at a point P due to this infinitesimal element is dE. This dE is resolved into
components.
(a) dE x  dE cos  (b) dE y  dE sin 
The net field E can be calculated b finding Ex and Ey from the above expressions. so
G   dx 
Ex   dE cos  , where dE 
r2
dx cos  dx cos 
 E x  G   G  2
r 2
 a  x2 
x
Also, we observe that tan  
a
 x  a tan 
 dx  a sec 2  d
a sec 2  cos  d
 E x  G 
a 2 sec2 
G 
 Ex   cos  d
a 
G
 Ex   sin   sin  
a
Similarly, let us calculate the value of Ey, given by
E y   dE y   dE sin 
dx sin 
 E y  G 
 a2  x2 
Since x  a tan 
 dx  a sec 2  d
a sec 2  sin  d
 E y  G 
a 2 sec 2 
G 
 Ey   sin  d
a 
G 
 Ey   cos   
a
So, the gravitational field due to a rod of length having uniform mass density  at a point P, that subtends
and angle  at one end and  at the other is given by

G
Ex   sin   sin  
a
G
Ey   cos   cos 
a
 A thin rod of mass M and length L is bent into a complete circle, then resultant force on a particle placed
at its centre is zero.
Work Done Against Gravity:

If the body of mass m is moved from the surface of earth to a point at distance h above the surface
of earth, then change in potential energy or work done against gravity will be
1 1 
W  U  GMm   
 r1 r2 
1 1 
 W  GMm    [As r1  R and r2  R  h ]
R R  h
GMmh mgh
W   GM
 2 h h [As  g]
R 1   1  R2
 R R
 When the distance h is not negligible and is comparable to radius of the earth, then we will use above formula.
 n 
 If h  nR then W  mgR  
n 1
1
 If hR then W 
2
mgR
 If h is very small as compared to radius of the earth then term h/R can be neglected
mgh  h 
From W 
1  h/R
 mgh  As R  0 
 
Gravitational Potential:
The amount of work done in bringing a unit mass from infinity to a certain point in the
gravitational field of another massive object is called gravitational potential at that point due to massive
object .
Let W is the work done and m0 is the test mass
W
then V  m
0
As this work is negative, the gravitational potential is negative

S.I unit : J/Kg
Dimensional formula :  M L T 
0 2 2
Potential due to a point mass:

M r
O P
The gravitational potential at a point p which is at a distance r from a point mass M is given by
GM
V 
r
In the system has a number of masses m1 , m2 , m3 .......... mn at distances r1 , r2 , r3 .......... rn
from the point ‘p’, the resultant gravitational potential at a point ‘p’ can be written as
V  V1  V2  V3  .........Vn
m m m m 
 V  G  1  2  3  ...........  n 
 r1 r2 r3 rn 
n
mi
 V  G 
i 1 ri
Potential due to Circular Ring:
Gravitational potential due to a circular ring, at a distance r from the centre and on the axis
of a ring of mass M and radius x is given by
GM O r
V R
P
R x
2 2
GM
At r=0, v   ,
R
GM
i.e., at the centre of the ring gravitational potential is 
R
Gravitational potential due to a spherical shell:
GM
Let M be the mass of spherical shell and R is its radius V 
r
GM
 At a point inside the spherical shell, (If r<R) Vinside 
R
R
r
O
GM
 At a point on the surface of the spherical shell, Vsurface  (If r=R)
R
GM
vcentre   (r=0 at centre)
R
GM
vinside  vsurface  vcentre   ,
R
GM M
 At a point outside the spherical shell, Voutside  (If r>R)
r R
 At infinity, V  0
graphical
GM

R r
R
Gravitational potential due to a solid sphere:
At a point inside the solid sphere,
GM
Vinside  (3R 2  x 2 )
2R3
 3 r2 
Vinside  GM   3  (if x<R)
 2R 2R 
GM
 At a point on the surface of the solid sphere, Vsurface  (If x=R)
R
GM
 At a point outside the solid sphere, Voutside  (If x>R)
x
3 GM 3
 At the centre, x=0  Vc    Vsurface .
2 R 2
 In case of solid sphere potential is maximum at centre.
3 GM Centre
2 R
V GM
 m surface
R
Inside
Outside
Newton’s Shell Theorem :

Gravitational potential at a point outside of a solid (or) hollow sphere of mass M is same as
potential at that point due to a point mass of M separated by same distance. Hence, the sphere can be
replaced by a point mass.
Gravitational potential difference :
The amount of work done in bringing a unit mass between two points in the
gravitational field is called as the gravitational potential difference between the two points.
 W  Wa 
V  Vb  Va    b 
 m0 
1 1
Wab  m0 Vb  Va   Gmm0   
 r0 ra 
Relation Between Gravitational Field and Potential:

 Gravitational field and the gravitational potential are related E  gradientV  gradV
  V  V  V 
E   i j k
 x y z 
V
Here, = Partial derivative of potential function V with respect to x,
x
i.e., differentiate V wrt x assuming y and z to be constant.
 The above equation can be written in the following forms.
 dV
 E , If gravitational field is along x-direction only..
dx
         
 dV   E .d r , (where d r  dx i  dy j  dz k and E  Ex i  E y j  Ez k )
Note :
r r  


a) If E is given V can be calculated by the formula V   dV    E.dr
 
b) The negative slope of V-r curve gives E

Gravitational Potential Energy:
The amount of work done by the gravitational force in bringing a body from infinity to any
point in the gravitational field is defined as the gravitational potential energy at that point.
dU
For a conservative field, F  
dr

 dU   F .d r

r  
u  
r
  dU    F .d r  U U0   F.dr
u0 r0 r0
We generally choose the reference point at infinity and assume potential energy to be zero
there. If we take r0   and U 0 = 0 then
r  
 r   
 U    F .d r  W  as  F .d r  W 
   
Potential energy of a body or system is the negative work done by the conservative forces in bringing it
from infinity to present position.
P.E=+mgh2
h2
Reference P.E=0
Level h1
P.E=-mgh1
 If a particle moves opposite to the field direction then work done by the field will be negative. So
potential energy will increase and change in potential energy will be positive.
 If a particle moves in the direction of the field work done is positive, so potential energy decreases and
change in potential energy is negative.
 potential energy exist for only conservative forces and it does not exist for non conservative forces.
 By the definition of gravitational potential,
W U
V    U  mV
m m
Gravitational Potential Energy of Two Particle System:
The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is
given by
Gm1m2
U 
r
Gravitational Potential Energy of Three Particle System
Z C
 m3
r13

m1 r23
A  m2
r12 B
Y
X
Consider a system consists of three particles of masses m1, m2 and m3 located at A, B and C respectively.
Total potential energy ‘U’ of the system is
m1m2 m2 m3 m1m2
U  G     
r12 r23 r13
If a body is moving only under the influence of gravitational force, from law of conservation
of mechanical energy
U 1  K1  U 2  K 2
Gravitational Potential Energy For a System of Particles:

The gravitational potential energy for a system of n particles is given by
 Gm1m2 Gm2 m3 
U  Ui      ............
 r12 r23 
n  n  1
For a n particle system there are pairs and the potential energy is calculated for
2
each pair and added to get the total potential energy of the system.
Gravitational Potential Energy of a Body in Earth’s Gravitational Field:
GMm
 If a point mass ‘m’ is at a distance r from the centre of the earth. U  
r
GMm  GM 
 On the surface of earth, U surface     mgR  g  2 
R  R 
GMm
 At a height ‘h’ above the surface of earth, U h  
Rh
 The difference in potential energy of the body of mass m at a height h and on the surface of earth is
U  Uh  Usurface
GMm  GMm 
  
Rh  r 
1 1 
 GMm   
 R Rh
GMmh GMmh
 
 R  h R  h
R 2 1  
 R
mgh
 U 
h
1
R
If h  R, U  mgh
 Work done in lifting a body of mass m from earth surface to a height h above the earths surface is
W  Uh Usurface ;
1 1 
W  GMm   
 R R h
mgh

h
1
R
3 GMm
 Gravitational potential energy at the centre of the earth is given by U c  mvc  
2 R
3 3GM
Here, Vc  Vs 
2 2R
(It is minimum but not zero. However ‘g’ at centre of earth is zero)
Self potential energy of a uniform sphere of mass ‘M’ and radius ‘R’ :
It is the amount of work done to bring identical massive particles to construct a sphere of
mass M radius R and density 
4 3
For a sphere of radius ‘x’, mass of the sphere   x  ,
3
where  =density of sphere
4
Gravitational potential on the surface   G x2
3
Gm G 4 3 4
(since gravitational potential     x    Gx 2  )
x x 3 3
Work done by the agent in increasing the surface from x to x+dx is
Gm  dm 
=Gravitational potential x dm
x
 4  16 r 2
   Gx 2    4 x 2dx    G  2 x 4 dx
 3  3
16 2 R 16 2G  2 R 5
Therefore, total work done  G  2  x 4 dx 
3 0 15
16 2 R 16 2G  2 R 5
Therefore, total work done  G  2  x 4dx 
3 0 15
2
 
16 GR 
2 5
M  3 GM 2
  
15  4 3  5 R
R
 3 
=Gravitational self potential energy of a sphere.
Gm 2
 Self potential energy of a thin uniform shell of mass ‘m’ and radius ‘R’ is 
2R
Change in the gravitational potential energy:
Change in the gravitational potential energy in lifting a body from the surface of the earth to
a height equal to ‘nR’ from the surface of the earth
GMmh G M m n R  GMmm m gRn

U    
R R  h  R R  nR  R n  1 n 1
PROBLEMS
1. Two bodies of masses 100Kg and 10,000Kg are at a distance of 1m apart. At what distance
from 100kg on the line joining them will the resultant gravitational field intensity be zero?
SOLUTION:
G  100 G  10,000

1  x 
2
x2
1
 100 x 2  1  x   x 
2
11
2. The potential energy of a body of mass ‘m’ is given by U=px+qy+rz. The magnitude of the
acceleration of the body will be
pqr p2  q2  r 2 p3  q3  r 3 p4  q4  r 4
1) 2) 3) 4)
m m m m
SOLUTION:
dU
F ; Fx   p, Fy  q , Fz   r
dr

F  p 2  q 2  r 2  ma  p 2  q 2r 2
3. Gravitational field intensity at the centre of the semi circle formed by a thin wire AB of mass
‘m’ and length ‘L’ is
X
A B
2 Gm  2 Gm 
Gm 2 
1) 2 i
L
 2)
Gm 2 
 L2
j  3)
L2
i  4)
L2
j 
SOLUTION:
m
 ; L   r ; dm   dl    rd 
L
G   


E   cos d i   sin  d j 
r 0 0 
4. The gravitational field due to a mass distribution is given by E=-K/x 3 in x-direction. Taking
the gravitational potential to be zero at infinity, find its value at a distance x.
SOLUTION:
The potential at a distance x is
x x
K  K  K
V    Edx   3
dx   2   2

x  2 x  2x
5. Consider two configurations in fig(i) and fig(ii)
3m
2m
a a a
m 2m m 3m
a a
fig(i) fig(ii)
The work done by external agent in changing the configuration from fig(i) to fig(ii) is
6Gm 2  1  6Gm 2  1  6Gm 2  1 
1) Zero 2)   1  3)   1  4)   2 
a  2 a  2 a  2
SOLUTION:
Gm1m2
GPE  ;W  GPE2  GPE1
r
6. A particle of mass m is placed at the centre of a uniform spherical shell of equal mass and
radius a. Find the gravitational potential at a point P at a distance a/2 from the centre.
SOLUTION:
Gm 2Gm
The gravitational potential at P due to particle at centre is V1  
a/2 a
Gm
The potential at P due to shell is V2 
a
3Gm
The net potential at P is V1  V2 
a
  
7.  
The gravitational field in a region is given by E    20 Nkg  i  j . Find the gravitational
1
potential at the origin (0, 0)

SOLUTION:
 
V    E.dr     Ex dx   E y dy   20 x  20 y
 V  0 at the origin (0, 0).

8. Calculate the gravitational potential at the centre of base of a solid hemisphere of mass M,
radius R.
SOLUTION:
dm
dr
M r
R
Consider a hemispherical shell of radius r and thickness dr. Its mass is given by
M 3Mr 2 dr
dm 
2 3
 2 r dr   R3
2
R
3
Since all points of this hemispherical shell are at the same distance r from centre O, potential at O due
to it is
Gdm 3GMrdr
dV  
r R3
3GM
R
V   dV  .
2R
0
9. A point mass M is at a distance S from an infinitely long and thin rod of linear density D. If G
is the gravitational constant then gravitational force between the point mass and the rod is
S
M
MGD MGD MGD 2 MGD

1) 2 2) 3) 4) ’
S S 2S 3 S
SOLUTION:
Sd
dm  D  dl  D 
cos 
GMdm
cos 
Gravitational force, DF   S 
 
 cos  
 /2
MGD 2 MGD
total force F 



/2
S
cos  d 
S
10. The gravitational field in a region is given by the equation E   5i  12 j N / kg . If a particle

of mass 2kg is moved from the origin to the point (12m, 5m) in this region, the change in the
gravitational potential energy is (2012 E)
SOLUTION:
 
dV   E.dr
   5i  12 j .12i  5 j    60  60   120
Change in gravitational potential energy
dU=mdV
=2(-120)=-240 J
11. A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is
located at a distance R/2 from the centre of sphere. The gravitational force on a particle of
mass m at a distance R/2 from the centre of the sphere on the line joining both the centres of
GM
sphere and cavity is (opposite to the centre of cavity) [here g  where M is the mass of
R2
the sphere].
mg 3mg mg mg
1) 2) 3) 4)
2 2 16 4
SOLUTION:
  R / 2
3
R
E1  , Ee 
6 0 3 0 R 2
M 1
Enet  E1  Ee ;  ; 0 
4 3
R 4 G
3
12. Find the gravitational potential energy of a system of four particles, each of mass m placed at
the vertices of a square of side l. Also obtain the gravitational potential at centre of the square.
SOLUTION:
m m
l
2 l
O
m m
The system has four pairs with distance l and two diagonal pairs with distance 2l .
Gm 2 Gm 2 2Gm 2  1 
U  4 2   2 
l 2l l  2
The gravitational potential the centre of the square is
V=Algebraic sum of potential due to each particle
4Gm 4 2Gm  2l 
V    r  
r l  2 
13. A mass m extends a vertical helical spring of spring constant k by x m at the surface of earth.
Extension in spring by the same mass at height h meter above the surface of earth is
GMm GMm  R  h
2 R2
1) k R  h 2) 3) x 4) R  h 2 x
  kR 2 R 2  
SOLUTION:
Let the extension at height h be x’ then
GMm  F
x 2 
 F  kx or x  
kR  k
x' R2 R2

then x k R  h 2 ; x '  x
   R  h
2
14. Two bodies of mass m and 4m are placed at a distance r. The gravitational potential at a point
on the line joining them where gravitational field is zero is (2011A)
SOLUTION:
Position of null point from mass m is
r r
x 
4m 3
1
m
r
3
Null point 4m
m
r
 3 12  9Gm
 potential V  Gm  r  2r   r
 
15. The change in potential energy, when a body of mass m is raised to a height nR from the earth’s surface
is (R = Radius of earth)
n n2 n
(a) mgR
n 1
(b) nmgR (c) mgR 2 (d) mgR
n 1
n 1
SOLUTION:
mgh mg nR nm gR
U   
h nR n  1
1  1 
R R
16. Four particles each of mass M are located at the vertices of a square with side L. The
gravitational potential due to this at centre of square is
GM GM GM
1)  32 2)  64 3) Zero 4)  16
L L2 L
GM GM
SOLUTION: U  4   32
L/ 2 L
17. A body of mass m rises to height h = R/5 from the earth’s surface, where R is earth’s radius. If g is
acceleration due to gravity at earth’s surface, the increase in potential energy is
4 5 6
(a) mgh (b) 5
mgh (c) 6
mgh (d) 7
mgh
SOLUTION:
mgh
U 
1  h/R
Substituting R  5 h
mgh 5
we get U  1  1/5  6 mgh
18. In a certain region of space, the gravitational field is given by -k/r, where r is the distance and
k is a constant. If the gravitational potential at r=r0 be V0, then what is the expression for the
gravitational potential V?
1) k log  r / r0  2) k log  r0 / r  3) V0  k log  r / r0  4) V0  k log  r0 / r 
SOLUTION:
dV dr
Here, I     k / r (or ) dV  k then integrate
dr r
19. Two masses 90kg and 160kg are 5 m apart. The gravitational field intensity at a point 3m from
90kg and 4m from 160kg is
1) 10G 2) 5 G 3) 5 2G 4) 10 2G
SOLUTION:
ER  E12  E22
20’. A spherical shell is cut into two pieces along a chord as shown in the figure. P is a point on the
plane of the chord. The gravitational field at P due to the upper part I1 and that due to the
lower part is I 2 . What is the relation between them?
1) I1  I 2 2) I1  I 2 3) I1  I 2 4) no definite relation
SOLUTION:
At the point P, we have I1-I2=0 (because the gravitational field inside a shell it is zero).
‘Hence I1=I2
21. Two concentric shells of different masses m1 and m2 are having a sliding particle of mass m.
The forces on the particle at position I, II and III are
II r2
m2 III
m1 r3
r1
Gm G  m1  m2  m Gm2 Gm1
2) r 2 ,0, r 2
1
1) 0, r 2 , r12
2 2 1
G  m1  m2  m Gm2 G  m1  m2  m G  m2  m
3) , 2 ,0 4) , ,0
r12 r2 r12 r22
SOLUTION:
G m  m1  m 2 
Position I. F  r12
(here the particle lies outside of both the shells)
Gm
Position II. F  r 2
2
here the particle lies outside of the shell of mass m1

Position III. Here the particle lies inside of both of the shells so F=0.
22. A ring has non-uniform distribution of mass having mass ‘M’ and radius ‘R’. A point mass m0
is moved from A to B along the axis of the ring. The work done by external agent against
gravitational force of ring is
R
A B
R R
GMm0 GMm0  1 1  GMm0  1 1  GMm0

1) 2)    3)    4)
2R R  2 5 R  5 2 5R
SOLUTION:
W  m VB  VA  ;
23. The gravitational field due to a mass distribution is E  K / x 3 in the x-direction. (K is a constant).
Taking the gravitational potential to be zero at infinity, its value at a distance x is
(a) K/x (b) K/2x (c) K / x2 (d) K / 2x 2
SOLUTION:
 K
Gravitational potential   I dx  x dx
x3
 
 x 3 1 
 K    K 
K
 3  1  2x 2 2x 2
 x x
24. The gravitational potential of two homogeneous spherical shells A and B of same surface
density at their respective centres are in the ratio 3:4. If the two shells coalesce into single one
such that surface density remains same, then the ratio of potential at an internal point of the
new shell to shell A is equal to :
1) 3:2 2) 4:3 3) 5:3 4) 5:6
SOLUTION:
4 r 2   4 r12   4 22   r 2  r12  r22
GM G 4 r 2 
V 
r r
V  4 rG   V  r
V1 r1 3 r2 9
    12 
V2 r2 4 r2 16
r12 : r22 : r 2  r12 : r22 :  r12  r22   9 :16 : 19  16 
 r1 : r2 r  3 : 4 : 5  V1 : V2 : V3
25. The masses and radii of the earth and moon are M 1 , R1 and M 2 , R2 respectively. Their centres are
distance d apart. The minimum velocity with which a particle of mass m should be projected from a
point midway between their centres so that it escapes to infinity is
G 2G Gm Gm (M 1  M 2 )
(a) 2 d
(M 1  M 2 ) (b) 2 d
(M 1  M 2 ) (c) 2 d
(M 1  M 2 ) (d) 2 d (R1  R 2 )
SOLUTION:
 GM 1 GM 2
Gravitational potential at mid point V  
d/2 d/2
2 Gm
Now, PE  m  V  (M 1  M 2 )
d
m = mass of particle
So, for projecting particle from mid point to infinity KE | PE|
1 2 Gm
 mv 2  (M 1  M 2 )
2 d
G (M 1  M 2 )
v  2 ’
d
26. A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R.
The work required to take a unit mass from point P on its axis to infinity is
4R
2GM
1)
7R
4 2 5   2) 
2GM
7R

4 2 5  3)
GM
2R
4)
2GM
5R
 2 1 
SOLUTION:
M  2 r  dr
dm 
 16 R 2  9 R 2  4 R
G  dm  4R
DV 
r 2  16 R 2
;V   dV ’
3R
W  m V  V 
27. The gravitational potential energy of a body of mass ‘m’ at the earth’s surface mgR e . Its
gravitational potential energy at a height R e from the earth’s surface will be (Here R e is the radius of
the earth)
1 1
(a) 2 mgR e (b) 2 mgR e (c) 2 mgR e (d)  2 mgR e
SOLUTION:
mgh mgR e mgR e
U  U 2  U1   
h Re 2
1 1
Re Re
mgR e
 U 2  (mgR e ) 
2
1
 U 2   mgR e
2
28. A body of mass m is placed on the earth’s surface. It is taken from the earth’s surface to a height
h  3 R . The change in gravitational potential energy of the body is
2 3 mgR mgR
(a) 3 mgR (b) 4 mgR (c) 2
(d) 4
SOLUTION:
mgh mg  3 R 3
 U    mgR
h 3 R 4
1  1 
R R

29. The gravitational force in a region is given by, F  ayi  ax j . The work done by gravitational
force to shift a point mass m from (0, 0, 0) to  x0 , y0 , z0  is
1) max 0 y0 z0 2) max 0 y0 3)  max 0 y0 4) 0
SOLUTION:
 x0 y0 z0

W   F .dr  ma  yi  x j dxi  dy j
0
 
‘  ma  d  xy   ma  xy 
30. Energy required to move a body of mass m from an orbit of radius 2R to 3R is
(a) GMm /12 R 2 (b) GMm /3 R 2 (c) GMm /8 R (d) GMm /6 R
SOLUTION:
Change in potential energy in displacing a body from r1 to r2 is given by
1 1  1 1  GMm
U  GMm     GMm   
 r1 r2   2 R 3 R 6R
31. Two concentric spherical shells A and B of radii R and 2R and masses 4M and M respectively
are as shown. The gravitational potential at point ’p’ at distance ‘r’ (R<r<2R) from centre of
shell is (r=1.5R)
B
A
2R R
r
P
4GM 9GM 4GM 19GM

1)  2)  3)  4) 
R 2R 3R 6R
SOLUTION:
G 4 M Mg
V 
3 2R
R
2
32. Four particles each of mass M, are located at the vertices of a square with side L. The gravitational
potential due to this at the centre of the square is
GM GM GM
(a)  32
L
(b)  64
L2
(c)Zero (d) 32
L
SOLUTION:
m m
L2
m m
GM
Potential at the centre due to single mass =
L/ 2
GM
Potential at the centre due to all four masses =  4
L/ 2
GM GM
4 2   32  .
L L
33. Suppose a vertical tunnel is dug along the diameter of earth assumed to be a sphere of uniform
mass having density  . If a body of mass m is thrown in this tunnel, its acceleration at a
distance y from the centre is given by
m
y
4 3 4 4
1) G  ym 2)  G y 3)  y 4)  G y
3 4 3 3
SOLUTION:
4 3
Mass of the sphere is given by M   y 
3
4 
G   y3  m
Gravitational force. F   3  a F ’
2
y m
34. The magnitudes of the gravitational force at distances r1 and r2 from the centre of a uniform sphere
of radius R and mass M are F1 and F2 respectively. Then
F1 r1 F r12
(a) F  if r1  R and r2  R (b) F1  if r1  R and r2  R
2 r2 2 r22
F1 r1 F r22
(c) F  if r1  R and r2  R (d) F1  if r1  R and r2  R
2 r2 2 r12
SOLUTION:
KEY: (a, b)
4
 g
3
Gr  gr if r  R
GM 1
 g
r 2  g
r2
if r  R
F1 g r
If r1  R and r2  R then  1  1
F2 g 2 r2
2
F1 g r 
If r1  R and r2  R then  1  2 
F2 g 2  r1 

35. A particle is placed in a field
characterized by a value of gravitational potential given V=-kxy,
where ‘k’ is a constant. If E g is the gravitational field then

 
1) E g  k xi  y j and is conservative in nature

 
2) E g  k yi  x j and is conservative in nature

 
3) E g  k xi  y j and is non conservative in nature

 
4) E g  k yi  x j and is non conservative in nature
SOLUTION:
   
Eg    i  j    kxy 
 x y 
36. Two identical thin rings each of radius ‘R’ are co-axially placed at a distance ‘R’. If the rings
have a uniform mass distribution and each has mass m1 and m2 respectively, then the work
done in moving a mass ‘m’ from the centre of one ring to that of the other is:
1) Zero 2)
Gm  m1  m2   
2 1
3)
Gm 2  m1  m2 
4)
Gm1m  
2 1
2R R m2 R
SOLUTION:
Gm1 Gm2 Gm2 Gm1
V1   and V2  
R 2R R 2R
Gm2 Gm1 Gm1 Gm2 1 1 

V  V2  V1      G  m1  m2    
R 2R R 2R R 2R 
Hence W  m  V  
mG  m1  m2   
2 1
2R
37. A mass m is placed in the cavity inside a hollow sphere of mass M as shown in the figure. The
gravitational force on m is
r
m
R
GMm GMm GMm

1) 2) 3) R  r 2 4) Zero
R2 r2  
SOLUTION: Gravitational force is zero due to symmetry
3 8 . A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates.
Two spheres of equal radii 1 unit with their centres at A(– 2, 0, 0) and B(2, 0, 0) respectively are taken
out of the solid leaving behind spherical cavities as shown in figure
Y
A B
O X
(a)The gravitational force due to this object at the origin is zero

(b)’The gravitational force at the point B (2, 0, 0) is zero
(c)The gravitational potential is the same at all points of the circle y 2  z 2  36
(d)The gravitational potential is the same at all points on the circle y2  z2  4
KEY:(a, c, d)
SOLUTION:
Since cavities are symmetrical w.r.t. O. So the gravitational force at the centre is zero.
The radius of the circle z 2  y 2  36 is 6.
For all points for r  6, the body behaves as if whole of the mass is concentrated at the centre.
So the gravitational potential is same.
Above is true for z2  y2  4 as well.

39. A homogeneous spherical heavenly body has a uniform and very narrow frictionless duct along
its diameter. Let mass of the body be M and diameter be D. A point mass m moves smoothly
inside the duct. Force exerted on this mass when it is at a distance s from the centre of the
body is
GMm  GMm 8GMm GMm

1) 2)  D / 2 3 s 3) s 4) R  s 2
s2 D3  
SOLUTION:
M Ms GmM s
 ; F 
4 D
3
4 S2
   S3
3  2 3
Escape Velocity:
The minimum velocity with which a body must be projected up so as to enable it to just
overcome the gravitational pull, is known as escape velocity.
The work done to displace a body from the surface of earth (r = R) to infinity ( r   ) is
 GMm 1 1 
W  R dx  GMm   
x 2
 R 
GMm
 W
R
This work required to project the body so as to escape the gravitational pull is performed
on the body by providing an equal amount of kinetic energy to it at the surface of the earth.
If v e is the required escape velocity,,
1
then kinetic energy which should be given to the body is mv e2
2
1 GMm
 2
mv e2 
R
2GM
 ve 
R
 v e  2 gR [As GM  gR 2 ]s
4
or ve  2  GR  R
3
8 4
 ve  R G  [As g GR ]
3 3
 Escape velocity is independent of the mass and direction of projection of the body.
 Escape velocity depends on the reference body. Greater the value of (M / R ) or (gR) for a planet,
greater will be escape velocity.
For the earth as g  9 .8 m / s 2 and R  6400 km
 ve  2  9 . 8  6 .4  10 6  11 . 2km / sec
 3 RT 
 A planet will have atmosphere if the velocity of molecule in its atmosphere v rms   is lesser
 M 
than escape velocity.

This is why earth has atmosphere (as at earth v rms  ve ) while moon has no atmosphere
(as at moon v rms  v e )
 If a body projected with velocity lesser than escape velocity ( v  v e ), it will reach a certain
maximum height and then may either move in an orbit around the planet or may fall down
back to the planet.
 Maximum height attained by body : Let a projection velocity of body (mass m ) is v , so that it
attains a maximum height h . At maximum height, the velocity of particle is zero, so kinetic
energy is zero.
By the law of conservation of energy
Total energy at surface = Total energy at height h.
GMm 1 GMm
  mv 2   0
R 2 R h
v2 1 1 
  GM  
2  R R  h 
GMh

R (R  h)
2 GM Rh R
  1
v2R h h
R R
h  2  v2 
 2 GM  v R  2
  2  1 e
1 2
 v R  v2  v e  v 
2 GM 2 GM
[As ve    v e2 ]
R R
) If a body is projected with velocity greater than escape velocity then by conservation of energy.
Total energy at surface = Total energy at infinite
1 GMm 1
mv 2   m (v )2  0
2 R 2
2 GM
i.e., (v )2  v 2 
R
2 GM
 v ' 2  v 2  v e2 [As R
 v e2 ]
 v   v 2  v e2
i.e, the body will move in interplanetary or inter stellar space with velocity v 2  v e2 .
Energy to be given to a stationary object on the surface of earth so that its total energy becomes zero, is
called escape energy.
GMm
Total energy at the surface of the earth  KE  PE  0 
R
GMm
 Escape energy  R
 If the escape velocity of a body is equal to the velocity of light then from such bodies nothing can
escape, not even light. Such bodies are called black holes.
2 GM
The radius of a black hole is given as R
C2
2 GM
[As C
R
, where C is the velocity of light]
Orbital Velocity of Satellite:
Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth.
Satellite
h
r
R
v
Earth
Fig. 8.26
For revolution of satellite around the earth, the gravitational pull provides the required centripetal force.
mv 2 GMm

r r2
GM
 v
r
As GM  gR 2 and r Rh
gR 2 g
v R
Rh Rh
 Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of
the orbit
i.e., satellites of diferent masses have same orbital velocity, if they are in the same orbit.
 Orbital velocity depends on the mass of central body and radius of orbit.
 For a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite
v  1 / r .
 Orbital velocity of the satellite when it revolves very close to the surface of the planet
GM GM
v 
r Rh
As h0 and GM  gR 2
GM
 v
R
 gR
For the earth v  9 . 8  6 . 4  10 6

 7 . 9 k m / s  8 km / sec
GM
 Close to the surface of planet v
R
2 GM
As ve 
R
ve
 v
2
i.e., v escape  2 v orbital
It means that if the speed of a satellite orbiting close to the earth is made 2 times (or
increased by 41%) then it will escape from the gravitational field.
 If the gravitational force of attraction of the sun on the planet varies as
1
F
rn
1
then the orbital velocity varies as v  .
r n 1
Time Period of Satellite:

It is the time taken by satellite to go once around the earth.
Circumfere nce of the orbit
 T 
orbital velocity
2r r GM
 T 
v
 2r
GM
[As v
r
]
r3 r3
 T  2  2 [As GM  gR 2 ]
GM gR 2
R  h 3
 T  2 g R2
3/2
R h
T  2 1  
g R
[As rRh]
r3
 From T  2 ,
GM
it is clear that time period is independent of the mass of orbiting body and depends on the
mass of central body and radius of the orbit
r3
 T  2
GM
2 4 2 3
 T  r
GM
i.e., T 2  r 3
This is in accordance with Kepler’s third law of planetary motion r becomes a (semi major axis)
if the orbit is elliptic.
 Time period of nearby satellite,

r3 R3 R
From T  2  2  2 [As h0 and GM  gR 2 ]
GM gR 2 g
For earth R  6400 km and g  9 .8 m / s 2

T  84 . 6 minute  1 . 4 hr
 Time period of nearby satellite in terms of density of planet can be given as
‘
3 3 
 
2 R 3
1/2

3
r R 1/2
G
T  2  2  4 3 
GM GM G. 3 R  
 
 If the gravitational force of attraction of the sun on the planet varies as then the time period varies
n 1
as T r 2
 If there is a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to
east, then for an observer, on the earth, angular velocity of satellite will be ( S   E ) .
The time interval between the two consecutive appearances overhead will be
2 T T  2 
T   S E  As T   
 s   E TE  TS  
If  S   E , T  
i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.
Height of Satellite:
r 3  2 (R  h)3
As we know, time period of satellite T  2
gR 2
GM
g R 2T 2
 R  h 
3
By squaring and rearranging both sides
4 2
1/3
 T 2g R2 
 h   
 R
 4
2

By knowing the value of time period we can calculate the height of satellite from the surface of the
earth.
Geostationary Satellite:
The satellite which appears stationary relative to earth is called geostationary or geosynchronous
satellite, communication satellite.
 A geostationary satellite always stays over the same place above the earth such a satellite is
never at rest. Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on earth.
 The orbit of a geostationary satellite is known as the parking orbit.
 It should revolve in an orbit concentric and coplanar with the equatorial plane.
 Its sense of rotation should be same as that of earth about its own axis
i.e., in anti-clockwise direction (from west to east).
 Its period of revolution around the earth should be same as that of earth about its own axis.
 T  24 hr  86400 sec
 Height of geostationary satellite
r3 (R  h) 3
As T  2  2  24 hr
GM GM
Substituting the value of G and M we get R  h  r  42000 km  7 R
height of geostationary satellite from the surface of earh h  6 R  36000 km
GM
 Orbital velocity of geo stationary satellite can be calculated by v
r
Substituting the value of and we get v  3 .08 km / sec
Energy of Satellite:
When a satellite revolves around a planet in its orbit, it possesses both potential energy (due to its
position against gravitational pull of earth) and kinetic energy (due to orbital motion).
 GMm  L2  GM 2 2 
(1) Potential energy : U  mV    As V  r , L  m GMr 
r mr 2  
1 GMm L2  GM 
(2) Kinetic energy : K mv 2    As v  
2 2r 2 mr 2  r 
 GMm GMm  GMm  L2

(3) Total energy : EUK    
r 2r 2r 2mr 2
 Kinetic energy, potential energy or total energy of a satellite depends on the mass of the satellite and
the central body and also on the radius of the orbit.
 From the above expressions we can say that
Kinetic energy (K) = – (Total energy)
Potential energy (U) = 2 (Total energy)
Potential energy (K) = – 2 (Kinetic energy)
 Energy graph for a satellite & Energy distribution in elliptical orbit
+
Satellite
r
En- K Perigee Focus
K.E. = max a Apogee
K.E. = min
ergO P.E. = min Semi major
P.E. = max
axis
y E r
U
r min r max
– (B)
(A) 8
 If the orbit of a satellite is elliptic then

GMm
(a) Total energy (E)  2a
 constant ;
where a is semi-major axis .

(b) Kinetic energy (K ) will be maximum when the satellite is closest to the central body (at
perigee) and minimum when it is farthest from the central body (at apogee)
(c) Potential energy (U) will be minimum when kinetic energy = maximum
i.e., the satellite is closest to the central body (at perigee)
Potential energy (U) will be maximum when kinetic energy = minimum
i.e., the satellite is farthest from the central body (at apogee).
Binding Energy :
The energy required to remove the satellite from its orbit to infinity is called Binding Energy of the
system,
Total energy of a satellite in its orbit is negative. Negative energy means that the satellite is bound to the
central body by an attractive force and energy must be supplied to remove it from the orbit to infinity.
GMm
i.e., Binding Energy (B.E.)  E 
2r
Weightlessness:
The weight of a body is the force with which it is attracted towards the centre of earth. When a body is
stationary with respect to the earth, its weight equals the gravity. This weight of the body is known as its static
or true weight.
We become conscious of our weight, only when our weight (which is gravity) is opposed by some other
object. Actually, the secret of measuring the weight of a body with a weighing machine lies in the fact that as we
place the body on the machine, the weighing machine opposes the weight of the body. The reaction of
the weighing machine to the body gives the measure of the weight of the body.
The state of weightlessness can be observed in the following situations.
(1) When objects fall freely under gravity :
For example, a lift falling freely, or an airship showing a feat in which it falls freely for a few
seconds during its flight, are in state of weightlessness.
(2) When a satellite revolves in its orbit around the earth :
Weightlessness poses many serious problems to the astronauts. It becomes quite difficult for
them to control their movements. Everything in the satellite has to be kept tied down. Creation of
artificial gravity is the answer to this problem.
(3) When bodies are at null points in outer space :
On a body projected up, the pull of the earth goes on decreasing, but at the same time the
gravitational pull of the moon on the body goes on increasing. At one particular position, the two
gravitational pulls may be equal and opposite and the net pull on the body becomes zero. This is zero
gravity region or the null point and the body in question is said to appear weightless.
Trajectories of a body projected with different velocities :
An object revolves around a planet only when it is projected with sufficient velocity in a
direction perpendicular to the gravitational force of attraction of the planet on the object.
point of projection
speed v part of ellipse

v  gr 
earth
circle
hyperbola v   gr   8 kms 1
v  2 gr  parabola
ellipse
v  2 gr   11km s 1
 2gr   v   gr 
If v  gr object falls on the surface of earth
If v  gr object revolve in a circular orbit.
If gr  v  2 gr object revolves in an elliptical orbit.
If v  2 gr object escapes from the field and follows parabolic path.
If v  2 gr object escapes from the field and follows hyperbolic path.
Special Cases :
V=0
V2 4
3 Escape
V1
2 h1
v 1 R h2
M
Case I:
Work done to lift an object at rest from the surface of a planet to a height h is
GMm GMm
TEi  TEsurface   0
R R
GMm GMm
TEi  TEheight   0
Rh Rh
GMm GMm
Work done W  TE f  TEi  
R Rh
GMmh mgh
W  
R  R  h 1  h
R
Case II:
Work done to shift an object at rest from the surface of planet in to an orbit in which object revolves
around the planet is
GMm GMm
TEi  TEsurface  0
R R
GMm 1 2 GMm
TE f  TEorbit   mv0 
Rh 2 2 R  h
GMm GMm
R 2  R  h
 R  2h 
W  GMm  
 2 R  R  h  
Case III:
Work done to shift an object revolving around the planet from one orbit in to another orbit is
GMm 1 2 GMm
TEi   TE  h   mv1 
1
R  h1 2 2  R  h1 
GMm 1 2 GMm
TE f  TE  h    mv2 
2
R  h2 2 2  R  h2 
GMm GMm
2 R  h1  2 R  h2 
GMm  h2  h1 
W   
2   R  h1  R  h2  
Case IV :
Work done (or) additional energy to be imparted for an object to just escape an object with is initially
revolving around the planet close the surface is
GMm GMm GMm
TEi   
R 2R 2R
TEf=O(object escapes only when its TE becomes zero (or) positive)
Work done (or) additional energy imparted to the object is
GMm
E  E  TE f  TEi   KE of the object
2R
Hence, an object (satellite) revolving around the planet escapes when
1) It’s KE is doubled (increases by 100%)
2) It’s velocity is increased to 2 times of present value (increases by 41.4%)
Additional velocity imparted to the body  ve  v0  2v0  v0
‘  
2  1 v0  3.2km / s  nearly 
Note :
In the above case if the object initially revolves around the planet at a highest h from the surface then it’s
GMm
TE 
2 R  h
GMm
Additional energy required to escape the object is 2 R  h ’’’
 
Quantities Variation Relation with r
1
Orbital Velocity Decreases v
r
Time period Increases T  r 3/ 2

1
Linear momentum Decreases p
r
Angular momentum Increases L r
1
Kinetic energy Decreases K
r
1
Potential energy Increases U 
r
1
Total energy Increases E 
r
1
Binding energy Decreases BE
r
::PROBLEMS::
1. Two bodies of masses m1 and m 2 are initially at rest at infinite distance apart. They are then allowed
to move towards each other under mutual gravitational attraction. Their relative velocity of approach
at a separation distance r between them is
1/2 1/2 1/2 1/2
 (m1  m 2 )   2G   r   2G 
(a) 2G  (b)  (m1  m 2  (c)  2G(m m  (d)  m 1m 2 
 r   r   1 )
2   r 
SOLUTION :
Let velocities of these masses at r distance from each other be v 1 and v 2
respectively.
By conservation of momentum m 1 v 1  m 2 v 2  0
 m 1v1  m 2 v 2 … (i)
By conservation of energy change in P.E.=change in K.E.
Gm 1 m 2 1 1
 m 1 v 12  m 2 v 22
r 2 2
m 12 v 12 m 22 v 22 2 Gm 1 m 2
   …(ii)
m1 m2 r
On solving equation (i) and (ii)
2 Gm 22
v1 
r(m 1  m 2 )
2 Gm 12
v2 
r(m 1  m 2 )
2G
 v app | v1 |  | v 2 |  (m 1  m 2 )
r
2. If v e and v o represent the escape velocity and orbital velocity of a satellite corresponding to a
circular orbit of radius R, then
(a) ve  vo (b) 2v o  ve (c) ve  v0 / 2 (d) and are not related
SOLUTION :
ve  2 gR
v0  gR
 2 v0  ve
3. A projectile is projected with velocity kv e in vertically upward direction from the ground into the
space. ( v e is escape velocity and k  1) . If air resistance is considered to be negligible then the
maximum height from the centre of earth to which it can go, will be : (R = radius of earth)
R R R R
(a) (b) (c) (d) k 1
k2 1 k 2 1 1k2
SOLUTION :
Kinetic energy = Potential energy
1 mgh
m (kv e ) 2 
2 h
1
R
1 mgh
mk 2 2 gR 
 2 h
1
R
Rk 2
h 
1k2
Height of Projectile from the earth’s surface = h
Rk 2
Height from the centre r  R  h  R 
1k2
R
By solving r 
1k2
4. If Earth has mass nine times and radius twice that of the planet mars, calculate the velocity
required by a rocket to pull out of the gravitational force of Mars. Take escape speed on
surface of Earth to be 11.2 km/s
SOLUTION :
Here, Me=9Mm, and Re=2Rm
ve (escape speed on surface of Earth)=11.2 km/s
Let Vm be the speed required to pull out of the gravitational force of mars.
We know that
2GM e 2GM m
ve  and vm 
Re Rm
vm 2GM m Re
Dividing, we get  
ve Rm 2GM e
M m Re 1 2
   2 
M e Rm 9 3
2
 vm  11.2km / s   5.3km / s
3
5. A rocket is fired with a speed v  2 gR near the earth’s surface and directed upwards.
(a) Show that it will escape from the earth.
(b) Show that it interstellar space its speed is v  2 gR .
SOLUTION :
(a) As PE of the rocket at the surface of the earth is (-GMm/R) and at infinity is zero,
 GMm   GM 
energy required for escaping from earth  0     mgR  g  R 2 
 R   
1 2
And as initial KE of the rocket mv  2mgR is greater than the energy required for escaping (=mg
2
R), the rocket will escape.
(b) If  is the velocity of the rocket in interstellar space (free from gravitational effects)
then by conservation of energy,
1
  1
  1 2
2 2
m 2 gR  m 2 gR  mv
2 2 2
v 2  4 gR  2 gR or v  2 gR
6. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10
times smaller. Given that escape velocity from the earth is 11km/s, the escape velocity from
the surface of the planet is
Re
SOLUTION : Given M P  10 M e ; RP   2008 A
10
2GM
We know that ve 
R
2GM P 100  2GM e

 vP    10ve
RP Re
=10 11 = 110 km/s
7 . The ratio of the K.E. required to be given to the satellite to escape earth’s gravitational field to the K.E.
required to be given so that the satellite moves in a circular orbit just above earth atmosphere is
(a) One (b) Two (c)Half (d)Infinity
SOLUTION :
K.E. required for satellite to escape from earth’s gravitational field
2
1 1  2 GM 
  GMm
mv e2  m 
2 2  R 
 R
K.E. required for satellite to move in circular orbit

2
1 1  GM  GMm
mv 02  m  
2 2  R  2R
The ratio between these two energies = 2
8. The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is v. For a
satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is
3 3 2 2
(a) 2
v (b) 2
v (c) 3
v (d) 3
v
SOLUTION :
GM
v
Rh
GM
For first satellite h  0 , v1  R
R 2 GM
For second satellite h
2
, v2 
3R
2 2
v2  v1  v
3 3
9. In a satellite if the time of revolution is T, then K.E. is proportional to

1 1 1
(a) T (b) T 2 (c) T 3 (d) T 2 / 3
SOLUTION :
GM
v 
r
1
 K . E.  v 2 
r
T 2  r3
 K . E.  T  2 / 3
10. Two identical satellites are at R and 7R away from earth surface, the wrong statement is (R = Radius of
earth)
(a) Ratio of total energy will be 4 (b)Ratio of kinetic energies will be 4
(c) Ratio of potential energies will be 4
(d) Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2
SOLUTION :
Orbital radius of satellites r1  R  R  2R
r2  R  7 R  8 R
GMm GMm
U1  and U2 
r1 r2
GMm GMm
K1  and K2 
2r1 2r2
GMm GMm
E1  and E2 
2r1 2r2
U1 K1 E
U 
K2
 1 4
E2
2
  for r  R
A spherically symmetric gravitational system of particles has a mass density    0 for
0
11. rR
.

Where  0 is a constant. A test mass can undergo circular motion under the influence of the
gravitational field of particles. Its speed v as function of distance r from the centre of system
is represented by (2008 I)
V V V V
1) 2) 3) 4)
r r r r
R R R R
SOLUTION :
mv 2 GmM
For r  R;   1
r r2
4 3
here, M    r   0
3 
substituting in Eq(1)
we get vr
i.e., v-r graph is a straight line passing through origin,
for r>R
3 
2Gm   R 3  0
mv 4  1
 2
or v
r r r
The corresponding v-r graph will be as shown in option (3)
12. In the following four periods

(i)Time of revolution of a satellite just above the earth’s surface (Tst )
(ii)Period of oscillation of mass inside the tunnel bored along the diameter of the earth (Tma )
(iii)Period of simple pendulum having a length equal to the earth’s radius in a uniform field of
9.8 N/kg (Tsp )
(iv)Period of an infinite length simple pendulum in the earth’s real gravitational field (Tis )
(a) T st  Tma (b) Tma  Tst (c) T sp  Tis (d) Tst  Tma  Tsp  Tis
SOLUTION :
(R  h)3 R
(i) T st  2  2 [As h <<R and GM  gR 2 ]
GM g
R
(ii) Tma  2
g
1 R
Tsp  2  2
(iii) 1 1  2g [As l = R]
g  
l R
R
(iv) Tis  2
g
13. A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the
radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is
(a) 10 hr (b) (6/ 2 ) hr (c)6 hr (d) 6 2 hr
SOLUTION :
Distances of the satellite from the centre are 7R and 3.5R respectively.
3/2 3/2
T2  R 2   3 .5 R 
   T 2  24    6 2 hr
T1  R 1 
  7R 
14. If the gravitational force between two objects were proportional to 1/R (and not as 1 / R 2 ) where R
is separation between them, then a particle in circular orbit under such a force would have its orbital
speed v proportional to
(a) 1 / R 2 (b) R 0 (c) R1 (d)1/R
SOLUTION :
Gravitational force provides the required centripetal force for orbiting the satellite
mv 2 K  1
 because F  
R R  R
 v  R
15. Potential energy of a satellite having mass ‘m’ and rotating at a height of 6 .4  10 6 m from the earth
surface is
(a) 0 . 5 mgR e (b) mgR e (c) 2 mgR e (d) 4 mgR e
SOLUTION :
GMm GMm GMm
Potential energy =  
r Re  h 2Re
gR e2 m
= 
2Re
1
 mgR e
2
 0 . 5 mgR e
16. Distance between the centre of two stars is 10a. the masses of these stars are M and 16M and
their radii a and 2a respectively. A body of mass m is fired straight from the surface of the
larger start towards the surface of the smaller star. What should be its minimum initial speed
to reach the surface of the smaller star?
GM 1 5GM 3 GM 3 5 GM
1) 2) 3) 4)
a 2 a 2 a 2 a
SOLUTION :
Let there are two stars 1 and 2 as shown below.
M 16M
C1 r1 P r2 C2
a
2a
1 2
Let P is a point between C1 and C2, where gravitational field strength is zero.
GM G 16 M  r2
Hence  ;  4, r1  r2  10a
r12 r22 r1
 4 
 r2    10a   8a
 4 1
r1  2a
Now, the body of mass m is projected from the surface of large star towards the smaller one. Between
C2 and P it is attracted towards 2 and between C1 and P it will be attracted towards 1. Therefore, the
body should be projected to just cross point P because beyond that the particle is attracted towards the
smaller star itself.
1 2
From conservation of mechanical energy mv = potential energy of the body at P
2
= potential energy at the surface of larger star.
1 2  GMm 16GMm   GMm 16GMm 

 mvmin      
2  r1 r2   10a  2a 2a 
1 2  45  GMm
mvmin   
2  8  a
3 5  GM 
vmin   
2  a 
1 1
17. If g (instead of ), then the relation between time period of a satellite near earth’s surface
R3 R2
and radius R will be
(a) T 2  R 3 (b) T  R 2 (c) T 2  R (d) T  R
SOLUTION :
Gravitational force provides the required centripetal force
GMm
m 2 R 
R3
4 2 GM
 
T2 R4
6
18. Gravitational acceleration on the surface of a planet is g , where g is the gravitational
11
acceleration on the surface of the earth. The average mass density of the planet is 2/3 times
that of the earth. If the escape speed on the surface of the earth is taken to be 11kms -1, the
escape speed on the surface of the planet in kms-1 will be (2010I)
1) 3 2) 6 3) 9 4) 12
SOLUTION :
4 
G   R3  
GM 3
g 2   2 
R R
g
g   R; R 

Now escape velocity, ve  2 gR
g g2
ve  gR ; ve g  
 
6 3
 ve  planet  11kms 1    3kms 1
121 2
19. A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of
the moon’s orbit. The satellite completes one revolution in
1 2
(a) lunar month (b) lunar month (c) 2 3 / 2 lunar month (d) 2 3 / 2 lunar month
2 3
SOLUTION :
Time period of revolution of moon around the earth
= 1 lunar month.
3/2 3/2
Ts  rs  1
   
Tm  rm 
 2
 Ts  2 3 / 2 lunar month.
20. Two spher ical planets P and Q have the same unif or m density  , masses MP and MQ and
surface areas A and 4A respectively. A spherical planet R also has uniform density  and its
mass is (MP+MQ). The escape velocities from the planets P, Q and R are V P, VQ and VR,
respectively. Then (2012I)
1
1) VQ  VR  VP 2) VR  VQ  VP 3) VR / VP  3 4) VP / VQ 
2
SOLUTION :
2GM 2G8M
VP  ,VQ   2VP
R 2R
2G9M
VR  1/3
 91/3VP
9 R
21. A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth,
is velocity must be increased
(a) 100% (b) 41.4% (c) 50% (d) 59.6%
v e  2 v 0  1 .414 v 0
 v 
Fractional increase in orbital velocity  v 

ve  v0
  0 . 414
v0
 Percentage increase = 41.4%
22. There is a crater of depth R/100 on the surface of the moon (radius R). A projectile is fired
vertically upward from the crater with velocity, which is equal to the escape velocity v from the
surface of the moon. Find the maximum height attained by the projectile. (2003A)
1) 90R 2) 95R 3) 99.5R 4) 50R
SOLUTION :
2GM
Speed of particle at A,vA = escape velocity on the surface of earth 
R
At highest point B, vB=0
Applying conservation of mechanical energy,
decrease in kinetic energy=increase in gravitational potential energy
1 2
 mv A  U B  U A  m VB  VA 
2
v A2
 VB  VA
2
GM  GM  R  
2
GM 
 1.5 R   0.5  R 
2
  
R R  h  R 3   100   
2
1 1 3  1  99  1
     
R R  h 2 R  2  100  R
Solving this equation,
we get h=99.5R
Kepler’s Laws :
Kepler’s first law or laws of orbits:
‘Every planet revolves around the sun in elliptical orbit with the sun is at one of its focii.
F1 F2
2b
2a
As shown in fig., sun may be at F1 or F2. Here a and b denote the lengths of semi major and
semi minor axes.
The nearest position of the planet from the sun is called perihelion.’
The farthest position of the planet from the sun is called aphelion.
A planet of mass m is moving in an elliptical orbit around the sun(S) of mass ‘M’, at one of its focii.
rp
rA
M m
2b
P S O A
C a
SO e
Eccentricity of the elliptical path e  e  c  ea
OA a
From fig, rp  a  c  a  ea  a 1  e 
Similarly ra  a  c  a  ea  a 1  e 
From conservation of angular momentum at A and P, we have mVP rP  mVA rA
VP rA 1  e
 
VA rP 1  e
GM  1  e  GM  1  e 
From conservation of energy, we have VA    and VP   
a 1 e  a 1 e 
If e>1 and total energy (K.E+P.E)>0,
the path of the satellite is hyperbolic and it escapes from its orbit.
If e<1 and total energy is negative, it moves in an elliptical path.
If e=0 and total energy is negative, it moves in an elliptical path.
If e=1 and total energy is zero, it will take parabolic path.
The path of the projectile thrown to lower heights is parabolic and thrown to greater heights is elliptical.
Kepler’s second law or Laws of Areas:
The radius vector joining the planet to the sun sweeps out equal areas in equal intervals of time.
 dA 
Areal Velocity of radius vector   joining the planet to sun remains constant.
 dt 
dA
Mathematically  constant
dt
dl
d p
sun r
1 1 1
But A   dl  r   ed  r  r 2d
2 2 2
d 1 2 
So,  r    constant
dt  2 
1 2 d 1 2
 r  r   constant
2 dt 2
1 mr 2 d I  L
    constant
2 m 2m 2 m
L=constant
As the gravitational force on planet by sun is central, torque is zero and hence angular momentum of
the planet is constant.
This law is consequence of law of conservation of angular momentum.
dA L mVr Vr
  
dt 2m 2m 2
Areal velocity of radius vector of the planet is independent of mass of the satellite.
As angular momentum is conserved,
m(Vmax )( rmin )  m(Vmin )(rmax )
Vmax 1  e
 
Vmin 1  e
Here V perihelion  Vmax
and Vapehelion  Vmin
Kepler’s laws can be applied to natural and artificial satellites as well.
Kepler’s third law or Law of periods :

The square of period of revolution of a planet around the sun is proportional to cube of the
average distance of planet (i.e., semi major axis of elliptical orbit) from the sun.
rmax  rmin 1  e  a  1  e  a
rmean   a
2 2
Hence T 2  a 3
where ‘a’ is length of semi major axis of ellipse
The gravitational force between the planet and the Sun provides the necessary centripetal force for the
planet to go round the Sun.
If M = mass of Sun,
m = mass of planet
‘r = average distance of the planet from the Sun
GmM
then, F 2
 mr 2
r
GM 4 2  2 
 2  as   
r3 T  T 
r3
T 2  4 2  T 2  r3
GM
NOTE: The gravitational force is a central force so torque on planet relative to sun is always zero, hence
angular momentum of a planet or satellite is always constant irrespective of shape of orbit.
PROBLEMS
1. An artificial satellite is in an elliptical orbit around the earth with aphelion of 6R and perihelion
of 2R where R is Radius of the earth = 6400Km. Calculate the eccentricity of the elliptical
orbit.
SOLUTION :
We know that
perigee (r p)=a(1-e)=2R ........(1)
apogee (ra)=a(1+e)=6R.......(2)
Solving (1) & (2),
eccentricity (e)=0.5
2. The mean distance of a planet from the sun is approximately 1/4 times that of earth from the
sun. Find the number of years required for planet to make one revolution about the sun.
SOLUTION :
1
Given rP  rE and TE  1yr
4
For Kepler’s third law, T 2  r 3
3
2 3
 TP   rP   rP  2

     TP  TE 
 TE   rE   rE 
3
3
 r 2  1  2
TP  1  E      0.125Yrs
 4rE  4
3. The speed of the planet at the perihelion P be VP and the Sun-planet distance SP be rP as shown
in Fig. Relate {tP,VP} to the corresponding quantities at the aphelion {rA, VA}. Will the planet
take equal times to traverse BAC and CPB ?
SOLUTION :
The magnitude of the angular momentum at P is LP=mPVPrP,
The magnitude of the angular momentum at A is LA=mAVArA

rp
rA
B
s
2b
P sun A
C
According to law of conservation of angular momentum,

VP rA
mP rPVP  mP rAVA or V  r
A P
Here rA  rP hence VP  VA .
The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig.
From Kepler’s second law, equal areas are swept in equal times.
Hence, the planet will take a longer time to traverse BAC than CPB.
4. Let us consider that our galaxy consists of 2.5x10 11 stars each of one solar mass. How long
will this star at a distance of 50,000 light years from the galastic entre take to complete one
revolution? Take the diameter of the Milky way to be 105ly.G=6.67x10-11 Nm2Kg-2. (1 ly =
9.46x1015m)
SOLUTION :
Here M=2.5x1011 solar mass=2.5x1011x(2x1030)kg=5.0x1041kg
r=50,000 ly=50,000x9.46x1015m=4.73x1020m
4 2 r 3
We know that, M 
GT 2
1
 4   22 / 7 2   4.73  1020 3  2
1
 4 r  3 3 2
T    
 GM    6.67  1011    5.0  10 41  
 
=3.53x1014s.
5. If the distance between the earth and the sun becomes half its present value, the number of days in a year
would have been
(a) 64.5 (b) 129 (c) 182.5 (d) 730
SOLUTION:
According to Kepler’s third law, the ratio of the squares of the periods of
any two planets revolving about the sun is equal to the ratio of the cubes
of their average distances from the sun
2 3 3
 T1  r   r 
i.e.  T   1
 r
  1  8
 1
 2   2   2 r1 
1 T
 T 2 2
2
T1 365 days
 T2    129 days
2 2 2 2
6. A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the
time period of a satellite orbiting a few hundred kilometres above the earth’s surface (R Earth  6400 km )
will approximately be
(a)1/2 h (b) 1 h (c)2 h (d)4 h
SOLUTION:
3/2 3/2
T 2  r2   6400 
   T 2  24    2 hour
T1  r1   36000 
7. The distance of neptune and saturn from sun are nearly 10 13 and 10 12 meters respectively. Assuming
that they move in circular orbits, their periodic times will be in the ratio
(a) 10 (b) 100 (c) 10 10 (d) 1 / 10
SOLUTION:
3/2 3/2
T1  R 1   10 13 
    12   (1000 )1 / 2
T2  R 2 
  10


 10 10
8. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a
period of revolution T. If the gravitational force of attraction between planet and star is proportional
5
to R

2 , then T2 is proportional to
(a) R 3 (b) R 7 / 2 (c) R 5 / 2 (d) R 3 / 2
SOLUTION:
For revolution of planet centripetal force is provided by gravitational force of attraction
m  2 R  R 5 / 2
1
 R 7 / 2
T2
 T 2  R7 /2
9. Suppose the gravitational force varies inversely as the n th power of distance. Then the time period
of a planet in circular orbit of radius R around the sun will be proportional to
 n 1   n 1   n2 
(a) R

2 

(b) R

2 

(c) Rn (d) R

2 

SOLUTION:
1
m 2 R 
Rn
 4 2 
 m  T 2 R  1
 Rn
 
 T 2  R n 1
 n 1 
 
2 

T  R
10. An artificial satellite revolves around earth in circular orbit of radius r with time period T. The
satellite is made to stop in the orbit which makes it fall onto earth. Time of fall of the satellite
on to earth is given by
T 2 T 2T
1) 3 2) T 3) 4)
6 8 3 3
SOLUTION:
On stopping, the satellite will fall along the radius r of the orbit which can be regarded as a limiting case
of an ellipse with semi major axis r/2
Using Kepler’s third law T 2r 3
T' T 2T
time of fall =  
2 2 8 8
11.The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit
of radius 4R is
(a) 4T (b) T/4 (c) 8T (d) T/8
SOLUTION:
3/2 3/2
T1  R 1   R 
     T 2  8 T1
T2  R 2 
  4R 
12. A planet moves around the sun. At a given point P, it is closest from the sun at a distance d1 and has
a speed v1 . At another point Q, when it is farthest from the sun at a distance d 2 , its speed will be
d12v1 d 2v1 d1v1 d 22v1
(a) d 22
(b) d1 (c) d2 (d) d12
SOLUTION:
Angular momentum remains constant
v1 d 1
mv 1 d 1  mv 2 d 2  v 2 
d2
13. A satellite moving in elliptical orbit around earth as shown. The minimum and maximum distance
of the satellite from earth are 3 units 5 units respectively. The distance of satellite from earth
when it is at ‘P’ is ______ (units)
1) 4 2) 3 3) 3.75 4) 6
S EO Q
SOLUTION:
1
Semi major axis = 4  ae  1  e 
4
Semi minor axis = b
1
b  a 1  e2  1  15  15
16
Required distance  b 2  1  4
14. The largest and the shortest distance of the earth from the sun are r1 and r2 , its distance from the
sun when it is at the perpendicular to the major axis of the orbit drawn from the sun
r1  r2 r1r2 2r1r2 r1  r2
(a) 4
(b) r1  r2 (c) r1  r2 (d) 3
SOLUTION:
The earth moves around the sun is elliptical path. so by using the properties of ellipse
r1  (1  e ) a
r2  (1  e ) a
r1  r2
a 
2
r1 r2  (1  e 2 ) a 2
where a = semi major axis
b = semi minor axis
e = eccentricity
b2
Now required distance = semi latusrectum 
a
a 2 (1  e 2 ) (r1 r2 ) 2r1 r2
  
a (r1  r2 ) / 2 r1  r2
15. A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the
orbit is R0 and mass of the earth M, the angular momentum about the centre of the earth is
GM GM
(a) m GMR 0 (b) M GmR 0 (c) m
R0 (d) M
R0
SOLUTION:
Angular momentum = Mass ×Orbital velocity ×Radius
 GM 
 m    R 0  m GMR 0
 R0 
 
16. The maximum and minimum distances of a comet from the sun are 8  10 12 m and 1 .6  10 12 m . If its
velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest
(a) 12 (b) 60 (c)112 (d)6
SOLUTION:
By conservation of angular momentum mvr = constant
v min  rmax  v max  rmin
60  1 . 6  10 12 60
 v min    12 m / s
8  10 12 5
17.A body revolved around the sun 27 times faster then the earth what is the ratio of their radii
(a) 1/3 (b) 1/9 (c)1/27 (d)1/4
SOLUTION:
 body  27  earth
1 1 1
T 2  r3   2    r 
r 3
r 3/2
 2/3
2/3
rbody  
 r   earth 
  body 
earth  
2/3
 1  1
  
 27  9
18. The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance
is increased to 16r, then the new angular momentum will be
L
(a) 16 L (b) 64 L (c) 4 (d)4 L
SOLUTION:
GM
L  mvr  m r  m GMr  L  r
r
19. The longest and the shortest distance of a planet from sun is R 1 and R2. Distance from sun
when it is normal to major axis of orbit is
R1  R2 R12  R22 R1R2 2R1R2

1) 2) 3) R  R 4) R  R
2 2 1 2 1 2
SOLUTION:
R1  1  e  a; R2  1  e  a
R1  R2
a ; R1R2  1  e 2  a 2
2
b2
semi-latus rectum =
a
a 2 1  e2  R1R2 2 R1 R2
  
a R1  R2 R1  R2
2
2 0 . The mass of a planet that has a moon whose time period and orbital radius are T and R respectively can
be written as
(a) 4  2 R 3 G 1 T 2 (b) 8  2 R 3 G 1 T 2 (c) 12 2 R 3 G 1T 2 (d) 16  2 R 3 G 1 T 2
SOLUTION:
2
GMm  2  GM
m 2 R    R 2
R2  T  R
4 2 R 3
M 
GT 2
21. A point mass is orbiting a significant mass M lying at the focus of the elliptical orbital having
major and minor axes given by 2a and 2b respectively. Let r be the distance between the mass
M and the end point of major axis. Velocity of the particle can be given as
2ab GM
ab GM ab GM ab GM 3
1) 2) 3) 4) r ab
r a3 r b3 2r r 3  
 2 
SOLUTION:
Gravitational force = Centripetal force
GM v2
 
r2 r'
here r is the radius of curvature. From Kepler’s law, time period is given by,
a3 2 ab
T  2 
GM rv
 dA vr 2 dA 2 
   dt  T    ab 
 dt 2 vr rv 
ab GM
v
r a3
22. A planet of mass m revolves in elliptical orbit around the sun so that its maximum and minimum
distances from the sun are equal to ra and rp respectively. The angular momentum of this
planet relative to the sun is
GMrP ra 2GMrP ra GmrP ra  rP  ra 

1) L  m 2) L  m 3) L  M 4) L  M
 rP  ra   rP  ra   rP  ra  GmrP ra
SOLUTION:
GMm 1 GMm 1
From conservation of energy  r 
2
mv P2 
ra
 mva2
2
P
L  mvP rP  mva ra
23. A satellite is orbiting just above the surface of a planet of average density D with period T. If
3
G is the universal gravitational constant, the quantity is equal to
G
1) T2D 2) 3 T 2 D 3) 3 D 2T 4) D 2T
SOLUTION:
4 2 R 3 3 3
R3 R3 T2     T 2D
Using T  2  GM
 2
4 : 4 3
G  R D DG G
G   R 3D
3 3
24. A planet revolves around sun in an elliptical orbit of eccentricity ‘e7’. If ‘T’ is the time period
of the planet then the time spent by the planet between the end of the minor axis and close to
sun is
1 e  Te e  T
1) T    2) 3)   1 4)
 4 2     e
SOLUTION:
dA
=constant;
dt
 ab 1
 b  ea    1 e
t AB  Area  SAB  4 2 1 e 
  4 2  T  4  2 
T  Area  ellipse  ab 
Polar Satellites:
These are low altitude (500 km to 800 km) satellites
They go round the poles of earth in north-south direction
Polar satellites have a time period of 100 minutes nearly
These satellites can view polar and equatorial regions at close distances with good resolution.
These satellites are useful for remote sensing, meteorology and environmental studies of earth.
Condition of Weightlessness in a Satellite:
GMm mv02
The force acting on the astronaut of mass ‘m’ is  FR  here FR is the reactional force
r2 r
The reactional force on the floor of the satellite is zero, hence there is the state of weight lessness
in a satellite.
GMm mv02
i.e., 
r2 r
As the frame of reference attached to the satellite is an accelerated frame, whose acceleration towards
v02 GM
the centre of the earth is a 
 2 g
r r
GAUSS THEOREM IN GRAVITATION IS:
 
g  4 G  M enclosed    E g .dA
:: THEORY BITS ::
1. If the earth is at one-fourth of its present distance from the sun, the duration of the year would
be
1) Half the present year 2) One-eight the present year
3) One-fourth the present year 4) One -sixteenth the present year
KEY :2
2. The radius vector drawn from the sun to a planet sweeps out ___ areas in equal time
1) equal 2) unequal 3) greater 4) less
KEY :1
3. Feeling of weightlessness in a satellite is due to
1) absence of inertia 2) absence of gravity
3) absence of accelerating force 4) free fall of satellite
KEY :4
4. According to Kepler’s second law, line joining the planet to the sun sweeps out equal areas in
equal intervals. This suggests that for the planet
1) radial acceleration is zero 2) tangential acceleration is zero
3) transverse acceleration is zero 4) All
KEY :3
5. If Fg and Fe are gravitational and electrostatic forces between two electrons at a distance 0.1 m
then Fg / Fe is in the order of
1) 1043 2) 10–43 3) 1035 4) 10–35
KEY :2
Gm1m2
6. F is valid
r2
1) Between bodies with any shape 2) Between particles
3) Between any bodies with uniform density 4) Between any bodies with same shape
KEY :2
7. Fg, Fe and Fn represent the gravitational, electro-magnetic and nuclear forces respectively,
then arrange the increasing order of their strengths
1) Fn,Fe, Fg 2) Fg,Fe, Fn 3) Fe,Fg, Fn 4) Fg,Fn, Fe
KEY :2
8. The time period of revolution of geo-stationary satellite with respect to earth is
1) 24 hrs 2) 1 year 3 ) Infinity 4) zero
KEY :3
9. Find the false statement
1) Gravitational force acts along the line joining the two interacting particles
2) Gravitational force is independent of medium
3) Gravitational force forms an action-reaction pair
4) Gravitational force does not obey the principle of superposition.
KEY :4
10. A relay satellite transmits the television programme from one part of the world to another part
continuously because its period
1) is greater than period of the earth about its axis
2) is less than period of rotation of the earth about its axis.
3) has no relation with the period of rotation of the earth about its axis.
4) is equal to the period of rotation of the earth about its axis.
KEY :4
11. Attractive Force is exists between two protons inside the Nucleus this is due to
1) Gravitational Forces 2) Electro magnetic Forces
3) Weak Nuclear Forces 4) Strong Nuclear Forces
KEY :4
12. Two equal masses separated by a distance d attract each other with a force (F). If one unit of
mass is transferred from one of them to the other, the force
1) does not change 2) decreases by (G/d2)
3) becomes d times
2
4) increases by (2G/d2)
KEY :2
13. When a satellite falls into an orbit of smaller radius its speed
1) decreases 2) increases 3) does not change 4) zero
KEY :2
14. Which of the following is the evidence to show that there must be a force acting on earth and
directed towards Sun?
1) Apparent motion of sun around the earth 2) Phenomenon of day and night
3) Revolution of earth round the Sun 4) Deviation of the falling body towards earth
KEY :3
15. Polar satellite go round the poles of earth in
1) South-east direction 2) north-west direction
3) east-west direction 4) north-south direction
KEY :4
16. If suddenly the gravitational force of attraction between earth and satellite revolving around it
becomes zero, then the satellite will (AIEEE 2002)
1) Continue to move in its orbit with same velocity
2) Move tangential to the original orbit with the same velocity
3) Becomes stationary in its orbit 4) Move towards the earth
KEY :2
17. A synchronous satellite should be at a proper height moving
1) From west to East in equatorial plan 2) From South to North in polar plane
3) From East to west in equatorial plan 4) From North to South in polar plane
KEY :1
18. If the speed of rotation of earth about its axis increases, then the weight of the body at the
equator will
1) increase 2) decrease
3) remains unchanged 4) some times decrease and sometimes increase
KEY :2
19. A body is dropped from a height equal to radius of the earth. The velocity acquired by it before
touching the ground is
1) V= 2 gR 2) V=3gR 3) V= gR 4) V=2gR
KEY :3
20. The ratio of acceleration due to gravity at a depth ‘h’ below the surface of earth and at a height
‘h’ above the surface for h<<R
1) constant only when h<<R 2) increases linearly with h
3) increases parabolically with h 4) decreases
KEY :2
21. A person will get more quantity of matter in Kg-Wt at
1) poles 2) a latitude of 600 3) equator 4) satellite
KEY :3
22. A geo-stationary satellite has an orbital period of
1) 2 hours 2) 6 hours 3) 24 hours 4) 12 hours
KEY :3
23. A pendulum clock which keeps correct time at the surface of the earth is taken into a mine,
then
1) it keeps correct time 2) it gains time 3) it loses time 4) none of these
KEY :3
24. A satellite is revolving in a elliptical orbit in free space; then the false statement is
1) its mechanical energy is constant 2) its linear momentum is constant
3) its angular momentum is constant 4) its areal velocity is constant
KEY :2
25. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very
small compared to the mass of the earth
1) the acceleration of S is always directed towards the centre of the earth
2) the angular momentum of S about the centre of the earth changes in direction, but its magnitude
remains constant
3) the total mechanical energy of S varies periodically with time
4) the linear momentum of S remains constant in magnitude
KEY :1
26. Assuming the earth to be a sphere of uniform density the acceleration due to gravity
1) at a point outside the earth is inversely proportional to the square of its distance from the centre
2) at a point outside the earth is inversely proportional to its distance from the centre
3) at a point inside is zero
4) at a point inside is inversely proportional to its distance from the centre.
KEY :1
27. An artificial satellite of mass m is revolving round the earth in a circle of radius R. Then work
done in one revolution is
mgR
1) mgR 2) 3) 2 R  mg 4) Zero
2
KEY :4
28. If earth were to rotate faster than its present speed, the weight of an object
1) increase at the equator but remain unchanged at poles
2) decrease at the equator but remain unchanged at the poles
3) remain unchanged at the equator but decrease at the poles
4) remain unchanged at the equator but increase at the poles
KEY :2
29. A satellite is revolving round the earth. lts kinetic energy is E k. How much energy is required
by the satellite such that it escapes out of the gravitational field of earth
Ek
1) 2 Ek 2) 3Ek 3) 4) infinity
2
KEY :1
30. If R=radius of the earth and g =acceleration due to gravity on the surface of the earth, the
acceleration due to gravity at a distance (r<R) from the centre of the earth is proportional to
1) r 2) r2 3) r–2 4) r–1
KEY :1
31. Earth is flattened at poles and bulged at equators this is due to
1) revolution of earth around the sun is an elliptical orbit
2) angular of velocity of spinning about its axis is more at equator
3) centrifugal force is more at equator than poles
4) more centrifugal force at poles than equator
KEY :3
32. Intensity of gravitational field inside the hollow spherical shell is
1) Variable 2) minimum 3) maximum 4) zero
KEY :4
33. If the universal gravitational constant increases uniformly with time, then a satellite in orbit
will still maintain its
1) weight 2) tangential speed 3) period of revolution 4) angular momentum
KEY :4
34. The work done by an external agent to shift a point mass from infinity to the centre of the
earth is ‘W’. Then choose the correct relation.
1) W=0 2) W>0 3) W<0 4) W  0
KEY :3
35. When a satellite in a circular orbit around the earth enters the atmospheric region, if encounters
small air resistance to its motion. Then
1) its angular momentum about the earth decreases 2) its kinetic energy decreases
3) its kinetic energy remains constant 4) its period of revolution around the earth increases
KEY :1
36. The intensity of the gravitational field of the earth maximum?
1) centre of earth 2) equator 3) poles 4) same everywhere
KEY :3
37. Let VG and EG denote gravitational potential and field respectively, then choose the wrong
statement.
1) VG  0, EG  0 2) VG  0, EG  0 3) VG  0, EG  0 4) VG  0, EG  0
KEY :3
38. The time period of an earth’s satellite in circular orbit is independent of
1) the mass of the satellite 2) radius of its orbit
3) both the mass and radius of the orbit 4) neither the mass of the satellite nor the radius of its orbit
KEY :1
39. A thin spherical shell of mass ‘M’ and radius ‘R’ has a small hole. A particle of mass ‘m’ is
released at the mouth of them. Then
1) the particle will execute S.H.M inside the shell
2) the particle will oscillate inside the shell, but the oscillations are not simple harmonic
3) the particle will not oscillate, but the speed of the particle will go on increasing
4) none of these
KEY :4
40. A hole is drilled through the earth along a diameter and a stone is dropped into it. When the
stone is at the centre of the earth, it has finite a) weight b) acceleration c) P.E. d) mass
1) a & b 2) b & c 3) a, b & c 4) c & d
KEY :4
41. A body has weight (w) on the ground. The work which must be done to lift it to a height equal to
the radius of the earth is
1) equal to WR 2) greater than WR 3) less than WR 4) we can’t say
KEY :3
42. The time period of a simple pendulum at the centre of the earth is
1) zero 2) infinite 3) less than zero 4) none of these
KEY :2
43. If the gravitational force of earth suddenly disappears, then,
1) weight of the body is zero 2) mass of the body is zero
3) both mass and weight become zero 4) neither the weight nor the mass is zero
KEY :1
44. A hallow spherical shell is compressed to half it radius. The gravitational potential at the
centre
1) increases 2) decreases 3 ) remains same
4) during the compression increases then returns to the previous value
KEY :2
45. A satellite is moving with constant speed ‘V’ is a circular orbit about earth. The kinetic energy
of the satellite is
1 3
1) mV 2 2) mV 2 3) mV 2 4) 2mV 2
2 2
KEY :1
46. For a satellite projected from the earth’s surface with a velocity greater than orbital velocity
the nature of the path it takes when its energy is negative, zero and positive respectively is
1) Elliptical, parabolic and hyperbolic 2) Hyperbolic, parabolic and elliptical
3) Elliptical, circular and parabolic 4) Parabolic, circular and Elliptical
KEY :1
47. If the mean radius of earth is R, its angular velocity is  and the acceleration due to gravity
at the surface of the earth is ‘g’ then the cube of the radius of the orbit of a satellite will be
Rg R2g R2g R 2
1) 2) 3) 4)
2  2 g
KEY :3
48. If satellite is orbiting in space having air and no energy being supplied, then path of that
satellite would be
1) circular 2) elliptical
3) spiral of increasing radius 4) spiral of decreasing radius
KEY :4
49. The earth retains its atmosphere, due to
1) The special shape of the earth
2) The escape velocity being greater than the mean speed of the molecules of the atmospheric gases.
3) The escape velocity being smaller than the mean speed of the molecules of the atmospheric gases.
4) The sun’s gravitational effect.
KEY :2
50. Ratio of the radius of a planet A to that of planet B is ‘r’. The ratio of accelerations due to
gravity for the two planets is x. The ratio of the escape velocities from the two planets is
1) rx 2) r / x 3) r 4) x/r
KEY :1
51. The ratio of the escape velocity and the orbital velocity is
1) 2 2) 1 / 2 3) 2 4) 1/2
KEY :1
52. The escape velocity from the earth for a rocket is 11.2 km/sec. Ignoring the air resistance, the
escape velocity of 10 mg grain of sand from the earth will be
1) 0.112 km/sec 2) 11.2 km/sec 3) 1.12 km/sec 4) None
KEY :2
53. The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/
s. If the body is projected at an angle of 450 with the vertical, the escape velocity will be
1) 11 2 km / s 2) 22 km / s 3) 11km / s 4 ) 11 2 km / s
KEY :3
54. Following physical quantity is constant when a planet that revolves around Sun in an elliptical
orbit.
1) Kinetic energy 2) Potential energy 3) Angular momentum 4) Linear velocity
KEY :3
55. A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and
potential energies is
1) Positive 2) Negative 3) Zero
4) May be positive or negative depending upon its initial velocity
KEY :2
56. The escape velocity of a body depends upon its mass as
1) m0 2) m1 3) m3 4) m 2
KEY :1
57. If the universal gravitational constant decreases uniformly with time, then a satellite in orbit
will still maintain its
1) weight 2) tangential speed 3) period of revolution 4) angular momentum
KEY :4
58. The magnitude of potential energy per unit mass of the object at the surface of earth is ‘E’.
Then escape velocity of the object is
1) 2E 2) 4E 2 3) E 4) E / 2
KEY :1
59. Tidal waves in the sea are primarily due to
1) the gravitational effect of the moon on the earth
2) the gravitational effect of the sun on the earth
3) the gravitational effect of the venus on the earth
4) the atmospheric effect of the earth itself
KEY :1
60. A space station is set up in space at a distance equal to earth’s radius from earth’s surface.
Suppose a satellite can be launched from space station. Let V1 and V2 be the escape velocities
of the satellite on earth’s surface and space station respectively. Then
1) V2  V1 2) V2  V1 3) V2  V1 4 ) No relation
KEY :2
61. The minimum number of geo-stationary satellites required to televise a programme all over
the earth is
1) 2 2) 6 3) 4 4) 3
KEY :4
62. When a satellite going around the earth in a circular orbit of radius r and speed v loses some
of its energy , then
1) r and v both increase 2) r and v both decrease
3) r will increase and v will decrease 4) r will decrease and v will increase
KEY :4
63. The satellite is orbiting a planet at a certain height in a circular orbit. If the mass of the planet
is reduced to half, the satellite would
1) fell on the planet 2) go to orbit of smaller radius
3) go to orbit of higher radius 4) escape from the planet
KEY :4
64. If R = radius of the earth and g = acceleration due to gravity on the surface of the earth, the
acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to
1) r 2) r 2 3) r–2 4) r–1
KEY :3
65. A satellite is revolving round the earth in an elliptical orbit. Its speed will be
1) same at all points of the orbit 2) different at different points of the orbit
3) maximum at the farthest point 4) minimum at the nearest point
KEY :2
66. A satellite is moving in a circular orbit round the earth. If any other planet comes in between
them, it will
1) Continue to move with the same speed along the same path
2) Move with the same velocity tangential to original orbit.
3) Fall down with increasing velocity.
4) Come to rest after moving certain distance along original path.
KEY :2
67. A space-ship entering the earth's atmosphere is likely to catch fire. This is due to
1) The surface tension of air 2) The viscosity of air
3) The high temperature of upper atmosphere
4) The greater portion of oxygen in the atmosphere at greater height.
KEY :2
68. An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth, gently
drops a ball from the space-ship. The ball will
1) Move randomly in space 2) Move along with the space-ship
3) Fall vertically down to earth 4) Move away from the earth
KEY :2
69. The energy required to remove an earth satellite of mass ‘m’ from its orbit of radius ‘r’ to
infinity is
GMm  GMm GMm Mm
1) 2) 3) 4)
r 2r 2r 2r
KEY :3
70. Pseudo force also called fictitious force such as centrifugal force arises only in
1) Inertial frames 2) Non-intertial frames
3) Both inertial and non-inertial frames 4) Rigid frames
KEY :2
71 A satellite launching station should be L
1) Near the equatorial region 2) Near the polar region
3) On the polar axis 4) At any place
KEY :1
72. The period of a satellite moving in circular orbit near the surface of a planet is independent of
1) mass of the planet 2) radius of the planet
3) mass of the satellite 4) density of planet
KEY :3
73. The motion of a planet around sun in an elliptical orbit is shown in the following figure. Sun is
situated on one focus. The shaded areas are equal. If the planet takes time ‘ t1 ’ and ‘ t2 ’ in
moving from A to B and from C to D respectively, then
D C
A B
1) t1 > t2 2) t1 < t2 3) t1 = t2 4) incomplete information

KEY :3
74. Two identical trains A and B move with equal speeds on parallel tracks along the equator. A
moves from east to west and B moves from west to east. Which train will exert greater force
on the track?
1) A 2) B 3) they will exert equal force
4) The mass and the speed of each train must be known to reach a conclusion.
KEY :1
75. Out of the following statements, the one which correctly describes a satellite orbiting about
the earth is
1) There is no force acting on the satellite
2) The acceleration and velocity of the satellite are roughly in the same direction
3) The satellite is always accelerating about the earth
4) The satellite must fall, back to earth when its fuel is exhausted.
KEY :3
76. Out of the following interactions weakest is
1) gravitational 2) electromagnetic 3) nuclear 4) electrostatic
KEY :1
77. When an astronaut goes out of his space-ship into the space he will
1) Fall freely on the earth 2) Go upwards
3) Continue to move along with the satellite in the same orbit. 4) Go spiral to the earth
KEY :3
78. If a satellite is moved from one stable circular orbit to a farther stable circular orbit, then the
following quantity increases
1) Gravitational force 2) Gravitational P.E.
3) linear orbital speed 4) Centripetal acceleration
KEY :2
79. When the height of a satellite increases from the surface of the earth.
1) PE decreases, KE increases 2) PE decreases, KE decreases
3) PE increases, KE decreases 4) PE increases, KE increases
KEY :3
80. Neutron changing into Proton by emitting electron and anti neutrino this due to
KEY :3
81. If S1 is surface satellite and S2 is geostationary satellite, with time periods T1 and T2, orbital
velocities V1 and V2,
1) T1 > T2; V1 >V2 2) T1 > T2; V1<V2
3) T1 < T2; V1 <V2 4) T1 < T2; V1 >V2
KEY :4
82. Among the following find the wrong statement is
1) Law of gravitation is framed using Newton’s third law of motion
2) Law of gravitation cannot explain why gravity exists
3) Law of gravitation does not explain the presence of force even when the particles are not in physical
contact
4) When the range is long, gravitational force becomes repulsive.
KEY :4
83. The following statement is correct about the motion of earth satellite.
1) It is always accelerating towards the earth
2) There is no force acting on the satellite
3) Move away from the earth normally to the orbit
4) Fall down on to the earth
KEY :1
84. Two satellites of masses m1 and m2 (m1 > m2) are revolving around earth in circular orbits of
radii r1 and r2 (r1 > r2) respectively. Which of the following statements is true regarding their
velocities V1 and V2.
V V
4) r  r
1 2
1) V1 = V2 2) V1 < V2 3) V1 > V2
1 2
KEY :2
85. If the area swept by the line joining the sun and the earth from Feb 1 to Feb 7 is ‘A’, then the
area swept by the radius vector from Feb 8 to Feb 28 is
1) A 2) 2A 3) 3A 4) 4A
KEY :3
86. An earth satellite is moved from one stable circular orbit to another larger and stable circular
orbit. The following quantities increase for the satellite as a result of this change
1) gravitational potential energy 2) angular velocity
3) linear orbital velocity 4) centripetal acceleration
KEY :1
87. Average density of the earth (2005A)
1) does not depend on ‘g’ 2) is a complex function of ‘g’
3) is directly proportional to ‘g’ 4) is inversely proportional to ‘g’
KEY :3
88. Two artificial satellites are revolving in the same circular orbit. Then they must have the same
1) Mass 2) Angular momentum
3) Kinetic energy 4) Period of revolution
KEY :4
89. Two identical spherical masses are kept at some distance. Potential energy when a mass ‘m’
is taken from the surface of one sphere to the other
1) increases continuously 2) decreases continuously
3) first increases, then decreases 4) first decreases, then increases
KEY :3
90. Six particles each of mass ‘m’ are placed at the corners of a regular hexagon of edge length
‘a’. If a point mass ‘ m0 ’ is placed at the centre of the hexagon, then the net gravitational force
on the point mass is
6Gm2 6Gmm0 6Gm
1) 2) 2 3) zero 4)
a 2
a a4
KEY :3
91. An artificial satellite of the earth releases a packet If air resistance is neglected, the point
where the packet will hit, will be
1) ahead 2) exactly below 3) behind 4) it will never reach the earth
KEY :4
92. A satellite in vacuum
1) is kept in orbit by solar energy 2) previous energy from gravitational field
3) by remote control 4) No energy is required for revolving
KEY :4
93. The gravitational field is a conservative field. The work done in this field by moving an object
from one point to another
1) depends on the end-points only
2) depends on the path along which the object is moved
3) depends on the end-points as well as the path between the points.
4) is not zero when the object is brought back to its initial position.
KEY :1
94. Two heavenly bodies s1 & s2 not far off from each other, revolve in orbit
1) around their common centre of mass
2) s1 is fixed and s2 revolves around s1
3) s2 is fixed and s1 revolves around s2
4) cannot say
KEY :1
95. A body of mass 5 kg is taken into space. Its mass becomes.
1) 5 kg 2) 10 kg 3) 2 kg 4) 30 kg
KEY :1
96. If V , T , L, K and r denote speed, time period, angular momentum, kinetic energy and radius
of satellite in circular orbit
a) V  r 1 b) L r 1/ 2 c) T  r 3/ 2 d) K r 2
1) a,b are true 2) b,c are true 3) a,b,d are true 4) a,b,c are true
KEY :2
97. Two satellites are revolving around the earth in circular orbits of same radii. Mass of one
satellite is 100 times that of the other. Then their periods of revolution are in the ratio
1) 100:1 2) 1:100 3) 1:1 4) 10:1
KEY :3
98. Two similar satellite s1 and s2 of same mass ‘m’ have completely inelastic collision while
orbiting earth in the same circular orbit in opposite direction then
1) total energy of satellites and earth system become zero
2) the satellites stick together and fly into space
3) the combined mass falls vertically down
4) the satellites move in opposite direction
KEY :3
99. For a planet revolving round the sun, when it is nearest to the sun is
1) K.E. is min and P.E. is max. 2) Both K.E. and P.E. are min
3) K.E. is max. and P.E. is min 4) K.E. and P.E. are equal
KEY :3
100. Consider earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist
B goes high up in a balloon. The gravitational field measured by
1) A goes on decreasing and that of B goes on increasing
2) B goes on decreasing and that of A goes on increasing
3) Each decreases at the same rate
4) Each decreases at different rates.
KEY :4
101. When projectile attains escape velocity, then on the surface of planet , its
1) KE  PE 2) PE  KE 3) KE  PE 4) KE  2 PE
KEY :3
102. The orbit of geo-stationary satellite is circular, the time period of satellite depends on (2008 E)
1) mass of the Earth 2) radius of the orbit
3) height of the satellite from the surface of Earth 4) all the above
KEY :4
103. A gravitation field is present in a region. A point mass is shifted from A to B, along different
paths shown in the figure. If W1 , W2 and W3 represent the work done by gravitational force for
respective paths, then
Path 1
Path 2
A B
Path 3
1) W1  W2  W3 2) W1  W2  W3 3) W1  W3  W2 4) none of these
KEY :1
104. The orbital angular velocity vector of a geo-stationary satellite and the spin angular velocity
vector of the earth are
1) always in the same direction 2) always in opposite direction
3) always mutually perpendicular 4) inclined at 23 1/20 to each other
KEY :1
105. Repulsive force exist between two protons out side the nucleus this due to
KEY :2
106. Radio activity decay exist due to
KEY :3
107. It is not possible to keep a geo-stationary satellite over Delhi. Since Delhi
1) is not present in A.P 2) is capital of India
3) is not in the equatorial plane of the earth 4) is near Agra.
KEY :3
108. If the area swept by the line joining the sun and the earth from Feb 1 to Feb 7 is ‘A’, then the
area swept by the radius vector from Feb 8 to Feb 28 is
1) A 2) 2A 3) 3A 4) 4A
KEY :3
109. The angle between the equatorial plane and the orbital plane of geo-stationary satellite is
1) 450 2) 00 3) 900 4) 600
KEY :2
110. The angle between the equatorial plane and the orbital plane of a polar satellite is
1) 450 2) 00 3) 900 4) 600
KEY :3
111. Law of gravitation is not applicable if
A) Velocity of moving objects are comparable to velocity of light
B) Gravitational field between objects whose masses are greater than the mass of sun.
1) A is true, B is false 2) A is false, B is true
3) Both A & B are true 4) Both A&B are false
KEY :3
112. Which of the following quantities remain constant in a planetary motion, when seen from the
surface of the sun.
1) K.E 2) angular speed
3) speed 4) Angular momentum
KEY :4
::PRACTICE BITS ::
1. A planet revolves round the sun in an elliptical orbit of semi minor and semi major axes x and
y respectively. Then the time period of revolution is proportional to
3 3 3 3
1)  x  y  2 2)  y  x  2 3) x 2 4) y 2
KEY:4
HINT:
From Kepler’s 3rd law,
T 2 r 3
2. The moon revolves round the earth 13 times in one year. If the ratio of sun-earth distance to
earth-moon distance is 392, then the ratio of masses of sun and earth will be
1) 365 2) 365x10-12 3) 3.56x105 4) 1
KEY:3
HINT
r3
From Kepler’s 3rd law, T 
2
M
3. An artificial satellite is revolving around the earth in a circular orbit. Its velocity is one-third of
the escape velocity. Its height from the earth’s surface is (in Km)
1) 22400 2) 12800 3) 3200 4) 1600
KEY:1
HINT
1 2 mgh
mv 
2 h
1
R
OA
4. The Earth moves around the Sun in an elliptical orbit as shown in the figure. The ratio  x.
OB
Then ratio of the speed of the Earth at B and at A is nearly
B Earth
o A
Sun
1) x 2) x 3) x x 4) x 2
KEY:2
HINT
From conservation of angular momentum
mvr=constant, v1r1=v2r2
5. The period of moon’s rotation around the earth is nearly 29 days. If moon’s mass were 2 fold
its present value and all other things remain unchanged, the period of moon’s rotation would
be nearly (in days)
1) 29 2 2) 29 / 2 3) 29 3 4) 29
KEY:4
HINT
Time period does not depend upon the mass of the satellite
6. A spaceship is launched into a circular orbit of radius ‘R’ close to the surface of earth. The
additional velocity to be imparted to the spaceship in the orbit to overcome the earth’s
gravitational pull is (g = acceleration due to gravity)
1) 1.414Rg 2) 1.414 Rg 3) 0.414Rg 4) 0.414 gR
KEY:4
HINT
V  Ve  V0  2 gR  gR  gR  
2 1
7. If the mass of earth were 2 times the present mass, the mass of the moon were half the present
mass and the moon were revolving round the earth at the same present distance, the time
period of revolution of the moon would be (in days)
1) 56 2) 28 3) 14 2 4) 7
KEY:3
HINT
r3
From Kepler’s 3rd law, T  2
GM
8. If the Earth shrinks such that its density becomes 8 times to the present value, then new
duration of the day in hours will be (2008M)
1) 24 2) 12 3) 6 4) 3
KEY:3
HINT
Given m1=m2  V1d1=V2d2
 R13d1  R23d 2  R1  2 R2
From law of conservation of angular momentum
2
T2 R 
I 1 1  I 2 2   2 
T1  R1 
9. If a graph is plotted between T2 and r3 for a planet then, its slope will be
T2
 r3
4 2 GM
1) 2) 3) 4 GM 4) Zero
GM 4 2
KEY:1
HINT
T 2 4 2
Slope   (From Kepler’s 3rd law)
R 3 GM
10. The ratio of Earth’s orbital momentum (about sun) to its mass is 4.4x1015m2s-1. The area enclosed
by the earth’s orbit is approximately
1) 1x1022m2 2) 3x1022m2 3) 5x1022m2 4) 7x1022m2
KEY:4
HINT
Areal velocity = Area swept/time for one revolution of earth about sun
So, Area = (Areal velocity)(Time period)
L
  365  86400
2m
11. Two identical spheres each of radius R are placed with their centres at a distance nR, where n
is integer greater than 2. The gravitational force between them will be proportional to
1) 1/R4 2) 1/R2 3) R2 4) R4
KEY:4
HINT
Gm1m2 4
F ; Here m   R
3
2
R 3
12. The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to
escape from the gravitational field of the earth is.
1) 2E 2) 4E 3) 2 2 E 4) 2E
KEY:1
HINT
K e 2 gR
  Ke  2K0
K0 gR
13. A satellite is orbiting round the earth. If both gravitational force and centripetal force on the
satellite is F, then, net force acting on the satellite to revolve round the earth is
1) F/2 2) F 3) 2F 4) Zero
KEY:2
HINT
Gravitational force provides centripetal force.
14. Mass M=1 unit is divided into two parts X and (1-X). For a given separation the value of X for
which the gravitational force between them becomes maximum is
1) 1/2 2) 3/5 3) 1 4) 2
KEY:1
HINT
G  m1  x mx 1
F 2 is maximum when x 
R 2
15. Gravitational force between two point masses m and M separated by a distance is F. Now if
point mass 3m is placed next to m, the total force on M will be
1) F 2) 2F 3) 3F 4) 4F
KEY:4
HINT
GMm GM  m  2m 
F 2
;F ' 
r r2
16. If the distance between the sun and the earth is increased by three times, then the gravitational
force between the two will
1) remain constant 2) decrease by 63% 3) increase by 63% 4) decrease by 89%
KEY:4
HINT
Gm1m2
F
R2
17. Energy required to shift a body of mass ‘m’ from an orbit of radius 2R and 3R is (2002A)
GMm GMm GMm GMm
1) 2) 3) 4)
12 R 3R 2 8R 6R
KEY:1
HINT
GMm  1 1 
W  T .E2  T .E1    
2  r2 r1 
18. The orbital speed of geostationary satellite is
1) 8km/sec from west to east 2) 11.2km/sec from east to west
3) 3.1km/sec from west to east 4) zero
KEY:3
HINT
V0  g  R  h 
19. Two lead balls of masses m and 5m having radii R and 2R are separated by 12R. If they
attract each other by gravitational force, the distance covered by small sphere before they
touch each other is
1) 10 R 2) 7.5 R 3) 9 R 4) 2.5 R
KEY:2
HINT
m2 d eff
x ; Here d eff  9 R
m1  m2
20. Two particles each of mass ‘m’ are placed at A and C are such AB=BC=L. The gravitational
force on the third particle placed at D at a distance L on the perpendicular bisector of the line
AC is
Gm2 Gm2 Gm 2 Gm 2
1) along BD 2) along DB 3) along AC 4) along BD
L2 2 L2 L2 E2
KEY:2
HINT
Gm2
F '  2F , F 
2 L2
21. If g on the surface of the earth is 9.8 m / s 2 , its value at a height of 6400 km is (Radius of the
earth = 6400km).
1) 4.9ms–2 2) 9.8ms–2 3)2.45ms–2 4)19.6ms–2
KEY:3
HINT
gh R2

g  R  h 2
22. If g on the surface of the earth is 9.8ms 2 , its value at a depth of 3200km (Radius of the earth
= 6400km) is
1) 9.8ms 2 2) zero 3) 4.9ms 2 4) 2.45ms 2
KEY:3
HINT
 d
g '  g 1  
 R
23. If mass of the planet is 10% less than that of earth and radius of the planet is 20% greater
than that of earth then the weight of 40kg person on that planet is
1) 10 kg wt 2) 25 kg wt 3) 40 kg wt 4)60 kg wt
KEY:2
HINT
GM M
g 2
 g 2
R R
24. The angular velocity of the earth with which it has to rotate so that the acceleration due to
gravity on 600 latitude becomes zero is
1) 2.5 103 rad s 1 2) 1.5 103 rad s 1 3) 4.5 103 rad s 1 4) 0.5 10 3 rad s 1
KEY:1
HINT
g  R 2 cos 2   0, given   600 , Find 
25. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is
launched into an orbit of radius 1.01 R. The time period of the second satellite is larger than
that of the first one by approximately
1) 0.5% 2) 1.5% 3) 1% 4) 3%
KEY:2
HINT
T 3 r
T  R 3/ 2   100   100
T 2 r
26. The value of acceleration due to gravity on the surface of earth is x. At an altitude of ‘h’ from
the surface of earth, its value is y. If R is the radius of earth, then the value of h is
 x   y  y x
1)  y  1 R 2)  x  1 R 3) R 4) R
    x y
KEY:1
HINT
GM GM
x ,y 
 R  h
2 2
R
27. The height at which the value of acceleration due to gravity becomes 50% of that at the
surface of the earth. (Radius of the earth=6400km) is
1) 2650 2) 2430 3) 2250 4) 2350
KEY:1
HINT
GM g
g'  g' 
 R  h
2 2
 h
1  
 R
28. The radius and density of two artificial satellites are R1, R2 and 1 ,  2 respectively. The ratio
of acceleration due to gravity on them will be
R2  2 R1 2 R1 1 R2 1
1) R  2) R  3) R  4) R 
1 1 2 1 2 2 1 2
KEY:3
HINT
4
g   GR   g  R 
3
29. The acceleration due to gravity at the latitude 45 0 on the earth becomes zero if the angular
velocity of rotation of earth is
2 2g 5R
1) 2) 2gR 3) 4)
gR R 2
KEY:3
HINT
2g
0  g  R 2 cos 2 450   
R
30. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the
ratio of densities of earth and moon is 5/3, then radius of moon in terms of radius of earth will
be
5 1 3 1
1) Re 2) Re 3) Re 4) Re
18 6 18 2 3
KEY:1
HINT
4
g   GR   g  R 
3
31. The point at which the gravitational force acting on any mass is zero due to the earth and the
moon system is. (The mass of the earth is approximately 81 times the mass of the moon and
the distance between the earth and the moon is 3,85,000km.)
1) 36,000km from the moon 2) 38,500km from the moon
3) 34500km from the moon 4) 30,000 from the moon
KEY:2
HINT
d
x
distance of null point
m2
1
m1
32. Two satellite M and N go around the earth in circular orbits at heights of RM and RN respectively
from the surface of the earth. Assuming the earth to be a uniform sphere of radius RE, the ratio
VM
of velocities of the satellite V is
N
2
 RM  RN  RE RN  RE RN
1)   2) RM  RE
3) R  R 4)
 RN  M E RM
KEY:2
GM 1
HINT V0  Rh
 V0 
Rh
33. Masses 2 kg and 8 kg are 18 cm apart. The point where the gravitational field due to them is
zero is
1) 6 cm from 8 kg mass 2) 6 cm from 2 kg mass
3) 1.8 cm from 8 kg mass 4) 9 cm from each mass
KEY:2
HINT
d
x
distance of null point
m2
1
m1
34. Particles of masses m1 and m2 are at a fixed distance apart. If the gravitational field strength
 
at m1 and m2 are I 1 and I 2 respectively. Then,
   
1) m1 I1  m2 I 2  0 2) m1 I 2  m2 I1  0
   
3) m1 I1  m2 I 2  0 4) m1 I 2  m2 I1  0
KEY:1
HINT
 Gm  Gm
I1  2 2 and I 2   2 1
d d
35. A satellite of mass ‘m’ revolves round the earth of mass ‘M’ in a circular orbit of radius‘r’ with
an angular velocity ‘  ’. If the angular velocity is  /8 then the radius of the orbit will be
1) 4r 2) 2r 3) 8r 4) r
KEY:1
HINT
1
From Kepler’s 3rd law, T  r   
2 3 2
r3
36. A body of mass ‘m’ is raised from the surface of the earth to a height ‘nR’ (R -radius of earth).
Magnitude of the change in the gravitational potential energy of the body is (g - acceleration
due to gravity on the surface of earth) (2007M)
 n   n 1  mgR mgR
1)  mgR 2)  mgR 3) 4) n 1

 n 1 

 n  n  
KEY:1
HINT
mgh
GPE  ; 27) W  m  V   KE
h
1
R
37. A person brings a mass 2 kg from A to B. The increase in kinetic energy of mass is 4J and work
done by the person on the mass is -10J. The potential difference between B and A is .......J/kg
1) 4 2) 7 3) -3 4) -7
KEY:4
HINT
38. The work done in shifting a particle of mass ‘m’ from the centre of earth to the surface of the
earth is
1
1) –mgR 2) mgR 3) zero 4) mgR
2
KEY:2
HINT
mgh
W  GPE2  GPE1 ; GPE 
h
1
R
39. The figure shows two shells of masses m1 and m2 . The shells are concentric. At which point, a
particle of mass m shall experience zero force?
m1
m2
1) A 2) B 3) C 4) D
KEY:4 D C B A
HINT
The gravitational field intensity at a point inside the spherical shell is zero.
40. Escape velocity of a body of 1kg mass on a planet is 100 m/s. Gravitational potential energy of
the body at the plane is
1) -5000J 2) -1000J 3) -2400J 4) 5000J
KEY:1
HINT
2GM GM
ve   100   5000
R R
GMm
 PE U     5000 J
R
41. If three particles, each of mass M are placed at the three corners of an equilateral triangle of
side a, the force exerted by this system on another particle of mass M placed (i) at the midpoint
of a side and (ii) at the centre of the triangle are, respectively.
4GM 2 4GM 2 3GM 2 GM 2
1) 0, 2) ,0 3) , 2 4) 0, 0
3a 2 3a 2 a2 a
KEY:2
HINT
Find individual forces and calculate resultant
Gm1m2
Use F 
R2
42. The ratio of escape velocities of two planets if g value on the two planets are 9.9m / s 2 and
3.3m / s 2 and their radii are 6400km and 3200km respectively is
1) 2.36 : 1 2) 1.36 : 1 3) 3.36 : 1 4) 4.36 : 1
KEY:1
HINT
Ve  2 gR  Ve  gR
43. The distance of Neptune and saturn from the Sun are respectively. 10 13 and 1012 meters and
their periodic times are respectively Tn and Ts. If their orbits are circular, the value of Tn/Ts is
1
1) 100 2) 10 10 3) 4) 10
10 10
KEY:2
HINT
3
3 T r 2
From Kepler’s 3rd law, T  r , 1   2 
2
T2  r1 
44. The escape velocity from the surface of the earth of radius R and density 
2  G 2  G R 2 G 
1) 2 R 2) 2 3) 2 4)
3 3 g R2
KEY:1
HINT
2GM 4 3
Ve  but M   R 
R 3
45. The escape velocity from the earth is 11 km/sec. The escape velocity from a planet having
twice the radius and same density as earth is (in km/sec)
1) 22 km/sec 2) 15.5 km/sec 3) 11 km/sec 4) 5.5 km/sec
KEY:1
HINT
Ve  R  ;
46. An object of mass ‘m’ is at rest on earth’s surface. Escape speed of this object is Ve . Same
object is orbiting earth with h  R then escape speed is Ve1 . Then
1) Ve1 2) Ve  2Ve1 3) Ve  2Ve1 4) Ve1  2Ve

KEY:3
HINT
1 2 GM
36) 2 mv0  R  h  O
 
1
47. A satellite revolves in a circular orbit with speed V  Ve . If satellite is suddenly stopped and
3
allowed to fall freely onto earth, the speed with which it hits earth’s surface is
gR 2
1) gR 2) 3) 2gR 4) gR
3 3
KEY:4
HINT
V  Ve2  2V02
48. A space station is set up in space at a distance equal to earth’s radius from the surface of
earth. Suppose a satellite can be launched from the space station also. Let v1 and v2 be the
escape velocities of the satellite on the earth’s surface and space station respectively. Then
1) v2  v1 2) v2  v1 3) v2  v1
4) 1, 2 and 3 are valid depending on the mass of satellite.
KEY:2
HINT
From the surface of earth
2GM
Escape velocity v1 
R
1 2 GMm
mv2  0
2 2R
49. A man weight ‘W’ on the surface of earth and his weight at a height ‘R” from surface of earth
is (R is Radius of earth)
W W
1) 2) 3) W 4) 4W
4 2
KEY:1
HINT
2
 R  W
W  mg ;W '  mg   
 Rh 4
50. A satellite moving in a circular path of radius ‘r’ around earth has a time period T. If its radius
slightly increases by 4%, then percentage change in its time period is
1) 1% 2) 6% 3) 3% 4) 9%
KEY:2
HINT:
T 3 R
T 2 r 3 ,  100   100
T 2 R
51. If a rocket is fired with a velocity, V  2 gR near the earth’s surface and goes upwards, its
speed in the inter-stellar space is
1) 4 gR 2) 2gR 3) gR 4) 4gR
KEY:2
HINT
According to the law of conservation of energy
T .E surface  T .E int er stellarspace
GMm 1 2 1
  mv  0  mv12
R 2 2
1
  1 2
2
 mgR  m 2 gR  mv1
2 2
1 2
 mv1  mgR  v12  2 R  V1  2 gR
2
52. A pr oj ectile is fir ed ver tically upwar ds fr om the sur face of the ear th with a velocity K v e, where
Ve is the escape velocity and K<1. If R is the radius of the earth, the maximum height to which
it will rise measured from the centre of the earth will be (neglect air resistance)
1 K2 R R
3) R 1  K 
2
1) 2) 4)
R 1 K2 1 K2
KEY:2
HINT
According to the law of conservation of energy
GMm 1 GMm
  mK 2ve2   0
R 2 Rh
2GM
ve2 
R
ve2 K 2ve2 v2  R 
   e
2 2 2 R  h
53. Two different artificial satellites orbiting with same time period around the earth having angular
moment 2:1. The ratio of masses of the satellites is
1) 2:1 2) 1:2 3) 1:1 4) 1:3
KEY:1
HINT
dA L
 =constant. (From Kepler’s 2nd law)
dt 2m
54. If the radius of earth shrinks by 0.2% without any change in its mass, escape velocity from the
surface of the earth
1) increases by 0.2% 2) decreases by 0.2%
3) increases by 0.1% 4) increases by 0.4%
KEY:3
HINT
2GM
ve 
R
1 ve 1  R 
 ve    100     100 
R ve 2 R 
55. If ‘A’ is areal velocity of a planet of mass M, its angular momentum is
1) M/A 2) 2MA 3) A2 M 4) AM 2
KEY:2
HINT:
dA L

dt 2 M
56. If d the distance between the centres of the earth of mass M1 and moon of mass M2, then the
velocity with which a body should be projected from the mid point of the line joining the earth
and the moon, so that it just escape is
G  M1  M 2  G  M1  M 2 
1) 2)
d 2d
2G  M 1  M 2  4G  M 1  M 2 
3) 4)
d d
KEY:4
HINT
Using law of conservation of energy,
1 2 2GM
mve   M1  M 2 
2 d
57. If ‘ve’ is the escape velocity of a body from a planet of mass ‘M’ and radius ‘R’. Then the
velocity of the satellite revolving at a height ‘h’ from the surface of the planet will be
R 2R Rh R
1) ve 2) ve 3) 4) ve 2 R  h
Rh Rh R  
KEY:4
HINT
2GM GM
ve  ,v 
R Rh
v R R
   v  ve
ve 2 R  h 2 R  h
58. A particle falls towards earth from infinity. The velocity with which it reaches earth’s surface is
1) v  2 gR 2) v  2 gR 3) v  gR 4) v  R / g
KEY:2
HINT
The projecting body having same final velocity to reach projecting place. So, v  2 gR
59. Two spheres of masses m and M are situated in air and the gravitational force between them
is F. The space between the masses is now filled with a liquid of specific gravity 3. The
gravitational force will now be
F F
1) 2) 3F 3) F 4)
9 3
KEY:3
HINT
Gravitational force does not depend upon the medium between the masses.
60. Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the
acceleration due to gravity on the surface of the earth. If R e is the maximum range of a
projectile on the earth's surface, what is the maximum range on the surface of the moon for
the same velocity of projection
1) 0.2 Re 2) 2 Re 3) 0.5 R e 4) 5 Re
KEY:4
HINT
u2 1
Rmax   Rmax 
g g
61. The orbital speed for an earth satellite near the surface of the earth is 7 km/sec. If the radius
of the orbit is 4 times the radius of the earth, the orbital speed would be
1) 3.5 km/sec 2) 7 km/sec 3) 7 2 km/sec 4) 14 km/sec
KEY:1
HINT
GM 1
Ve   Ve 
r r
62. Two masses ‘M’ and ‘4M’ are at a distance ‘r’ apart on the line joining them, ‘P’ is point where
the resultant gravitational force is zero (such a point is called as null point). The distance of ‘P’
from the mass ‘M’ is
r r 2r 4r
1) 2) 3) 4)
5 3 3 5
KEY:2
HINT
d
x
m2
1
m1
63. Let ‘A’ be the area swept by the line joining the earth and the sun during Feb 2012. The area
swept by the same line during the first week of that month is
1) A/4 2) 7A/29 3) A 4) 7A/30
KEY:2
HINT:
For 29 days - A, For 1 day - A/29,
For 1 week -7A/29,
64. Three identical particle each of mass “m” are arranged at the corners of an equilateral triangle
of side “L”. If they are to be in equilibrium, the speed with which they must revolve under the
influence of one another’s gravity in a circular orbit circumscribing the triangle is
3Gm Gm Gm 3Gm
1) 2) 3) 4)
L L 3L L2
KEY:2
HINT
mv 2 Gm1m2 L
3F  ;F  2
and r 
r L 3
65. Two satellites are revolving round the earth at different heights. The ratio of their orbital
speeds is 2 : 1. If one of them is at a height of 100km, the height of the other satellite is
1) 19600km 2) 24600km 3) 29600km 4) 14600km
KEY:1
HINT
GM 1
Ve   Ve 
Rh Rh
66. A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g
is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite
is [AIEEE-2004]
1/ 2
 gR 2  gR 2 gR
1) gx 2)   3) 4)
Rx Rx Rx
KEY:2
HINT
GM gR 2
V0  
Rh Rx
67. A body is projected vertically up from surface of the earth with a velocity half of escape velocity.
The ratio of its maximum height of ascent and radius of earth is
1) 1 : 1 2) 1 : 2 3) 1 : 3 4) 1 : 4
KEY:1
HINT
R
h Here n=2
n 1
2
68. The PE of three objects of masses 1kg, 2kg and 3kg placed at the three vertices of an
equilateral triangle of side 20cm is
1) 25G 2) 35G 3) 45G 4) 55G
KEY:4
HINT
Gm1m2
GPE U   Use, U net  U1  U 2  U 3
r12
69. If the mass of one particle is increased by 50% and the mass of another particle is decreased
by 50% the force between them
1) decreases by 25% 2) decreases by 75% 3) increases by 25% 4) does not change
KEY:1
HINT
 m1  m2 
mm  m1   m2  
F1  G 1 2 2 and F2  G  2  2 
d d2
70. An astronaut orbiting in a spaceship round the earth has a centripetal acceleration of 2.45m / s 2 .
The height of spaceship from earth’s surface is (R= radius of earth)
1) 3R 2) 2R 3) R 4) R / 2
KEY:3
HINT
gR 2
a ;
 R  h
2
71. Two satellites P, Q are revolving around earth in different circular orbits. The velocity of P is
twice the velocity of Q. If the height of P from earth’s surface is 1600 km. The radius of orbit
of Q is (radius of earth R =6400 km).
1) 1600 km 2) 20000 km 3) 32000 km 4) 40000 km
KEY:3
HINT
Given, vP  2vQ
GM GM
2  rQ  4rP
rP rQ
72. A planet is revolving around the sun. Its distance from the sun at apogee is r A and that at
perigee is rp. The masses of planet and sun are ‘m’ and M respectively, V A is the velocity of
planet at apogee and VP is at perigee respectively and T is the time period of revolution of
planet round the sun, then identify the wrong answer.
2 3
 rA  rP   rA  rP 
3 3
2) T 
2
1) T 2 
2GM 2GM
3) VA rA  VP rP 4) VA  VP ; rA  rP
KEY:1
HINT
3
4 2  rA  rP   rA  rP 
T   r 
2

GM  2   2 
By the law of conservation of angular momentum VA rA  VP rP
73. Suppose the gravitational force varies inversely as the n th power of distance, then the time
period of a planet in circular orbit of radius ‘R’ around the sun will be proportional to
(2004A)
 n 1   n2   n 1 
1) 
 2 
 2) 
 2 
 3) Rn 4) 
 2 

R R R
KEY:1
HINT
k
F n
, F  mr 2
r
74. A satellite moves around the earth in a circular orbit with speed ‘V’. If ‘m’ is mass of the
satellite then its total energy is
1 1 3
1) mv2 2) mv2 3) – mv2 4) mv2
2 2 2
KEY:3
HINT
1
TE   KE   mv 2
2
75. The mass of a planet is half that of the earth and the radius of the planet is one fourth that of
earth. If we plan to send an artificial satellite from the planet, the escape velocity will be,
(Ve=11kms-1)
1) 11kms-1 2) 5.5kms-1 3) 15.55kms-1 4) 7.78kms-1
KEY:3
HINT
2GM M
ve   ve 
R R
76. Two satellites of masses 400 kg, 500 kg are revolving around earth in different circular orbits
of radii r1, r2 such that their kinetic energies are equal. The ratio of r 1 to r2 is
1) 4 : 5 2) 16 : 25 3) 5 : 4 4) 25 : 16
KEY:1
HINT
GMm m
KE   KE   m  r
2r r
77. The kinetic energy needed to project a body of mass m from earth’s surface (radius R) to
infinity is
mgR mgR
1) 2) 2mgR 3) mgR 4)
2 4
KEY:3
HINT
1
KE  mVe2
2
78. The work done to increase the radius of orbit of a satellite of mass ‘m’ revolving around a
planet of mass M from orbit of radius R in to another orbit of radius 3R is
2GMm GMm GMm GMm
1) 2) 3) 4)
3R 3R 6R 24 R
KEY:2
HINT
Workdone = change in TE
GMm GMm
Workdone= 
2R 6R
79. A stone is dropped from a height equal to nR, where R is the radius of the earth, from the
surface of the earth. The velocity of the stone on reaching the surface of the earth is
2 g  n  1 R 2 gR 2 gnR
1) 2) 3) 4) 2gnR
n n 1 n 1
KEY:3
HINT
1 2 mgh
mv 
2 h
1
R
80. Three particles of equal mass ‘m’ are situated at the vertices of an equilateral triangle of side
‘L. The work done in increasing the side of the triangle to 2L is
2G 2 m Gm 2 3Gm 2 3Gm 2
1) 2) 3) 4)
2L 2L 2L L
KEY:3
HINT
3Gm2
Initial potential energy U i 
L
3Gm 2
Final potential energy U f  
2L
3Gm 2  3Gm 2  3Gm 2

W  U f  Ui    
2L  L  2L
81. The time of revolution of planet A round the sun is 8 times that of another planet B. The
distance of planet A from the sun is how many times greater than that of the planet B from the
sun
1) 2 2) 3 3) 4 4) 5
KEY:3
HINT:
3
3 T r 2
From Kepler’s 3rd law, T  r , 1   2 
2
T2  r1 
82. A small body is at a distance ‘r’ from the centre of mercury, where ‘r’ is greater than the radius
of Mercury. The energy required to shift the body from r to 2r measured from the centre is E.
The energy required to shift it from 2r or 3r will be
E E E
1) E 2) 3) 4)
2 3 4
KEY:3
HINT
E  U 2  U1
GMm GMm GMm
  
2R R 2R
GMm GMm GMm
E'    
3R 2R 6R
83. By what percent the energy of the satellite has to be increased to shift it from an orbit of
3r
radius ‘r’ to
2
1) 66.7% 2) 33.3% 3) 15% 4) 20.3%
KEY:2
HINT
GMm GMm GMm
E W  E2  E1  , E2 
2r 2r1 2r2
84. The gravitational force between two bodies is 6.67x10 -7N when the distance between their
centres is 10m. If the mass of first body is 800 kg, then the mass of second body is
1) 1000 kg 2) 1250 kg 3) 1500 kg 4) 2000 kg
KEY:2
HINT
Gm1m2 Fg  R 2
Fg   m2 
R2 Gm1
85. At what height from the surface of earth, the total energy of satellite is equal to its potential
energy at a height 2R from the surface of earth (R=radius of earth)
1) 2R 2) R/2 3) R/4 4) 4R
KEY:2
HINT
GMm  GMm 
 r  R  h
2r  3R 
86. A geo-stationary satellite is orbiting the earth at a height 6R above the surface of the earth,
where R is the radius of earth. The time period of another satellite revolving around earth
at a height 2.5R from earth’s surface is
1) 12 2hr 2) 12 hr 3) 6 2hr 4) 6 hr
KEY:3
HINT
7R
3
T1 R13
 3
 3
T2 R2  7R 
 
 2 
87. A planet moves around the sun. At a given point P, it is closest from the sun at a distance d1 and
has a speed V1. At another point Q, when it is farthest from the sun at a distance d 2, its speed
will be
d12V1 d 2V1 d1V1 d 22V1
1) 2) d 3) d 4) 2
d2 1 2 d1
KEY:3
HINT
From conservation of angular momentum vr=Constant.
88. A small body is initially at a distance ‘r’ from the centre of earth. ‘r’ is greater than the radius
of the earth. If it takes W joule of work to move the body from this position to another position
at a distance 2r measured from the centre of earth, how many joules would be required to
move it from this position to a new position at a distance of 3r from the centre of the earth.
1) W/5 2) W/3 3) W/2 4) W/6
KEY:2
HINT
Gm1m2
W  GPEe  GPE1 Here, GPE  r12
89. The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is
1) 3ve 2) 3ve 3) 2v 4) 2ve
KEY:1
HINT
vP M P Re

ve M e RP
90. The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to
escape from the gravitational field of the earth is.
1) 2E 2) 4E 3) 2 2 E 4) 2E
KEY:1
HINT
K e 2 gR
  Ke  2K0
K0 gR
‘
PREVIOUS JEE MAINS QUESTIONS & SOLUTIONS
Keplers Laws of Planetary Motion:
1. Ifthe angular momentum ofa planet ofmass 𝐦, moving around the Sun in a circular orbit is 𝐋,
about the center of the Sun, its areal velocity is: [9 Jan. 2019 I]
𝐋 𝟒𝐋 𝐋 𝟐𝐋
(a) 𝐦 (b) (c) 𝟐𝐦 ‘ (d)
𝐦 𝐦
dA
SOLUTION : (c) Areal velocity; dt
1 2
dA = r d𝜃
2
dA 1 L
=
dt 2 m
𝟏
2. Figure shows elliptical path abcd of a planet around thesun 𝐒 such that the area oftriangle csa is
𝟒
the area ofthe ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis.
𝐈𝐟𝐭 𝟏 is the time taken for planet to go over path abc and 𝐭 𝟐 for path taken over cda then:
(a) 𝐭 𝟏 = 𝟒𝐭 𝟐 (b) 𝐭 𝟏 = 𝟐𝐭 𝟐 (c) 𝐭 𝟏 = 𝟑𝐭 𝟐 (d) 𝐭 𝟏 = 𝐭 𝟐
SOLUTION : . (c)
Let area ofellipse abcd = x
x x
Area of SabcS = 2 + 4 (i. e., arc of abca +SacS)
3x
(Area ofhalfellipse + Area oftriangle)=
4
a
3x x
Area of SadcS = x − =4
4
AreaofSabcS 3x/4 t1
= =
AreaofSadcS x/4 t2
t1
= 3 or, t1 = 3t 2
t2
3. India’s Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the
sun. It leaves the earth at 𝐄 and meets Mars at M. Ifthe semi‐major axis of Earth’s orbit is
𝐚𝐞 = 𝟏. 𝟓 × 𝟏𝟎𝟏𝟏 𝐦, that ofMars orbit 𝐚𝐦 = 𝟐. 𝟐𝟖 × 𝟏𝟎𝟏𝟏 𝐦, taken Kepler’s laws give the estimate
oftime for Mangalyan to reach Mars from Earth to be close to: [Online April 9, 2014]
(a) 500 days (b) 320 days (c) 260 days (d) 220 days
SOLUTION :’ (𝑏)
4. The time period of a satellite of earth is 5 hours. If the separation between the earth and the
satellite is increased to 4 times the previous value, the new time period will become [2003]
(a) 10 hours (b) 80 hours (c) 40 hours (d) 20 hours
SOLUTION : ‘(c)
According to Kepler’s law ofperiods 𝑇 2∝ 𝑅 3
𝑇2 2 𝑅2 3
= = 5 × 23 = 40 hours
𝑇1 𝑅1
Newton’s Universal Law of Gravitation :
5. A straight rod oflength 𝐋 extends from 𝒙 = 𝐚 tox = 𝐋 + 𝐚. The gravitational force it exerts on
point mass 𝐦’ at 𝒙 = 𝟎, ifthe mass per unit length ofthe rod is 𝐀 + 𝐁𝒙𝟐, is givenby:
[12 Jan. 2019 I]
𝟏 𝟏 𝟏 𝟏
(a) 𝐆𝐦 𝐀 − 𝐚 − 𝐁𝐋 (b) 𝐆𝐦 𝐀 − 𝐚+𝑳 − 𝐁𝐋
𝐚+𝑳 𝐚
𝟏 𝟏 𝟏 𝟏
(c) 𝐆𝐦 𝐀 − 𝐚 + 𝐁𝐋 (d) 𝐆𝐦 𝐀 − 𝐚+𝑳 + 𝐁𝐋
𝐚+𝑳 𝐚
SOLUTION : ‘. (d)
Given = A + Bx 2 ,
Taking small element dm oflength dx at a distance x fromx = 0
a+L
Gm
⇒𝛤= A + Bx 2 dx
a x2
a+L
A
= Gm − + Bx
x a
1 1
= Gm A − + BL
a a+L
6. Take the mean distance ofthe moon and the sun from the earth to be 𝟎. 𝟒 × 𝟏𝟎𝟔 km and 𝟏𝟓𝟎 ×
𝟏𝟎𝟔 km respectively. Their masses are 𝟖 × 𝟏𝟎𝟐𝟐 𝐤𝐠 and 𝟐 × 𝟏𝟎𝟑𝟎 kg respectively. The radius ofthe
earth is 6400 km. Let 𝜟𝜞𝟏 be the difference in the forces exerted bythe moon at the nearest and
farthest points on the earth and 𝜟𝜞𝟐 be the difference in the force exerted by the sun at the
𝜟𝑭
nearest and farthest points onthe earth. Then, the number closest to 𝜟𝑭𝟏 is:[Online Apri115, 2018]
𝟐
(a) 2 (b) 6 (c) 𝟏𝟎−𝟐 (d) 𝟎. 𝟔
SOLUTION : (a)
GMm
As we know, Gravitational force ofattraction, 𝛤 = R2
GM e m GM e M s
𝛤1 = and 𝛤2 =
r 21 r 22
2GM e m GM e M s
𝛥𝛤1 = 𝛥r1 and 𝛥𝛤2 = 𝛥r2
r 31 r 32
𝛥𝛤1 m𝛥r1 r23

= 3
𝛥𝛤2 r1 Ms 𝛥r2
Using 𝛥r1 = 𝛥r2 = 2R earth ; m = 8 × 1022 kg;
Ms = 2 × 1030 kg
r1 = 0.4 × 106 km and r2 = 150 × 106 km
𝛥𝛤1
=2
𝛥𝛤2
7. Four particles, each ofmass 𝐌 and equidistant from each other, move along a circle ofradius 𝐑
under the action of their mutual gravitational attraction. The speed of each particle is: [2014]
𝐆𝐌 𝐆𝐌 𝐆𝐌 𝟏 𝐆𝐌
(a) (b) 𝟐 𝟐 (c) 𝟏+𝟐 𝟐 (d) 𝟐 𝟏+𝟐 𝟐
𝐑 𝐑 𝐑 𝐑
SOLUTION : (d)
𝑀𝑣 2
2𝐹 cos 45∘ + 𝐹 ′ = (From figure)
𝑅
𝐺𝑀 2 𝐺𝑀 2
Where 𝐹 = 2 and 𝐹 ′ =
2𝑅 4𝑅 2
2×𝐺𝑀 2 𝐺𝑀 2 𝑀𝑣 2
⇒ 2 + =
2 𝑅 2 4𝑅 2 𝑅
so, dm = 𝜆dx : dm = A + Bx 2 dx
Gmdm
d𝛤 =
x2
𝐺𝑀 2 1 1
⇒ + = 𝑀𝑣 2
𝑅 4 2
𝐺𝑀 2+4 1 𝐺𝑀
𝑣= = 1+2 2
𝑅 4 2 2 𝑅
𝐑
8. From a sphere ofmass 𝐌 and radius 𝐑, a smaller sphere of radius 𝟐 is carved out such that the
cavity made in the original sphere is between its centre and the periphery (See figure). For the
configuration in the figure where the distance between the centre ofthe original sphere and the
removed sphere is 𝟑𝐑, the gravitational force between thetwo sphere is: [Online April 11, 2014]
𝟒𝟏𝐆𝐌𝟐 𝟒𝟏𝐆𝐌𝟐 𝟓𝟗𝐆𝐌𝟐 𝐆𝐌𝟐

(a) 𝟑𝟔𝟎𝟎𝐑𝟐 𝐛 𝐜 𝐝
𝟒𝟓𝟎𝐑𝟐 𝟒𝟓𝟎𝐑𝟐 𝟐𝟐𝟓𝐑𝟐
SOLUTION : (a)
4 𝑅 3 4 1
Volume ofremoved sphere 𝑉remo = 3 𝜋 = 3 𝜋𝑅 3
2 8
4 4 1 4 7
Volume ofthe sphere(remaining) 𝑉remain = 3 𝜋𝑅 3 − 3 𝜋𝑅 3 = 3 𝜋𝑅 3
8 8
1 7
Therefore mass of sphere carved and remaining sphere are at respectively 8 M and 8 M.
Therefore, gravitational force between these two sphere,
7𝑀 1
𝐺M𝑚 𝐺 8 × 8 𝑀 7𝐺𝑀2 41GM 2
𝐹= 2 = = =
𝑟 3𝑅 2 64 × 9𝑅 2 3600R2
9. Two particles ofequal mass ‘m’ go arounda circle ofradius 𝑹 under the action oftheir mutual
gravitational attraction. The speed ofeach particle with respect to their centre of mass is [2011 ]
𝑮𝒎 𝑮𝒎 𝑮𝒎 𝑮𝒎
(a) (b) (c) (d)
𝟒𝑹 𝟑𝑹 𝟐𝑹 𝑹
SOLUTION : (a)
As two masses revolve about the common centre of mass 𝑂.
Mutual gravitational attraction = centripetal force
Ifthe velocity ofthe two particles with respect to the centre of gravity is v then
𝑣= w R
𝐺𝑚 𝐺𝑚
𝑣= ×𝑅 =
4𝑅 3 4𝑅
10. Two spherical bodies ofmass 𝑴 and 5M& radii 𝑹 & 𝟐𝑹 respectively are released in free space with
initial separation between their centres equal to 12 𝑹. If they attract each other due to gravitational
force only, then the distance covered bythe smaller body just before collision is [2003]
(a) 𝟐. 𝟓𝑹 (b) 𝟒. 𝟓𝑹 (c) 𝟕. 𝟓𝑹 (d) 𝟏. 𝟓𝑹
SOLUTION : (c)
We know that Force =mass ×acceleration.
The gravitational force acting on both the masses is the same. 𝐹1 = 𝐹2
𝑚𝑎1 = 𝑚𝑎2
9𝑀 5𝑀 9𝑀 1
⇒ 95𝑀 = =5 ⇒ 95𝑀 = 5
𝑀
Let 𝑡 be the time taken for the two masses to collide and 𝑥5𝑀 , xM be the distance travelled by the mass 5𝑀
and 𝑀 respectively.
1 1
For mass 5𝑀 𝑢 = 0, 𝑆 = 𝑢𝑡 + 2 𝑎𝑡 2 ; 𝑥5𝑀 = 2 𝑎5𝑀 𝑡 2 (ii)
For mass 𝑀 𝑢 = 0, 𝑠 = 𝑥𝑀 , 𝑡 = 𝑡, 𝑎 = 𝑎𝑀
1 1
𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2 ⇒ 𝑥𝑀 = 2 𝑎𝑀 𝑡 2 ... (iii)
1
𝑥 5𝑀 𝑎 5𝑀 𝑡 2 𝑎 5𝑀 1
Dividing (ii) by(iii) = 21 = =5 [From (i)]
𝑥𝑀 𝑎 𝑡2 𝑎𝑀
2 𝑀
5𝑥5𝑀 = 𝑥𝑀
From the figure it is clear that
𝑥5𝑀 + 𝑥𝑀 = 9𝑅
Where 𝑂 is the point where the two spheres collide.
From(iv) and(v)
𝑥𝑀
+ 𝑥𝑀 = 9𝑅
5
6𝑥𝑀 = 45𝑅
45
𝑥𝑀 = 𝑅 = 7.5𝑅
6
Asseleration due to Gravity :
𝑹
11. The value ofacceleration due to gravity is 𝒈𝟏 at a height 𝒉 = (𝑹 = radius of the earth) from the
𝟐
surface of the earth. It is again equal to 𝒈𝟏 and a depth 𝒅 below the surface ofthe earth. The ratio
𝒅
equals: [5 Sep. 2020 (I)]
𝑹
𝟒 𝟓 𝟏 𝟕
(a) 𝟗 (b) 𝟗 (c) 𝟑 (d) 𝟗
SOLUTION : (b)
According to question, 𝑔ℎ = 𝑔𝑑 = 𝑔1
∙:: (R‐d):
𝐺𝑀 𝐺𝑀 𝑅−𝑑
𝑔ℎ = 𝑅 2
and 𝑔𝑑 =
𝑅+ 𝑅3
2
𝐺𝑀 𝐺𝑀 𝑅 − 𝑑 4 𝑅−𝑑
2 = 3
⇒ =
3𝑅 𝑅 9 𝑅
2
⇒ 4𝑅 = 9𝑅 − 9𝑑 ⇒ 5𝑅 = 9𝑑
𝑑 5
=
𝑅 9
12. The acceleration due to gravity on the earth’s surface at the poles is 𝒈 and angular velocity ofthe
earth about the axis passing through the pole is 00. An object is weighed at the equator and at a
height 𝒉 above the poles by using a spring balance. Ifthe weights are found to be same, then 𝒉 is:
(𝒉 << 𝑹, where 𝑹 is the radius ofthe earth) [𝟓 Sep. 2020 (II)]
𝑹𝟐 𝖈𝟎𝟐 𝑹𝟐 𝖈𝟎𝟐 𝑹𝟐 𝖈𝟎𝟐 𝑹𝟐 𝖈𝟎𝟐

(a) (b) (c) (d)
𝟐𝒈 𝒈 𝟒𝒈 𝟖𝒈
SOLUTION : . (b)
Value of 𝑔 at equator, 𝑔𝐴 = 𝑔 ⋅ −𝑅o)2
Value of 𝑔 at height ℎ above the pole,
2ℎ
𝑔𝐵 = 𝑔 ⋅ 1 −
𝑅
As object is weighed equally at the equator and poles, it means 𝑔 is same at these places.
𝑔𝐴 = 𝑔𝐵
2 2
2ℎ 2 2𝑔ℎ 𝑅 (0
⇒ 𝑔 − 𝑅oo2 = 𝑔 1 − ⇒ 𝑅0) = ⇒ℎ=
𝑅 𝑅 2𝑔
13. The height ↑ 𝒉′ at which the weight of a body will be the same as that at the same depth ↑
𝒉′ 𝐟𝐢𝐢 𝐨𝐦 the surface ofthe earth is (Radius ofthe earth is 𝑹 and effect ofthe rotation ofthe earth is
neglected) : [2 Sep. 2020(II)]
𝟓 𝑹 𝟓𝑹−𝑹 𝟑𝑹−𝑹
(a) 𝑹−𝑹 (b) 𝟐 (c) (d)
𝟐 𝟐 𝟐
SOLUTION : (c)
𝐺𝑀
The acceleration due to gravity at a height ℎ is given by 𝑔 = 𝑅+ℎ 2
Here, 𝐺 = gravitation constant
𝑀 = mass of earth
The acceleration due to gravity at depth ℎ is
𝐺𝑀 ℎ
𝑔↑ = 2
1−
𝑅 𝑅
Given, 𝑔 = 𝑔′
𝐺𝑀 𝐺𝑀 ℎ
= 1 −
𝑅+ℎ 2 𝑅2 𝑅
𝑅3 = 𝑅 + ℎ 2
𝑅 − ℎ = 𝑅 2 + ℎ2 + 2ℎ𝑅 𝑅 − ℎ
⇒ 𝑅 3 = 𝑅 3 + ℎ2 𝑅 + 2ℎ𝑅 2 − 𝑅 2 ℎ − ℎ3 − 2ℎ2 𝑅
⇒ ℎ3 + ℎ2 2𝑅 − 𝑅 − 𝑅 2 ℎ = 0
⇒ ℎ3 + ℎ2 𝑅 − 𝑅 2 ℎ = 0
⇒ ℎ2 + ℎ𝑅 − 𝑅 2 = 0
−𝑅 ± 𝑅 2 + 4 1 𝑅 2 −𝑅 + 5𝑅 5−1
⇒ℎ= = = 𝑅
2 2 2
14. A box weighs 196 𝐍 on a spring balance at the north pole. Its weight recorded on the same
balance if it is shifted to the equator is close to (Take 𝒈 = 𝟏𝟎 ms −𝟐 at the north pole and the
radius ofthe earth = 𝟔𝟒𝟎𝟎 km): [7 Jan. 2020 II]
(a) 𝟏𝟗𝟓. 𝟔𝟔𝐍 (b) 𝟏𝟗𝟒. 𝟑𝟐𝐍 (c) 𝟏𝟗𝟒. 𝟔𝟔𝐍 (d) 𝟏𝟗𝟓. 𝟑𝟐𝐍
SOLUTION : . (d)
Weight at pole, 𝑤 = 𝑚𝑔 = 196𝑁
⇒ 𝑚 = 19.6kg
2
Weight at equator, 𝑤’ = 𝑚𝑔’ = 𝑚 𝑔−0) 𝑅
2𝜋 2 2𝜋
= 19.6 10 − × 6400 × 103 N ⋅.⋅ 𝑖 =
24×3600 𝑇
= 19.6 10 − 0.034 = 195.33𝑁
15. The ratio ofthe weights ofa body on the Earth’s surface to that on the surface of a planet is 9:4.
𝟏
The mass of theplanet is 𝟗 th ofthat ofthe Earth. If‘ 𝐑’ is the radius ofthe Earth, what is the radius
ofthe planet? (Take the planets to have the same mass density). [12 April 2019 II]
𝑹 𝑹 𝑹 𝑹
(a) 𝟑 (b) 𝟒 (c) 𝟗 (d) 𝟐
SOLUTION : (d)
𝑊𝑒 𝑚𝑔 9 𝑔 9 𝐺𝑀/𝑅 2 9
= 𝑚 𝑔𝑒 = 4 or 𝑔 𝑒 = 4 or 𝐺 =4
𝑊𝑝 𝑝 𝑝 𝑀/9 /𝑅𝑝2
𝑅
𝑅𝑝 =
2
16. The value ofacceleration due to gravity at Earth’s surface is . 𝟖 𝐦𝐬−𝟐 . The altitude above its
surface at which the acceleration due to gravity decreases to 𝟒. 𝟗𝐦𝐬 −𝟐 , is close to : (Radius
ofearth = 𝟔. 𝟒 × 𝟏𝟎𝟔 𝐦) [10 April 2019 I]
(a) 𝟐. 𝟔 × 𝟏𝟎𝟔 𝐦 (b) 𝟔. 𝟒 × 𝟏𝟎𝟔 𝐦 (c) 𝟗. 𝟎 × 𝟏𝟎𝟔 𝐦 (d) 𝟏. 𝟔 × 𝟏𝟎𝟔 𝐦
SOLUTION : . (a)
h −2
Given Acceleration due to gravity at a height h from earth’s surface is g h = g 1 + Re
h −2
4.9 = 9.8 1 + Re [as h <<< R e ]h = R e 2−1
h = 6400 × 0.414km = 2.6 × 106 m

17. Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence.
(a) There will be no change in weight anywhere on the earth
(b) Weight ofthe object, everywhere on the earth, wild decrease
(c) Weight ofthe object, everywhere on the earth, will increase
(d) Except at poles, weight ofthe object on the earth will decrease
SOLUTION : (d)
With rotation of earth or latitude, acceleration due to gravity vary as g ′ = g − w 2 Rcos2 𝜑
Where 𝜑 is latitude, there will be no change in gravity at poles as 𝜑 = 90∘
At all other points as w increases g ′ will decreases hence, weight, W = mg decreases.
18. The variation ofacceleration due to gravity gwith distance 𝐝 from centre of the earth is best
represented by (𝐑 =Earth’s radius): [2017, Online May 7, 2012]
(a) (b) (c) (d)
SOLUTION : (b)
Variation of acceleration due to gravity, 𝑔 with distance 𝑑 ′ from centre ofthe earth
𝐺𝑚
If 𝑑 < 𝑅, 𝑔 = .𝑑
𝑅2
i.e., 𝑔 ∝ 𝑑(straight line)
𝐺𝑚
If 𝑑 = 𝑅, 𝑔𝑠 = 𝑅2
𝐺𝑚
If 𝑑 > 𝑅, g = 𝑑2
1
i.e., 𝑔 ∝ 𝑑 2
𝒌
19. The mass density of a spherical body is given by 𝐩 𝒓 = 𝒓 for 𝒓 < 𝐑 and 𝐩 𝒓 = 𝟎 for 𝒓 >
𝑹,where 𝒓 is the distance from the centre.The correct graph that describes qualitatively the
acceleration, 𝐚, ofa test particle as a function ofr is: [Online April 9, 2017]
a) (c)
SOLUTION : . (b)
mass k
Given that, mass density of a spherical body p r =
volume r
M k
= r for inside r ≤ R
V
kv
M= − (i)
r
GMr 𝛤
Inside the surface of sphere Intensity I = I=m
R3
GMr mg G kv
g inside = or I = = g = R3 . . r = constant From eq. (i),
R3 m r
GM
Similarly, g out = r2
Hence, option (2) is correct graph.
20. Ifthe Earth has no rotational motion, the weight ofa person on the equator is W. Determine the
speed with which the earth would have to rotate about its axis so that the person at the equator
𝟑
will weight 𝟒 W. Radius of the Earth is 6400 km and 𝐠 = 𝟏𝟎𝐦/𝐬𝟐 . [Online April 8, 2017]
(a) 𝟏. 𝟏 × 𝟏𝟎−𝟑 𝐫𝐚𝐝/𝐬 (b) 𝟎. 𝟖𝟑 × 𝟏𝟎−𝟑 𝐫𝐚𝐝/𝐬 (c) 𝟎. 𝟔𝟑 × 𝟏𝟎−𝟑 𝐫𝐚𝐝/𝐬 (d) 𝟎. 𝟐𝟖 × 𝟏𝟎−𝟑 𝐫𝐚𝐝/𝐬
SOLUTION : . (c)
We know, g ′ = g − 𝑤 2 Rcos2 𝜃
3g
= g − 𝔠02 R
4
3
Given, g ′ = 4 g
g
w2 R =
4
g 10 1
w= = 3
= = 0.6 × 10−3 rad/s
4R 4 × 6400 × 10 2 × 8 × 100
21. The change in the value ofacceleration of earth towards sun, when the moon comes 𝐟𝐢𝐢 𝐨𝐦 the
position of solar eclipse to the position on the other side of earth in line with sun is:(mass ofthe
moon = 𝟕. 𝟑𝟔 × 𝟏𝟎𝟐𝟐 𝐤𝐠, radius ofthe moon’s orbit = 𝟑. 𝟖 × 𝟏𝟎𝟖 𝐦). [Online April 22, 2013]
(a) 𝟔. 𝟕𝟑 × 𝟏𝟎−𝟓 𝐦/𝐬𝟐 (b) 𝟔. 𝟕𝟑 × 𝟏𝟎−𝟑 𝐦/𝐬𝟐
(c) 𝟔. 𝟕𝟑 × 𝟏𝟎−𝟐 𝐦/𝐬𝟐 (d) 𝟔. 𝟕𝟑 × 𝟏𝟎−𝟒 𝐦/𝐬𝟐
SOLUTION : (a)
22. Assuming the earth to be a sphere ofuniform density, the acceleration due to gravity inside the
earth at a distance of 𝒓𝐟𝐢𝐢 𝐨𝐦 the centre is proportional to [Online May 12, 2012]
(a) 𝒓 (b) 𝒓−𝟏 (c) 𝒓𝟐 (d) 𝒓−𝟐
SOLUTION : (a)
Acceleration due to gravity at depth d from the surface ofthe earth or at a distance r from the centre ’ 𝑂’ of
4
the earth 𝑔′ = 3 𝜋p𝐺𝜈 Hence 𝑔↑ ∝ 𝑟
r= 𝑅−𝑑
𝒈
23. The height at which the acceleration due to gravity becomes 𝟗 (where 𝒈 = the acceleration due to
gravity on the surface of the earth) in terms 𝐨𝐟𝑹, the radius of the earth, is [2009]
𝑹
(a) (b) 𝑹𝒍𝟐 (c) 𝟐𝑹 (d) 𝟐𝐑
𝟐
SOLUTION : . (d)
𝐺𝑀
On earth’s surface 𝑔 = 𝑅2
𝐺𝑀
At height above earth’s surface 𝑔ℎ = 𝑅+ℎ 2
𝑔𝑛 𝑅2 𝑔/9 𝑅 2 𝑅 1
= 2
⇒ = ⇒ 𝑅+ℎ = 3
𝑔 𝑅+ℎ 𝑔 𝑅+ℎ
ℎ = 2𝑅
24. The change in the value of ‘g’ at a height ‘h’ above the surface ofthe earth is the same as at a
depth 𝒅’ below the surface ofearth. When both 𝒅’ and 𝒉’ are much smaller than the radius
ofearth, then which one ofthe following is correct? [2005]
𝟑𝒉 𝒉
(a) 𝒅 = (b) 𝒅 = 𝟐 (c) 𝒅 = 𝒉 (d) 𝒅 = 𝟐𝒉
𝟐
SOLUTION : . (d)
2ℎ
Value of 𝑔 with altitude is, 𝑔ℎ = 𝑔 1 − ;
𝑅
𝑑
Value of 𝑔 at depth d below earth’s surface, 𝑔𝑑 = 𝑔 1 − 𝑅
Equating 𝑔ℎ and 𝑔𝑑 , we get 𝑑 = 2ℎ
25. Average density ofthe earth [2005]
(a) is a complex function ofg (b) does not depend on 𝒈
(c) is inverselyproportional to 𝒈 (d) is directly proportional to 𝒈
SOLUTION : . (d)
4
𝐺𝑀 𝐺p×𝑉 𝐺×p× 𝜋𝑅 3
Value ofg on earth’s surface, 𝑔 = = ⇒𝑔= 3
𝑅2 𝑅2 𝑅2
4
𝑔 = 3 p𝜋𝐺. 𝑅
where p →average density
3𝑔
p= ⇒ p is directly proportional to 𝑔.
4𝜋𝐺𝑅
Gravitational Field and Potential Energy :
26. Two planets have masses 𝑴 and 16 𝑴 and their radii are 𝒂 and 𝟐𝒂, respectively. The separation
between the centres ofthe planets is 𝟏𝟎𝒂. A body ofmass 𝐦 is fired from the surface ofthe larger
planet towards the smaller planet along the line joining their centres. For the body to be able to
reach the surface of smaller planet, the minimum firing speed needed is : [6 Sep. 2020 (II)]
𝑮𝑴 𝑮𝑴 𝑮𝑴𝟐 𝟑 𝟓𝑮𝑴
(a) 𝟐 (b) 𝟒 (c) (d) 𝟐
𝒂 𝒂 𝒎𝒂 𝒂
SOLUTION : (d)
Let 𝐴 be the point where gravitation field ofboth planets cancel each other i.e. zero.
𝐺𝑀 𝐺 16𝑀
=
𝑥2 10𝑎 − 𝑥 2
1 4
⇒𝑥= ⇒ 4𝑥 = 10𝑎 − 𝑥 ⇒ 𝑥 = 2𝑎 ... (i)
10𝑎 −𝑥
𝐺𝑀𝑚 𝐺 16𝑀 𝑚 𝐺𝑀𝑚 𝐺 16𝑀 𝑚

Using conservation of energy, we have − − + 𝐾𝐸 = − −
8𝑎 2𝑎 2𝑎 8𝑎
1 16 1 16
𝐾𝐸 = 𝐺𝑀𝑚 + − −
8𝑎 2𝑎 2𝑎 8𝑎
1 + 64 − 4 − 16
⇒ 𝐾𝐸 = 𝐺𝑀𝑚
8𝑎
1 45 90𝐺𝑀 3 5𝐺𝑀
⇒ 2 𝑚𝑣 2 = 𝐺𝑀𝑚 ⇒𝑣= ⇒𝑣=2
8𝑎 8𝑎 𝑎
27. On the 𝒙‐axis and at a distance 𝒙 from the origin, the gravitational field due to a mass distribution
𝑨𝒙
is given by 𝟑/𝟐 in the 𝒙‐direction. The magnitude of gravitational potential on the 𝒙‐axis at a
𝒙𝟐 +𝒂𝟐
distance 𝒙, taking its value to be zero at infmity, is: [4 Sep. 2020 (I)]
𝑨 𝑨
(a) (b) (c) 𝑨 𝒙 + 𝒂 𝟐𝟐𝟏 /𝟐 (d) 𝑨 𝒙 + 𝒂 𝟐𝟐𝟑 /𝟐
𝒙+𝒂 𝟐𝟐𝟏 /𝟐 𝒙+𝒂 𝟐𝟐𝟑 /𝟐
SOLUTION : . (a)
𝐴𝑥
Given : Gravitational field,𝐸𝐺 = ,𝑉 =0
𝑥 2 +𝑎 2 3/2 ∞
𝑉𝑥 𝑥 𝑥 𝐴𝑥
𝑉∞
𝑑𝑉 = − ∞
𝐸𝐺 ⋅ 𝑑𝑥 ⇒ 𝛻𝑥 − 𝛻∞ = − ∞ 𝑥 2 +𝑎 2 3/2
𝑑𝑥
𝐴 𝐴
𝑉𝑥 = −0=
𝑥2 + 𝑎2 1/2 𝑥2 + 𝑎2 1/2
28. The mass density of a planet of radius 𝑹 varies with thedistance 𝒓𝐟𝐢𝐢 𝐨𝐦 its centre as 𝐩 𝒓 =
𝒓𝟐
𝐩𝟎 𝟏 − 𝑹𝟐 . Then the gravitational field is maximum at: [3 Sep. 2020 (II)]
𝟑 𝟏 𝟓
(a) 𝒓 = 𝑹 (b) 𝒓 = 𝑹 (c) 𝒓 = 𝑹 (d) 𝒓 = 𝑹
𝟒 𝟑 𝟗
SOLUTION : (d)
Mass ofsmall element ofplanet ofradius 𝑥 and thickness 𝑑𝑥.
x2
𝑑𝑚 = p × 4𝜋𝑥 2 𝑑𝑥 = p0 1 − × 4𝜋𝑥 2 𝑑𝑥
R2
𝑟 𝑥4
Mass of the planet 𝑀 = 4𝜋p0 0
𝑥 2 − 𝑅 2 𝑑𝑥
𝑟3 𝑟5
⇒ 𝑀 = 4𝜋p0 | − 2 |
3 5𝑅
𝐺𝑀 𝐺
Gravitational field, 𝐸 = = 𝑟 2 × 4𝜋p0
𝑟2
𝑟 𝑟3
⇒ 𝐸 = 4𝜋𝐺p0 −
3 5𝑅 2
𝑑𝐸
𝐸 is maximum when 𝑑𝑟 = 0
𝑑𝐸 1 3𝑟 2
⇒ = 4𝜋𝐺p0 − =0
𝑑𝑟 3 5𝑅 2
5
⇒𝑟= 𝑅
3
29. Consider two solid spheres ofradii 𝑹𝟏 = 𝟏𝒎, 𝑹𝟐 = 𝟐𝒎 and masses 𝑴𝟏 and 𝑴𝟐 , respectively. The
𝒎𝟏
gra vitational field due to sphere 𝐎𝐢 and 𝐎𝟐 are shown. The value of 𝒎𝟐
is: [8 Jan. 2020 I]
𝟐 𝟏 𝟏 𝟏
(a) 𝟑 (b) 𝟔 (c) 𝟐 (d) 𝟑
SOLUTION : (b)
𝐺𝑚
Gravitation field at the surface𝐸 = 𝑟2
𝐺𝑚 1 𝐺𝑚 2
𝐸1 = and 𝐸2 =
𝑟12 𝑟22
From the diagram given in question,
𝐸1 2
= 3(𝑟1 = 1m, 𝑅2 = 2𝑚 given)
𝐸2
m1 1
⇒ =
m2 6
30. An asteroid is moving directly towards the centre of the earth. When at a distance of10 𝑹(𝑹 is the
radius of the earth) from the earths centre, it has a speed of12 𝐤𝐦/𝐬. Neglecting the effect of
earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth
(escape velocity 𝐟𝐢𝐢𝐨𝐦 the earth is 𝟏𝟏. 𝟐𝐤𝐦/𝐬)? Give your answer to the nearest integer in
kilometer/s [NA8 Jan. 𝟐𝟎𝟐𝟎𝐈𝐈]
SOLUTION : . (16.00)
Using law of conservation of energy
Total energy at height 10 R = total energy at earth
𝐺𝑀𝐸 𝑚 1 𝐺𝑀𝐸 𝑚 1
− + 𝑚𝑉02 = − + 𝑚𝑉 2
10𝑅 2 𝑅 2
𝐺𝑀𝑚
[⋅.⋅ Gravitational potential energy = − ]
𝑟
𝐺𝑀𝐸 1 𝑉02 𝑉2 9
⇒ 1 − 10 + = ⇒ 𝑉 2 = 𝑉02 + 5 𝑔𝑅
𝑅 2 2
9
⇒𝑉= 𝑉02 + 5 𝑔𝑅 ≈ 16 km 𝑙𝑠 [V0 = 12km/s given]
31. A solid sphere ofmass 𝐌’ and radius 𝐚’ is surrounded by a uniform concentric spherical shell
ofthickness 𝟐𝐚 and mass 𝟐𝐌. The gravitational field at distance 𝟑𝐚’ from thecentre will be:
[9 April 2019 I]
𝟐𝐆𝐌 𝐆𝐌 𝐆𝐌 𝟐𝐆𝐌
(a) (b) 𝟗𝒂𝟐 (c) 𝟑𝒂𝟐 (d)
𝟗𝒂𝟐 𝟑𝒂𝟐
𝐺𝑀 𝐺 2𝑀 𝐺𝑀
SOLUTION : (c) 𝐸𝑔 = + = 3𝑎 2
3𝑎 2 3𝑎 2
32. Four identical particles ofmass 𝐌 are located at the corners of a square of side 𝒂’. What should be
their speed ifeach of them revolves under the influence of others’ gravitational field in a circular
orbit circumscribing the square? [8 April 2019 I]
𝐆𝐌 𝐆𝐌 𝐆𝐌 𝐆𝐌
(a) 𝟏. 𝟑𝟓 (b) 𝟏. 𝟏𝟔 (c) 𝟏. 𝟐𝟏 (d) 𝟏. 𝟒𝟏
𝒂 𝒂 𝒂 𝒂
SOLUTION : (b)
𝐴𝐶 𝑎 2 𝑎
𝐴𝐶 = 𝑎 2 𝑟 = = =
2 2 2
𝐺𝑀 2 𝐺𝑀 2 𝐺𝑀 2
Resultant force on the body𝐵 = î+ 𝑗+ 2 cos 45o 𝑖 + sin 45o 𝑗
𝑎2 𝑎2 𝑎 2
𝑀𝑣 2
= Resultant force towards centre
𝑟
𝑀𝑣 2 𝐺𝑀2 1
= 2 2+
𝑎 𝑎 2
2
𝐺𝑀 1
⇒ 𝑣2 = 1+
𝑎 2 2
𝐺𝑀 1 𝐺𝑀
⇒ 𝑣= 1+ = 1.16
𝑎 2 2 𝑎
33. Atest particle is moving in circular orbit in the gravitational field produced by a mass density
𝑲
𝒓 𝒓 = 𝒓𝟐 𝐈𝐝𝐞𝐧𝐭𝐢𝖍 the correct relation between the radius 𝐑 ofthe particle’s orbit and its period 𝐓:
[8 April 2019 II]
(a) 𝐓/𝐑 is a constant (b) 𝐓 𝟐 /𝐑𝟑 is a constant
(c) 𝐓/𝐑𝟐 is a constant (d) TR is a constant
SOLUTION : . (a)
R 𝑅 𝑅
𝐺𝑀𝑚 p 𝑑𝑉 𝑚 𝑘4𝜋𝑟 2 𝑑𝑟 1 4𝜋𝑘𝐺𝑚
𝐹= = 𝑎 = 𝑚𝐺 = −4𝜋𝑘𝐺𝑚 =−
𝑟 0 𝑟2 0
2
𝑟 𝑟 2 𝑟 0 𝑅
Using Newton’s second law, we have

𝑚 𝑣02 4𝜋𝑘𝐺𝑚
= or 𝑣0 = C (const.)
𝑅 𝑅
2𝜋𝑅 2𝜋𝑅 𝑇
Time period, 𝑇 = = or = 𝑅 =constant.
𝑣0 𝐶
34. Abody ofmass 𝐦 is moving in a circular orbit ofradius 𝑹 about a planet of mass 𝑴. At some
𝑹
instant, it splits into two equal masses. The first mass moves in a circular orbit of radius 𝟐 , and
𝟑𝑹
the other mass, in a circular orbit of radius . The difference between the final and initial total
𝟐
energies is: [Online Apri115, 2018]
𝑮𝑴𝒎 𝑮𝑴𝒎 𝑮𝑴𝒎 𝑮𝑴𝒎

(a) − (b)+ (c) − (d)
𝟐𝑹 𝟔𝑹 𝟔𝑹 𝟐𝑹
SOLUTION : . (c)
𝐺𝑀𝑚
Initial gravitational potential energy, 𝐸𝑗 = − Final gravitational potential energy,
2𝑅
𝐺𝑀𝑚 /2 𝐺𝑀𝑚 /2 𝐺𝑀𝑚 𝐺𝑀𝑚

𝐸𝑓 = − 𝑅 − 3𝑅 =− −
2 2 2𝑅 6𝑅
2 2
𝐺𝑀𝑚 2 1 𝐺𝑀𝑚
Difference between initial and final energy, 𝐸𝑓 − 𝐸𝑗 = −3 + 2 = −
𝑅 6𝑅
35. From a solid sphere of mass 𝐌 and radius 𝐑, a spherical portion ofradius 𝐑/𝟐 is removed, as
shown in the figure. Taking gravitational potential 𝐕 = 𝟎 at 𝐫 = ∞, the potential at the centre
ofthe cavity thus formed is: (𝑮 = gravitational constant) [2015]
−𝟐𝐆𝐌 −𝟐𝐆𝐌 −𝐆𝐌 −𝐆𝐌

(a) (b) (c) 𝐝
𝟑𝐑 𝐑 𝟐𝐑 𝐑
SOLUTION : . (d)
−GM R 2
Due to complete solid sphere, potential at point 𝑃Vsphere = 3R2 −
2R 3 2
−GM 11R2 GM
= 3
= −11
2R 4 8R
3GM
Due to cavity part potential at point 𝑃 Vcavity = −
8R
So potential at the centre of cavity
11GM 3 GM −GM
= Vsphere − Vcavity = − − −8 =
8R R R
36. Which of the following most closely depicts the correct variation ofthe gravitational potential V(r)
due to a large planet of radius 𝐑 and uniform mass density? (figures are not drawn to scale)
SOLUTION : . (c)
𝐺𝑀
As, 𝑉 = − 2𝑅 3 3𝑅 2 − 𝑟 2
Graph (c) most closely depicts the correct variation of𝑣 𝑟 .
37. The gravitational field in a region is given by 𝐠 = 𝟓𝐍/ kgî+12N/ kgî. The change in the
gravitational potential energy of a particle ofmass 1 kg when it is taken from the origin to a point
𝟕𝐦, −𝟑𝐦 is: [Online Apri119, 2014]
(a) 71 𝐉 (b) 𝟏𝟑 𝟓𝟖𝐉 𝐜 − 𝟕𝟏𝐉 (d) 1 𝐉

SOLUTION : . (d)
Gravitational field, I=(5î+12j) N/kg
dv
I=−
dr
x y
v=− 0 x
I dx + 0 y
I dy = − Ix . x + Iy . y
= − 5 7 − 0 + 12 −3 − 0 = − 35 + −36 = 1J/kg
i. e., change in gravitational potential 1 J/kg.
Hence change in gravitational potential energy 1 J
38. Two hypothetical planets ofmasses 𝐦𝟏 and 𝐦𝟐 are at rest when they are infinite distance apart.
Because of the gravitational force they move towards each other along the line joining their
centres. What is their speed when their separation is 𝐝’? [Online Apri112, 2014]
(Speed 𝐨𝐟𝐦𝟏 is 𝐯𝟏 and that 𝐨𝐟𝐦𝟐 is 𝐯𝟐 )
𝟐𝐆 𝟐𝐆
(a) 𝐯𝟏 = 𝐯𝟐 (b) 𝐯𝟏 = 𝐦𝟐 𝐯𝟐 = 𝐦𝟏
𝐝 𝐦𝟏 +𝐦𝟐 𝐝 𝐦𝟏 +𝐦𝟐
𝟐𝐆 𝟐𝐆 𝟐𝐆 𝟐𝐆
(c) 𝐯𝟏 = 𝐦𝟏 𝐯𝟐 = 𝐦𝟐 (d) 𝐯𝟏 = 𝐦𝟐 𝐯 = 𝐦𝟐
𝐝 𝐦𝟏 +𝐦𝟐 𝐝 𝐦𝟏 +𝐦𝟐 𝐦𝟏 𝟐 𝐦𝟐
SOLUTION : (b)
We choose reference point, infinity, where total energy ofthe system is zero.
So, initial energy ofthe system = 0
1 1 Gm 1 m 2
Final energy = 2 m1 v12 + 2 m2 v22 − d
From conservation of energy,
Initial energy = Final energy
1 1 Gm1 m2
0 = m1 v12 + m2 v22 −
2 2 d
1 1 Gm 1 m 2
or 2 m1 v12 + 2 m1 v22 = (1)
d
By conservation oflinear momentum
v m m
m1 v1 + m2 v2 = 0 or v 1 = − m 2 ⇒ v2 = − m 1 v1
2 1 2
Putting value ofv2 in equation (1), we get
m1 v1 2 2Gm1 m2
m1 v12 + m2 − =
m2 d
m1 m2 v12 + m12 v12 2Gm1 m2

=
m2 d
2Gm22 2G
v1 = = m2
d m1 + m2 d m1 + m2
2G
Similarly v2 = −m1
d m 1 +m 2
39. The gravitational field, due to the ↑left over part’ ofa uniform sphere (from which a part as
shown, has been ‘removed out’), at a very far offpoint, 𝐏, located as shown, would be(nearly) :
𝟓 𝑮𝑴 𝟖 𝑮𝑴 𝟕 𝑮𝑴 𝟔 𝑮𝑴
(a) 𝟔 (b) 𝟗 (c) 𝟖 (d) 𝟕
𝒙𝟐 𝒙𝟐 𝒙𝟐 𝒙𝟐
SOLUTION : (c)
R
Let mass ofsmaller sphere(which has to 𝑘 removed) is m Radius = (from figure)
2
M m M
4 =4 R 3
⇒m=
𝜋R 3 𝜋 8
3 3 2
M 7
Mass of the left over part of the sphere M′ = M − = 8M
8
GM ′ 7 GM
Therefore gravitational field due to the left over part ofthe sphere = =8
x2 x2
40. The mass of a spaceship is 1000 𝐤𝐠. It is to be launched from the earth’s surface out into free
space. The value of 𝒈 and 𝑹 (radius ofearth) are 10 𝐧𝒚𝐬𝟐 and 6400 km respectively. The required
energy for this work will be [2012]
(a) 𝟔. 𝟒 × 𝟏𝟎𝟏𝟏Joules (b) 𝟔. 𝟒 × 𝟏𝟎𝟖 Joules (c) 𝟔. 𝟒 × 𝟏𝟎𝟗 Joules (d) 𝟔. 𝟒 × 𝟏𝟎𝟏𝟎 Joules
SOLUTION : (d)
∞ ∞ 𝐺𝑀𝑚
The work done to launch the spaceship 𝑊 = − 𝑅
𝐹 ⋅ 𝑑𝑟 = − 𝑅 𝑟2
𝑑𝑟
𝐺𝑀𝑚
𝑊=+ … … …. .(i)
𝑅
The force ofattraction ofthe earth on the spaceship, when it was on the earth’s surface
𝐺𝑀𝑚
𝐹=
𝑅2
𝐺𝑀𝑚 𝐺𝑀
⇒ 𝑚𝑔 = ⇒𝑔= ... (ii)
𝑅2 𝑅2
Substituting the value of 𝑔 in (i) we get
𝑔𝑅 2 𝑚
𝑊= ⇒ 𝑊 = 𝑚𝑔𝑅
𝑅
⇒ 𝑊 = 1000 × 10 × 6400 × 103 = 6.4 × 1010 Joule
41. Apoint particle is held on the axis ofa ring ofmass 𝒎 and radius 𝒓 at a distance 𝒓𝐟𝐢𝐢 𝐨𝐦 its centre
𝑪. When released, it reaches 𝑪 under the gravitational attraction ofthe ring. Its speed at 𝑪 will be
𝟐𝑮𝒎 𝑮𝒎 𝟐𝑮𝒎 𝟏 𝟐𝑮𝒎

(a) 𝟐−𝟏 (b) (c) 𝟏− (d)
𝒓 𝒓 𝒓 𝟐 𝒓
SOLUTION : . (c)
Let 𝑀 be the mass of the particle
Now, Einitia 1 = Ef 1 na 1
GMm GM 𝑚 1 1 𝐺𝑀𝑚 1
i. e. +0= + 2 𝑀V 2 or , 𝑀𝑉 2 = 1−
2𝑟 𝑟 2 𝑟 2
1 2
𝐺𝑚 1 2𝐺𝑚 1
⇒2 𝑉 = 1− or , 𝑉= 1−
𝑟 2 𝑟 2
42. Two bodies ofmasses 𝒎 and 4 𝒎 are placed at a distance 𝒓. The gravitational potential at a point
on the line joining them where the gravitational field is zero is: [2011]
𝟒𝐆𝐦 𝟔𝐆𝐦 𝟗𝐆𝐦

(a) − (b) − (c) − (d) zero
𝐫 𝐫 𝐫
SOLUTION : . (c)
Let 𝑃 be the point where gravitational field is zero.
𝐺𝑚 4𝐺𝑚
2
= 2
𝑥 𝑟−𝑥
1 2 𝑟
⇒ 𝑥 = 𝑟−𝑥 ⇒ 𝑟 − 𝑥 = 2𝑥 ⇒ 𝑥 = 3
𝐺𝑚 4𝐺𝑚 9𝐺𝑚
Gravitational potential at 𝑃, 𝑉=− r − 2r =− 𝑟
3 3
43. This question contains Statement‐l and Statement‐2. Of the four choices given after the
statements, choose the one that best describes the two statements. [2008]
Statement‐l : For a mass 𝑴 kept at the centre ofa cube of side 𝒂’, the flux ofgravitational field
passing through its sides 4 𝝅𝑮𝑴. and
Statement‐2: Ifthe direction ofa field due to a point source is radial and its dependence on the
𝟏
distance ‘r’ 𝐟𝐫𝐨𝐦 the source is given as 𝒓𝟐 , its flux through a closed surface depends only on
the strength of the source enclosed by the surface and not on the size or shape ofthe surface.
(a) Statement‐l is false, Statement‐2 is true
(b) Statement‐l is true, Statement‐2 is true; Statement ‐𝟐 is a correct explanation for Statement‐ 1
(c) Statement‐l is true, Statement‐2 is true; Statement2 is not a correct explanation for
Statement‐l
(d) Statement‐l is true, Statement‐2 is false
SOLUTION : (b)
𝐺𝑀
Gravitational field, 𝐸 = − 𝑟2
Flux, 𝜑 = 𝐸𝑔 ⋅ 𝑑 𝑆 = |𝐸 ⋅ 4𝜋𝑟 2 | = −4𝜋𝐺𝑀
where, M = mass enclosed in the closed surface
1
This relationship is valid when 𝐸𝑔 ∞ 𝑟 2
44. A particle ofmass 10 𝐠 is kept on the surface ofa uniform sphere ofmass 100 kg and radius 10 cm.
Find the work to be done against the gravitational force between them to take the particle far
away from the sphere(you may take 𝑮 = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏𝐍𝐦𝟐 /𝐤𝐠 𝟐 ) [2005]
(a) 𝟑. 𝟑𝟑 × 𝟏𝟎−𝟏𝟎𝐉 (b) 𝟏𝟑. 𝟑𝟒 × 𝟏𝟎−𝟏𝟎𝐉 (c) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟎 𝐉 (d) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟗 𝐉
SOLUTION : (c)
𝐺𝑀𝑚
Initial P.E. 𝑈𝑖 = − 𝑅
When the particle is far away from the sphere, the P.E. of the system is zero.
𝑈𝑓 = 0
−𝐺𝑀𝑚
𝑊 = 𝛥𝑈 = 𝑈𝑓 − 𝑈𝑗 = 0 −
𝑅
6.67×10−11 ×100 10
𝑊= × 1000 = 6.67 × 10−10 J
0.1
45. If ‘g’ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of
an object of mass ‘m’ raised from the surface ofthe earth to a height equal to the radius 𝑹′ ofthe
earth is [2004]
𝟏 𝟏
(a) 𝟒 𝒎𝒈𝑹 𝐛 𝒎𝒈𝑹 (c) 2 𝒎𝒈𝑹 (d) 𝒎𝒈𝑹
𝟐
SOLUTION : . (b)
𝐺𝑚𝑀
On earth’s surface potential energy, 𝑈 = 𝑅
𝐺𝑚𝑀
At a height 𝑅 from the earth↑ s surface, P.E. of system = − 2𝑅
−𝐺𝑚𝑀 𝐺𝑚𝑀
𝛥𝑈 = + ;
2𝑅 𝑅
𝐺𝑚𝑀
⇒ 𝛥𝑈 =
2𝑅
𝐺𝑀 𝐺𝑀
Now = 𝑔; .⋅. = 𝑔𝑅
𝑅2 𝑅
1
𝛥𝑈 = 𝑚𝑔𝑅
2
46. Energyrequired to move abodyofmass 𝒎𝐟𝐢𝐢 𝐨𝐦 an orbit of radius 𝟐𝑹 to 𝟑𝑹 is [2002]
(a) GMml 𝟏𝟐𝑹𝟐 (b) 𝑮𝑴𝒎𝒍𝟑𝑹𝟐 (c) 𝑮𝑴𝒎𝒍𝟖𝑹 (d) 𝑮𝑴𝒎𝒍𝟔𝑹.
SOLUTION : . (d)
𝐺𝑀𝑚
Gravitational potential energy ofmass 𝑚 in an orbit ofradius R 𝑢=− 𝑅
−𝐺𝑀𝑚 −𝐺𝑀𝑚
Energy required = potential energy at 3R − potential energy a2R = −
3𝑅 2𝑅
−𝐺𝑀𝑚 𝐺𝑀𝑚 −2𝐺𝑀𝑚 +3𝐺𝑀𝑚 𝐺𝑀𝑚

= + = =
3𝑅 2𝑅 6𝑅 6𝑅
47. A satellite is in an elliptical orbit around a planet P. It is observed that the velocity ofthe satellite
when it is farthest from the planet is 6 times less than that when it is closest to the planet. The
ratio of distances between the satellite and the planet at closest and farthest points is:
[NA 6 Sep. 2020 (I)]
(a) 1: 6 (b) 1: 3 (c) 1: 2 (d) 3: 4
SOLUTION : . (a)
By angular momentum conservation 𝑟m ln 𝑣 max = 𝑚𝑟max 𝑣 min

𝑣 max
Given, 𝑣 min = 6
𝑟 min 𝑣 min 1
= =
𝑟max 𝑣 max 6
48. Abody is moving inalow circular orbit about aplanet of mass 𝑴 and radius 𝑹. The radius ofthe
orbit can be taken to be 𝑹 itself. Then the ratio ofthe speed of this body in the orbit to the escape
velocity from the planet is: [4 Sep. 2020 (II)]
𝟏
(a) (b) 2 (c) 1 (d) 𝟐
𝟐
SOLUTION : (a)
Orbital speed ofthe body when it revolves very close to the surface of planet
𝐺𝑀
𝑉0 = (i)
𝑅
Here, 𝐺 = gravitational constant
Escape speed from the surface ofplanet
2𝐺𝑀
𝑉𝑒 = (ii)
𝑅
Dividing (i) by(ii), we have
𝐺𝑀
V0 𝑅 1
= =
Ve 2𝐺𝑀 2
𝑅
49. Asatellite is moving inalow nearly circular orbit around the earth. Its radius is roughly equal to
that ofthe earth’s radius 𝑹𝒆 . By firing rockets attached to it, its speed is instantaneously increased
𝟑
in the direction ofits motion so that it become times larger. Due to this the farthest distance
𝟐
𝐟𝐢𝐢 𝐨𝐦 the centre of the earth that the satellite reaches is 𝑹. Value 𝐨𝐟𝑹 is: [3 Sep. 2020 (I)]
(a) 𝟒𝑹𝒆 (b) 𝟐. 𝟓𝑹𝒆 (c) 𝟑𝑹𝒆 (d) 𝟐𝑹𝒆

SOLUTION : .(c)
𝐺𝑀
Orbital velocity, 𝑉0 = 𝑅𝑒
2
𝐺𝑀𝑚 1 3 𝐺𝑀𝑚 1 2
From energy conversation, − +2𝑚 𝑉 =𝑅 + 2 𝑚𝑉 min (1)
𝑅𝑒 2 max
3
From angular momentum conversation 𝑉𝑅𝑒 = 𝛻 min 𝑅 max (2)
2
Solving equation (1) and (2) we get, 𝑅 max = 3𝑅𝑒
𝑲
50. The mass density ofa spherical galaxy varies as over a large distance’r’ from its centre. In that
𝒓
region, a small star is in a circular orbit of radius 𝑹. Then the period of revolution, 𝑻 depends on
𝑹 as: [2 Sep. 2020 (I)]
𝟏
(a) 𝑻𝟐 ∝ 𝑹 𝐛 𝑻𝟐 ∝ 𝑹𝟑 𝐜 𝑻𝟐 ∝ 𝑹𝟑 𝐝 𝑻∝𝑹
SOLUTION : (a)
𝑘
According to question, mass density of a spherical galaxy varies as 𝑟
𝑟−−𝑅0 𝑘 𝑅0
Mass, 𝑀 = p 𝑑𝑉 ⇒ 𝑀 = 0
4𝜋𝑟 2 𝑑𝑟 ⇒ 𝑀 = 4𝜋𝑘 0
𝑟 𝑑𝑟
𝑟
4𝜋𝑘 𝑅02
or, 𝑀 = = 2𝜋𝑘𝑅 2
2
4𝜋𝑘 𝑅 2
𝐺𝑀𝑚 𝐺 2𝜋𝐾𝐺 2𝜋
𝐹𝐺 = = 𝑚𝔠020 𝑅 = 𝐹𝐶 ⇒ 2
= 0)2 0𝑅 ⇒ 0)0 = ⋅.⋅ t𝑖J =
𝑅02 𝑅2 𝑅 𝑇
2𝜋 2𝜋 𝑅 2𝜋𝑅 2𝜋𝑅
𝑇= )= = ⇒ 𝑇2 =
00 2𝜋𝐾𝐺 𝐾𝐺 𝐾𝐺
2𝜋, 𝐾 and 𝐺 are constants
𝑇 2 ∝ 𝑅.
51. Abody A ofmass 𝒎 is moving in a circular orbit ofradius𝐑 about a planet. Another body 𝐁
𝒎 𝒗
ofmass collides with A with a velocity which is half the instantaneous velocity 𝒗 or A. The
𝟐 𝟐
collision is completely inelastic. Then, the combined body: [9 Jan. 2020 I]
(a) continues to move in a circular orbit
(b) Escapes 𝐟𝐢𝐢 𝐨𝐦 the Planet’s Gravitational field
(c) Falls vertically downwards towards the planet
(d) starts moving in an elliptical orbit around the planet
SOLUTION : . (d)
From law ofconservation ofmomentum, 𝑃𝑖 = 𝑃𝑓
𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑀𝑉𝑓
𝑚𝑣
𝑚𝑣 + 4 3𝑚 5𝑣
⇒ 𝑣𝑓 = =
2 6
Clearly, 𝑣𝑓 < 𝑣𝑗 Path will be elliptical

52. The energy required to take a satellite to a height 𝐡′ above Earth surface (radius of Eareth
= 𝟔. 𝟒 × 𝟏𝟎𝟑 km) is 𝐄𝟏 and kinetic energy required for the satellite to be in a circular orbit at this
height is 𝐄𝟐 . The value of 𝐡 for which 𝐄𝟏 and 𝐄𝟐 are equal, is: [9 Jan. 2019 II]
(a) 𝟏. 𝟔 × 𝟏𝟎𝟑 km (b) 𝟑. 𝟐 × 𝟏𝟎𝟑 km (c) 𝟔. 𝟒 × 𝟏𝟎𝟑 km (d) 𝟐𝟖 × 𝟏𝟎𝟒 km
SOLUTION : (b)
K.E. ofsatellite is zero at earth surface and at height h from energy conservation
Usu 𝛤 + E =Uface 1h
GMe m GMe m
− + E1 = −
Re Re + h
GMe m h
⇒ E1 = GMe m −
R e R e +h ⇒ E1 = ×
Re + h Re
Gravitational attraction
mv 2 GMe m
𝛤G = mac = =
Re + h Re + h 2
GMe m
mv 2 =
Re + h
mv 2 GMe m
E2 = =
2 2 Re + h
E1 = E2
h 1 Re
Clearly, R = 2 ⇒ h = = 3200km
e 2
53. Planet A has mass 𝐌 and radius R. Planet 𝐁 has halfthe mass and halfthe radius ofPlanet A. Ifthe
𝒗𝐀 𝒏
escape velocities from the Planets A and 𝐁 are 𝒗𝐀 and 𝒗𝐁 , respectively, then = 𝟒 The value of n
𝒗𝐁
is: [9 Jan. 2020 II]
(a) 4 (b) 1 (c) 2 (d) 3
SOLUTION : (a)
2𝐺𝑀𝐴
Escape velocity of the planet 𝐴 is 𝑉𝐴 = 𝑅𝐴
where 𝑀𝐴 and 𝑅𝐴 be the mass and radius ofthe planet𝐴.
𝑀𝐴 𝑅𝐴
According to given problem𝑀𝐵 = , 𝑅𝐵 =
2 2
⇒ 𝑛=4
54. A satellite of mass 𝒎 is launched vertically upwards with an initial speed 𝒖 from the surface of
𝒎
the earth. After it reaches height 𝑹 (𝑹 = radius of the earth), itejects a rocket of mass 𝟏𝟎 so that
subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (𝑮 is the
gravitational constant; 𝑴 is the mass ofthe earth): [7 Jan. 2020 I]
𝟐
𝟐𝑮𝑴 𝟐
𝒎 𝟏𝟏𝟑𝑮𝑴 𝟏𝟏𝟗𝑮𝑴 𝟑𝒎 𝟓𝑮𝑴 𝒎 𝒖− )
(a) 𝟐𝟎 𝒖𝟐 + (b) 𝟓𝒎 𝒖𝟐 − (c) (d) 𝟐𝟎 ( 𝟑𝑹
𝟐𝟎𝟎𝑹 𝟐𝟎𝟎𝑹 𝟖 𝟔𝑹
SOLUTION : .(b)
1 −𝐺𝑀𝑚 1 −𝐺𝑀𝑚
𝑚𝑢2 + = 𝑚𝑣 2 +
2 𝑅 2 2𝑅
1 −𝐺𝑀𝑚 GM
⇒ 2 𝑚 𝑣 2 − 𝑢2 = ⇒V= V = u2 − … .. (i)
2𝑅 R
𝐺𝑀 𝑚×𝑣
𝑣0 = 𝑣𝑟𝑎𝑑 = 𝑚 = 10𝑣
2𝑅
10
𝑚
Ejecting a rocket ofmass 10
9𝑚 𝐺𝑀 𝑚 𝐺𝑀
× = × 𝑣𝜏 ⇒ 𝑉𝜏2 = 81
10 2𝑅 10 2𝑅
1𝑀
Kinetic energy ofrocket, KE𝑟𝑜𝑐𝑘𝑒𝑡 = 2 10 𝑉𝑇2 + 𝛻𝑟2
1 𝑚 𝐺𝑀 𝐺𝑀 𝑚 𝐺𝑀 81𝐺𝑀
= 2 × 10 × 𝑢2 − 100 + 81 = 20 × 100 𝑢2 − +
𝑅 𝑅 𝑅 200𝑅
119𝐺𝑀
= 5𝑚 𝑢2 −
200𝑅
55. A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only
gravitational field of the planet acts on the spaceship, what will be the number ofcomplete
revolutions made by the spaceship in 24 hours around the planet? [Given : Mass ofPlanet
= 𝟖 × 𝟏𝟎𝟐𝟐𝐤𝐠, Radius of planet = 𝟐 × 𝟏𝟎𝟔 𝐦, Gravitational constant𝐆 = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏𝐍𝐦𝟐 /𝐤𝐠 𝟐 ]
[10 April 2019 II]
(a) 9 (b) 17 (c) 13 (d) 11
SOLUTION : (d)
2𝜋𝑟
Time period ofrevolution of satellite, 𝑇 =
v
𝐺𝑀
𝑣=
𝑟
𝑟 𝑟3
𝑇 = 2𝜋𝑟 = 2𝜋
𝐺𝑀 𝐺𝑀
202 3 ×1012
Substituting the values, we get 𝑇 = 2𝜋 sec
667×10−11 ×8×1022
T = 7812.2s
T = 2.17hr ⇒ 11 revolutions.
56. A rocket has to be launched from earth in such a waythat it never returns. 𝐈𝐟𝐄 is the minimum
energy delivered by the rocket launcher, what should be the minimum energy that the launcher
should have ifthe same rocket is to be launched 𝐟𝐢𝐢 𝐨𝐦 the surface ofthe moon? Assume that the
density ofthe earth and the moon are equal and that the earth’s volume is 64 times the volume
ofthe moon. [8 April 2019 II]
𝐄 𝐄 𝐄 𝐄
(a) 𝟔𝟒 (b) 𝟑𝟐 (c) 𝟒 (d) 𝟏𝟔
SOLUTION : . (d)
Escape velocity,
2𝐺𝑀 2𝐺p𝑉 2𝐺𝑆×4𝜋𝑅 3 8

𝑣𝑐 = = = = 𝜋p𝐺𝑅 2
𝑅 𝑅 𝑅 3
8
For moon, 𝑣 𝑐 = 3
𝜋 p𝐺 𝑅 2𝑚
4 4 𝑅
Given, 3 𝜋 𝑅 3 = 64 × 3 𝜋 𝑅 3𝑚 or 𝑅 𝑚 = 4
2
8 𝑅 𝑣𝑐
𝑣𝑒 = 𝜋 p𝐺 =
3 4 4
1
𝐸 2 𝑚𝑣 2𝑒 𝑣 2𝑒 𝑣𝑒
= = 2 = 𝑣 = 16
𝐸 ’ 1 𝑚𝑣 2 𝑣 𝑐 ′ 𝑒
2 𝑒 4
𝐸
or 𝐸 ’ = 16
57. A satellite ofmass 𝐌 is in a circular orbit ofradius Rabout the centre ofthe earth. A meteorite
ofthe same mass, falling towards the earth collides with the satellite completely in elastically. The
speeds ofthe satellite and the meteorite are the same, Just before the collision. The subsequent
motion ofthe combined bodywill be [12 Jan. 2019 I]
(a) such that it escape to infinity (b) In an elliptical orbit
(c) in the same circular orbit ofradius 𝐑 (d) in a circular orbit ofa different radius
SOLUTION : (b)
v v
mvî+ mvj = 2mv ⇒ v = 2 î+ 2 j
v 2 v 2 v 1 GM
⇒v= + = = ×
2 2 2 2 R
58. Two satellites,AandB, have masses m𝐚𝐧𝐝𝟐𝐦𝐫espectively. A is in a circular orbit ofradius 𝐑, and
𝐁 is in a circular orbit of radius 𝟐𝐑 around the earth. The ratio of their kinetic energies, 𝐓𝐀 /𝐓𝐁 ,
is: [12 Jan. 2019 II]
𝟏 𝟏
(a) 𝟐 (b) 1 (c) 2 (d) 𝟐
SOLUTION : (b)
GM
Orbital, velocity, v = r
Kinetic energy of satellite A, T = mA V2A
Kinetic energy ofsatellite B, T = mB V2B
GM
TA m × R
⇒ = 1
TB 2m × GM
2R
59. Asatellite is revolving in acircular orbit at aheight hfrom the earth surface, such that 𝐡 <<
Rwhere 𝐑 is the radius of the earth. Assuming that the effect ofearth’s atmosphere can be
neglected the minimum increase in the speed required so that the satellite could escape from the
gravitational field ofearth is: [11 Jan. 2019 I]
𝒈𝑹
(a) 𝟐𝒈𝑹 (b) 𝒈𝑹 (c) (d) 𝒈𝑹 𝟐−𝟏
𝟐
SOLUTION : . (d)
For a satellite orbiting close to the earth, orbital velocity is given by v0 = g R+h ≈ gR
Escape velocity ve is ve = 2g R + h ≈ 2gR [h << 𝑅 ]

𝛥 v = ve − v0 = 2−1 gR
60. A satellite is moving with a constant speed 𝒗 in circular orbit around the earth. An object ofmass
‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth.
At the time of ejection, the kinetic energy of the object is: [10 Jan. 2019 I]
𝟏 𝟑
(a) 2 𝐦𝒗𝟐 (b) 𝐦𝒗𝟐 (c) 𝟐 𝐦𝒗𝟐 (d) 𝟐 𝐦𝒗𝟐
SOLUTION : (b)
GM
At height rfii om center ofearth, orbital velocity𝑣 = 𝑟
GMm
By principle of energy conservation KE of m’ + − =0+0
r
(At infinity, PE = KE = 0)
m 2
GMm GM
or KE of m’ = = = mv2
r r
61. Two stars of masses 𝟑 × 𝟏𝟎𝟑𝟏 kg each, and at distance 𝟐 × 𝟏𝟎𝟏𝟏 𝐦 rotate in a plane about their
common centre of mass O. A meteorite passes through 𝐎 moving perpen‐dicular to the star’s
rotation plane. In order to escape from the gravitational field of this double star, the minimum
speed that meteorite should have at 𝐎 is: (Take Gravitational constant 𝐆 = 𝟔𝟔 × 𝟏𝟎−𝟏𝟏 𝐍𝐦𝟐 𝐤𝐠 −𝟐 )
[10 Jan. 2019 II]
(a) 𝟐. 𝟒 × 𝟏𝟎𝟒 𝐦/𝐬 (b) 𝟏. 𝟒 × 𝟏𝟎𝟓 𝐦/𝐬 (c) 𝟑. 𝟖 × 𝟏𝟎𝟒 𝐦/𝐬 (d) 𝟐. 𝟖 × 𝟏𝟎𝟓 𝐦/𝐬
SOLUTION : (d)
Let M is mass ofstar m is mass ofmeteroite By energy convervation between 0 and ∞.
GMm −GMm 1
− + + mV2ese = 0 + 0
r r 2
4GM 4×667×10−11 ×3×1031

v= = = 2.8 × 105 m/s
r 1011
62. Asatellite is revolving in acircular orbit at aheight ‘h’from the earth’s surface (radius ofearth
𝐑; 𝐡 << 𝑹). The minimum increase in its orbital velocityrequired, so that the satellite could
escape from the earth’s gravitational field, is close to: (Neglect the effect ofatmosphere.) [2016]
(a) 𝐠𝐑/𝟐 (b) 𝐠𝐑 𝟐−𝟏 (c) 𝟐𝐠𝐑 (d) 𝐠𝐑
SOLUTION : (b)
For h << 𝑅 , the orbital velocity is gR Escape velocity = 2gR
The minimum increase in its orbital velocity = 2gR − gR = gR 2−1
63. An astronaut ofmass 𝐦 is working on a satellite orbiting the earth at a distance 𝐡𝐟𝐢𝐢 𝐨𝐦 the
𝐞𝐚𝐫𝐭𝐡↑ 𝐬 surface. The radius ofthe earth is 𝐑, while its mass is M. The gravitational pull 𝜞𝐆 on the
astronaut is: [Online Apri110, 2016]
𝐆𝐌𝐦 𝐆𝐌𝐦
(a) Zero since astronaut feels weightless (b) < 𝜞𝐆 <
𝐑+𝐡 𝟐 𝐑𝟐
𝐆𝐌𝐦 𝐆𝐌𝐦
(c) 𝜞𝐆 = (d) 𝟎 < 𝜞𝐆 <
𝐑+𝐡 𝟐 𝐑𝟐
SOLUTION : . (c)
According to universal law ofGravitation,
2
GMm Gravitational force F = R + h
64. A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius
𝐑(𝐑 << 𝑳). 𝑨 star outside the galaxy is orbiting the galaxy in a plane perpendicular to the
galaxy and passing through its centre. If the time period of star is 𝐓 and its distance from the
galaxy’s axis is 𝒓, then: [OnlineApri110, 2015]
(a) 𝐓 ∝ 𝐫 (b) 𝐓 ∝ 𝐫 (c) 𝐓 ∝ 𝐫 𝟐 (d) 𝐓 𝟐 ∝ 𝐫 𝟑
SOLUTION : ( a )
65. What is the minimum energyrequired to launch a satellite of mass 𝐦 from the surface of a planet
of mass 𝐌 and radius 𝐑 in a circular orbit at an altitude of 𝟐𝐑? [2013]
𝟓𝐆𝐦𝐌 𝟐𝐆𝐦𝐌 𝐆𝐦𝐌 𝐆𝐦𝐌
(a) 𝐛 𝐜 (d)
𝟔𝐑 𝟑𝐑 𝟐𝐑 𝟐𝐑
SOLUTION : . (a)
−𝐺𝑀𝑚
As we know, Gravitational potential energy U = 𝑟
and orbital velocity, 𝑉0 = GM/R + h
1 GMm 1 GM GMm
Ef = mv20 − = m −
2 3R 2 3R 3R
GMm 1 −GMm
= −1 =
3R 2 6R
−GMm
Ei = +K
R
5GMm
Therefore minimum required energy, K = 6R
66. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10
times smaller. Given that the escape velocity from the earth is 11 km 𝐬−𝟏 , the escape velocity
𝐟𝐢𝐢 𝐨𝐦 the surface ofthe planet would be [2008]
(a) l.lkm 𝐬−𝟏 (b) llkm 𝐬−𝟏 (c) 𝟏𝟏𝟎𝐤𝐦𝐬−𝟏 (d) 𝟎. 𝟏𝟏 km 𝐬−𝟏
SOLUTION : (c)
67. Suppose the gravitational force varies inversely as the nth power ofdistance. Then the time period
ofa planet in circular orbit ofradius ‘R’ around the sun will be proportional to
𝒏−𝟏 𝒏+𝟏 𝒏−𝟐

(a) 𝑹𝒏 (b) 𝑹 [2004] (c) 𝑹 (d) 𝑹
𝟐 𝟐 𝟐
SOLUTION : (c )
68. The time period of an earth satellite in circular orbit is independent of [2004]
(a) both the mass and radius ofthe orbit (b) radius of its orbit
(c) the mass ofthe satellite (d) neither the mass ofthe satellite nor the radius of its orbit.
SOLUTION : (c)
Time period of satellite is given by
𝑅+ℎ 3
𝑇 = 2𝜋
𝐺𝑀
Where 𝑅 + ℎ = radius of orbit of satellite
𝑀 = mass ofearth.
Time period is independent ofmass ofsatellite.
69. A satellite ofmass 𝒎 revolves around the earth ofradius R at a height 𝒙𝐟𝐢𝐢 𝐨𝐦 its surface. Ifg is the
acceleration due to gravity on the surface ofthe earth, the orbital speed ofthe satellite is [2004]
𝒈𝑹𝟐 𝒈𝑹 𝒈𝑹𝟐
(a) 𝑹+𝒙 (b) 𝑹−𝒙 (c) 𝒈𝒙 (d) 𝑹+𝒙
SOLUTION : (d)
70. The escape velocity for a body projected vertically upwards 𝐟𝐢𝐢 𝐨𝐦 the surface ofearth is 11 𝐤𝐦/𝐬.
Ifthe bodyis projected at an angle of 𝟒𝟓𝐨 with the vertical, the escape velocitywillbe [2003]
𝟏𝟏
(a) 𝟏𝟏 𝟐𝐤𝐦/𝐬 (b) 22 𝐤𝐦/𝐬 (c) 11 𝐤𝐦/𝐬 (d) 𝐤𝐦/𝐬
𝟐
SOLUTION : (c)
71. The kinetic energyneeded to project a 𝐛𝜶𝐥𝐲 ofmass 𝒎𝐟𝐢𝐢 𝐨𝐦 the earth surface (radius 𝑹) to
infinity is [2002]
(a) 𝒎𝒈𝑹𝒍𝟐 (b) 𝟐𝒎𝒈𝑹 (c) 𝒎𝒈𝑹 (d) 𝒎𝒈𝑹𝒍𝟒.
SOLUTION : (c)
72. If suddenly the gravitational force of attraction between Earth and a satellite revolving around it
becomes zero, then the satellite will [2002]
(a) continue to move in its orbit with same velocity
(b) move tangentially to the original orbit in the same velocity
(c) become stationary in its orbit
(d) move towards the earth
SOLUTION : ( b )
Due to inertia ofmotion it will move tangentially to the original orbit with the same velocity.
73. The escape velocity of a body depends upon mass as [2002]
(a) 𝒎𝟎 (b) 𝒎𝟏 (c) 𝒎𝟐 (d) 𝒎𝟑
SOLUTION : (a ) ‘
MECHANICAL PROPERTIES OF SOLIDS
Deforming Force :
External force which try to change in the length, volume or shape of the body is called deforming
force.
Elasticity:
Elasticity is that property of the material of a body by virtue of which the body oppose any
change in its shape or size when deforming forces are applied to it, and recover its original
state as soon as the deforming force are removed.
Perfectly Elastic Body :
The body which perfectly regains its original form on removing the external deforming force,
is defined as a perfectly elastic body. Ex. : quartz – Very nearly a perfect elastic body.
Plastic Body:
(a) The body which does not have the property of opposing the deforming force, is known
as a plastic body.
(b) The bodies which remain in deformed state even after removed of the deforming force
are defined as plastic bodies.
Internal restoring force :
When a external force acts at any substance then due to the intermolecular force there is
a internal resistance produced into the substance called internal restoring force.
At equilibrium the numerical value of internal restoring force is equal to the external force.
STRESS:
The internal restoring force acting per unit area of cross-section of the deformed body is
called stress.
Internal restoring force Fint ernal Fexternal
Stress  = =
Area of cross section A A
SI UNIT : N-m–2 Dimension : M1L–1T–2
There are three types of stress –

(I) Longitudinal stress
(II) Volume stress
(III) Shear stress
Longitudinal Stress :
When the stress is normal to the surface of body, then it is known as longitudinal stress.
there are two types of longitudinal stress
Tensile Stress
The longitudinal stress, produced due to increase in length of a body, is defined as tensile
stress.
1
F ll
wa
 
F
tensile stress
tensile stress
Compressive Stress :
The longitudinal stress, produced due to decrease in length of a body, is defined as
compressive stress.
Volume Stress :
If equal normal forces are applied every one surface of a body, then it undergoes change
in volume. The force opposing this change in volume per unit area is defined as volume
stress.
Shear Stress :
When the stress is tangential or parallel to the surface of a body then it is known as shear
stress. Due to this stress, the shape of the body changes or it gets twisted.
Difference between pressure and stress
Pressure Stress
Pressure is always normal to the Stress can be either normal or tangential.
area.
Pressure on a body is always Stress can be compressive or tensile
compressive. or shear.
Pressure is a scalar Stress is a tensor
If deforming force is applied tangential to the surface, such that tangential stress is developed
in a body, then the shape of the body may change.
F  er
n
A
 F
F
When a force ‘F’ acts at an angle '  ' with outward normal n to the area A as shown in figure. In
this case, the stress will have the normal and tangential components.
To find the linear (or) longitudinal stress, take the component of the force perpendicular to the
plane of a given area A, then divide this component ( Fer ) by the area ‘A’.
2
F cos 
Fer
Longitudinal stress = 
A A
To find the shearing stress, take the component of force parallel to the plane of the given area
and then devide Fel by the area ‘A’.
FPal F sin 
Shearing stress = 
A A
The total stress = longitudinal stress + shearing stress
But not F/A.
Breaking Stress :
The stress required to cause actual facture of a material is called the breaking stress or
ultimate strength.
F
Breaking stress =
A
Dependence of breaking stress : (i) Nature of material (ii) Temperature (iii) Impurities.
Independence of breaking stress : (i) Cross sectional area or thickness ( i i )
Applied force.
Note :The stress required to cause actual fracture of a material is called the breaking stress
or the ultimate strength.
Breaking force = Breaking stress x area of cross section

Ø Breaking force is independent of length of the wire, but it depends on the nature of material
and area of cross section.
F  A, F  r 2 [in case of cylindrical wire]
Ø The maximum length of the wire that can be hanged without breaking under its own weight
breaking stress

dg
Ø If we cut a wire that can support a maximum load W into two equal parts, then each part of the
wire can support a maximum load W.
Punching a hole :
Ø The force required to punch a hole of radius 'r' in a metal plate of thickness 't' is
F = Maximum shearing stress x 2rt
STRAIN L
change in size of the body

Strain = original size of the body
3
There are three types of strain :
Type of strain depends upon the directions of applied force.
Longitudinal strain =
Longitudinal strain is possible only in solids
change in volume of the body V

Volume strain = original volume of the body =
V
Shear strain
When a deforming force is applied to a body parallel to its surface then its shape (not size) changes.
The strain produced in this way is known as shear strain.
The strain produced due to change of shape of the body is known as shear strain.


tan =
L F
A
 A' B B'
 displacement of upper face L

or  = = 
L distance between two faces
F
D C
Relation Between angle of twist and Angle of shear
fixed
r B
When a cylinder of length 'l' and radius 'r' is fixed at one end 
'

O 
tw is A A'
ted
and tangential force is applied at the other end, then the

cylinder gets twisted. Figure shows the angle of shear
ABA' and angle of twist AOA'.
Arc AA' = r  and Arc AA' = l 
r
so r  l r 

4
 = angle of twist,  = angle of shear
Youngs modulus
Longitudinal stress and Longitudinal strain are in constant ratio is called Young's modulus.
longitudinal stress
Young's modulus=
longitudinal strain
F
A F . F. 
or Y= e
= Þ e = A.Y
e A

1. Two wires made of same material having lengths l1 & l2 and radii r1 and r2 are subjected to the same
force. Then the ratio of their elongations is
e1 l1 r22 l
  (since e  2 )
e 2 l 2 r12 r
2.Two wires made of same material having lengths l1 and l2 and masses m1 and m2 are subjected to the
same force. Then the ratio of their elongations is
e1 l12 m 2 l2
 2 (since e  )
e 2 l 2 m1 m
3. Two wires of same material and same volume having areas of cross section A1 & A2 are subjected
to the same force. Then the ratio of their elogations is
e1 A 22 1
 (since e  )
e 2 A12 A2
4. Two wires of same material and same volume having radii r1 and r2 are subjected to the same
force. Then the ratio of their elongations is
e1 r24 1
 (since e  )
e 2 r14 r4
5.When a body of mass 'm' and density 'dB' is suspended from a wire its elongation is 'e' when it is in air.
If it is completely immersed in a non-viscous liquid of density dL then its new elongation is
dL
e1 = e(1– d )
B
PROBLEMS
1.A steel wire of 2mm in diameter is stretched by applying a force of 72N. Find the stress
in the wire.
Solution : r  1103 m; F=72N
F F 72
The stress =  2 
A r  1103 
2
72
=  2.292107 Nm2
 106
5
2:A uniform rope of mass M and Length L, on which a force F is applied at one end,
then find stress in the rope at a distance x from the end where force is applied?
Solution :
M T F
= mass per unit length
L
X
F
From F  Ma  a 
M
M
Tension, T  L  xa
L
M F F
 L  x   L  x
L M L
T F x
Stress   1  
A A L
Where tension T and area A must be perpendicular for tensile stress.
3.A force F is required to break a wire of length L and radius r. What is the force required
to break a wire of the same material, twice the length and 4 times the radius?
F
Solution : breaking stress =
 r2  F  r
2
F1 r12 1
 
F2 r22 16
 F 1  16 F
So the force required is 16 F in the second case
4.The breaking stress of steel is 7.9×109 Nm–2 and density of steel is 7.9 × 103 kgm–3 and
g = 10ms–2. The maximum length of steel wire that can hang vertically without breaking
is
F
solution: Breaking Stress   lρ g
A
Breaking stress 7.9 109
L   105 m
g 7.9 10  10
3
.
5: A body of mass “m” is connected to an inextensible thread of length “L” is whirled in
horizontal circle. Find the maximum angular velocity with which it can be whirled
without breaking the thread (Breaking stress of thread =S).
6
T  F centripetal  mL2
solution: force
Tmax mLmax
2
S 
A A
SA
max 
mL
6: In the given graph extension (e) of a wire of length 1m is suspended from the top of a
roof at one end and with a load “w” connected to other end , if the cross sectional
area of the wire is 10-6m2, the Young’s modulus of the material of wire is
 e  10  4 m 
4
3
2
wN
20 40 60 80
e 1
Sol: From graph slope,  104
w 20
w F
  20  104
e e
F  w 
from Y     
A e e A
1
 20 10 4   2 1011 N / m 2
106
7. The length of a metal wire is l1 when the tension in it is F1 and l2 when the tension is F2.
find the natural length of wire
Solution: Let l be the natural length of the wire for the force F1 if elongation is e1 .
F1l Fl
e1  Y  1
AY Ae1
Similarly for the force F2, if elongation is e2
Fl Fl
e2  1
Y  1
AY Ae2
l1  l  e1 and l2  l  e1
Fl F2l
Y 1

A l1  l  A l2  l 
7
 F1  l2  l   F2  l1  l  and F1l2  F1l  F2l1  F2l
l1 F2  l2 F1
 F2  F1  l  l1F2  l2 F1 or l  F  F .
 2 1
8 Find the pressure that has to be applied to the ends of a steel wire of length 10 cm,
to keep its length constant when its temperature is raised by 1000C is
[2014, Jee-main]
Ysteel  2  1011 N / m 2  wire  1.1 10 5 / K
F
Solution: Pressure = P 
A
F/A F 
From, Y   Y  Y t
 A 

 P  Y t  2 1011 1.1105 102  2.2 108 Pa
9. A slightly conical wire of length  and radii r1and r2 is stretched by two forces of magnitude
F applied parallel to length in opposite directions and normal to end faces. If Y denotes
the Youngs’s modulus, then find the elongation of the wire (r1>r2).
dx
r1 r r2
F F
Solution: Consider an element of length dx at distance x as shown in the figure. The radius of the
 r2  r1 
section rx  r1   x
  
F  dx 
The extension of the element d  
AxY
Fdx

 rx 2Y
elongation of wire is

Fdx F
   dl  
 r2  r1 
2
 rr
1 2Y.
0
 r1  x Y
  
10. A Steel rod of cross-sectional area 1m2 is acted upon by forces as shown in the Fig.
Determine the total elongation of the bar. Take Y  2.0 1011 N / m2
8
A B C D
6 0 kN 10kN 20k N 50kN
1 .5 m 1 m 2m
Solution: The action of forces on each part of rod is shown in

60kN 10kN 20kN
60kN 50kN 50kN

60kN 50kN
60kN
60kN 50kN 50kN
70kN 70kN
We know that the extension due to external force F is given by

F
e
AY
 eAB 
 60 10  1.5  4.5 10
3
7
m
1 2  1011
eBC 
 70 10  1.5  3.5 10
3
7
m
1 2  10 11
and eCD 
 50 10  1.5  5.0 10
3
7
m
1 2  10 11
The toal extension e  e AB  e BC  eCD

 4.5  10 7  3.5  10 7  5.0  10 7
 13107 m  1.3 m
11 .A uniform rod of radius “R” and Length “L” is rotated with some angular velocity  in
a horizontal plane about a vertical axis passing through one of its ends, then find
tension in the rod?
Solution:

r
T  dT T

dr
L
Tension in small element of mass dm is

dT  dm r  2  dT    Adr r 2
this tension is only due to centripetal force due to all elements between x=L to x=r
r r
T   dT     A r dr   A  rdr 2 2
L L
1
T   AW 2  L2  r 2 
2
9
dy T  stress 
if ‘dy’ is the elongation in the element ‘dr’ then   Strain  Y 
dr Ay
T
 dy   AY dr
1  A 2 2
L L
T
y
AY
dr  
2 AY
 L  r 2  dr
O O
y = total elongation in the rod
1   2 L3
y
3 Y
12. A uniform rod of length “L” and mass “M” is pulled horizontally on a smooth surface
with a force “F” elongation of the rod of a material of Young’s modulus Y is
x L-x
F
solution: dx
Let a small element dx from the free end of the rod, the magnitude of force at this section is
F
F1   x ,
L
F
elongation on this differential element is dl   x  dx
YAL
 Total elongation,
L
F  x2 
L
F 1 FL
l   dl   x dx    =l 
0
YAL YAL  2 0 2 YA
13. The following four wires of length L and radius r are made of the same material. Which of
these will have the largest extension when the same tension is applied?
(a) L = 40 cm, r = 0.20 mm (b) L = 100 cm, r = 0.5 mm
(c) L = 200 cm, r = 1 mm (d) L = 300 cm, r = 1.5 mm.
FL L
solution: Y or L  2 .
 r L
2
r
Here L/r2 is maximum when L = 40 cm and r = 0.20 mm as compared to other cases.
14.. If the ratio of lengths, radii and Young’s modulii of steel and brass wires in the Fig. are a, b,
c respectively. Then the corresponding ratio of increase in their lengths would be
10
2ac 3a 3c 2a 2 c
(a) (b) (c) (d)
b2 2b2 c 2ab2 b
solution:
Fl l F l A Y  3M  1 1 3a
As Dl = AY ; l  F  l  A  Y   2M   a  b2  c  2b2 c
s s s B B
B B B s s  
15.The magnitude of the force developed by raising the temperature from 0ºC to 100ºC of the
iron bar 1.00 m long and 1 cm2 cross section when it is held so that it is not permitted to expand
or bend is (  10 50 C 1 and Y  1011 N m2 )
(a) 103 N (b) 104 N (c) 105 N (d) 109 N.
solution: key a l   l t and F  YA l/l  Y A  t
16. A lift is tied with thick iron wires and its mass is 1000 kg. The minimum diameter of wire if
the maximum acceleration of lift is 1.2 ms–2 and the maximum safe stress is 1.4 ´ 108 N m–
2
is (g = 9.8 ms–2)
(a) 0.00141 m (b) 0.00282 m (c) 0.005 m (d) 0.01 m.
. solution:
When the lift is accelerated upwards with acceleration a, then tension in the rope is
T = m(g + a) = 1000 (9.8 + 1.2) = 11000 N.
F T T
Now, stress = A   r 2 or r2 
  series
11000  7 1 1 1
  or r ;so D  2r   0.01m
22  1.4  10 8
4  104 200 100
SO F = 1011 ´ (10–4) ´ 10–5 ´ 100 = 104 N

17. A wire of length L and of area of cross-section A is stretched through a certain length l. If Y is
Young’s modulus of the material of the wire, then the force constant of the wire is
YL Yl YA YA
(a) (b) (c) (d) .
A A l L
F L F YA
solution: Y  or force constant = 
A l l L
18. A uniform bar of length L with an elastic modulus Y and thermal coefficient a is held between
two rigid planes, one at each end of the bar. In this way the bar is prevented from expansion
in these directions when it is heated. When the temperature of the bar is raised by DTºC, the
stress developed in the bar is
Y  T Y YL
(a) Y  T (b) (c) (d)
L T T .
F Y l  Y[l0 (1   T)  l0 ]  Y  T
solution: Stress = 
A l l0
11
19.A wire (Y = 2 ´ 1011 N/m2) has length 1 m and area 1 mm2. The work required to increase its
length by 2 mm is
(a) 400 J (b) 40 J (c) 4 J (d) 0.4 J.
F l YAx
solution: Y   or F
A x l
1 1 YAx 2 1  2  1011  (106 )  (2  103 )2
Workdone, W Fx    0.4 J
2 2 l 2 1
20. A rod PQ of length 1.05 m having negligible mass is supported at its ends by two wires
one of steel (wire A), and the other of aluminium (wire B) of equal lengths as shown
in Fg. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0mm2
respectively. At what point along the rod a load W be suspended in order to produce
(a) equal stress
(b) equal strains in both steel and aluminium.(Ysteel=200 GPa,Yaluminium=70 GPa)
Ysteel YAl
P Q
W
solution:e end P of the rod PQ, then for rotational equilibrium of the rod about O ,
F1 F2
O
P
Q
x
m
w= mg
x=0.7 m
b) If F1 and F2 are the tensions in the wires A and B respectively to produce equal strain in
both wires, i.e, for  L  A   L  B ,
 1mm2   200GPa  10
 2  
 2mm   700GPa  7
If mass m is now placed at a distance x’ from the end P of the rod PQ, for the rotational
equilibrium of the rod about O,
F1 F 2
O
P
Q
x
m
w = m g
12
F1 1.05  x
F1 x  F2 1.05  x  or 
F2 x
7.35
x  0.43m  7.35  7 x  10 x
17
7.35
x  0.43m
17
STRESS - STRAIN GRAPH

Proportion Limit :The limit in which Hook's law is valid and stress is directly
proportional to strain is called proportion limit.
Stress  Strain
The limit in which Hook's law is valid and stress is directly

proportional to strain is called proportion limit.
Elastic limit
That maximum stress which on removing the
deforming force makes the body to recover
completely its original state.
Yield Point
The point beyond elastic limit, at which the length of wire starts increasing without increasing
stress, is defined as the yield point.
Breaking Point
The position when the strain becomes so large that the wire breaks down at last, is called
breaking point. At this position the stress acting in that wire is called breaking stress and
strain is called breaking strain.
When the deforming force, applied on a body, is changed rapidly then it temporarily loses
its property of elasticity. This is known as elastic fatigue.
Elastic Fatigue
When the deforming force, applied on a body, is changed rapidly then it temporarily loses
its property of elasticity. This is known as elastic fatigue.
13
Elastic Hysteresis
The strain persists even when the stress is removed.
g
This lagging behind of strain is called elastic hysteresis.
sin
load or stress
g
sin
re a
r ea
nc
di
ec
l oa
dd
loa
extension or strain
This is the reason why the values of strain for same stress
are different while increasing the load and while decreasing the load.
Breaking stress also measures the tensile strength.
Metals with small plastic deformation are called brittle.
Metals with large plastic deformation are called ductile.
Elasticity restoring forces are strictly conservative only when the elastic hysteresis is zero.
i.e. the loading and unloading stress - strain curves are identical.
The material which have low elastic hysteresis have also low elastic relaxation time.
HOOKE'S LAW
Within elastic limit, the extension of an elastic body is directly proportional to the force that
is producing it.
Thomus Young, an English scientist modified the law to a general form.
This modified law is true for all kinds of deformation such as bending, compression, stretching,
twisting etc.
This modified form, given below, is now the accepted form of Hooke's Law.
Within elastic limit, stress is proportional to strain.
This constant is known as modulus of elasticity or coefficient of elasticity.

The modulus of elasticity depends only on the type of material used.
It does not depend upon the value of stress and strain.
21. Two rods of different metals, having the same area of cross - section A, are placed
end to end between two massive walls as shown in fig. If the temperature of both
the rods are now raised by t 0C then
a) Find the force with which the rods acts on each other at heigher
temperature.
1 
Y
 Y 2
A 1 1 2 2
B
14
solution:a) Due to heating the increases in length of the composite rod will be
   I  11t   2 2 t    11   2 2  t ... (1)

due to compressive force ‘F’ from the walls, due to elasticity, the decrease in length will be
F 1 F  2 F   1  2 
  D      
AY1 AY2 A  Y1 Y2  .... (2)
As the length of the composite rod remains unchanged the increase in length due to heating
must be equal to decrease in length due to compression.
F  1  2 
 11   2 2  t    
A  Y1 Y2 
shift in the joint of a rod
If is 1   2 then final length of first rod

elong ation + l compression
Fl
 l  l 1   
Ay1
l1  l  l1 
1   2  
1 1

y1 y2
l    y1 1  y 2 2 
Shift in joint = l  l  x 
1
y1  y 2
22.The edge of an aluminium cube is 10 cm long. one face of cube is firmly fixed to a
vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The
shear modulus of aluminium is 25 GPa. What is the displacement of upper face relative
to lower face.
solution. Here; L  10cm  10 1 m
A  L2  102 m2
F  100kg wt  100  9.8 N  9.8  102 N .
  25 GPa  25  109 pa  25  109 N / m2
As
shear stress F / A FL
  
Shear strain x / L Ax
15
FL  9.810 10 
2 1
x   2
A 10  25109 
 0.4 106 m  4 107 m.

23 The rubber cord catapult has a cross-sectional area 1 mm2 and total unstretched length 10.0
cm. It is stretched to 12.0 cm and then released to project a missile of mass 5.0 g. Taking
Young’s modulus for rubber as 5.0108 N m–2, the tension in the cord is
(a) 1000 N (b) 100 N (c) 10 N (d) 1 N.
F l l (5.0  108 )  (106 )  (2  102 )
solution. Y  or F  YA =  100 N
A l l (10  102 )
24Two wires of equal length and cross-section area suspended as shown in Fig. in Their Young’s
modulus are Y 1 and Y-2 respectively. The equivalent Young’s modulus will be
F l
solution. Y  
A l
F YA
Spring constant of wire, k   , keq = k1 + k2
l l
Y (2 A ) Y1 A Y2 A Y1  Y2
or   or Y 
l l l 2
25.fig. shows the stress-strain graph of a certain substance. over which region of the graph is
IIooke’s Law obeyed?
(a) BC (b) CD (c) AB (d) OD.

For Hooke’s law ; stress µ stain i.e. the graph between stress and strain is a straight line,
which is so for portion O to D.
16
26. The potential energy U between two molecules as a function of the distance X between
them has been shown in the adjoining figure. The two molecules are
(a)attracted when X lies between A and B and repelled whenX lies between B and C
(b) attracted when X lies between B and C and are repelled when X lies between A and B
(c) attracted when they reach B. (d) repelled when they reach B.
When there is attraction between molecule, the potential energy decreases with decrease in
distance, in case of force of repulsion between the molecules the potential energy increases with
the decrease in distance between the molecules
Bulk modulus (B) :

volume stress
Bulk modulus 
bulk strain
F
V F  PV
B A  
V = Constant V V A V
V
* negative sign indicates the decrease in volume with increase in pressure
Bsolids  Bliquids  Bgases
* If a block of coefficient of cubical expansion  is heated through a rise in temperature of  ,
the pressure to be applied on it to prevent its expansion = K , where K is its bulk modulus.
* When a rubber ball of volume V, bulk modulus K is taken to a depth 'h' in water decrease in its
hdgV
volume V  ; (d = density of material)
K
* For an incompressible material, V  0 ,
so bulk modulus is infinity.
Solid possesses y, n and k
Liquids and gases possess only K.
* Isothermal bulk modulus of the gas = P(pressure)
Adiabatic bulk modulus of the gas = p
Cp
(where   )
CV
adiabatic change in volume Va
 
isothermal change in volume Vi
1
* The reciprocal of bulk modulus is called compressibility. C 
K
17
Density of compressed liquid :
If a liquid of density '  ', volume V and bulk modulus 'K' is compressed, then its density in-
creases
m
density  
V
1  V
   -------- (1)
V  V
1  V
   -------- (1)
V  V
But by definition of bulk modulus
V  P V P
B   ----------------(2)
V V B
 P
from (1) and (2)   K
1   P P  P 
  1      1   1 
K 
 K K 
Also    1  CP where ‘C’ is the compressibility..

1
27.A 8m long string of rubber, having density 1.5 x 10 3 kg/m3 and young's modulus
N/m2 is suspended from the ceiling of a room. The increase in its length due to its own
weight will be (g=10m/s2)
Solution: The increase in its length due to its own weight

l2g 82 1.5103 10
e   9.6102 m
2Y 2510 6
28.What is the density of water at a depth where the pressure is 80.0 atm, given its density
at the surface is 1.03 103 kg / m3 ? Compressibility of water  45.8  1010 pa 1.
Solution.
P V P
B  
V V B
V
  79  1.013  105    45.8  10 10 
 36.65  103
V  m /     m /  
1

but   1 1
V m /   
 V
 1
 V
18
1.03  103 1.03  103
or   
1
1  36.65  103 0.964

 1.07  103 kg / m 3
29.Determine the volume contraction of a solid copper cube, 10 cm on an edge, when
subjected to a hydraulic pressure of 7.0  106 pa. (Bulk modulus for copper=140 GPa)
solution: Here;
L  10cm  101 m,
V  L3 103 m3,
P  7.0  106 Pa, B  140 GPa
 140  109 Pa.
P V
As B  ,
V
P V  7.0  10 10 
6 3
V    5  108 m 3
B 140 10 
9
30.A solid sphere of radius 'R' made of a material of bulk modulus B is surrounded by a
liquid in a cylindrical container. A massless piston of area 'A' floats on the surface of the
liquid. Find the fractional change in the radius of the sphere , when a mass M is placed on
the piston to compress the liquid.
solution: As for a spherical body
4 V R
V  R 3 , 3
3 V R
Now by definition of bulk modulus
P V P Mg  Mg 
B  V i.e    as P 
V V B AB  A 
dR 1 V dR Mg
  
R 3 V R 3 AB
(iii) Bending of Beam :

Beam is the structural member which can carry transverse load. A simply supported beam is
19
supported at its ends. A cantilever beam is fixed at one end.
(iv) Deflection of beam : Deflection of beam at its centre due to load placed as shown in Figure.
W 3 W 3
 for simply supported beam and   for cantilever beam where I is called
48YI 3YI
geometric moment of area.
bd 3
i) For rectangular cross - section I  d
12
b
r 4 r
ii) For circular cross - section I 
4
Ø A force F is applied tangentially on the upper face of a cube of side length L by fixing its lower
face. If l is the displacement of the upper face and A is the area of the upper face, A  L2 .
FL FL F
 Rigidity modulus,   Al  2  Ll .
Ll
Ø One end of the rod is fixed the other free end is
twisted through an angle  by applying a torque
 . The work done on the rod is
1
W 
2
wires in series
Two wires of different length 1 ,  2 and of same radii are joined end to end and loaded.
Young’s modulii Y1 , Y2 respectively ,and combination behaves as a single wire
stress produced in two wires is same
e1 Y1
***
e2 Y2
20
Total elongation is e  e1  e2
F leff F l1 F l2
 
A Yeff A Y1 A Y2
1   2 1  2
 
Yeff Y1 Y2
if two wires are having same length, 1   2 then
2 1 1 2 Y1Y 2
   Y eff 
Y eff Y1 Y 2 Y1  Y 2
TWO WIRES CONNECTED IN PARALLEL

Two wires of same length but different area of cross section A1 , A2 are joined in parallel and
loaded. If the youngs modulii of the materials of the wires are Y1 & Y2 and the combination
behaves as a single wire
The strain produced in the two wires is same
Y 1 Y 2
Ø Elongation produced in the two wires is same but stress is shared between them
(stress)1+(stress)2=stress
F  F1  F2
Yeff Aeff Y1 A1 Y2 A2
 
l l l
Y1 A1  Y2 A2  Ycff  A1  A2 
Y1 A1  Y2 A2
 Ycff , if two wires are of same area of cress section, A  A
A1  A2 1 2
Y1  Y2
Yeff 
2
31.A 40 kg boy whose legs are 4cm2 in area 50cm long falls through a height of 2m
without breaking his leg bones. If the bones can with stand a stress of 0.9  108 N / m 2 .
Calculate the Young’s modulus of material of the bone.
Sol:
 1 stress 2 
mgh  2    volume ,
 2 Y 
0.9108  4104 50102

2
Y
409.8 2
 2.0510 N / m 29
21
32. A copper wire 2m long is stretched by 1mm. If the energy stored in the stretched wire
is converted into heat, calculate the rise in temperature of the wire.
( Y  12.5 1010 N / m 2 ;
 = 9 103 kg / m3 ; s  385 J / Kg / K )
1 m
Sol: m s  t  2 Y  stra in   
2
2
1 Y e
t     
2  
2
1 12.5 1010  1 
t      0.045 0C
2 9 103 1000  2 
So the rise in temperature of the wire is 0.0045OC
33. A catapult consists of two parallel rubber cords each of length 20 cm and
cross- sectional area 5 cm2 when stretched by 8 cm, it can throw a stone of mass
4gm to a vertical height 5 m, the Young’s modulus of elasticity of rubber is
 g  10 m / sec 2 
Sol: The total elastic potential energy is converted into gravitational potential energy
1 YAe2 2mghL
  mgh  Y 
2 L Ae 2
mghL
for a single string, Y 
Ae 2
4  103  10  5  20 10 2 4 102
 
5  104   8  10 2 
2
5  64 108
 1.25  10 4 N / m 2
Factors effecting Elasticity:

* Annealing decreases elasticity while hammering and rolling increases it.
* The impurity having higher elasticity than the sample to which it is added increases the elasticity
while the impurity with smaller elasticity decreases the elasticity of the sample.
* Normally, elasticity of the material gets decreased with rise in temperature. However, INVAR
STEEL is a material whose elastic behaviour is not affected by rise in temperature.
Work done in stretching a wire (Potential energy of a stretched wire)
For a wire of lenght Lo stretched by a distance x, the restoring elastic force is
L0 L0
x
:
F
22
 x 
F = stress × area = Y  L  A
 o
The work has to be done against the elastic restoring forces.

YA
dW = F . dx = L x. dx
o
The total work done in stretching the wire from x = 0 to x = Dl is, then
 
YA YA  x2  YA( )2
W 
0 Lo
x.dx   
L o  2 0
or W
2L o
1
w= 2
x stretching force x extension.
1 YAe 2 1 F 2  1 F2 
ii) w = 1
Fe = = 
2 2 l 2 AY 2 r 2 y
iii) w = 12 x stress x strain x volume of the wire
iv) Area under F-e graph gives the work done or the strain energy stored in the
.
previous jeemains questions
Topic:1_Hooke’s Law & Young’s Modulus
A B
1. If the potential energy between two molecules is given by U    , then at
r 6 r12
equilibrium, separation between molecules, and the potential energy are:
sep 6 mains 2020
1 1
 B 6 A2  B 6
(1)   ,  (2)   , 0
 2A  2B  A
1 1
2 2
 2B  6 A  2B  6 A
(3)   , (4)   ,
 A  4B  A  2B
2. A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The
maximum angular speed (in rad s-1) with which it can be rotated about its other end
in space station is (Breading stress of wire = 4.8 107 Nm2 and area of
crosssection of the wire = 102 cm2 ) is______ 9 jan mains 2020
3. A uniform cyclindrical rod of length L and radius r, is made from a material whose
Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T
and simultaneously subjected to a net longitudinal compressional force F, its
length remains unchanged. The coefficient of volume expansion, of the material of
the rod, is (nearly) equal to:
 2
(1) 9 F / r YT  2
(2) 6 F / r YT  12 apr mains2020
(3) 3F /  r YT  (4) F /  3r YT 

2 2
4. In an environment, brass and steel wires of length 1 m each with areas of cross
section 1 mm2 are used. The wires are connected in series and one end of the
combined wire is connected to a rigid support and other end is subjected to
23
elongation. The stress required to produce a net elongation of 0.2 mm is,
[Given, the Young’s modulus for stell and brass are, respectively, 120 109 N / m2
and 60 109 N / m2 ]
(1) 1.2 106 N / m2 (2) 4.0 106 N / m2 10 apr mains 2020
(3) 1.8 106 N / m2 (4) 0.2 106 N / m2
5. The elastic limit of brass is 379 MPa. What should be the minimum diameter of a
brass rod if it is to support a 400 N load without exceeding its elastic limit?
(1) 1.00 mm (2) 1.16 mm 10APR MAINS 2019
(3) 0.90 mm (4) 1.36 mm
6. A steel wire having a radius of 2.0 mm, carrying a load of 4kg, is hanging from a ceiling.
Given tha g = 3.1 A ms–2, what will be the tensile stress tha would be developed in the
wire? 9 APR MAINS 2019
(1) 6.2 106 Nm2 (2) 5.2 106 Nm2
(3) 3.1106 Nm2 (4) 4.8 106 Nm2
7. A steel wire having a radius of 2.0 mm, carrying a load of 4kg, is hanging from a ceiling.
Given that g = 3.1 A ms–2, what will be tha tensile stress tha twould be developed in
the wire? 9 APR MAINS 2019
(1) 6.2 10 Nm
6  2 (2) 5.2 10 Nm
6 2
(3) 3.1106 Nm2 (4) 4.8 106 Nm2

8. Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has
radius R. Wire B is 1.5 m long and has radius 2mm. If the two wires stretch by the
same length for a given load, then the value of R is close to:8 APR MAINS 2019
(1) 1.5 mm (2) 1.9 mm
(3) 1.7 mm (4) 1.3 mm
9. As shown in the figure, forces of 105N each are applied in opposite directions, on the
upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself
by 0.5cm. If the side of another cube of the same material is, 20 cm, then under similar
conditions as above, the displacement will be APR 15 MAINS 2018
(1) 1.00 cm (2) 0.25 cm

(3) 0.37 cm (4) 0.75 cm
10. A thin 1m long rod has a radius of 5 mm. A force of 50 kN is applied at one end to
determine its Young’s modulus. Assume that the force is exactly known. If the least
count in the measurement of all lengths is 0.01 mm, which of the following
statements is false ? APR MAINS 2016
(1) The maximum value of Y that can be determined is 2 1014 N / m2
Y
(2) gets minimum contribution from the uncertainty in the length
Y
Y
(3) gets its maximum contribution from the uncertainty in strain
Y
(4) The figure of merit is the largetst for the length of the rod
11. A uniformly tapering conical wire is made from a material of Young’s modulus Y
and has a normal, unextended length L. the radii, at the upper and lower ends of
24
this conical wire, have values R and 3R, respectively. The upper end of the wire is
fixed to a rigid support and a mass M is suspended from its lower end. The
equilibrium extended length, of this wire, would equal.. APR9 MAINS 2016
 2 Mg   1 Mg 
(1) L 1  2  (2)
L 1  
 9 YR   9 YR 2 
 1 Mg   2 Mg 
(3) L  1  2  (4)
L 1  
 3 YR   3 YR 2 
12. The pressure that has to be applied to the ends of a stell wire of elngth 10 cm to
keep its length constant when its temperature is raised by 1000C is
(1) 2.2 108 Pa (2) 2.2 109 Pa MAINS 2014
(3) 2.2 107 Pa (4) 2.2 106 Pa
13. Two blocks of masses m and M are connected by means of a metal wire of cross-
sectional area A passing over a frictionless fixed pulley as shown in the figure. The
system is then released. If M = 2m, then the stress produced in the wire is
APR MAINS2013
2mg 4mg mg 3mg

(1) (2) (3) (4)
3A 3A A 4A
14. A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-
sectional areas are joined end to end. The composite wire is stretched by a certain
load which stretches the copper wire by 1 mm. If the Young’s modulii of copper
and steel are respectively 1.0 1011 Nm2 and 2.0 1011 Nm2 , the total extension of
the composite wire is APR23 MAINS 2013
(1) 1.75 mm (2) 2.0 mm
(3) 1.50 mm (4) 1.25 mm
15. A uniform wire (Young’s modulus 2 1011 Nm2 ) is subjected to longitudinal tensile
stress of 5 1011 Nm2 . If the overall volume change in the wire is 0.02%, the
fractional decrease in the radius of the wire is close to:
(1) 1.0  104 (2) 1.5 104 APR22 MAINS 2013
(3) 0.25 104 (4) 5  104
16. If the ratio of lengths, radii and Young’s moduli of stell and brass wires in the figure
are a, b and c respectively, then the corresponding ratio of increase in their
lengths is:
25
APR9 MAINS 2013
3c 2a 2 c 3a 2ac
(1) 2 (2) (3) 2 (4) 2
2ab b 2b c b
17. A steel wire can sustain 100 kg weight without breaking. If the wire is cut into two
equal parts, each part can sustain a weight of MAY19 MAINS 2012
(1) 50 kg (2) 400 kg
(3) 100 kg (4) 200 kg
18. A structural steel rod has a radius of 10 mm and length of 1.0 m. A 100 kN force
stretches it along its length. Young’s modulus of structural steel is 2 1011 Nm2 .
The percentage strain is about MAY7 2012
(1) 0.16% (2) 0.32% (3) 0.08% (4) 0.24%
19. The load versus elongation graphs for four wires of same length and made of the
same material are shown in the figure. The thinnest wire is represented by the line
MAY7 2012
(1) OA (2) OC (3) OD (4) OB

20. Two wires are made of the same material and have the same volume. However
wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the
length of wire 1 increases by  x on applying force F, how much force is needed to
stretch wire 2 by the same amount? MAY7 2009
(1) 4F (2) 6F (3) 9F (4) F
21. A wire elongates by l mm when a load W is hanged from it. If the wire goes over a
pulley and two weights W each are hung at the two ends, the elongation of the
wire will be (in mm)
(1) l (2) 2l (3) zero (4) l/2 MAY 2006
Topic-2: Bulk and Rigidity Modulus and Work Done in Stretching a Wire
22. Two steel wires having same length are suspended from a ceiling under the same
load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their
diameters is: 9JAN20120 MAINS
(1) 2 :1 (2) 1: 2 (3) 2 :1 (4) 1: 2
23. A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of
cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on
it and stretches the cord by 20 cm by applying a constant force. When released,
the stone flies off with a velocity of 20 ms–11. Neglect the change in the area of
cross-section of the cord while stretched. The Young’s modulus of rubber is
closest to: 8APRMAINS2019
(1) 106 N/m–2 (2) 104 N/m–2
26
(3) 108 N/m–2 (4) 103 N/m–2
24. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by
a liquid in a cylindrical container. A massless piston of area a floats on the surface of
the liquid, covering entire cross-section of cylindrical container. When a mss m is
placed on the surface of the piston to compress the liquid, the fractional decrement
 dr 
in the radius of the sphere   , is 2018 MAINS
 r 
Ka Ka mg mg
(1) (2) (3) (4)
mg 3mg 3Ka Ka
25. A bottle has an opening of radius a and length b. A cork of length b and radius
 a  a where  a  a  is compressed to fit into the opening completely (see
figure). If the bulk modulus of cork is B and frictional coefficient between the bottle
and cork is  then the force needed to push the cork into the bottle is
APR12 2014 MAINS
(1)  Bb  a (2)  2Bb  a

(3)  Bb  a (4)  4Bb  a
26. Steel ruptures when a shear of 3.5 108 Nm2 is applied. The force needed to punch
a 1 cm diameter hole in a steel sheet 0.3 cm thick is nearly APR12 2014 MAINS
(1) 1.4 104 N (2) 2.7 104 N
(3) 3.3 104 N (4) 1.1104 N
27. The bulk moduli of ethanol, mercury and water are given as 0.9, 25 and 2.2 respectively
in units of 109 Nm–2. For a given value of pressure, the fractional compression in
V V
volume is . Which of the following statements about for these three liquids is
V V
correct ? APR11 2014 MAINS
(1) Ethanol > Water > Mercury
(2) Water > Ethanol > Mercury
(3) Mercury > Ethanol > Water
(4) Ethanol > Mercury > Water
28. In materials like aluminium and copper, the correct order of magnitude of various
elastic modului is APR9 2014 MAINS
(1) Young’s modulus < shear modulus < bulk modulus
(2) Bulk modulus < Shear modulus <Young’s modulus
(3) Shear modulus < Young’s modulus < Bulk modulus
(4) Bulk modulus < Young’s modulus < Shear modulus
29. If ‘S’ is stress and ‘Y’ is young’s modulus of material of wire, the energy stored in
the wire per unit volume is APR9 2005 MAINS
S2 S 2Y
(1) (2) 2S 2Y (3) (4)
2Y 2Y S2
27
30. A wire fixed at the upper end stretches by length  by applying a force F. The work
done in stretching is
F F
(1) 2F (2) F (3) (4) APR2004 MAINS
2 2
HINTS & SOLUTIONS

A B
1. (3) Given: U  
r 6 r12
For equilibrium,
F
dU
dr
   
  A 6r 7  B 12r 13  0
6 A 12 B 6A 1
 0  7  13   6
r r 12 B r
1/6
 2B 
 Separation between molecules, r   
 A 
Potential energy,
  2B  
1/6
A B
U r      2 2
  A   2B / A 4B / A

 A2 A2  A 2
 
2B 4B 4B
2. (4) Given: wire length, l  0.3m
Mass of the body, m  10kg
Breaking stress,   4.8 107 Nm2
Area of cross-section, a  102 cm2
Maximum angular speed   ?
T  Ml 2
T ml 2
 
A A
ml 2 7 2
 48 10   
48 107 A 
A ml
 2

 48 10 10   16
7 6
10  3
 max  4rad / s
3. (3)  temp   force
FL
or L  T  
AY
FL F
  
AYT r 2YT
Coefficient of volume expression
28
3F
r  3 
r 2YT
4. (Bonus)
Stress
Y
Young modulus,  l 
 
 L
Let  be the stress
L1 L1 L2
Total elongation lnet  Y  Y  Y
1 1 2
1 1 
lnet       L1  L2  1m
 Y1 Y2 
 YY 
  l  1 2 
 Y1  Y2 
 120  60  6 N
 0.2  103   9
  10  8  10 2
 180  m
F  120  60  9 6 N
5. (2) Stress     10  8  10 2
A  180  m
400  4
 d2 
379 106 
d  1.15mm
6. (3) Goven,
Radius of wire, r = 2 mm
Mass of the load m = 4 kg
F mg
 
Stress A  r 2
 
4  3.1
  3.1106 N / m 2
 
2
 2 103
7. (3) Given,
Radius of wire, r = 2 mm
Mass of the load m = 4 kg
F mg 4  3.1
  
A r 
 
2 2
Stress  2  103
 3.1106 N / m2
8. (3) 1   2
29
Fl1 Fl2 2 1.5
or r 2 y  r 2 y or 2  2
1 1 2 2 R 7 2 4
 R  1.75 mm
9. (2) For same material the ratio stress to strain is same
For first cube
force1 105
Stress1  
area1  
0.12
change in length1 0.5 102

Strain1  
original length1 0.1
For second block,
force2 105
Stress 2  
area 2 
0.2 2 
cahnge in length 2 x
strain 2  
original length 2 0.2
x is the displacement for second block
stress1 stress 2
For same material, strain  strain
1 2
10.5 105
 0.12 
 0.2 
2
or, 0.5  102 x

0.1 0.2
Solving we get, x  0.25cm
F 
10. (1) Young’s modulus Y  /
A 
F
Y
r 2 
Given, radius r = 5mm, force F = 50 F  50kN ,

 0.01 mm

F 
Y  2  2  1014 N / m2
r  
2R
11. (3) Consider a small element dx of radius r, r  xR
L
30
At equilibrium change in length of the wire
1
Mgdx
 dL   2R 
2
0
 x  R y
 L 
Taking limit from 0 to L
 
 
Mg  1 L  MgL
L    
y   2 Rx 
L
2 R  R 2 y
   R 
  L  
The equilibrium extended length of wire L  L
MgL  1 Mg 
 L 2
 L 1  
3R Y  3 YR 2 
stress
12. (1) Young’s modulus Y 
strain
stress = Y x strain
Stress in steel wire = Applied pressure
Pressure = stress = Y x strain
L
Strain =  T
L
(As length is constant)
 2 1011 1.1105 100  2.2 108 Pa
 2mM 
13. (2) Tension in the wire, T   g
mM 
Force / Tension 2mM
Stress   g
Area A m  M
2  m  2m  g
  M  2m given 
A  m  2m 
4m 2 4mg
 g
3mA 3A
14. (4) Yc   Lc / Lc   Ys   Ls / Ls 
 1 103 
11 11  Ls 
 110     2  10   
 1   0.5 
31
0.5  103
Ls   0.25m
2
Therefore, total extension of the composite wire
 Lc  Ls
 1mm  0.25m  1.25m
15. (3) Given, y  2  1011 Nm 2
F 7 2
Stress    5  10 Nm
 A
V  0.02%  2 104 m3
r
?
r
stress    
  strain    ...(i)
strain   0  stress
V  2x 0 r  r 2  ....(ii)
From equation (i) and (ii) putting the value of ,  0 and V and solving we get
r
 0.25  104
r
16. (3) According to questions,
s r y 
 a , s  b, s  c , s  ?
b rb yb  b
F F
As, y   
A Ay
3mg  s
 s   Fs   M  2M  g 
rs2 . ys
2mg  b
 b   Fb  2Mg 
rb2 . yb
3Mg  s
 r 2 . y 3a
 s  s s  2
 b 2 Mg . b 2b C
2
rb . yb
17. (3) Breaking force  area of cross section of wire Load hold by wire is
independent of length of the wire.
18. (1) Given: F  100kN  105 N
Y  2 1011 Nm2
 0  1.0m
radius r = 10 mm = 10–2 m
Stress
From formula, Y 
Strain
32
Stress F
 Strain  
Y AY
105 105 1
 2
 4 11

r Y 3.14 10  2 10 628
1
There fore%strain =  100  0.16%
638
19. (1) From the graph, it is clear that for the same value of load, elongation is maximum
for wire OA. Hence OA is the thinnest wire among the four wires
20. (3)
For wire 1
Length L1  1
Area, A1  A
Fore wire 2

Length, L2 
3
Area, A2  3 A
As the wires are made of same material, so they will have same young’s modulus
For wire 1,
F/A
Y ....(i)
x / 
For wire 2,
F '/ 3 A
Y ....(ii)
x /   / 3
From (i) and (ii) we get,
F  F' 
    F '  9F
A x 3 A 3x
21. (1) Case (i)
At equilibrium, T  W
W/A
Young’s modules, Y  ....(1)
/L
W L 33
Elongation,   
A Y
Case (ii) At equilibrium T = W
W/A
Y
 Young’s moduls, /2
L/2
W/A W L
Y   
/L A Y
 Elongation is the same
22. (1) If force F acts a long the length L of the wire of crosssection A, then energy stored
in unit volume of wire is given by
1
Energy density  stress  strain
2
1 F F
  
2 A AY
F X
( stress  and strain  )
A AY
1 F2 1 F 2  16 1 F 2  16
  
 
2 A2Y 2 d 2 2 Y 2 d 4Y
If u1 and u2 are the densities of two wires, then

4
u1  d 2  d d
    1   4   1  2 :1
1/4
u2  d1  d2 d2
23. (1) When a catapult is stretched up to length l , then the stored energy in it  k .E 
1  YA  1 mv 2 L
.    I   mv 2  y 
2
  l 
2
2  L  2
m = 0.02 kg
v = 20 ms–1
L = 0.42 m
 
2
A = d / 4
d = 6 103 m
l  0.2 m
0.02  400  0.42  4
y 6
 2.3 106 N / m2
 36  10  0.04
So, order is 106
volumetric stress
24. (3) Bulk modulus, K  volumetric strain
mg
K
 dV 
a 
 V 
dV mg
 
V Ka
34
4 3
volume of sphere, V  R
3
dV 3dr
Fractional change in volume  .....(ii)
V r
3dr mg
Using eq. (i) & (ii) 
r Ka
dr mg
  (fractional decrement in radius)
r 3Ka
Normal force N N
25. (4) Stress  Area
 
A  2a  b
Stress = B x strain
N 2aa  b
B
 2a  b a 2b
N B
 2a  ab 2
2
a 2b
Force needed to push the cork
f  N  4baB   4Bb  a
26. (3)
Shearing strain is created along the side surface of the punched disk. Note that the
forces exerted on the disk are exerted along the circumference of the disk, and the
total force exerted on its center only
Let us assume that the shearing stress along the side surface of the disk is uniform,
then
F  max dAmax  max  dA
surface surface
D
   max . A   max .2   h
2
1 
 3.5  108    102   0.3  10 2  2
2 
 3.297 104  3.3 104 N
1
27. (1) Compressibility  Bulk mod ulus
As bulk modulus is least for ethanol (0.9) and maximum for mercury (25) among
V
ehtanol, mercury and water. Hence compression in volume .
V
Ethanol > Water > Mercury
35
lateral strain   
28. (3) Poisson’s ratio,   longitudinal strain 
 
For material like copper,   0.33
And, Y  3k 1  2 
9 1 3
Also,  
Y k 
Y  2 1   
Hence,   Y  k
29. (1) Energy stored in the wire per unit volume,
1
E   stress  strain ....(i)
2
We know that,
stress
Y
strain
stress
 strain 
Y
On substituting the expression of strain in equation (i) we get
1 stress 1 S 2
E   stress   .
2 Y 2 Y
30. (4) Let A and L be the area and length of the wire. Work done by constant force in
displacing the wire by the distance  .
= change in potential energy
1
  stress  strain  volume
2
1 F  F
    A L 
2 A L 2
36
Fluid Mechanics ( Viscosity )
Fluid is the name given to a substance which begins to flow when external force is applied on it.
Liquids and gases are fluids.
Fluids do not have their own shape but take the shape of the containing vessel.
Fluids flow from one place to other because of pressure difference.
The branch of physics which deals with the study of fluids at rest is called hydrostatics
The branch which deals with the study of fluids in motion is called hydrodynamics.
Pressure:.
¨The normal force exerted by liquid at rest on a given surface in contact with it is called thrust of
liquid on that surface.
¨The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it,
is called pressure of liquid or hydrostatic pressure.
If F be the normal force acting on a surface of area A in contact with liquid,
F
then pressure exerted by liquid on this surface is P 
A
F
¨The average pressure on the surface area A due to a normal force F is Pavg  
A
F dF
¨Pressure at a point is given by P  ALt0 
A dA
¨ Units : N / m2 or Pascal (S.I.)
 Dyne/cm2 (C.G.S.)
1Nm-2=10dyne/cm2
[F] [MLT 2 ]
¨ Dimension : [P ] 
[ A]

[L2 ]
 [ML1T  2 ]
¨ At a point pressure acts in all directions and a definite direction is not associated with it.
So pressure is a tensor quantity.
Atmospheric pressure :
¨The gaseous envelope surrounding the earth is called the earth’s atmosphere and the
pressure exerted by the atmosphere is called atmospheric pressure.
Ïts value on the surface of the earth at sea level is nearly 1 .013  10 5 N / m 2 or Pascal
¨The practical units of pressure are atmosphere, bar and torr (mm of Hg)
¨ 1 atmospheric pressure = 1.01325 105 pascal ,
1 bar = 760mm of Hg = 76cm of Hg = 0.76m of Hg,
1 torr = 1 mm of Hg.’
1
The atmospheric pressure is maximum at the surface of earth and goes on decreasing as we move
up into the earth’s atmosphere.
The pressure at the bottom of the container due to liquid column of height ‘h’ is P  h  g ,
where '  ' is the density of the liquid
If atmospheric pressure( Po ) is considered, then net pressure at the bottom of the container is
P  Po  hg
Gauge pressure :
The pressure difference between hydrostatic pressure P and atmospheric pressure P0 is called
gauge pressure.
P  P0  h  g
Pabsolute  Patm  Pgauge
Absolute pressure is always positive and is never equal to zero.
Gauge pressure may be positive, negative or zero.’nn
NOTE :
Hydrostatic pressure depends on the depth of the point below the surface (h), nature of liquid (  ) and
acceleration due to gravity (g) while it is independent of the amount of liquid, shape of the container or
cross-sectional area considered. So if a given liquid is filled in vessels of different shapes to same height,
the pressure at the base in each vessel’s will be the same, though the volume or weight of the liquid in
different vessels will be different.
PA  PB  PC but WA  WB  WC
nnNN
(A) (B) (C)
 In a liquid at same level, the pressure will be same at all points, if not, due to pressure difference the
liquid cannot be at rest. This is why the height of liquid is the same in vessels of different shapes
containing different amounts of the same liquid at rest when they are in communication with each other.
Pressure is isotropic i.e., the pressure exerted by a liquid at a point is same in all directions.
Pressure is uniform on a horizontal plane for a liquid at rest or moving with uniform velocity or vertical
acceleration.’
2
Pressure varies with depth and height:
P0
C
y PA
dy
dw
 P  dp  A
up wards force =  P  dP   PA  AdP

downward force = PA  dw  PA  A gdy
at equilibrium PA  AdP  PA  A gdy
dP P y
   g   dP   g  dy
dy P0 0
 P  P0   gy  dP   gy
Pressure increases with depth linearly
Pressure decreases with height linearly
The average pressure of a liquid on the walls of the container filled up to height ‘h’
1
with the liquid is  gh
2
Density:
In a fluid, at a point, density  is defined as:
m dm
  lim 
V  0 V dV
 In case of homogenous isotropic substance, it has no directional properties, so is a scalar.
 S.I. unit kg/m3
C.G.S. unit g/cc
1 g / cc  10 3 kg / m 3
Dimensions [ML3 ]
 Density of substance means the ratio of mass of substance to the volume occupied by the
substance while density of a body means the ratio of mass of a body to the volume of the
body. So for a solid body.
Density of body = Density of substance
While for a hollow body, density of body is lesser than that of substance [As Vbody  Vsub. ]
When immiscible liquids of different densities are poured in a container, the liquid of highest
density will be at the bottom while that of lowest density at the top and interfaces will be plane.
 Sometimes instead of density we use the term relative density or specific gravity
which is defined as :
3
Density of body
RD 
Density of water
 If m1 mass of liquid of density 1 & m 2 mass of density  2 are mixed,
then as m  m1  m 2 and V  (m1 / 1 )  (m 2 /  2 ) [As V  m /  ]
m m1  m 2  mi
  
V (m 1 / 1 )  (m 2 /  2 ) (m i / i )
2 1  2
If  
m1  m 2
1   2 Harmonic mean
 If V1 volume of liquid of density 1 and V2 volume of liquid of density 2 are mixed,
then as: m  1 V1   2 V2 and V  V1  V2 [As   m / V ]
If V1  V2  V   (1   2 ) / 2 = Arithmetic Mean
 With rise in temperature due to thermal expansion of a given body, volume will increase while
mass will remain unchanged, so density will decrease,
 (m / V ) V0 V0
i.e.,     [As V  V0 (1   ) ]
0 (m / V0 ) V V0 (1   )
0 ~–  0 (1   )
or 
(1   )
With increase in pressure due to decrease in volume, density will increase,

 (m / V ) V0 m
i.e.,    [As   ]
0 (m / V0 ) V V
p
But as by definition of bulk-modulus B  V0 V
 p 
i.e., V  V0 1 
 B 
1
 p  ~  p 
So   0 1    0  1  
 B   B 
MEASUREMENT OF ATMOSPHERIC PRESSURE:
Mercury Barometer :
To measure the atmospheric pressure experimentally, torricelli invented a mercury baromer in 1643.
Torricelli
Vaccum
h Mercury
A Trough
Pa=hrg
The pressure exerted by a mercury column of 1mm high is called 1 Torr.
‘Open tube Manometer :
Open-tube manometer is used to measure the pressure gauge. When equilibrium is reached, the pressure
at the bottom of left limb is equal to the pressure at the bottom of right limb.
4
pa
y=y2-y1
y2
p
y1
p a  y1g p a  y 2 g
i.e. p + y1rg = pa+ y2rg
p - pa = rg(y2-y1) = rgy
p = absolute pressure, p-pa = gauge pressure
Thus, knowing y and r (density of liquid), we can measure the gauge pressure.
Water Barometer:
Let us suppose water is used in the barometer instead of mercury.
1.013  105
hrg = 1.013 x 10 or h 
5
pg
The height of the water column in the tube will be 10.3m. Such a long tube cannot be managed
easily, thus water barometer is not feasible.
Pressure difference when liquid is accelerating in vertical direction:
i) When liquid column is uniform acceleration upwards, P  h  g  a 
ii) When liquid column is in uniform acceleration downwards, P  h  g  a 

Pressure difference when liquid is accelerating in horizontal direction:
Consider a liquid placed in beaker which is accelerating horizontally with an acceleration a0.
Let A and B be two points in the liquid at a seperation  in the same horizontal line along the acceleration
a0.We shall first obtain the pressure difference between the points A and B. Construct a small vertical
area S around A and an equal area around B. Consider the liquid contained in the horizontal cylinder
with two areas as the flat faces. Let the pressure at A be P1and the pressure at B be P2. The forces along
the line AB are
D
h1 C
l 
h2
A B
a0
P1S P2 S
 P1 S towards right due to the liquid on the left

 P2 S towards left due to the liquid on the right .
Under the action of these forces, the liquid contained in the cylinder is accelerating towards right.
From Newton’s second law.
P1 S -P2 S =ma0 or, (P1-P2) S =( S ) ρa 0
or P1-P2= ρa 0 ---------------(1)
The two points in the same horizontal line do not have equal pressure, if the liquid is accelerated
horizontally.
As there is no vertical acceleration, the equation is valid. If the atmospheric pressure is P0, the
pressure at A is P1  P0  h1 g and the pressure at B is P2  P0  h2  g , where h1 and h2 are the depths
5
of A and B from the free surface. Substituting values in (1).
h1  h2 a0 a
h1ρg-h 2ρg=lρa 0 or  or , tan   0
 g g
Where  is the inclination of the free surface with the horizontal.
Here PA  PC   PA  PB    PB  PC 
  la   gh2    la  gh2 
If a U shape tube is moving horizontally with an accleration ‘a’ as shown in the fig. then
l
h a h  
tan    a
l g
If container is accelerated ‘a’ at some angle with the horizontal,

ay
a
ax
tan  
ay  g
 ax
ax = a cos   horizontal component

a y = a sin   vertical component
 = angle of inclination of free surface of the liquid with horizontal
Pascal’s Law:
The pressure applied to an enclosed incompressible liquid is transmitted undiminished to every
point of the liquid and the walls of the container.
The pressure in a liquid at rest is same at all points if we ignore gravity.
 F2 
Mechanical Gain : It is the ratio of output force to input force (or) Mechanical gain =  
 F1 
Hydraulic lift :
in p u t F1 A2
A1
h2
h1
o u tp u t f 2
F1 F2 A
P    F2  F1 2
A1 A2 A1
6
As A2  A1 , F 2  F1
As the same volume of fluid is displaced at both pistons
A1h1  A2 h2
 h2  h1
Pressure Energy:
The energy possessed by a fluid by virtue of its pressure is called the pressure energy.
Pressure energy is equal to the work done in keeping an elementary mass of a fluid at a point against
the pressure existing at that point.
Pressure energy=Pressure x Volume = P(A x x).
Where P = pressure,
A = Area of cross section,
x = distance through which liquid is moved
P A x
Pressure energy per unit volume = P
A x
P  A  x P Pr essure
Pressure energy per unit mass=  
  A x  density
Pressure energy has same units and dimensions as that of energy.
Buoyancy:
When a body is partly or wholly dipped in a fluid, the fluid exerts force on the body due to
hydrostatic pressure.
At any small portion of the surface of the body, the force exerted by the fluid is perpendicular to
the surface and is equal to the pressure at that point multiplied by the area.
The resultant of all these constant forces is called upthrust or buoyancy.
F1 h1
h2
F2
To determine the magnitude and direction of this force consider a body immersed in a fluid of
density  as shown in figure. The forces on the vertical sides of the body will cancel each other..
The top surface of the body will experience a downward force
F1  AP1  A(h1g  P0 ) [As P  h g  P0 ]
While the lower face of the body will experience an upward force.
F2  AP2  A(h2g  P0 )
As h2  h1 , F2 will be greater than F1 ,
so the body will experience a net upward force
F  F2  F1  A g(h2  h1 )
If L is the vertical height of the body F  A gL  Vg [As V  AL  A(h2  h1 )]
i.e., F = Weight of fluid displaced by the body.
7
This force is called upthrust or buoyancy and acts vertically upwards (opposite to the weight of
the body) through the centre of gravity of displaced fluid (called centre of buoyancy).
Though we have derived this result for a body fully submerged in a fluid, it can be shown to hold
good for partly submerged bodies or a body in more than one fluid also.
When a body is partly or wholly immersed in a fluid the upward force exerted by fluid on thebody is
called buoyancy.
Force of buoyancy is equal to the hydrostatic pressure at the point multiplied by area of cross section
of the body.
Laws of floatation:
Let W is weight of a body
W1 is the the buoyant force .
 If W > W1 body sinks
 W = W1 body is just submerged (body floats with its volume completely under the liquid)
 W < W1 body floats (a part of the body lies outside the liquid)
A body of volume V and density b is floating with a volume Vin inside the fluid of density  l ,
then V  b g  Vin 1 g
weight of the body = weight of the liquid displaced (due to body submerged in the liquid).
A body of mass M and volume V is floating in a liquid of density  l with some volume in air. To make it to
just sink, the mass ‘m’ to be placed on it is given by
mg  Vg l ,
where V is the volume of body that was initially outside the liquid.
Floatation:
When a body of density b and volume V immersed in a liquid of density  then forces acting are
 The weight of body acting vertical downwards through the center of gravity of the body
W = mg  V b g
’ The upthrust(force of buoyancy) acting upwards through center of gravity of displaced liquid called center
of buoyancy FB  V l g
1) If b    W  FB  body sinks
2) If b  l  W  FB  body just floats
3) If b  l  V b g  Vin 1 g  V b  Vin 1
 body is partially immersed or floats partially..
Archimede’s Principle:
When a body is immersed partly or wholly in a fluid it appears that it loses some weight, which
is equal to the weight of the liquid displaced (which is equal to the force of buoyancy).
FB  Buoyancy
mg
FB=mg
8
Apparent loss of weight of a body or weight of fluid displaced = Vin l g
Vin = Volume of body immersed or volume of fluid displaced
Note :
Upthrust or buoyancy is independent of mass, size, density, shape etc. of the body.
It depends only on the volume of the body immersed inside the fluid, nature (density) of the fluid and
acceleration due to gravity
 FB eff  Vin  l  g  a 
Relative density (Specific gravity) of a solid:
density of the body
RD = density of water at 4 0 C
weight of the body

‘ = upthrust exerted by water
weight of the body in air


loss of weight of body in water
w1
R.D. = w  w ;
1 2
w1 = weight of the body in air

w2 = weight of the body in water
w1  w2 = Loss of weight of body in water
Relative density of a liquid:
If loss of weight of a body in water is ‘a’ and that of in liquid is ‘b’ then,
V  w g  a; V  L g  b
 L Loss of weight in liquid b Wair  Wliquid

RD of liquid =  = = =
w Loss of weight in water a Wair  Wwater
Volume of a cavity in a body:
Consider a metal piece of mass M and density  m
M
The volume of metal V  
m
Weight of the metal in air = W1
Weight of the metal in water = W2
Loss of weight of the metal = W1 - W2 = Fb
If V ' is the geometric volume of the body immersed in the liquid then
V '  w g  W1  W2
W1  W2
V'
w g
9
 W1  W2  M
Volume of cavity = V ' V   
 w  m
 g
Where
V ' is the total volume of the metallic body..
V is the volume of the material in the metal piece.
The amount of impurity in a given metal:
Let w 1 be the weight of an alloy in air
w2 be the weight in water.
Let the alloy consists of two metals having masses m1 and m2
Total mass m = (m1 + m2). The buoyant force on the alloy is,
Fb  w1  w2  V  w g
w1  w2 m m
 V  1  2
w g 1  2
‘= Volume of the first metal in the alloy + Volume of the second metal in the alloy
= Volume of the alloy ( 1 ,  2 are the densities of the metals)
w1  w2 m1  m  m1 
 
w g 1 2
m  m1  m2  total mass
Fraction of Volume of the Body outside the Liquid:
Weight of the body = Weight of the displaced liquid
Vtotal b g  Vin 1 g Vtotal  volume of the body 
  
Vout  Vtotal  Vin  V 1  b 
 l 
Vout  b 
The fraction of volume outside the liquid is f out   1  
Vtotal  
Floating of Ice:
When a block of ice, floats in a liquid of density dl , melts completely, the level of (liquid + water)
(i) Rises, if 1   w
(ii) Falls, if 1   w
(iii) Remains unchanged, if 1   w
A piece of solid is embedded inside an ice block which floats in water.
When ice melts completely, the level of water
(i) Remains same, if  s   w
(ii) Falls, if  s   w
10
Tension in the string connected to a submerged body :
When the body hangs by light string and  b   l
T FB
Case-I : When the system is at rest or moving with uniform velocity (a=0) in vertical direction
The tension in the string is T = Apparent weight of the body
T  W  FB  V b g  V l g  V  b  l  g
Case-II: When the system is acclerating vertically upward with acceleration a. Tension
T  V  b  l  g  a 
Case-III : If the system is accelerating vertically downward with an acceleration a (a<g)
T  V  b  l  g  a 
11
::PROBLEMS::
1. The pressure at the bottom of a lake, due to water, is 4.9x 106 N/m2. What is the depth of lake?
SOLUTION :
Pressure P  h  g  4.9  106 N / m2
 density of water = 1000 kg/m3; g=9.8m/s2
P 4.9  106
hence h    500m
 g 1000  9.8
2. Equal masses of water and a liquid of density 2 are mixed together, then the mixture has a
density of
(a) 2/3 (b) 4/3 (c) 3/2 (d) 3
SOLUTION :
If two liquid of equal masses and different densities are mixed together then density of mixture
2 1  2 2 1  2 4
  
1   2 12 3
3. If pressure at half the depth of a lake is equal to 2/3 pressure at the bottom of the lake then what is the
depth of the lake
(a) 10 m (b) 20 m (c)60 m (d) 30 m
SOLUTION :
Pressure at bottom of the lake = P0  h g
h
Pressure at half the depth of a lake  P0  g
2
According to given condition
1 2
P0  h g  (P0  h g)
2 3
1 1
 3
P0  h g
6
2 P0 2  10 5
 h  3
g 10  10
 20 m .
4. In air, a metallic sphere with an internal cavity weighs 40g and in water it weighs 20g. The
volume of cavity if the density of material with cavity be 8g/cm3 is
1) zero 2) 15 cm3 3) 5 cm3 4) 20 cm3
SOLUTION :
Weight of sphere in air = 40 g
Weight of sphere in water = 20 g
Loss in weight = (40-20)g=20g
Weight of water displaced = loss in weight = 20g
Volume of water displaced = 20 cm3’
12
Actual volume of sphere=volume of water
displaced = 20 cm3
40
Volume of material in sphere =  5cm3
8
Volume of cavity = (20-5) = 15 cm3
5. A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The
specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float
with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust
will be [AIIMS 1995]
(a) 8 (b) 4 (c)3 (d) Zero
SOLUTION :
Let specific gravities of concrete and saw dust are 1 and  2 respectively..
According to principle of floatation
weight of whole sphere = upthrust on the sphere
4 4 4
 (R 3  r 3 )  1 g  r 3  2 g  R 3  1  g
3 3 3
R 3 1  r 3 1  r 3  2  R 3
R 3 (1  1)  r 3 (1   2 )
R3   2
 1
r3 1  1
R 3  r3    2  1  1
 1
r3 1  1
(R 3  r 3 )1  1   2  1
  

r3 2  1  1   2
Mass of concrete  1  0 .3  2 .4
  4
Mass of saw dust  2 .4  1  0 . 3
6. Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass
of one body is 36 g and its density is 9 g / cm3. If the mass of the other is 48 g, its density in g /
cm3 is
4 3
(a) 3 (b) 2 (c)3 (d)5
SOLUTION :
m
Apparent weight  V (    )g  (    )g

where m  mass of the body,,
density of the body
  density of water
If two bodies are in equilibrium then their apparent weight must be equal.
m1 m2
  (  1   )   ( 2   )
1 2
13
36 48
(9  1)  ( 2  1)
9 2
By solving we get 2  3 .
7. What is force on the base of a tank of base area 1.5m2 when it is filled with water upto a height
of 1m   water  103 kg / m 3 , P0  105 Pa and g  10m / s 2 
SOLUTION :
Absolute pressure at the bottom of the container is P  P0  h  g  105  1  103  10  1.1  105 Pa
Then force due to water on the base is Fbase  PA  1.1  10  1.5   1.65  10 N
5 5
8. A tank accelerates upwards with acceleration a=1 m/s 2 contains water. A block of mass 1kg
and density 0.8 g/cm3 is held stationary inside the tank with the help of the string as shown in
figure. The tension in the string is: (Given:density of water=1000kg/m3)
1) T=2.2 N 2) T=2.75 N 3) T=3N 4) T=2.4N
SOLUTION :
F=upthrust force = V  w  g  a 
 mass of block 
  w  g  a 
 density of block 
1
 1000 11  13.75 N
800
F-T-W=ma; 13.75-T-10=1 (1)
T=2.75N
9. An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The
bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric
pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3)
(a) 350 cm3 (b) 300 cm3 (c) 250 cm3 (d) 22 cm3
SOLUTION :
P2 V 2

h
(P1 V1)
14
According to Boyle’s law, pressure and volume are inversely proportional to each other
1
i.e. P
V
 P1 V1  P2 V2
 (P0  h w g)V1  P0 V2
 h g 
 V2   1  w V1
 P0 
 47 . 6  10 2  1  1000 
 V2   1   V1

 70  13 . 6  1000 
 V2  (1  5 )50 cm 3  300 cm 3 .
[As P2  P0  70 cm of Hg  70  13 .6  1000 ]
1 0 . A closed rectangular tank is completely filled with water and is accelerated horizontally with an
acceleration a towards right. Pressure is (i) maximum at, and (ii) minimum at
A D
a
B C
‘
(a) (i) B (ii) D (b) (i) C (ii) D (c) (i) B (ii) C (d) (i) B (ii) A
SOLUTION :
Due to acceleration towards right, there will be a pseudo force in a left direction.
So the pressure will be more on rear side (Points A and B) in comparison with front side (Point D
and C).
Also due to height of liquid column pressure will be more at the bottom (points B and C) in comparison
with top (point A and D).
So overall maximum pressure will be at point B and minimum pressure will be at point D.
11. A body of density d1 is counterpoised by Mg of weights of density d 2 in air of density d.

Then the true mass of the body is
 d   d  M (1  d / d 2 )
(a)M (b) M  1  d 
 (c) M  1  d 
 (d) (1  d / d 1 )
 2   1 
SOLUTION :
Let M 0  mass of body in vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air
Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air
 M0  M
 M 0 g   d  dg  Mg    dg
 d 
 1   2
15
 d 
M 1  
d
M 0  
2 
 d 
1  
 d1 
12. When equal volumes of two metals are mixed together, the specific gravity of alloy is 4. When
equal masses of the same two metals are mixed together, the specific gravity of the alloy now
becomes 3. Find specific gravity of each metal?
 density of substance 
 specific gravity= 
 density of water 
SOLUTION :
m1  m2
In case of mixture,  m ix  V  V
1 2
V 1  V  2 1   2
When equal volumes are mixed, 4   ....  i 
V V 2
3
 m  m   2 1 2 ..... ii 
When equal masses are mixed, m

m 1   2
1 2
Therefore, from (i) and (ii)
specific gravity of the metals are 2 and 6.
13. When a polar bear jumps on an iceberg, its weight 240 kg.wt is just sufficient to sink the
iceberg. What is the mass of the iceberg? (specific gravity of ice is 0.9 and that of sea water is
1.02)
SOLUTION :
If M is the mass of iceberg in kg
M
volume V  m3  density   specific gravity  103 kg / m 3 
0.9  10 3
The weight of displaced sea water = V  1.02  10  g  N .

3
 M 
3  
Mg  240 g    1.02  103   g
 0.9  10 
 1.02  12
 240  M   1  M
 0.9  90
90
or M   240  1800kg .
12
14. Fig. Shows a U-tube of uniform cross-sectional area A accelerated with acceleration ‘a’ as
shown. If d is the separation between the limbs, then the difference in the levels of the liquid in
the U-tube is
16
h1

a
h2
d
ad g
1) g 2) 3) adg 4) ad  g
ad
SOLUTION :
h
ta n  
d
ad
h
g
15. The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of
density of mercury to that of air is 104. The height of the hill is
(a) 250 m (b) 2.5 km (c) 1.25 km (d) 750 m
SOLUTION :
Difference of pressure between sea level and the top of hill
P  (h1  h2 )   Hg  g  (75  50)  102  Hg  g …(i)
pressure difference due to h meter of air
P = h   air  g …(ii)
By equating (i) and (ii) we get
h   air  g  (75  50 )  10 2   Hg  g
  Hg 
 h  25  10  2    25  10 2  10 4  2500 m

  air 
 Height of the hill = 2.5 km.
16. Four-fifths of a cylindrical block of wood, floats in a liquid. Asuming the relative density of
wood be 0.8, find the density of the liquid.
SOLUTION :
Let volume of wooden block = V
4
Volume of liquid displaced = V
5
Weight of the block = V  0.8  103 g
As the block floats,
weight of the body = weight of the liquid displaced
4V
V  0.8  103  g   l  g ; l  103 kg / m3
5
17
17. A cubical block of wood edge 3 cm floats in water. The lower surface fo the cube just touches the
free end of a vertical spring fixed at the bottom of the pot. The maximum weight that can be put on
the block without wetting it is (density of wood = 800 kg/m 3 and spring constant of the spring = 50 N/
m. Take g=10 m/s2)
1) 1.35 N 2) 1.55 N 3) 1.65 N 4) 1.75 N

SOLUTION :’
In equilibrium total weight W  Kx  Fb
 M  m  g  Kx  V w g
18. A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10 3kg/m3. If outer
diameter and the density of the bowl are 1 m and 2 × 104 kg/m3 respectively, then the inner
diameter of the bowl will be
(a) 0.94 m (b) 0.97 m (c)0.98 m (d) 0.99 m
SOLUTION :
4  D   d  
3 3
Weight of the bowl = mg = Vg 3  2   2   g

      
 
where D = Outer diameter ,

d = Inner diameter
 = Density of bowl
3
4  D
Weight of the liquid displaced by the bowl  Vg     g
3 2
where  is the density of the liquid.

4  D   d  
3 3 3
4 D
For the flotation    g         g
3 2 3  2   2  
 
1
3  1  3  d  3 
   1 . 2  10 3        2  10 4
2  2   2  
By solving we get d = 0.98 m.

19. Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass
of one body is 28g and its density is 5.6g/c.c. If the mass of the other body is 36g. Find its
density d
SOLUTION :
18
 28 
Apt wt. of 1st body = m1 g  Fb   28  g
 5.6 
 36 
Apt wt. of 2nd body = m2 g  Fb   36   l  g
 d 
 28   36  36
 28   g   36   g ;  28  5   36 
 5.6   d  d
36 36
  36  23  13; d   2.8 g / cc
d 13
20. A certain block weighs 15N in air. But it weighs only 12N when completely immersed in water.
When immersed completely in another liquid, it weighs 13N. Calculate the relative density of
(i) the block and (ii) the liquid.
SOLUTION :
Wair
i) Relative density of body = W  W
air water
where Wair  15 N  weight of the body in air 

and Wwater  12 N  weight of the body in water 
15 N
 R.Dblock  5
15 N  12 N
loss of weight in liquid 15  13 2
ii) R.DL  loss of weight in water  15  12  3
21.. Figure shows a hydraulic press with the larger piston of diameter 35 cm at a height of 1.5 m
relative to the smaller piston of diameter 10 cm. The mass on the smaller piston is 20 kg. What
isthe force exerted on the load by the larger piston. The density of oil in the process is 750 kg/m3,
(Take g=9.8 m/s2).
1.5 m
20 kg
1) 5 x 103 N 2) 1.3 x 103 N 3) 3.7 x 103N 4) 4.8 x 103N

SOLUTION :
F1 F2
  h g
A1 A2
19
m1 g m2 g
  h g
 r12  r22
1  m1 m2 
   
h  r12 r22 
22. A cubical block of iron of side 5cm is floating in mercury taken in a vessel. What is the height
of the block above mercury level.   Hg  13.6 g / cm ,  Fe  7.2 g / cm 
3 3
5cm
SOLUTION :
From the law of flotation, Vb b g  Vin  L g
  5    7.2    52 x   13.6  ; x  2.65cm

3
Then, the height of the block above mercury level = 5cm - x = 2.35 cm
23. A solid sphere of radius ‘R’ has a concentric cavity of radius R/3 inside it. The sphere is found to
just float in water with the highest point of it touching the water surface. Find the specific gravity
of the material of the sphere.
SOLUTION :
Vcavity VS  Vmetal V
  1  metal  1
VS VS VS
(VS = Total volume of the sphere)
According to Archimede’s principle Weight of body = Weight of displaced liquid
mg  VS d w g  d SVmetal g  VS d w g
Vmetal d w
(ds = Density of solid material, dw = Density of water) V  d   2  ;
SA S
From equation 1 and 2

Vcavity dw 1 1
  1  1  1
VS d d S .G
dw
3
4 R
 
3 3 1 1 1
 1   1
4 3 S .G 27 S .G
R
3
1 1 27
 1  S .G 
S .G 27 26
20
mA 2
24. In the arrangement shown in the figure m  3 and the ratio of density of block B and the
B
liquid is 2:1. The system is released from rest. Then
A
B
1) block B will oscillate but not simple harmonically

2) block B will oscillate simple harmonically
3) the system will remain in equilibrium
4) None of the above
SOLUTION :
Let mA  2m, mB  3m
mA g   mB g  upthrust on B  g
When block-B is inside the liquid a1  
m A  mB 10
mB g  mA g 3mg  2mg g
When block-B is outside the liquid a2  m  m  5m

5
A B
Since a1 & a2 are constants, motion is periodic, but not simple harmonic
25. A ball of relative density 0.8 falls into water from a height of 2m. Find the depth to which the
ball will sink (neglect viscous focus)
SOLUTION :
Gravitational potential energy = Apparent weight of the body  Displacement of the body inside
liquid
mgh  mg 1h1
 1 
(where g  g  d /   1 )
1
 w 
d
Relative density    0.8
w
 1   0.2  g
g1  g   1  g   ’
 0.8   0.8  4
g g
h1  1
h  2  8m
g g/4
26. A log of wood of mass 120 Kg floats in water. The weight that can be put on the raft to make it
j ust sink , should be (density of wood = 600 Kg/m3)
(a) 80 Kg (b) 50 Kg (c) 60 Kg (d) 30 Kg
21
SOLUTION :
mass 120
Volume of log of wood V  =0.2 m3
density 600
Let x weight that can be put on the log of wood.

So weight of the body = (120  x )  10 N
Weight of displaced liquid = Vg  0 .2  10 3  10 N
The body will just sink in liquid if the weight of the body will be equal to the weight of displaced
liquid.
 (120  x )  10  0 .2  10 3  10
 120  x  200
 x = 80 kg
2 7 . A vertical U-tube of uniform inner cross section contains mercury in both sides of its arms. A
glycerin (density = 1.3 g/cm3) column of length 10 cm is introduced into one of its arms. Oil of
density 0.8 gm/cm3 is poured into the other arm until the upper surfaces of the oil and glycerin
are in the same horizontal level. Find the length of the oil column, Density of mercury = 13.6 g/
cm3
Glycerine
Oil h
10 cm
Mercury
(a) 10.4 cm (b) 8.2 cm (c) 7.2 cm (d) 9.6 cm

SOLUTION :
At the condition of equilibrium
Pressure at point A = Pressure at point B

PA  PB
10  1 .3  g  h  0 .8  g  (10  h)  13 .6  g
By solving we get h = 9.7 cm
28. The hydraulic press shown in the figure is used to raise the mass M through a height of 5.0 mm
by performing 500 J of work at the small piston. The diameter of the large piston is 10 cm while
that of the smaller one is 2 cm. The mass M is
M W=500J
2 cm
10 cm
22
1) 104 kg 2) 103 kg 3) 100 kg 4) 105kg
SOLUTION :
Since, M moves 5.0 mm and volume of liquid remains constant
    5  102   5  103     1  102  x

2 2
 x  125  10 3 m
Suppose force F does 500 J work on small piston.
Fx=500
500
F 3
N  4  103 N
125  10
F  5  102 
2
Mg 
Using Pascal’s law,
1  10 
2 2
Mg  100  103  105 N  M  10 4 kg

29. An ice berg of density 900 Kg/m3 is floating in water of density 1000 Kg/m3. The percentage of
volume of ice-cube outside the water is
(a) 20% (b) 35% (c) 10% (d) 25%
SOLUTION :
Let the total volume of ice-berg is V and its density is r.
If this ice-berg floats in water with volume Vin inside it
then Ving  Vg

Vin    V [   density of water]
 
   
or Vout  V  Vin   V
  
Vout      1000  900 1

  
V    1000 10
\ Vout  10 % of V
30. A rubber ball of mass m and density  is immersed in a liquid of density 3  to a depth h and
released. To what height will the ball jump up above the surface due to buoyancy force of liquid
on the ball? (neglect the resistance of water and air).
SOLUTION :
m
Volume of ball V 

Acceleration of ball moving in upward direction inside the liquid
Fnet upthrust  weight Vtotal l g  mg
a  
m m m
23
m
    3  g   mg
a   2 g  upwards 
m
 velocity of ball while crossing the surface v  2ah  4 gh
 The ball will jump to a height
v 2 4 gh
H   2h
2g 2g
31. A small ball of density  is immersed in a liquid of density      to a depth h and released.
The height above the surface of water upto which the ball will jump is
       
1)   1 h 2)   1 h 3)   1 h 4)   1 h
       
SOLUTION :
2
H
2g
   
  2ah and a     g
 
2     gh


2    
H   h
2g   
32. Two spheres of volume 250cc each but of relative densities 0.8 and 1.2 are connected by a string
and the combination is immersed in a liquid. Find the tension in the string. (g=1m/s2).
SOLUTION :
The tension on denser sphere is upwards and on lighter sphere is downards.
F.B.D of F.B.D of
heavier body lighter body
Lighter T FB
FB
T
T
Heavier T
mg mg
Vb1 1 g  T  Vb1 1 g
  250  106  800 g   T  250  106  liquid g   i 
Vb2  2 g  T  Vb2 l g
24
 250  10 6
 1200 g   T  250  10 6  liquid g   ii 
subtract Eqs. (ii) from (i), we get
2T  250  106  400 g  T  0.5 N
33. A bowl of soap water is at rest on a table in the dining compartment of a train, if the
acceleration of the train is g/4 in forward direction, the angle made by its surface with
horizontal is
1  1  1  1  1  1  1  1 
1) tan   2) tan   3) tan   4 ) tan  
2 4 5 3
SOLUTION :
mg
ma
ma a
 tan   
mg g
34. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, when its
lengths vertical from a fixed point by a massless spring such that it is half submerged in a liquid
of density  at equilibrium position. The extension x0 of the spring when it is in equilibrium is
(AIEEE-13)
SOLUTION :
FRestoring
FB
Mg
In equilibrium, upward force = downward force

L 
Fspring  FB  mg  kx0   A   g  mg
2 
 LAg
mg 
2 mg   LA 
 x0   1  
k k  2m 
25
35. A block is fully submerged in a vessel filled with water by a spring attached to the bottom of
the vessel. In equilibrium position spring is compressed. If the vessel now moves downwards
with an acceleration a(<g). What happens to the length of the spring?
SOLUTION :
When the vessel moves downwards, with acceleration, block experiences net pseudo force
upwards. Hence apparent weight of the block decreases and the block moves upwards. Hence length
of the spring increases. Additional force on the block in upward direction=pseudo force on the block-
decrease in buoyance force
3 6 . A solid sphere of density  ( > 1) times lighter than water is suspended in a water tank by a
str ing tied to its base as shown in f ig. I f the mass of the spher e is m then the tension in the
string is given by
 1  mg
(a)   mg (b) mg (c)   1 (d) (  1) mg
  
SOLUTION :
Tension in spring T = upthrust – weight of sphere
 Vg  Vg
 Vg  Vg (As    )
 (  1)Vg
= (  1)mg .
37. A hollow sphere of volume V is floating on water surface with half immersed in it. What should
be the minimum volume of water poured inside the sphere so that the sphere now sinks into the
water
(a) V /2 (b) V /3 (c) V /4 (d) V
SOLUTION :
When body (sphere) is half immersed, then
upthrust = weight of sphere
26
V  liq
2
  liq  g  V    g  
2
When body (sphere) is fully immersed then,
Upthrust = wt. of sphere + wt. of water poured in sphere
V   liq  g  V    g  V '  liq  g
V   liq
V   liq   V '  liq
2
V
V' 
2
38. Two communicating vessels contain mercury. The diameter of one vessel is n times larger
than the diameter of the other. A column of water of height h is poured into the left vessel.
The mercury level will rise in the right-hand vessel (s = relative density of mercury and 
= density of water) by
Water
h
Mercury
n 2h h h h
(a) (n  1)2 s
(b) 2
(n  1) s (c) (n  1)2 s
(d) n2s
SOLUTION :
Water
h
h2
h1
A B
Mercury
If the level in narrow tube goes down by h1 then in wider tube goes up to h2,
Now, r 2 h1   (nr )2 h2
 h1  n 2 h2
Now, pressure at point A = pressure at point B
h g  (h1  h2 ) ' g
 ' 
 h = (n 2 h2  h2 )sg  As s   
 
h
 h2  (n 2  1)s
39. A uniform rod of density  is placed in a wide tank containing a liquid of density  0 ( 0   ) .
The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with
its lower end resting on the bottom of the tank. In this position the rod makes an angle 
with the horizontal
1 1 0
(a) sin   0 /  (b) sin   . (c) sin    / 0 (d) sin    0 / 
2 2 
SOLUTION :
27
Q
R
FB S
L
h=L/2
P W

L
Let L = PQ = length of rod SP  SQ 
2
Weight of rod, W  Alg , acting At point S

And force of buoyancy, FB  Al 0 g , [l = PR] which acts at mid-point of PR.
For rotational equilibrium,
l L
Al 0 g  cos   AL g  cos 
2 2
l2  l 
 L 2

0
 L

0
h L 1 0
From figure, sin    
l 2l 2 
Comprehension :
A solid hemisphere of radius R is made to just sink in a liquid of density  . Find the
40. vertical thrust on the curved surface

 R3  g  R3  g
A) B) C) 0 D)  R 3  g
3 2
SOLUTION :
Vertical thrust of the liquid is equal to weight of the liquid column above the curve (spherical ) surface.
FV  v  g .
2  R3
Volume of the shaded portion =  R    R =
3 3
3 3
R
R
(I) (II)
Substituting V in the equation Fv  V  g , we have
 R3
FV   g  down 
3
28
41. Vertical thrust on the flat surface
 R3  g  R3  g
A) B) C) 0 D)  R 3  g
3 2
SOLUTION :
The upward thrust on the base of the hemisphere is Fv'  V  g ,
where, V = volume of the liquid column above the base   R  R   R
2 3
Then, we have Fv'   R 3  g  up 

42. Side thrust on the hemisphere
 R3  g  R3  g
A) B) C) 0 D)  R 3  g
3 2
SOLUTION :
Let the horizontal and vertical thrusts on the tortoise be Fh and Fv , respectively..
we know that Fh   gyC Av
where yC  R
Av   R 2 .
This gives Fh   g  R 3 towards right
43. A block of ice floats on a liquid of density 1.2in a beaker then level of liquid when ice completely
melt
(a) Remains same (b) Rises (c) Lowers (d)(a), (b) or (c)
SOLUTION :
M
The volume of liquid displaced by floating ice VD  
L
M
Volume of water formed by melting ice, VF 
W
M M
If  1   W , then, 
L W
i.e. VD  VF
i.e. volume of liquid displaced by floating ice will be lesser than water formed and so the level if
liquid will rise.
44. A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The
volume of the remaining cylinder is V and mass M. It is suspended by a string in a liquid of
density  , where it stays vertical. The upper surface of cylinder is at a depth h below the liquid
surface. The force on the bottom of the cylinder by the liquid is
29
h
A)  g V   R h  C) Mg  V  g
2
B) Mg D) Mg   R 2 h g
SOLUTION :
Upthrust = F2  F1 ;
F2  F1 +Upthrust
=  gh  R   V  g ;
2
=  g V   R h 
2
45. The top view of closed compartment containing liquid is moving with an acceleration along x -
axis as shown. Find the incorrect statement.
2a
Y
E O1
O a
X
D A
C B
A) The pressure at A and O is same B) The pressure at O and O1 is same

C) The pressure at B and C is same D) The pressure at D and E is same
SOLUTION :
dP
   x acceleration
dx
The pressure decreases in positive x - direction.
The pressure is lower in front side.
The pressure at B and C can not be same.
Comprehension :
A tortoise is just sinking in water of density  . The tortoise is assumed to be a hemisphere of
radius R.
30
R
46. Find Vertical thrust

1 2
A)  g R 3 B)  g R 3 C)  g R 3 D) 0
3 3
SOLUTION :
Similarly using the formula Fv  Vg
2
where V = volume of the tortoise =  R3
3
2
we have Fv   g R 3 (up)
3
FV yC
Fh
47. Fint the total hydrostatic force

13 2 16
A)  g R 3 B)  g R3 C)  g R 3 D)  g R 3
3 3 3
SOLUTION :
Hence the net hydrostatic force on the tortoise is F  Fh2  Fv2

2
2 
   g R 
3 2
   g R 3 
3 
13
  g R 3
3
48. A vessel of height H and length L contains a liquid of density  upto height H/2. The vessel
starts accelerating horizontally with acceleration ‘ a ’ towards right. If A is the point at the surface
31
of the liquid at right end while the vessel is accelerating and B is the point at bottom of the
vessel on the other end, the difference of pressures at B and A will be
A
H
H/2
a
B
L
  3
A)  gH  aL  B)  gH  aL  C) 2   gH  aL  D)  gH  aL 
2 2 2
SOLUTION :
Let force  P  A acts on the surface of the liquid while vessel is accelerating.
( P.A)
 y
C
H/2 L/2 A
B h1
The surface can not sustain tangential force.

 P. A cos  mg ;
 P. A  sin   ma
L aL
tan   a / g and y  tan  
2 2g
H aL
 h1 
2 2g
Considering a fluid element at distance x from left side of the vessel, then
dx
area A'
pressure 
+dp
 p  dp  A ' pA '   A ' dx   a

C C
 dp    adx ; pC  p A  a  L
A A
32
If points A and C are at same level
 H aL 
pB  pC   gh1   g   
 2 2g 
 H aL 
pB    gL  p A    g   
 2 2g 

pB  p A   gH  aL 
2
49. A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the
right contains mercury (density 13.6 g/cm3). The level of mercury in the narrow limb is at a
distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury
in the right limb if the left limb is filled to the top with water
Water
Mercury
(a) 1.2 cm (b) 2.35 cm (c) 0.56 cm (d) 0.8 cm

SOLUTION :
x
A B A B
4x
A' B'
If the rise of level in the right limb be x cm.

the fall of level of mercury in left limb be 4x cm
( because the area of cross section of right limb is 4 times as that of left limb.}
 Level of water in left limb is (36 + 4x) cm.
Now equating pressure at interface of Hg and water (at A’B’)
(36  4 x )  1  g  5 x  13 .6  g
By solving we get x = 0.56 cm.
50. A homogeneous solid cylinder of length L (L  H / 2) . Cross-sectional area A / 5 is immersed
such that it floats with its axis vertical at the liquid-liquid interface with length L / 4 in the
denser liquid as shown in the fig. The lower density liquid is open to atmosphere having
pressure P0 . Then density D of solid is given by
H/2 d
3L/4
L
H/2 2d
33
5 4 d
(a) d (b) d (c) d (d)
4 5 5
SOLUTION :
Weight of cylinder = upthrust due to both liquids
A 3   A L
V  Dg    L   d  g      2d  g
5 4  5 4
A  A Ld  g
  L D  g 
 5  4
D d

5 4
5
D  d
4
51. A vessel contains oil (density = 0.8 gm/cm3) over mercury (density = 13.6 gm/cm3). A
homogeneous sphere floats with half of its volume immersed in mercury and the other half in
oil. The density of the material of the sphere in gm/cm3 is
(a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8
SOLUTION :
Oil
Mercury
As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it
4
Weight of sphere  R 3 g …(i)
3
2 3 2
Upthrust due to oil and mercury  R   oil g  R 3 Hg g …(ii)
3 3
Equating (i) and (ii)
4 2 2
R 3 g  R 3 0 . 8 g  R 3  13 . 6 g
3 3 3
 2   0 .8  13 . 6  14 . 4    7 .2
34
Fluid Dynamics:
It is the study of behaviour of fluids in motion. Fluid dynamics is caused by the difference in
pressures of a fluid between two points.
The Rate Of Flow Of A Liquid:
The rate of flow of a liquid means the volume of a liquid that flows across any cross section in unit time
and is given by
Volume V l
Q   A    Av  v  vel.of the fluid 
time t t
 m3 
SI unit :   ;
 sec 
D.F: L3T-1
M Volume l 
Mass of the liquid that flows per unit time   density  A     Av 
t time t 
Where
A is the area of cross section of the tube,
V is the velocity of the liquid
 is the density of the liquid.
Stream line flow :
A streamline may be defined as the path, straight or curved, the tangent to which at any point
gives the direction of the flow of liquid at that point.
 B
v1 C
 
v2 v3
The two streamlines cannot cross each other and the greater is the crowding of streamlines at
a place, the greater is the velocity of liquid particles at that place.
Path ABC is streamline as shown in the figure and v1 , v 2 and v 3 are the velocities of the
liquid particles at A, B and C point respectively.
Characteristics of fluid flow:
A stream line may be a straight line or a curve.
The tangent drawn at any point of curved stream line gives the direction of velocity of the particle at
that point.
Two stream lines never intersect, if they intersect, at the point of intersection the fluid may have two
directions of velocity which is impossible.
An imaginary tube consisting of a number of stream lines is called tube of flow.
There is no radial flow of liquid.
Types of flow of liquid:
There are two types of liquid flow.
35
1) Stream line flow
2) Turbulent flow.
Stream Line Flow:
If the velocity of all the fluid particles crossing a point remains constant both in magnitude and
direction then the flow of the fluid is known as Stream Line Flow
ATurbulent Flow:
If the velocity of the different fluid particles crossing a point doest not remain constant in ‘’
magnitude and direction then the flow of the fluid is known asTurbulent Flow
Eddies and whirl pools are formed in turbulent flow.
Critical Velocity:
Critical Velocity is the velocity beyond which stream line flow is gradually changed to turbulent flow.
R
Critical Velocity Vc 
D
DVc
R

where  =coefficient of viscosity,,
R=Reynolds number,
D=diameter of the tube,
 =density of the liquid
Reynold’s number depending upon the diameter of the pipe, the density and coefficient of viscosity of
the liquid.
 If 0<R<1000, the fluid flow is said to be stream line.
 If the value of R > 2000 then the liquid flow becomes turbulent.
 If 1000 < R < 2000 then the flow in unsteady
Equation of Continuity:
When an incompressible fluid flows steadily through a tube of non-uniform cross section, the
rate of mass of fluid entering the tube is equal to rate of mass of the fluid leaving the tube.
a2
v2
B
a1
v1
m
   constant
l
m1 m2
  
t1 t2
l  l 
  A1  1  1  A2  2  2
 t1   t2 
36
   A1v1 1  A2 v2  2
  leas, aasss 1   2
A1v1  A2 v2  constant
1
V
A
1
v
r2
Equation of continuity represents the law of conservation of mass in case of moving fluids.
Types of Energies in fluid flow:
A fluid in motion prossesses three types of energies namely
1 2
(i) KE  mv kinetic energy,,
2
(ii) PE=mghpotential energy
(iii) Pressure energy = P x V
Bernoulli's theorem:
Bernoulli's Theorem states that the sum of the pressure energy, kinetic energy and potential energy at
any point in steady flow calculated per unit mass or per unit volume is constant.
(or)
Bernoulli's Theorem can also be stated as follows: "In a stream line flow of fluid, the sum of
gravitational head, pressure head and velocity head at any point in the path of the flow is constant"
1
 P   v 2   gh  constant (per unit volume)
2
P 1
  v 2  gh  Constant (per unit mass)
 2
2
 P  v  h = Constant (per unit weight)
g 2g
P
Here  is called pressure head,
g
v2
 2 g is called Velocity head
 h is called gravitational head.
A1 V1
h1 A2
h2 V2
Ground
1 2
Here   v is called dynamic pressure.
2
( P   gh ) is called static pressure
37
Bernoulli's theorem represents Law of conservation of energy.
When the flow is horizontal, h is same and hence sum of pressure head and velocity head is
constant.
1 2 1
P1   v1  P2   v22
2 2
1
P1  P2    22   12 
2
where P1 and P2 are pressures at two points,
v1, v2 are velocities at two points

 is the density of the liquid.
For horizontal flow of liquid, maximum pressure corresponds to minimum velocity and vice versa
 1 2 
 P   v  constant 
 2 
Torricelli's theorem:
The velocity of efflux of a liquid through an orifice is equal to that of the velocity acquired by a freely
falling body from a height which is equal to that of the liquid level from the orifice.
v  2 gh
Time taken by the efflux liquid to reach the ground is given by
2 H  h
t
g
Where H = height of the liquid in the container
h = the distance between the free surface of the liquid and centre of the hole
H
v  2 gh
 H  h
Horizontal range of the liquid is given by R = V  t

2  H  h
R  2 gh.
g
R = 2 h  H  h
The volume of the liquid coming out of the orifice per second is
Q  Av  A 2 gh   r 2 2 gh
Horizontal range is maximum when the orifice is at the middle of liquid level and bottom.
H
i.e., if h 
2
then R M a x  H  2 h
38
If the level of free surface, in a container is at height H from the base.There are two holes at height
‘h’above the bottom and other at depth ‘y’ below the free surface, then
h
x  x'
x  2 h  H  h  and x1  2 y  H  y 
now if x  x1
i.e. h( H  h)  y ( H  y )
1
from above equation y  [ H  ( H  2h)]
2
i.e. y  h or ( H  h)
so the range is same for liquid coming out of holes at same distance below the top and above the bottom.
A tank having an area of cross-section A is filled with water upto height ‘H’ and ‘A0’ is the area of
cross-section of hole at the bottom of tank. If A0 is the area of orifice at a depth ‘y’ below the free
surface and A is that of container, then the volume of liquid coming out of the orifice per second will be
 dV / dt   vA0  A0 2 gy  as v  2 gy 
yH
A0
Due to this, the level of liquid in the container will decrease and so, if the level of liquid in the
container above the hole changes from y to y-dy in time t to t+dt then -dV=Ady.
Substituting the value of dV in the above equation,
dy A 1
 A0 2 gy ;  dt  
2g 
A y 1/ 2 dy
dt A0
So the time taken for the level to fall from H to H1
A 1 H1 A 2
t  y 1/ 2dy  H  H1 
A0 2g H A0 g  
39
H
Time after which level of water falls from H to is
2
A 2 H 
t1   H  
A0 g  2 
H
Time after which water level falls from to ‘0’ is
2
A 2 H 
t2    0
A0 g 2 
t1 2 1

t2 1
In a cylindrical vessel containing liquid of density '  ' , there are two holes in the side wall at
heights h1 and h2 respectively such that the range of efflux at the bottom of vessel is same. If
v1, v2 are the velocities of efflux and t1, t2 are the times taken by the efflux liquid to reach the
floor respectively from holes at heights h1, h2 then
2h1
v1  2g  H  h1  ; t1 
g
2h2
v2  2g  H  h2  ; t2 
g
since x1  x2  v1t1  v2t2
2h1 2h2
2 g  H  h1   2 g  H  h2 
g g
on solving H  h1  h2
H h2
h1
Note :
In the above case, the height of a hole, in terms of h 1 and h2 for which the range of efflux would be
maximum will be
H  h1  h2 
h  
2  2 
40
A tank is filled upto a height, 2H with a liquid and is placed on a platform of height H from the
ground. The density ‘y’ from the ground where a small hole is made in the tank, to get the
maximum horizontal range R
2H
y
Horizontal range will be maximum when the hole in the tank lies at the middle of total height of
water surface from the ground.
H  h1  h2 
For maximum range h   
2  2 
 2 H  H  3H
i.e y   
 2  2
A water tank is kept on the top of a table of height h. If a small hole is punched in the side of
the tank at its base it is found that the resultant stream of water strikes the ground at a
R2
horizontal distance R from the tank then the depth of water in the tank x 
4h
2h
R  ut  2 gx
g
R  2 xh
R 2  4 xh
R2
x
4h
A tank is filled with water of density 1 and oil of density  2 . The height of water column is h1 and that
of the oil is h2. The velocity of efflux through a hole at the bottom of the tank is obtained as follows
According to Bernoulli’s theorem
41
v2 P0
oil
h2
2
1
h1
v1
Water
P0
Water
1 1
P1  1v12  0  P2  2v22   1 gh1  2 gh2 
2 2
But P1  P2  P0
v2  0  v2  v1 
1
1v12  1 gh1   2 gh2
2
2g  1h1   2 h2 
v1 
1
Applications of Bernoulli's theorem:

Dynamic lift:
The upward lift experienced by a body in motion in a fluid is called dynamic lift.
The dynamic lift experienced by a body when it is in motion in air is called aerodynamic lift.
Aeroplanes get the dynamic lift because of the shape of their wings.
The upper surface of the wing is more curved than the lower surface. Air flows with greater speed
above the wing and so pressure above the wing will be less than that at the bottom. This difference in
pressures produces the aerodynamic lift and allows aeroplane to fly.
P1 V2 is more; P1 is less 

V2
P2
V1
V1 is less; P2 is more
Pupper  Plower
1
Dynamic lift =  P2  P1  A =  . V12  V22   A
2
42
Spinning ball :
VR V
V VR
The plane of motion of a spinning ball gets changed due to an effect called Magnus effect.
Resultant velocity on the top  V  VR  V  R
Resultant velocity on the bottom  V  VR  V  R
Pressure at the top of the ball will be less than that of at the bottom.
Atomiser, paintgun and Bunsen burner, work onthe principle of Bernoulli's Theorem.
Venturimeter:
It is an ideal device of measuring rate of flow of a liquid through pipe.It is also known as venturi tube
or flow meter.
The decrease in cross-sectional area of the flow passage causes increase in pressure
The measurement of the pressure difference enables the determination of the rate of flow through the
pipe.
2  P1  P2  2  h1  h2   g 2hg
Q  A1 A2  A1 A2  A1 A2
A 1
2
 A22   A 1
2
 A22   A
1
2
 A22 
 h  h1  h2 
P1 P2
h1
h2
V1 V2
A2
A1
Aspirator pump:
In aspirator pump when air is pressed inside the tube, it comes out rapidly so that pressure at
A reduces whereas pressure at B is more. For this pressure difference liquid rises till the barrel and
sprayed with the expelled air.
43
Pitot tube:
V V
a b
It is used to measure the speed of fluid flowing through a pipe. Here, the pressure in the left
arm of the manometer whose opening is parallel to the direction of flow, is equal to the pressure in the
fluid stream while pressure in right arm can be computed by using Bernoulli’s theorem. It is obvious that
velocity of fluid is zero at point b.
v 2
Pa  2  P0 , where  is the density of fluid flowing through tube.
2
Pb  Pa  0 gh where is the density of liquid in the manometer tube.
v 2 20 gh
So,  0 gh  v 
2 
Siphon Tube:
C
H
A
h
B
y
D
E
Siphon tube is used to empty the tanks etc, which are either very heavy or cant’t be lifted. In
short, we can say siphon is used to remove liquid from containers without using pumps etc.
The siphon tune is of uniform cross-section and to operate it successfully, it must be intially filled with
liquid. For the situation show in figure.
vA = 0 and vB = vC = vD = v
Because siphon tube is of uniform cross section so from equation of continuity, flow speed is
same at all points within the siphon tube.
Applying Bernoulli’s equation at A, B, C, D and E (consider water level in the tank as reference for
gravitational potential energy)
v 2
P0  0  0  PB   gh
2
v 2 v 2
 PC   gH  PD   g  h  y 
2 2
v 2E
 P0  g  h  y 
2
From above equation, we have PA = PE = P0
VE  2g  h  y 
so for liquid to come out h+y>0
44
::PROBLEMS::
1. What are the dimensions of Reynolds number?
SOLUTION :
 v0 D
R

 is density of the fluid,
v0 is the critical velocity,
D is diameter through which the fluid is flowing

 is the coefficient of viscosity of the fluid)
 ML3   LT 1   L   ML1t 1 
R  1 1    M L T 
0 0 0
 ML1T 1   ML T 
 R is dimensionless.
2. In the syphon as shown, which of the option is not correct, if h2  h1 and h3  h1 ?
B C
h1
A
h2
E
h3
D
A) pE  pD B) pE  pC C) pB  pC D) pB  pE
SOLUTION :
If p0  atmospheric pressure PA  P0  PB   h1 g ;
PD  P0  PE   h3 g
Since, h3  h1 , So, PE  PB
3. What should be the average velocity of water in a tube of diameter 2cm so that the flow is (i)
laminar (ii) turbulent? The viscosity of water is 0.001 Pa-s. (for water pipe R<2000 steam line
flow; R>3000 turbulent flow)
SOLUTION :
Here, d=2cm=0.02m;
  0.001Pa  s;   103 kg / m3
i) For stream line flow; Reynolds no. R=2000, V=?
R 2000  0.001
Now, v    0.1ms 1
d 103  0.02
ii) For turbulent flow, Reynolds number,
3000  0.001
R  3000; v  ? v   0.15ms 1
10  0.02
3
45
4. A cylindrical tank has a hole of 2 cm2 at its bottom. If the water is allowed to flow into the tank from
a tube above it at the rate of 100 cm3/s, then find the maximum height upto which water can rise in
the tank (Take g=10 ms-2)
1) 2.5x10-2 m 2) 1.25x10-2m 3) 5.5x10-2 m 4) 3.5x10-2m
SOLUTION :
Qin  Qout Q  a 2 gh
5. A siphon tube is used to remove liquid from a container as shown in fig. In order to operated
the siphon tube, it must initially be filled with the liquid.
C
H
A
B
y
(i) Determine the speed of the liquid through the siphon

(ii) Determine the pressure at the point C.
SOLUTION :
(i) Applying Bernoulli’s equation at points A and D,
1 2 1 2
we get PA  rv A  rgy A  PD  rv D  rgy D
2 2
Assuming datum for potential energy at the free surface,
we have y A  0; y D    h  y  ;
PA  PD  Patm v A2  0; v D  v
1
Patm  0  0  Patm  rv 2  rg    h  y   or v  2g  h  y 
2
(ii) Applying Bernoulli’s equation at A and C,
1 2 1 2
we get PA  rv A  rgy A  PC  rv C  rgy C
2 2
Here, YC= + H; vC = v (according to the continuity equation)
1
Patm  0  0  PC  rv 2  rgh
2
or, Patm  PC  rg(h  y)  rgh
or, PC  Patm  rg(h  H  y)
6. In a vessel as shown, the opening has a cross - sectional area A. F1 is the net force applied on
the plate by liquid and air, which is kept to close the opening. The plate is now displaced a short
distance away from the opening in which case liquid strikes the plate inelastically with a force
F2 . Find F2 / F1
46
A
SOLUTION :
F1  P  A    ghA
F2   AV 2   A  2 gh  ;
F2
2
F1
7. A vessel has water to a height of 50 cm. It has three horizontal tubes of same diameter each of
length 15cm coming out at heights 10 cm, 15 cm, 20 cm. The length of a single tube of same
diameter as that of the three tubes which can replace them when placed horizontally at the
bottom of the vessel is:
1) 45 cm 2) 5 cm 3) 8 cm 4) 16 cm
SOLUTION :
h is total height of water, h1, h2, h3 are the heights at which tubes are fitted
  h  h1  dgr 4   h  h2  dgr 4
Q1  ; Q2 
8l 8 l
  h  h3  dgr 4
Q3  and Q  Q1  Q2  Q3
8 l
 hdgr 4 h h1 h2 h3
with Q  ;   
8l l l1 l2 l3
8. Air is streaming past a horizontal aeroplane wing such that the speed of air is 120 m/s over the
upper surface and 90 m/s at the lower surface, with respect to the plane. If the density of air is 1.3
kg/m3, find the difference in pressure between the top and bottom of the wing. If the wing is 10m
long and has an average width of 2m, calculate the gross lift of the wing.
SOLUTION :
1 2 1 2
According to Bernoulli’s equation, P1   v1  P2   v2
2 2
1 1
i.e., P1  P2    v2  v1   1.3 120  90   4.1  10 N / m
2 2 2 0 3 2
2 2
The net uplift due to the difference in pressure is
Fup  P  A   4.1  10   10  2   8.2  104 N
9. A sphere falls from rest into water from a height of 2m. The relative density of the sphere is 0.80.
Find the depth to which ball will sink (in m)
47
SOLUTION :
upthrust  weight  V 1 g  V  0.8  g  g / 4
Retardation in water, a  V  0.8 
mass
Now, V22  V12  2  a  h0 ;
0  2 gh  2ah0 ;
gh  g  2 
h0   8m
a  g / 4
10. The velocity of the liquid coming out of a small hole of a vessel containing two different liquids
of densities 2  and  as shown in fig is
2h
A) gh B) 2 gh C) 2 2gh D) 6gh
SOLUTION :
1
P0   g  2h    2   gh  P0   2   v 2
2
liquid of density 2  is coming out
11. A cylindrical tank of height H is open at the top end and it has a radius R. water is filled in it up to
a height of h. The time taken to empty the tank through a hole of radius r at its bottom is
2h R 2 2H R 2 2H R
A) B) C) hH D)
g r2 g r2 g r
SOLUTION :
dh dh r2
 R 2   r 2 2 gh ;   2 2 gh
dt dt R
R2 1
dt   dh
r2 2 gh
t R2 0
  dt   h 1/ 2 dh
0
r2 2g h
R2  h
0
t  2  h
r2 2g
2h  R2 
t
g  r 2 
48
12. A cylindrical vessel of height 500 mm has an orifice at its bottom. The orifice is initially closed and
water is filled in it upto a height H. Now the top is sealed with a cap and the orifice at the bottom
is opened. Some water comes out from the orifice and the water level in the vessel becomes
steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due
to opening of the orifice. (Take atmospheric pressure  1.0 105 N / m 2 . denisty of water
1000kg / m3 and g  10m / s 2 . Neglect any effect of surface tension.)
SOLUTION :
500 mm
H
500 mm
200 mm
P  P0   gh  98  103 N / m 2 ;
P0V0  PV
105  A  500  H    98  103  A  500  200  
H  206mm ;
level fall  206  200  6mm
13. A horizontal pipe line carries water in a streamline flow. At a point along the pipe where the
cross-sectional area is 10cm2, the velocity of water is 1 m/s and the pressure is 2000 Pa. What
is the pressure at another point where the cross-sectional area is 5cm2.
SOLUTION :
10
According to equation of continuity A1v1  A2v2 ; v2   1  2m / s
5
49
1 2 1
Now, according to Bernouli’s equation, P1   v1  P2   v22 ( horizontal pipe)
2 2
1 1
2000  103 1  P2  103  2 2  ; P2  500 Pa
2
2 2
14. A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks
in water. The ratio of radius of cavity to that of outer radius of the sphere must be
‘
71/3 51/3 91/3 31/3
A) B) C) D)
2 2 2 2
SOLUTION :
Let  be the density of the material,  0 be the density of water..
When the spehere has just started sinking,
the weight of the spehre= weight of water displaced (approx)
4 4
   R3  r 3   g   R3 0 g
3 3
  R3  r 3    R 3 0 
R 3
 r3 

0
R 3

r 7
1/3
 
R 2
15. A block of ice (specific gravity Si- = 0.90) is floating in a container having two immiscible liquids
(one of specific gravity S = 0.50 and other is water) as shown in the figure. (H , H are heights
1 1 2
of water, other liquid columns respectively.) Now the ice block melts completely, then
H2 S = 0.50
‘
Ice Block
H1 Water
1. H will decrease 2. H will increase

2 1
3. H +H will remains unchanged 4. H +H decreases

1 2 1 2
SOLUTION :
Density of ice is in between that of water and liquid.
50
16. Two unequal blocks of densities  1 and  2 placed over each other are immersed in fluid of
density  . The block of density  1 is fully submerged and the block of density  2 is partly
submerged so that ratio of their masses is 1/2 and  /  1  2 and  /  2  0.5 . Find the degree
of submergence of the upper block of density  2 .
A) 50% submerged B) 25% submerged C) 75% submerged D) Fully submerged
SOLUTION :
 m2   m1 m2 
 m1  g     g
 K    1 K 2 
dividing by m2 g ,
m1 1 m 
  1 
m2 K m2 1 K 2
m1 1
 .2  .0.5 ;
m2 K
m1 0.5
 ; K  1.0 , fully submerged
m2 K
17. In a cylindrical container water is filled up to a height of h0  1.0m . Now a large number of small
iron balls are gently dropped one by one into the container till the upper layer of the balls
touches the water surface. If average density of the contents is   4070kg / m3 , density of iron
is  i  7140 kg / m 3 and density of water is 0  1000kg / m3 , find the height h of the water
level (in S.I units) in the container with the iron balls.
h
h0
SOLUTION :
total mass m  m2
  1
total volume Ah
m1 m2
0  ; i 
Ah0 A  h  h0 
 0 h0   h  h0  i  i  0 
h   2m
  i   
18. A spherical solid ball of volume V is made of a material of density 1 . It is falling through a
liquid of density  2  2  1  . Assume that the liquid applies a viscous force on the ball that is
proportional to the square of its speed v, i.e., Fviscous  kv 2  k  0  . The terminal speed of the
ball is (AIEEE-2008)
51
Vg 1 Vg  1   2  Vg  1   2  Vg 1
1) 2) 3) 4)
k k k k
SOLUTION :
mg  Fb  Fv
19. Calculate rate of flow of glycerin of density 1.25x10 3kg/m3 through the conical section of a ‘’’
horizontal pipe, if the radii of its ends are 0.1m and 0.04m and pressure drop across its length
is 10 N/m2.
SOLUTION :
 0.1  25
2
v2 A1 r12
According to equation of continuity v  A  r 2 
 0.04  4
2
1 2 2
1 2 1
and, according to Bernoulli’s equation for a horizontal tube, P1   v1  P2   v22
2 2
v22  v12  2
 P1  P2   16  103 m2 / s 2

25
but v2  v1  6.25v1
4
  6.25   12  v12  16  103 m2 / s 2

2
 
or v1  0.0205m / s .
the rate of volume flow  A1v1    0.1   0.02   6.28  10 4 m3 / s
2
dm
And the rate of mass flow is   Av .  1.25  103    6.28  10 4   0.785kg / s
dt
20. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating
vertically in water in half-submerged state. If  c is the relative density of the material of the
shell with respect to water, then the correct statement is that the shell is
A) more than half-filled if  c is less than 0.5 B) more than half-filled if  c is more than 1.0
C) half-filled if  c is more than 0.5 D) less than half-filled if  c is less than 0.5
SOLUTION :
Vm  Va  Vw
 w g  Vm c  w g  Vw  w g
2
Vw  Vm 1  2 c   Va
1
If  c   Vw  Va
2
52
1
if  c   Vw  Va
2
where Vw = volume occupied by water in the shell
Va = volume occupied by air in the shell
Vm = volume of the material in the shell
21. The range of water flowing out of a small hole made at a depth 10m below water surface in a
large tank is R. Find the extra force applied on water surface so that range becomes 2R (in atm,
an approximate value)
10
‘
R
SOLUTION :
V   2 gh 
1/2
Range will be twice, if efflux velocity becomes twice

or h becomes four times or 40m.
Extra pressure = 30m of water head
But 1 atm = 0.7 13.6m of water = 10.336m of water,,
30m of water=3.0 atm
22. Equal volumes of two immiscible liquids of densities  and 2  are filled in a vessel as shown in
h 3h
figure. Two small holes are punched at depth and from the surface of lighter liquid. If
2 2
v1
v1 and v2 are the velocities of efflux at these two holes, then v is
2
h  V1
h V2
1 1 1 1
A) B) C) D)
2 2 2 4 2
53
SOLUTION :
h 1
P0   g  P0   v12
2 2
h h 1
P0   g   2   g  P0   2   v22
2 2 2
v1 gh 1
 
v2 2 gh 2
23. A slit is cut at the bottom, along the right bottom edge of a rectangular tank. The slit is closed by
a wooden wedge of mass m and apex angle  as shown in figure. The vertical plane surface of
the wedge is in contact with the right vertical wall of the container. Coefficient of static friction
between these two surfaces is  . To what maximum height, can water be filled in the tank
without leakage from the slit? The width of tank is b and density of water is  .
‘
h
 ‘’’
2m 4m 2m 2m cos 
A)  b  tan     B)  b  tan     C)  b  sin    cos   D)  b  tan    cos  
SOLUTION :
mg   N  Fpressure
 h  1 
 mg     g  bh     g   h tan   h 
 2  2 
 gh 2b  gh 2 tan 
 mg  
2 2
bh 2 
m  tan    
2
2m
h 
 tan     b
54
24. A vessel with a symmetrical hole in its bottom is fastened on a cart. The mass of the vessel and
the cart is 1.5kg. With what force F(in 10 2 N) should the cart be pulled so that the maximum
amount of water remains in the vessel. The dimensions of the vessel are as shown in figure.
Given that b=50cm, c=10cm, area of base A=40 cm2 , L=20cm, g =10 m / s 2 .
b
‘ C
F
SOLUTION :
As the cart is drawn by a force F, the water in the vessel takes up a slant position rising upward at the
back of the vessel.
To prevent water from flowing out of the hole H, the acceleration of the vessel should have such a
value that it occupies a face area DBH and a width of vessel given by A/L.
D
L
ma b
mg 
B H F
1
Area of DBH  bc
2
1 A
volume of the liquid retained  bc 
2 2
bcA
Mass of the cart and water  M 
2L
ma
tan   ;
mg
b
a  g tan   g 
c
 bcA  gb
Required force is  M    1.5  0.5  50  2.0  50  100N
 2L  c
55
25. A thin V-shaped glass tube is fixed in the vertical plane as shown. Initially, the left part of the
tube contains a column of water of length d = 2 m. A valve at the bottom of the tube prevents
the water from moving to right part. At some time, the valve is quickly opened. Neglecting
friction, find the time (in seconds) it takes for the water to move completely into the right part of
the tube. (Take g=  2 m/s2)
45 45
Fixed horizontal surface

SOLUTION :
The water of density  will execute SHM inside the fixed v-shape glass tube with time period
m
T  2 2  Ag cos 45
where m = mass of water
A = cross-sectional area of tube.
d
T  2 2g
Hence the required time is half the time period of oscillation.
26. A square box of water has a small hole located in one of the bottom corners. When the box is
full and sitting on a level surface, complete opening of the hole results in a flow of water with a
speed v0, as shown. When the box is still half empty, it is tiled by 450 so that the hole is ta the
lowest point. Now the water will flow out with a speed of
v0
v0 v
v0 v0 v0
1) v0 2) 3) 4) 4
2 2 2
SOLUTION :
h V0
V0  2 gh  V '  2 g  4
2 2
27. A cylindrical vessel contains a liquid of density  upto a height h. The liquid is closed by a
piston of mass m and area of cross-section A. There is a small hole at the bottom of the vessel.
Find the speed v with which the liquid comes out of the hole.
56
m, A
1 2
P0
SOLUTION :
Applying Bernoulli’s theorem at 1 and 2;
difference in pressure energy between 1 and 2 = difference in kinetic energy between 1 and 2
 mg  1 2
 P0   gh    0   p0    v
 A  2
mg 1
p0   gh   p0   v 2
A 2
1 2mg  mg 
 gh   v 2  v  2 gh   2  gh 
2 A   A 
28. Two balls of same size but different masses m1 and m2  m2  m1  are attached to the two ends
of a thin light thread and dropped from a certain height. It is known that the viscous drag of air
depends on the size and velocities of the balls. Other than the gravitational pull from the earth
and the viscous drag, the buoyant force from air also act on the balls. The buoyant force on a
ball equals to the weight of air displaced by the ball. After sufficiently long time from the instant
the balls were dropped both of them acquire uniform velocity known as terminal velocity. When
the balls have acquired terminal velocity, the tension in the thread is
A) Zero B)  m2  m1  g C) 0.5  m2  m1  g * D) 0.5  m2  m1  g
SOLUTION :
For vertical translational equilibrium, use freebody diagram for single and two body system seperately.
29. A gas flows with a velocity v along a pipe of cross sectional area ‘s’ and bent an angle of 90 0 at
point A. What force does the gas exert on the pipe at ‘A’. If it’s density is  ?
2SV SV 2 
1) 2) 2SV 2
3) 3 4) 3SV 2 
 2
SOLUTION :
Take x-axis along the flow and y-axis perpendicular to it
  
Vinitial  V i,V final  V j; V  V 2  V 2  2V
v v l
F m m    S  2V  2 SV 2 
t t t
57
30. A pump draws water from a reservoir and sends it through a horizontal pipe with speed v. Find
the relation between power of the pump and velocity of liquid.
SOLUTION :
From work - energy theorem
KE  imparted to water  KE volume of water
P  
time volumeof water time
1 
   v 2   Av  or P  v 3
2 
31. A cylinder stands vertical in two immiscible liquids of densities  and 2  as shown. Find the
difference in pressure at point A and B:
Air
h
2h A 
h 2
B
A) 2  gh B) 3 gh C) 4  gh D) None
SOLUTION :
Difference in pressure at A and B = 4  gh   gh  3 gh
Comprehension :
An open rectangular tank of dimensions 6m x 5m x 4m contains water upto a height of 2m. The
vessel is accelerated horizontally with an acceleration of a m / s 2 as shown. Take
 water  103 kg / m 3 , g  10m / s 2 , atmospheric pressure = 105 N / m 2 . Base on above information
answer the following questions :
a
4m
2m
1m
6m
58
32. Determine the maximum value of a so that no water comes out from tank.
2g g
A) g B) C) D) 2g
3 3
SOLUTION :
1  a 
The angle made by free surface of liquid with horizontal is   tan  
g
4m
y
6m
For no water to spill out, limiting case is shown in the fig.

As volume of water inside the beaker remains the same,
1
so 2  6  5   4  y   6  5
2
 y0
4 y 2 a
tan    
6 3 g
2g
 a
3
33. Determine the height to which the water should be filled in the tank so that when a  5m / s 2 , no
water comes out from the tank
A) 2m B) 3m C) 2.5m D) 3.5m
SOLUTION :
For no water to come out, the situation is as shown in fig
a
4m
y
6m
4 y a
tan   
6 g
59
4 y 5
 
6 10
 y  1m
1
So, volume of water inside the tank is, 4  y 6 5
2
Let h be the initial height, then
1
5  6  h    4  1  6  5 ;
2
h  5 / 2  2.5m
34. Instead of open top if the vessel is closed, then absolute pressure at point A would be
20
[ Take a  m / s 2 and initially height of water in tank is 2m]
3
A) 1.33 105 N / m 2 B) 1.0  105 N / m 2 C) 3.33  10 45 N / m 2 D) None
SOLUTION :
20 2 g
For a   the situation would be as shown in fig.
3 3
Q A P
pressure P  p0
Pressure at A, PA  p0   a  5
20
 105  103  5
3
 1.33  105 N / m 2
35. A liquid of density   0 1   y  is stored in a container where y is the distance from the the
2 1
liquid surface and   m . A small hole is made at the bottom of the container. Find nearest
3
integer of velocity of efflux (in m/s) when the liquid height is 1m. Assume flow is laminar .
 g  10 m / s 
2
SOLUTION :
dp
   g   0 (1   y) g
dy
  dp    0 (1   y) g
after applying Bernoulli’s principle,

the velocity is 4 m/s.
60
36. A light cylindrical vessel is kept on a horizontal surface. It’s base are is A. A hole of cross-
sectional area ‘a’ is made just at it’s bottom side. The minimum coefficient of friction necessary
for sliding of vessel due to the impact force of the emerging liquid
a 2a
1) Varying 2) 3) 4) None
A A
SOLUTION :
dm
 
2
F v  va  v  v 2 a   2 gy a   2a  gy
dt
Force of friction f=F=2a  gy
2a
  N    A gy    2a  gy;  
A
37. There are two identical small holes, each of area of cross-section ‘a’ on the opposite sides of a
tank containing a liquid of density  . The difference in height between the holes is ‘h’. Tank is
resting on a smooth horizontal surface. Find the horizontal force which will have to be applied on
the tank to keep it in equilibrium.
SOLUTION :
F  F1  F2   av12   av22
V2 h1
h2
h
V1
F2 F1
  a  2 gh1    a  2 gh2   2  ag  h1  h2   2 agh

38. A uniform vertical cylinder is released from rest with its lower end just touching the liquid
surface of a deep lake. Calculate the maximum displacement of the cylinder in meters. Take
 1
  4m and 
 2
 l
61
SOLUTION :
Al  g  Ax  g  Al  a
 gx
ag ;
l
v x
 gx
 v.dv   g 
0 0

dx
v 2
 g x2
  gx 
2  2
At maximum displacement,
2  1
x   2   4  4m
 2
39. A tank is filled with two immiscible liquids of densities 2  and  each of height h. Two holes
h 3h
are made to the side wall at and from upper surface of the liquid, then find the ratio of
2 2
velocity of efflux of the liquids through the holes
SOLUTION :
According to Bernoulli’s theorem,
 v1
h
3h
2
h 2
v2
h 1
For v1 , p   g    P     v1   v1  gh
2
2  2
h 1
For v2 , P    gh    2   g    P   2   v2
2
2  2
v1 1
 v2  2 gh  v  2
2
62
40. A hose shoots water straight up to a height of 2.5 m. The opening end of the hose has an area
of 0.75cm2. What is the speed of the water as it leaves the hose? How much water will come
out in one minute?
SOLUTION :
Kinetic energy at bottom = Potential energy at the top
1 2
v  gh  v  2 gh  v  2  9.8  2.5
2
 5  9.8  49  7 m / s; v  7 m / s or 700cm/s.
The rate of flow of water = Av. So in one minute the volume of water that flows out
 Av  60   0.75  700  60  0.75  42  103
 3.15  104 cm 3  31.5 litre.
41. A wooden block is floating in a water tank. The block is pressed to its bottom. During this
process work done is equal to
A) Work done against upthrust exerted by the water
B) Work done against upthrust plus loss of gravitational potential energy of the block
C) Work done against upthrust minus loss of gravitational potential energy of the block
D) all the above
SOL UT I ON :
Initially the wooden block floats with partially immersed in water. Initially, upthrust exerted by
water is equal to weight of the block.
But when it is pressed down more water is displaced. Hence upthrust exerted by water
increases.
The force required to press down the block is equal to F  upthrust  mg .
Hence, work done by the force F will be equal to work done against (upthrust - mg).
It means, work done by the external force is equal to work done against upthrust loss of gravitational
potential energy of the block.
42. A small hole is made at the bottom of a symmetrical jar as shown in figure. A liquid is filled into
the jar upto a certain height. The rate of dissension of liquid is independent of level of the
liquid in the jar. Then the surface of jar is a surface of revolution of curve
y
1) y=kx4 2) y=kx2 3) y=kx3 4) y=kx5

SOLUTION :
63
y
Let y be the height of liquid at same instant then

 dy
(a=area of hole =constant given)
dt
 dy 
a 2 gy   r 2   
 dt 
  dy 
 ,  and g are constant
 dt 
squaring the equation, we get y  kx 4
43. A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the
top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is
completely filled with water, the quantities of water flowing out per second from both holes are
the same. Then, find the value of R. (EAMCET-2011)
SOLUTION :
Velocity of efflux at a depth h is given by v  2 gh .
Volume of water flowing out per second from both the holes are equal
 V 
 a1v1  a2v2   av 
 t 
L
L 
2
2g  y    R2 2g  4 y   R 
2
44. The fig shows a semi-cylindrical massless gate of unit length perpendicular to the plane of the
page and is pivoted at the point O holding a stationary liquid of density  . A horizontal force F
is applied at its lowest position to keep it stationary. The magnitude of the force is :
h=R
 O (Pivot)
R

F
3 9
A)  gR 2 B)  gR 2 C)  gR 2 D) 2  gR 2
2 2
64
SOLUTION :
Torque about ‘O’ due to water pressure is given by    d
where d  dF .R sin 
after integrating from 0 to 1800
we will get   4R 3  g
this should be balanced by applied torque.
  4 R 3  g  F  2 R 
 F  2  gR 2
45. In figure-I is shown a sphere of mass m and radius r resting at the bottom of a large container
filled with water. Depth of the container is h. Density of material of the sphere is the same as
that of water. Now the whole sphere is slowly pulled out of water as shown in figure-II
I II
Work done by the agent in pulling the sphere is equal to
A) mgr B) 0.5mgr C) mg(0.5r+h) D) mg(r+h)
SOLUTION :
3dr

r R
Inside
net force on inner side body's
r
cos    r  R cos   dr   R sin  .d
R
net force on immersed body’s part=0  body   water 
 dw  workdone in shifting above part through dr height  Vabove  g  dr   1

determination of Vabove :-   2 1  cos  
 1 3 
 4  3  R 
R3  
 volume related to surface area above water level is    4  
3    R3  
 3  4 
65
R3
Vabove  2 1  cos     2 
3
R3
 dw   2  2 cos    g   R  sin  d
3
R4  g R4  g
 dw   2 sin  d  2 cos  .sin  .d
3 3
R4  g
 sin 2 d  2sin d 
3
R 4  g
w 4
3
[ after integration limits of  : 0 to 1800 ]
4 3
w R  gR  mgR
3
Comprehension :
Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above
the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There
is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long
vertical tube is connected as shown.
h2
2P0 h0
h1
46. Find the height h2 of the water in the long tube above the top initially
3 p0 h0 2 p0 h0 p0 p0
A)  B)  C) h D)  2h0
g 3 g 2 g 0 2 g
SOLUTION :
2P0   h2  h0   g  p0
(since liquids at the same level have the same pressure)
P0  h2  g  h0  g ;
h2  g  P0  h0  g
P0 h0  g P0
h2    h
g g g 0
66
47. Find the speed with which water comes out of the hole.
1/2 1/ 2 ½
1 2  3  4 
A)   p0   g  h1  2h0  B)    p0   g  h1  h0    C)    p0   g  h1  h0  
1/2
D)    p0   g  h1  h0  
     
SOLUTION :
1 P
KE of the water = Pressure energy of the water at that layer mV 2  m 
2 
2P 2
V2    P   g  h1  h0  
  0
1/ 2
2 
V    P0   g  h1  h0 
 
48. Find the height of the water in the long tube abvoe the top when the water stops coming out of
the hole.
A) 2h0 B) h0 C) h2 D)  h1
SOLUTION :
we know 2P0   g  h1  h0   P0   gX
P0
X   h1  h0   h2  h1
g
i.e, X is h1 meter below the top or X is - h1 above the top.
49. A rod is of length 6 m and of specific gravity   25 / 36 . One end of the rod is tied to a 5 m long
light rope which in turn is tied to the floor of a pool 10 m deep as shown. Find the length of the
part of rod in metres which is out of water.
10m
5m
SOLUTION :
Taking moments about A
 6  x  A w g / 2  18 A r g
2
Solving, we get x  1 m
67
68
VISCOSITY :
 In case of steady flow of a fluid when a layer of fluid slips or tends to slip on adjacent layers in contact,
the two layer exert tangential force on each other which tries to destroy the relative motion between
them.
 The property of a fluid due to which it opposes the relative motion between its different layers is called
viscosity (or fluid friction or internal friction) and the force between the layers opposing the relative
motion is called viscous force.
Viscosity in gases due to collisions between gas molecules and transfer of momentum.
Viscosity in liquids is due to cohesive force between molecules in succesive layers
Waves in sea water subside due to viscosity.
Rain drops fall to the ground with terminal velocity due to viscosity.
It is only due to viscosity, a liquid flow becomes orderly.
Viscous Force (Newton’s Formula):

v + dv
M N
x + dx
C D
v
x
A B
Rest
Consider the two layers CD and MN of the liquid at distances x and x + dx from the fixed
dv
surface AB, having the velocities v and v + dv respectively. Then dx
denotes the rate of change of
velocity with distance and is known as velocity gradient.
According to Newton’s hypothesis, the tangential force F acting on a plane parallel layer is
dv
proportional to the area of the plane A and the velocity gradient dx
in a direction normal to the
layer, i.e.,
dv
F A and F
dx
dv
 F  A dx
dv
or F  A
dx
Where  is a constant called the coefficient of viscosity.

dv
The viscous force acting between two adjacent layers of a liquid is given by F  A
dx
The viscous force acts tangential to the liquid layer
Negative sign indicates the force is opposite in direction to the relative velocity of flow of the liquid.
69
dv
velocity gradient = . It is a vector
dx
 is coefficient of viscosity. It may be defined as the tangential force per unit area required to maintain
unit velocity gradient (or) it is the ratio between tangential stress and velocity gradient.
It is also called coefficient of dynamic viscosity.
Dimensional formula of   M 1 L1T 1
The SI unit of  is Pa-s. Its CGS unit is poise.
1pa-s=10 poise.
Note : The value of  changes from liquid to liquid and for ideal liquid  =0
Coefficient of kinematic viscosity:
The ratio between the coefficient of viscosity and density of the liquid is called Coefficient of Kinematic
viscosity.

Coefficient of Kinematic viscosity =

Its S.I. unit is m s
2 1
Its practical unit is stoke. 1 stoke = 10-4m2s-1.

Its dimensional formula is  L2T 1  , same as that of areal velocity..
Effect of temperature:
In the case of liquids, coefficient of viscosity decreases with increase of temperature as the cohesive
forces decrease with increase of temperature.
In the case of gases, coefficient of viscosity increases with increase of temperature because number of
collisions between the molecules of the gas increases.
Effect of pressure:
For liquids the value of  increases with increase of pressure.
Above 330C the viscosity of water increases with pressure, and that below this temperature, initially
the pressure effect is negative
For gases, value of  increases with increase of pressure at low pressure. But at high pressure,  is
independent of pressure.
The machine parts get jammed in winter, because the viscosity oil (used as lubricants in machine parts)
increases due to fall in temperature
A viscous fluid tends to cling to a solid in contact with it. That is why dust particles cling to a fan blade
even when it is rotating rapidly.
Poiseuille's equation:
According to Poiseuille’s Formula the rate of flow of a liquid through a horizontal capillary tube is
 Pr 4
V
8 l
Where V is volume of the liquid flowing out per second,
‘p’ is the pressure difference across the capillary pipe,
‘r’ is the radius of the pipe,
‘ l ’ is the length of the capillary pipe,
' ' is the coefficient of viscosity of the liquid,
 / 8 is a proportionality constant.
70
Poiseuille's equation is valid for
a) flow through a horizontal capillary tube.
b) steady and laminar flow
c) the liquid the contact with the walls of the capillary tube must be at rest.
d) the pressure at any cross section must be same
Arrangement of capillary tubes:
a) Capillary tubes in series :
 Pr 4 P P
V  or V  8l 
From Poiseuille’s equation, 8 l   R
 4
r 
Total volume of the liquid flowing out per sec remains constant
8 l
Where R  is called fluid resistance.
 r4
l1 l2
When two capillaries are connected in series across a constant pressure difference P,
the fluid resistance R=R1+R2.
8 l1 8 l2 8  l1 l2 
R     
 r14  r24   r14 r24 
Volume of liquid flowing per second is same through both capillaries
P P P
V   
R1  R2 R   
8  14  42 
 r1 r2 
Here P  P1  P2   P1  P0    P0  P2 
 P1  P0  and  P0  P2  are pressure differences across individual capillaries.

P1  P0 R1 1 r24
  
P0  P2 R2  2 r14
b) Capillary tubes in parallel :
l1
l2
When two capillary tubes are connected in parallel across constant pressure difference P,
1 1 1
then fluid resistance R is given by R  R  R
1 2
8 l1 8 l2
 R1  and R2   r 4
 r14 2
71
1   r14 r24 
   
R 8  l1 l2 
Here volume of the fluid flowing per second in each tube is different but pressure difference
P   P1  P2  is same.
Total volume flowing per second is V=V1+V2
P P  P  r14 r24 
V   V    
R1 R2 or 8  l1 l2 
4
V1 R2 l2  r1 
   
V2 R1 l1  r2 
Stoke's law - Terminal velocity:
When a spherical body is dropped in a fluid, the fluid layer in contact with the body is dragged
with the body. The remaining layers of the fluid excerts viscous force on the body to oppose its motion.
When the sum of the viscous force and upthrust of the liquid on the body is equal to its weight,
the body then begins to fall with a constant velocity, known as terminal velocity
According to stoke,
i) the viscous force (F)  6 r vt .
Where
 =coeff. of viscosity of the fluid,
r=radius of the body
vt terminal velocity..
4 3
ii) Upward of the body(W) = mg=(volume of sphere x density)  g   r  g
3
iii) Upward thrust(T) = weight of the fluid displaced
=(volume of the fluid displaced x density of the fluid) g
4
  r3 g
3
When the body attains terminal velocity the net force acting on the body is zero
W-R-F=0;
F=W-T
2r 2 g     
 vt 
9
 vt  (  )
If    , body will rise instead of falling.
For example air bubble in water
1
 vt   ,
greater the viscosity, smaller is the terminal velocity.
For example a lead shot attains less terminal velocity in glycerine than in water.
72
2
vt1r 
 vt r   1 
2
vt2  r2 
When ‘n’ identical droplets are falling down with their terminal velocity ‘v’ are combined to form a big
drop, then terminal velocity of big drop is
vbig R2
 2  vbig  n 2/3v  R  n1/3 r 
v r
 Two spherical bodies having masses m1, m2 respectively falling in viscous medium the ratio of terminal
velocities is
2/ 3
v1  m1 
 
v2  m2 
73
:: PROBLEMS ::
1. A boat of area 10 m floating on the surface of a river is made to move horizontally with a
2
speed of 2m/s by applying a tangential force. If the river is 1m deep and the water in contact
with the bed is stationary, find the tangential force need to keep the boat moving with constant
speed. (coefficient of viscosity of water = 10-2 poise)
SOLUTION :
dv 2  0

Velocity gradient =  2 s 1
dx 1  0
From, Newton’s law of viscous force,
dv
Fv   A  10 2  101  10  2   0.02 N
dx
2. A spherical ball falls through viscous medium with terminal velocity v. If this ball is replaced by
another ball of the same mass but half the radius, then the terminal velocity will be (neglect the
effect of buoyancy.)
A) v B) 2v C) 4v D) 8v
SOLUTION :
4
m   r3 g
3
v0 r 2  ;
r2
.  8    2r 2 
v0 
,
4
3. A spherical solid ball of volume V is made of a material of density 1 . It is falling through a
liquid of density  2   2  1  . Assume that the liquid applies a viscous force on the ball that is
proportional to the square of its speed v , i.e., Fviscous  kv 2  k  0  . The terminal speed of the
ball is:
Vg   1   2  Vg  1 Vg 1 Vg   1   2 
A) B) C) D)
k k k k
SOLUTION :
The force acting on the ball are gravity force, buoyancy force and viscous force.
When ball acquires terminal speed, it is in dynamic equilibrium,
let terminal speed of ball is vT .
So, v2 g  kvT 2  V 1 g ;
V  1   2  g
vT 
k
4. Water flows through a capillary tube of radius ‘r’ and length at a rate of 40 ml per second, when
connected to a pressure difference of ‘h’ cm of water. Another tube of the some length but
radiud . r/2 is connected in series with this tube and the combination is connected to the same
pressure heaD) [density of water is  ]
74
 gh 16
A) The pressure difference accross each tube is p1  and P2  gh
17 17
 gh 17
B) The pressure difference accross each tube is p1  and P2  gh
16 16
40
C) The rate of flow of water through the combination is c.c / sec.
17
17
D) The rate of flow of water through the combination is c.c / sec.
40
KEY : A , C
SOLUTION :
According to Pissiulle’s equation.
P
Rate of flow = 8 L
 
 4
 r 
Making an analogy with current flow,
R1
R2
P2
P1
8 L
Let R1  ;
 r4
8 L
R2  4
 16 R
g
 
2
R 1
P1  P2  P   gh and P1 / P2  R  16
1
P P 40
Also Rate of flow = P  P  17 R  17  mL/s
1 2 1
5. A 16cm3 of water flows per second through a capillary tube of radius r cm and of length 1cm,
when connected to a pressure head of h cm of water. If a tube of the same length and radius r/
s is connected to the same pressure head, find the mass of water flowing per minute through
the tube.
SOLUTION :
Pr 4 V P r4 l
V  2  2  24  1
l V1 P1 r1 l2
r / 2
4 4
l 1 1 16
     ;V 2   1cm 3 / s
r4 l 2 16 16
Volume of water flowing per minute = 1  60  60cm3 / min
 Mass of water flowing per minute  V  d  60  1  60 gram / min.
75
6. Water is flowing in a river. If the velocity of a layer at a distance of 10 cm from the bottom is 20
cm/s. Find the velocity of layer at a height of 40 cm from the bottom
A) 10 m / s B) 20 m / s C) 30 m / s D) 80 m / s
SOLUTION :
dv
F   A ;
dx
dv v
F or F
dx x
v v
 or  x
F
x F
but F is constant
 x v
7. Water flows in a streamline manner through a capillary tube of radius a. The pressure difference
being P and the rate of flow is Q. If the radius is reduced to a/2 and the pressure difference is
increased to 2P, then find the rate of flow
SOLUTION :
P
Rate of flow Q 
8 l /  r 4 
4
Q1 P  r 
 Q  P r 4    1  1 
Q2 P2  r2 
4
Q1 P  a  Q
     Q2 
Q2 2P  a / 2  8
Rate of flow will become 1/8 times
8. If the terminal speed of a sphere of gold (density = 19.5 kg/m 3) is 0.2 m/s in a viscous liquid
(density 1.5 kg/m3), find the terminal speed of a sphere of silver (density = 10.5 kg/m3) of the
same size in the same liquid.
A) 0.2 m/s B) 0.4 m/s C) 0.133 m/s D) 0.1m/s
SOLUTION :
2r 2
Terminal velocity of spherical body in a viscous liquid is given by: vT      g
9
Where r  radius of the sphere,
 =density of the sphere,
 =coefficient of viscosity,,
 = density of liquid.
 v    
vg  g   19.5  1.5
  2
vs  s   10.5  1.5
vg 0.2
 vs    0.1 m / s
2 2
76
9. Capillary tubes of lengths l and 2l are connected in series. Their radii are r and 2r respectively.
If stream line flow is maintained and pressure difference across first and second capillary
P1
tubes are P1 and P2 respectively, then find the ratio P
2
SOLUTION :
Equating the rate of flow of liquid
 r 4P
Here Q1  Q2 and Q 
8l
  r  P1   2r  P2
4 4
P
   1 8
8 l 8  2l  P2
10. A liquid of density 900 kg / m3 is fillled in a cylindrical tank of upper radius 0.9m and lower
radius 0.3m A capillary tube of length l is attached at the bottom of the tank as shown in fig. The
capillary has outer radius 0.002m and inner radius A) When pressure P is applied at the top of
the tank volume flow rate of liquid is 8  106 m3 / s and if capillary tube is detached, the liquid
comes out from the tank with a velocity 10m / s. Then the coefficient viscosity of liquid is
(  a 2  106 m 2 a 2 / l  2  106 m.)
0.9m
0.3m
A)   1.25  10 3 N  s / m 2 . B)   2.50  10 3 N  s / m2
C)   5.00 10 3 N  s / m2 D)   7.25 10 3 N  s / m2
SOLUTION :
Applying Bemoullie’s equation P   gh 

1
2
 
 v2 2  v12 ;
but v1   A2v2 / A1 
dV  a 4 P
P  P   gh;  
dt 8l
77
11. Three capillary tubes of same radius 1 cm but of lengths 1m, 2m and 3m are fitted horizontally
to the bottom of a long vessel containing a liquid at constant pressure and flowing through
these. What is the length of a single tube which can replace the three capillaries.
SOLUTION :
 Pr 4  Pr 4  Pr 4
. V1  ,V2  , V3 
8 l1 8 l 2 8 l3
and V   Pr
4
8 l
1 1 1 1
Now V  V1  V2  V3    
 1  2  3
Substituting the values, we get
l1l2l3 1 2  3 6
l   m
l1l2  l1l3  l2l3 1 2  1  3  2  3 11
12. A cylindrical vessel of area of cross-section A and filled with liquid to a height of h1 has a
capillary tube of length l and radius r protuding horizontally at its bottom. If the viscosity of
liquid is  and density  . Find the time in which the level of water in the vessel falls to h2.
8 lA 8 lA A 8 lA
h1
A)  gr 4 ln h
2
B)  gr 4 C) g h1  h2  h2
D)  gr 4 ln h
1
SOLUTION :
Let h be the height of water level in the vessel at instant t which decreases by dh in time dt.
 dh 
 Rate of flow of water through the capillary tube, V   A  dt  ....................... (1)
 
 Pr 4
Further, the rate of flow from Poiseuille formula V  ....................... (2)
8 l
The hydrostatic pressure at depth h is P   gh
From eqns (1) and (2), we have
dh  hr 4
A 
dt 8 l
8 lA dh
dt   ;
 r 4 h
h
8 lA 2 dh
 gr 4 h1 h
t
13. Two identical drops of water are falling through air with a steady speed of ‘v’ each. If the drops
coalesce to form a single drop, what is the new terminal velocity? (EAMCET-2013)
SOLUTION :
4 4 4
From conservation of mass  R3     r 3     r 3  
3 3 3
 1
or R   2 3  r and vt r 2 (stokes law)
 
78
v1 R 2
  22 / 3
v r2
 v1  22 / 3 v .
14. A spherical steel ball stretched at the top of a long column of glycerin of length l falls through
a distacne l / 2 with accelerated motion and the remaining distance l / 2 with uniform velocity..
Let t1 and t2 denote the times taken to cover the first and second half and w 1 and w2 are the
work done against gravity in the two halves, then compare times and work done.
SOLUTION :
Average velocity in first half of the distance <v,
while in the second half the average velocity is v.
Therefore, t1>t2.
The work done against gravity in both halves is mg l / 2
 t1  t2  w1  w2
15. A volume V of a viscous liquid flows per unit time due to a pressure head P along a pipe of
diameter d and length l . Instead of this pipe, a set of four pipes each of diameter d / 2 and
length 2l is connected to the same pressure head P . Now the volume of liquid flowing per
unit time is:
A) V /16 B) V / 8 C) V / 4 D) V
SOLUTION :
 Pr 4
V ;
8l
r4
V
l
4
V1  r1  l2
     32 ;
V2  r2  l1
V
V2 
32
For 4 pipes,
1 V V
V2  4V2  4  
32 8
16. A small steel ball falls through a syrup at a constant speed of 1.0 m/s. If the steel ball is pulled
upwards with a force equal to twice its effective weight, how fast will it move upward?
SOLUTION :
6 rv  mg1 (effective weight)
Let We = effective weight and 6 r  K
In equilibrium, we=kv...(1)
Again, 2we-we=kv1.....(2)
From Eqs (1) and (2) v1  v  1.0 m / s
79
KV 2We
V V1
We
We
KV 1
17. A marble of mass x and diameter 2r is gently released in a tall cylinder containing honey. If the
mar ble displaces mass. y (< x) of the liquid, the terminal velocity is proportional to
x y x y
A) x  y b. x  y C) D)
r r
SOLUTION :
2 2     g
1
v0  r ;
9 
4
x   r 3   or 
3
x y
 3 ;
 1 3 ;
r r
x y
v0
r
18. Between a plate of area 100 cm and another plate of area 100 m2 there is a 1 mm, thick layer of
2
water, if the coefficient of viscosity of water is 0.01 poise, then the force required to move the
smaller plate with a velocity 10 cm/s with reference to large plate is.
A) 100 dyn B) 104 dyn C) 106 dyn D) 109 dyn
SOLUTION :
dv
F  A
dx
where A=100cm2;
dv 10cm / s
  100 s 1
dx 1m / m
19. A tube of length l and radius R carries a steady flow of fluid whose density is  and viscos ity
 . The velocity v of flow is given by v  v0 (1  r 2 / R 2 ), Where r is the distance of flowing fluid
from the axis.
A) The volume of fluid, flowing across the section of the tube, in unit time is 2 v0 ( R 2 / 4)
B) The kinetic energy of the fluid within the volume of the tube is K .E.   lv0 2 ( R 2 / 6)
C) The frictional force exerted on the tube by the fluid is F  4 kv0
4 lv0
D) The pressure difference at the ends of tube is P 
R2
80
KEY : ALL
SOLUTION :
The volume of fluid flowing through this section per second dv   2 rdr  v0 1  r / R 
2 2
 2 r dr  v0 1  r 2 / R 2 
R
total volume V  
0
 2 v0  R 2 / 4 
(ii). The kinetic energy of the fluid within the volume element of thickness dr
K.E of fluid within the tube is
1
  2 l   v0  1  r 2 / R 2  r dr
2 R
2 0
we get K.E   l v0  R / 6 
22
(iii). The viscous drag exerts a force on the

 dv 
tube F   A  
 dr  r  R
 dv 
Here    v0  2 r / R r  R  2v0 / R  F  4 lv0
2
 dr r  R
(iv) P  P2  P1  P
where P1  0 and P2  P
force  F 
4 l v0
P 
area  R 2 R2
20. A ball moves successively through three liquids, at rest as shown, of densities  1 ,  2 and  3 nd
viscosity coefficient 1 ,2 and 3 and respectively with the same (terminal) velocity. Then
h1
h2
h3
          
B)      C)    D)         
1 2 3 1 3 2 1 1 2 2 1
A) 3   2  1
1 2 3 3 2 3 1 1 3 3 1
KEY : C, D
SOLUTION :
  1    2    3
 
1 2 3
 1   2   3  1   2  3
1    1 3    3
Also,      3  1      1   C 
3 2 2
Eliminating  gives (D)
81
21. Two capillary tubes of same radius r but of lengths l1 and l2 are fitted in parallel to the bottom
of a vessel. The pressure head is P . What should be the length of a single tube of same radius
that can replace the two tubes so that the rate of flow is same as before ?
1 1 l1l2 1
A) l1  l 2 B) l  l C) l  l D) l  l
1 2 1 2 1 2
SOLUTION :
8 L
Fluid resistance R 
 r4
1 1 1
in parallel, R  R  R
1 2
 r4  r4  r4
or 8 l  8 l  8 l
eq 1 2
l1l2
leq 
l1  l2
22. When water flows at a rate Q through a tube of radius r placed horizontally, a pressure difference
p develops across the ends of the tube. If the radius of the tube is doubled and the rate of flow
halved, the pressure difference will be
A) 8 p B) p C) p / 8 D) p / 32
SOLUTION :
 pr 4
Poiseuille’s formula gives the quantity of liquid flowing through a ckapillary, Q 
8 l
8 nl
i.e., p  Q. ;
 r4
Q
if Q '  , r '  2r
2
p 8 Q nl 8 Q.nl 1
 .  
n  2  2r  4 p r 4 32  n  32 
p
i.e., pressure p ' 
32
23. L, L / 2 and L / 3 are connected in series. Their radii aree r , r / 2 and r / 3 respectively. Then,
if stream-line flow is to be maintained and the pressure across the first capillary is P , then:
A) the pressure difference across the ends of second capillary is 8P
B) the pressure difference across the third capillary is 43P
C) the pressure difference across the ends of second capillary is 16P
D) the pressure difference across the third capillary is 59P
SOLUTION :
82
dQ  Pr 4

dt 8 L
As capillaries are joined in series, so  dQ / dt  will be same for each capillary..
 Pr 4  P '  r / 2   P "  r / 3
4 4
Hence,  
8 L 8  L / 2  8  L / 3
So pressure difference across the ends of 2nd capillary P '  8 p
across the ends of 3rd capillary P "  27 p
83
Surface Tension
The attractive forces between molecules of same substance are called cohesive forces.
The attractive forces between molecules of different substances are called adhesive forces.
The maximum distance upto which the cohesive force between two molecules exists is called the
molecular range and is of the order of 10-9m.
An imaginary sphere drawn around a molecule with molecular range as radius is called the sphere of
influence of that molecule.
The force per unit length normal to any imaginary line drawn on the free surface of a liquid is known as
surface Tension.
F
T ;
l
SI Unit : N/m,
CGS Unit : dyne/cm.
Dimensional formula : ML0T-2
Surface tension is due to cohesive force between the molecules of a liquid.
Surface tension is a molecular phenomenon.
Surface tension is independent of surface area.
It decreases with increase of temperature.
Applications of force due to surface tension :
Force required to pull a wire of length ' l ' from the surface of water of surface tension T is
F=2  T
Force required to pull a circular ring of radius R from the surface of water of surface tension T is
F  2 rT
Force required to pull a thin circular ring of radius r from the surface of water of surface tension T is
F= 4 r T
84
r
Force required to pull a rectangular plate of length ' l ' and breadth 'b' from the surface of water of
surface tension T is
F  2l  b T
Force required to pull a circular disc of radius R with hole of radius r from the surface of water of
surface tension T is
F  2 R  r T
I
Between two glass plates a water drop is squeezed to form a thin film of thickness (d) and surface area
(A). The force required to seperate the two plates is,
F = 2TA/d.
Surface Energy:
Work done to increase surface area of a film by one unit is known as surface energy. It is numerically
equal to surface tension. (or) The additional potential energy possessed due to increase in
surface area by one unit is called surface energy
W
S ;
A
SI Unit : J/m2;
D.F=MT-2
85
Applications of Surface Energy :
Work done in forming a liquid drop is,
W = Change in surface Area x Surface tension
W = AT  4 r 2T .
Work done in increasing the size of liquid drop from radius r1 to r2 is
W = 4  T(r22 - r12)
Work done in blowing a soap bubble of radius r is
W=8  r2 T. (Soap bubble has two free surfaces)
Work done in increasing the size of a soap bubble from radius r1 to r2 is,
W = 8  T (r22 - r12)
Work done in forming a circular liquid film of radius 'r' is,
w = 2  r2 T..
Work done in increasing the area of circular soap film from radius r1 to r2 is
W = 2  T (r22 - r12).
When a big liquid drop splits into 'n' identical droplets, then
a) Surface area increases,

work is done on the system against surface tension and potential energy increase.
b) Energy is absorbed by the system.
c) Temperature of each droplet decreases.
d) As mass is constant
4 4 
M  nm   R 3   n   r 3    R  n1/ 3r
3 3 
Increase in surface area FA  n 4 r 2  4 R 2
Workdone = T A  T 4  nr  R 
2 2
 13 
W  T 4 R  n  1
2
 
When ‘n’ small droplets each of radius ‘r’ are merged to form a big drop of radius ‘R’, then
r
a) Work is done by surface tension and total surface area decreases,

b) Energy is released by the system.
c) Temperature of big droplet increases.
d) In this case workdone by the system is
86
W  4 r 2T  n1/ 3  1
W  4 r 2T  n  n 2/ 3 
4 3
If V   R = volume of big drop
3
3VT 1/3  n1/3 1 
W
R
 n  1  3VT  R  R
 
1 1 
W  3VT   
r R
87
::PROBLEMS::
1. A small piece of wire of length 4 cm is floating on the surface of water. If a force of 560 dynes
in excess of its apparent weight is required to pull it up from the surface, find the surface
tension of water.
SOLUTION :
Length of wire l  4cm . Contact length of solid with liquid surface L  2l  8cm .
F 560
Surface tension T  T   70dyne / cm
L 8
Twater  70 dyne cm 1  0.07 Nm 1
2. An annular metal ring of inner radius 7cm and outer radius 14cm and negligible weight is
f loating on the sur f ace of a liquid. I f sur f ace tension of liquid is 0.08Nm -1, calculate the force
required to detach it from liquid surface.
SOLUTION :
The contact length of annular ring with liquid surface
22
L  2 r1  2 r2  L  2   14  7  cm
7
L  132cm  1.32m .
The force required.
F=T(L)=0.08x1.32=0.1056N
3. A wire is bent in the form of a ‘U’-shape and a slider of negligible mass is connecting the two
vertical sides of the U-shape. This arrangement is dipped in a soap solution and lifted, a thin
soap film is formed in the frame. It supports a weight of 2.0 x 10-2N. If the length of the slider
is 40cm, what is the surface tension of the film?
SOLUTION :
W=2.0 x 10-2N
l =40cm = 0.4 m
Upward force due to surface tension = T  2l
In equilibrium, W = T  2l
W 2  10 2
T    2.5  10 2 Nm 1 .
2l 2  0.4
4. When a wire of length l  l  r  and cross sectional radius r is kept floating on surface of a
liquid. Maximum radius of wire such that it may not sink is
SOLUTION :
Weight = maximum force of Surface Tension
Mg  T  2l     r 2  lg  T  2l 
2T
rmax  
 g ( is the density of the wire)
5. If the surface tension of soap solution is 35 dynes/cm, calculate the work done to form an air
bubble of diameter 14mm with that solution.
SOLUTION :
Surface tension T=35 dynes cm-1=0.035Nm-1
88
14mm
Radius of the bubble r   7 mm  7103 m
2
22
W  A  T  8 r 2T  8   49  106  0.035
7
W  4.312  105 J
6. A soap bubble is blown to a radius of 3cm. If it is to be further blown to a radius of 4cm what is
the work done? (Surface tension of soap solution = 3.06 x 10 -2 Nm-1)
SOLUTION :
Initial radius of soap bubble R1=3cm=3x10-2m
Final radius of sopa bubble R2=4cm=4x10-2m
Work done in blowing soap bubble from radius R1 to R2 is
 W  8  R22  R12  T
22
 8  3.06  102 16  9   104
7
 176  3.06  106 J  539.6  10 6 J
7. A water drop of diameter 2mm is split up into 109 identical water drops. Calculate the work
done in this process. (The surface tension of water is 7.3x10-2Nm-1).
SOLUTION :
Let a water drop of radius R be split up into 109 identical water drops each of radius r.
D 2
R   1mm  1  103 m
2 2
No. droplets n=109; T=7.3x10-2Nm-1
W  4 R 2T  n1/ 3  1
 4 10 3   7.3  102 109   1  9.17  10 4 J

2 1/ 3
 
8. 1000 drops of a liquid each of diameter 4mm coalesce to form a single large drop. If surface
tension of liquid is 35 dyne cm-1. Calculate the energy evolved by the system in the process.
SOLUTION :
No. of drops n=1000  n1/3  10; n 2/ 3  100
Surface tension of liquid T=35 dyne cm-1
Radius of each small drop r=2mm=0.2cm
Energy evolved in merging W  4 r T  n  n 
2 2/3
22
  2  10 1  351000  100
2
 W  4
7
88  35  4  102
W  900
7
 15840ergs.  1.58  103 J
89
9. If n drops of a liquid, each with surface energy E, join to from a single drop, then
A) Some energy will be released in the process
B) Some energy wil be released in the process
C) the energy released or ansorbed wil be E  n  n 
2/3
D) the energy released or absorbed will be nE  2  1

/3
KEY : A , C
SOLUTION :
Let S = surface tension
= surface energy per unit drop
r = radius of each small drop
R radius of a single drop
4 4
n   r 3   R3
3 3
R  rn1/ 3
Initial surface energy Ei  n  4 r 2  S  nE
Final surface energy, E f  4 R 2 S  4 r 2 n 2 / 3 S  n 2 / 3 E
Therefore, energy released  Ei  E f  E  n  n 

2/3
10. A thin liquid film formed between a U-shaped wire and a light slider supports a weight of
1.5x10-2N. The length of the slider is 30 cm and its weight negligible. The surface tension of
the liquid film is (AIEEE-2012)
Film
1) 0.0125Nm-1 2) 0.1Nm-1 3) 0.05Nm-1 4) 0.025Nm-1

SOLUTION :
At equilibrium, weight of the given block is balanced by force due to surface tension,
i.e., 2L.T=W
W 1.5 102 N
T   0.025Nm1
2L 2  0.3m
11. Work W is required to from a bubble of volume V from a given solution. What amount of work
is required to be done to form a bubble of volume 2V ?
A) W B) 2W C) 21/ 3W D) 41/ 3W
SOLUTION :
4 4 4 4
. V   R 3 ; 2V   R 3 ; 2   R 3   R 3
3 3 3 3
R'  2 R
1/
2
W '    21/ 3 R    2 2 / 3  2  4 R 2  41/ 3W
90
12. A large number of liquid drops each of radius “r” merge to form a single spherical drop of
radius “R”. If the energy released in the process is converted into the kinetic energy of the
big drop formed. Find the speed of the big drop (d is density of the liquid)?
SOLUTION :
Energy released is W  3VT    ..... 1

1 1
r R
If V is volume of big drop, M the mass of the drop and  the density then
1 1
Kinetic energy = Mv 2   V  v 2 ...  2 
2 2
As (1) = (2)
1 1  1
3VT      V  v 2
r R 2
6T  1 1 
or v  
  r R 
13. A drop of liquid of density  is floating half immersed in a liquid of density d. If  is the
surface tension, then what is the diameter of the drop of the liquid.
3 6 4 12
1) g  2  d  2) g  2  d  3) g  2  d  4) g  2  d 
SOLUTION :
In equilibrium
force due to surface tension + Force of buoyancy = Weight of the spherical liquid drop
2 4
2 rT   r 3d 2 g   r 3d1 g
3 3
T   , d1   , d 2  d
2 4
2 r   r 3dg   r 3  g
3 3
3 3
r2  r
g  2   d  or g  2  d 
12
Diameter  2r  g  2  d 
14. A large number of droplets, each of radius a , Coalesce to from a bigger drop of radius b .
Assume that the energy released in the process is converted into the kinetic energy of the drop.
The velocity of the drop is (  =surface tension,  = density)
1/ 2 1/ 2 1/ 2 1/ 2
  1 1   2  1 1   3  1 1   6  1 1 
A)      B)     C)     D)    
   a b     a b    a b     a b 
91
‘
SOLUTION :
Energy released  n  4 a  4 b  
2 2
4 3 4 3 b3
Now, 4   r   b or n  3
3 3 a
Therefore, enery released is
 b3 2 2b 
 a 3  4 a  4 b    4 b  a  1 
2
   
14 2 2 2 b 
Now,   b   v  4 b   1 
23  a 
1/ 2
 6  1 1  
or v      
   a b 
15. A drop of radius R is split under isothermal conditions into ‘n’ droplets, each of radius ‘r’, the
ratio of surface energies of big and each small drop is
SOLUTION :
 4 R T  R n r 
2 1/ 3 2
U big 2
   n 2/ 3 :1
U small  4 r T r
2 2
r 2
16. Eight spherical droplets, each of radius ‘r’ of a liquid of density '  ' and surface tension ‘r’
coalesce to form one big drop. If ‘s’ in the specific heat of the liquid. Then the rise in the temeprature
of the liquid in this process is
2T 3T 3T T
1) 2) 3) 4)
3r  s rs 2r  s rs
SOLUTION :
Q  mst  T A  mst
4
T 16 r 2    R 3  S T
3
4 3T
T 16 r 2  
 2   S T  t 
3
3 2r  s
17. Number of droplets are combined isothermally to form a big drop, the ratio of initial and final
surface energies of the system is
SOLUTION :
ui n  4 r 2  T nr 2 nr 2
1
    n 3
:1
uf  4 R 2  T R  n 13 r 
2
 
18. Two mercury drops (each of radius r) merge to form a bigger drop. The surface energy of the
bigger drop, if T is the surface tension, is (AIEEE-2011)
1) 2  r T
5 /3 2 2) 4 r T
2 3) 2 r T
2 4) 2  r T
8/ 3 2
92
SOLUTION :
Let R be the radius of the bigger drop,
then volume of bigger drop = 2 x volume of small drop
4 3 4
 R  2   r 3 ; R  21/ 3 r
3 3
Surface energy of bigger drop,
E  4 R 2T  4  2 2/ 3  r 2T  28/ 3  r 2T
19. When a big drop of water is formed from n small drops of water, the energy loss is 3E, where,
E is the energy of the bigger drop. If R is the radius of the bigger drop and r is the radius of the
smaller drop, then number of smaller drops (n) is (EAMCET-2014)
SOLUTION :
The energy of n small drops-the energy of the bigger drop=Energy loss by bigger drop
n  4 r 2  T  4 R 2  T  3  4 R 2  T
R2
n  4 r 2  12 R 2  4 R 2  n  4
r2
20. Find the maximum possible mass of a greased needle floating on water surface. T is the surface
tension of water, l is the length of the needle
2Tl g 2Tg Tl
A) mmax  B) mmax  C) mmax  D) mmax 
g 2Tl l g
SOLUTION :
Let the mass of the needle be m.Asthe liquid surface is distorted,the surface tensionforces acing on
both sides of the needle make an angle  ,say,with vertical.Since the forces acting on the needle are
F,F and mg, resolving the forces vertically for its equilibrium, we have
F   F
mg
F y  F cos   F cos   mg  0
2 F cos 
This gives m 
g
where F  Tl
2Tl cos 
Then m 
g
For m to be maximum, cos   1
2Tl
Hence, mmax 
g
93
21. A straw 6 cm long floats on water. The water film on one side has surface tension of 50 dyne/
cm. On the other side, camphor reduces the surface tension to 40 dyne/cm. The resultant force
acting on the straw is
1) 60 dyne 2) 10 dyne 3) 30 dyne 4) 0 dyne
SOLUTION :
  
F  F1  F2  T1  T2  l
22. A drop of water of mass m and density  is placed between two well cleaned glass plates, the
distance between which is d. What is the force of attraction between the plates? (T=surface
Tension)
Tm 4Tm 2Tm Tm
1) 2) 3) 4)
2 d 2 d 2 d 2 d 2
SOLUTION :
F  2TA / d
m  V   Ad
m
A
d
2Tm
F
d 2
23. Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is nearly (Surface
tension of soap solution = 0.03Nm-1) (AIEEE-2011)
1) 0.2 mJ 2) 2 mJ 3) 0.4 mJ 4) 4 mJ
SOLUTION :
Work done = change in surface energy
 W  2T  4  R22  R12 
 2  0.03  4  5    3   104 J  0.4 mJ
2 2
 
24. A film of soap solution is trapped between a vertical frame and a light wire ab of length 0.1m .
If g  10m / s 2 . Then the load W that should be suspended from the wire to keep it in equilibrium
is
a a
94
A) 0.2 g B) 0.3g C) 0.4 g D) 0.5g
SOLUTION :
25 103  2  0.1  m 10
5  103
m kg  0.5 g
10
95
Angle of contact:
Angle of contact of a liquid with respect to a solid is the angle between the tangent drawn to the liquid
surface at the point of contact and the surface of the solid, measured inside the liquid.
The angle of contact depends on solid-liquid pair, temperature and impurities.
The angle of contact may assume any value between 00 to 1800.
If the angle of contact is less than 900 then the liquid wets the solid
If the angle of contact is greater than 900 then the liquid does not wet the solid.
The angle of contact between pure water and glass = 00
The angle of contact between Hg and glass=1400.
The angle of contact is not changed by the angle of inclination of solid object in the liquid.
Water proofing agents increase the angle of contact.
Wetting agents decrease the angle of contact.
Shape of a liquid surface in a tube:
When a glass capillary tube is dipped in water, water rises into the tube.
When a glass capillary tube is dipped in mercury, mercury depresses into the tube.
When a silver capillary tube is dipped in water, water neither rises nor falls.
FA FA FA
 ‘
F FC FC F FC
F
FA  Fc FA  Fc FA  Fc
Case(i) Case(ii) Case(iii)
(FA:adhesive forces, FC=Cohesive forces, PA:Pressure at A, PB : Pressure at B)
Case (i) : - When glass capillary tube is dipped in water, observations are
 B
1) Capillary rise
2) FA  FC (adhesive > Cohesive)
3) Concave meniscus
4) Water wets the glass
5)   90o
6) PA > PB
96
Case (ii) : - When glass capillary tube is dipped in Mercury, observations are
 A
1) Capillary fall
2) FA  FC (adhesive < Cohesive)
3) Convex meniscus
4)   90o
5) PB > PA
6) Mercury does not wet the glass
Case (iii) : - When silver capillary tube is dipped in water, observations are
A
 B
1) water neither rises nor falls

2) FA  FC (adhesive = Cohesive)
3) Flat surface (not curved)
4) Critical wetting
5)   90o
6) PA = PB
Capillarity:
The rise or fall of a liquid column in a capillary tube dipped in a liquid is known as capillarity.
Capillarity is due to relative strengths of cohesive and adhesive forces.
In a gravity free space, a liquid in a capillary tube will rise to full length of the tube but not over flow.
The weight of the liquid column in the capillary tube is balanced by the force due to surface Tension.
97
T cos T cos
T   T
T sin  A B T sin 
D C T cos
h
r B
A
r
D
C
(a) (b)
2  r T cos  = Mg
h mgh 2T 2 cos2 
U  weight   
2 2 dg
r  h  r / 3 dg
T
2cos 
If r is very small compared to h,
hrdg
then T 
2 cos 
where
r = radius of the capillary tube
h = height of liquid column,
d = density of the liquid,
g = acceleration due to gravity,
 = angle of contact
Gravitational Potential energy of a liquid rises in a tube is
h mgh 2 T 2 cos 2 
U  weight    (when r is very less than h)
2 2 dg
When diameter of capillary tube increases twice, the height of liquid column falls down to half.
(r1h1 = r2h2).
1
Since h  , the graph between h and r is a rectangular hyperbola.
r
l 
 h
For a given radius, the capillary rise in a capillary tube does not depend either on the angle of inclination
or on the shape of the tube.
h
cos    h  l cos 
l
98
l1 cos 1  l2 cos  2
h = height of water in the tube
l = length of water in the tube
If the radii of the two limbs of a U-tube are r1 and r2 , the difference between the levels of a liquid in
‘u’ tube is
2T  1 1 
h      if r1  r2  h 
dg  r1 r2 
If two parallel plates with the spacing ‘t’ are placed in water reservoir, then height of rise
2T   mg  V  g  ht  g (  density of the liquid)
2T
h
 tg
l
If two concentric cylinders of radii r1 & r2 (inner one is solid are placed in water reservoir,
T cos  l1  l2   mg
T  2 r1  2 r2    r22 h   r12 h   g    00 
2T
h  if r  r  h 
 r2  r1   g 1 2
r1 r2
A drop of liquid of density d1 are floating half immersed in a liquid of density d 2 . If T is the surface
tension of the liquid, then the radius of the drop is (if   00 );
Fsurface tension  Fbuoyancy  mg
2 4
2 rT cos    r 3d 2 g   r 3 d1 g
3 3
3T
r
 2d1  d 2  g
99
A capillary tube is vertically dipped in a liquid. The height of the liquid in the
tube is ‘h’ and the total set up is kept in a lift.
 If the lift is moving up with an acceleration ‘a’ then the height of the liquid in the tube is given by
 g 
h'  h  
g a
 If the lift is moving down with an acceleration ‘a’ then the height of the liquid in the tube is given by
 g 
h'  h  
g a
If the lift is falling freely the height of the liquid raised in the tube is equal to full length of the tube
available, but not over flow.
 g 
h'  h    (is not true in such situations)
g  g
The relation between radius of tube and radius of meniscus is:
R C
A  B
r

AB = radius of tube (r)

AC = radius of meniscus (R)
AB r
In ABC cos   
AC R
 r  R cos 
If  = 0o then r = R;
r
If  = 90o, R 
cos900
i.e., The liquid meniscus is plane.
Capillary tube of insufficient length:
2T
Theoretically the rise of liquid in the tube is h= .
Rpg
If the length of the tube above liquid is  , if less than h, then the liquid will rise to full length of the tube
and the free surface of the liquid will acquire larger radius of curvature in such a way, that the product
h1 R1  h2 R2  l2 R2
100
R1  h2  l2 
 R2
l1 
h1 l2
h2
2T
We have, h  R  g
Thus, When a capillary tube of insufficient length is dipped in the liquid, the liquid will not overflow but
stays at the top with adjustable meniscus.
Effect of temperature on surface tension:
Over small ranges of temperature, the surface tension of a liquid decreases linearly with the
rise of temperature due to increase in inter molecular distances according to the relation
St = S0 (1- t )
where st = surface tension at t0C
S0 = surface tension at 00C
t = Change in temperature.
 = temperature coefficient of surface tension.
Dependence of Surface Tension -On temperature :
In the case of molten copper and molten cadmium,surface tension increases with increase of temperature.
The surface tension of any liquid at its critical temperature is zero.

Effect of impurities on Surface Tension:
With the addition of impurities surface tension may increase or decrease, depending on the
type of impurity.
There are two types of impurities.
They are
(i) weakly soluble impurities,
(ii) highly soluble impurities.
If the added impurities are weakly soluble in liquid its surface tension decreases.
Eg: When soap is mixed with water surface tension decreases.
If the added impurities are highly soluble in liquid, its surface tension increases.
Eg: When highly soluble salts like NaCl, ZnSO4, etc. are mixed with water, its surface tension
increases.
101
Excess Pressure inside a curved liquid surface :
The pressure on the concave side of curved liquid surface is greater than that on the convex side. This is
the reason why pressure difference exists across two sides of a curved surface.
Case : A
Excess pressure inside a soap bubble
= Pressure inside the bubble - outside pressure soap bubble has two free surface
A
2  T  2
B
(A) one is exposed to outside air
(B) another to inside air.
Balancing atmospheric pressure = surface tension at 2 free surface soap surface has [2free surfaces]
4T
P  pr 2  2  T  2pr  P 
r
Case : B
Excess pressure in air bubble inside a liquid.
Here only one free surface which is exposed to inside air of bubble.
A
P  2 T  2
B
Balancing liquid pressure = surface tension at one free surface
P  pr 2  T  2pr[one free surface]
2T
P
r
Case : C
Excess pressure in a liquid drop
102
r
r
liquid
here, one free surface exposed to air

Balancing atm pressure = surface tension at one free surface
2T
P  pr 2  T  2pr  P 
r
Case :D
If curvature are in same direction
r1 r2
1 1 
P T  
 r1 r2 
Case : E
If the curvature are in same direction
r1 r2
1 1 
P T   
 r1 r2 
From above
T
For cylindrical surface P  \ r1  r, r2  
r
2T
For spherical surface P  \ r1  r2  r
r
Note: 1
(A) Always for liquid surface one free surface is taken
(B) Always for liquid film two free surface is taken.
Note : 2
For liquid surface, pressure on concave side is always high than convex side.
Low Pressure
High High
Pressure Pressure
Low Pressure
103
When the soap bubble coalesce, then the radius of curvature of common surface
4T
Excess pressure in first bubble P1  r
1
4T
Excess pressure in second bubble P2  r
2
Excess pressure on common surface P = P1- P2

if radius of curvature of common surface is r,
4T
P
r
P = P 1- P 2
4T 4T 4T
 
r r1 r2
1 1 1
 
r r1 r2
r1r2
r
r2  r1 , where r1 < r2
When two soap bubble of radii r1 & r2, combine to form a new bubble in vaccum under isothermal
condition,
At isothermal condition SPV=0
P1V1 +P2V2 = PV
4T 4 3 4T 4 3 4T 4 3
. pr1  . pr2  . pr
r1 3 r2 3 r 3
r  r12  r22
Liquid betwen two plates:
When a small drop of water is placed between two glass plates put face to face. it forms a thin
film which is concave outward along its boundary. Let ‘R’ and ‘r’ be the radii of curvature of the
enclosed film in two perpendicular directions.
Hence the pressure inside the film is less than the atmospheric pressure outside it by an amount P given
by
1 1  d 
P  T     R1  R  ; R2   
 R1 R2   2 
T T 2T
  
R d /2 d
Force required to separate glass plates is
104
2T
F A
d
2T  A 2TA2
  V  A  d 
VA V
Detergents And Surface Tension:
We clean dirty clothes containing grease and oil stains sticking to cotton (or) other fabrics by adding
detergents (or) soap to water.
Adding detergent or soap to water makes the angle of contact less than 90 0 and there by wets the
clothes.
The kind of process using surface active detergents or surfactants is important not only for cleaning,
but also in recovering oil, mineral ores, etc.
Wetting Agents:
Wetting agent is a material, mixed with liquid, to decrease the angle of contact with the given solid.
Eg: Soaps and detergents
Water Proofing Agent:
Water proofing agent is a material applied on the surface of solid to increase the angle of contact with water.
Eg : Wax
105
::PROBLEMS ::
1. Expression for the height of capillary rise between two parallel plates dipping in a liquid of
density  separated by a distance D) The surface tension of the liquid is T. [Take angle of
contact to be zero]
T T
A l
h
B C
2T 2d T 2T 2
A) h   dg B) h  C) h  D) h 
T d d
SOLUTION :
The meniscus between two plates is cylindrical in shape. Pressure at A(the lowestpoint of the
meniscus)
p A  p0  T / r
Pressure at B = Pressure at C  p0 = Pressure at A   gh
T T 2T
PB  p0  p0    gh, h  
r  gr  gd
Alternative method:
Force upward  2lT cos   2lT   0 
0
Gravitational pull = (Volume  Density)g  lhd g

2T
 2lT  lhd  g  h 
d g
2. Find the weight of water supported by surface tension in a capillary tube with a radius of 0.2mm.
Surface tension of water is 0.072Nm-1 and angle of contact of water is 00.
SOLUTION :
Assume the weight of water to be ‘F’
Weight of water in capillary tube = upward force due to surface tension
i.e., F  2 r  T cos 
Surface tension of water T=0.072Nm-1
106
0.2
Angle of contact   00 , Radius of capillary tube (r) = m  0.2  10 3 m
1000
22
F  2 r T cos    2   0.2  10 3  0.072  1
7
 90.51  10 6 N  F  90.51  10 6 N
3. A capillary tube of radius ‘r’ is immersed in water and water rises to a height of ‘h’. Mass of
water in the capillar y tube is 5x10-3 kg. The same capillary tube is now immersed in a liquid
whose surface tension in 2 times the surface tension of water. The angle of contact between
the capillary tube and this liquid is 450. The mass of liquid which rises into the capillary tube now
is, (in kg) (EAM-13)
SOLUTION :
Height of water rise in a capillary tube
2T cos 2T cos 1 2T cos 2
h ; h1  1 ; h2  2
rdg rdg rdg
1
Given, T2  2T ,  45 ,cos 45 
0 0
2
1
2 2T 
 h2  2
rdg
From Eqs(i) and (ii), we observe h2=h
Hence, same mass of liquid rises into the capillary as before 5 x 10-3 kg.
4. A U-tube is supported with its limbs vertical and is partly filled with water. If internal diameters
of the limbs are 1x10-2m and 1x10-4m respectively, what will be the difference in heights of
water columns in the two limbs (Surface tension of water is 0.07Nm-1)
SOLUTION :
Surface tension, T=0.07Nm-1;
Density, d=1000kgm-3;
g=9.8ms-2
Angle of contact   00 ;
Radius, r1=0.5x10-2m;
Radius, r2=0.5x10-4m
2T  1 1 
Then, h    
dg  r1 r2 
2  0.07  1 1 
  2
 4 
 0.283m
10  9.8  0.5  10
3
0.5  10 
5. A glass capillary sealed at the upper end is of length 0.11m and internal diameter 2 105 m. The
tube is immersed vertically into a liquid of surface tension 5.06  102 N / m. To what length the
capillary has to be immersed so that liquid level inside and outside the capillary becomes the
same.?
A) 5cm B) 3cm C) 1cm D) 7cm
SOLUTION :
Let P0 be atmospheric pressure and P1 the pressure of air within the sealed tube.
107
Therefore equilibrium pressure just below the meniscus should be equal to atmospheric pressure
because levels of water inside and outside the tube is same.
 2T  2T
i.e.,  P1    p0orP1  p0 
 r  r
If L=0.11m is the length of tube and x the
L-x
P0 P1 P0
P x
 2T 
length of immersed part, then from Boyle’s law p1V1  p2V2 ; P0 La   P0    L  x a
 r 
where a is the cross-sectional area of tube,
 2T 
i.e., P0  L   P0   L  x
 r 
2T
P0  L  P0  L  x    L  x
r
4TL
x
Po d  4T
6. What would be the pressure inside a small air bubble of 1.0mm radius situated just below the
surface of water. T=72 x 10-3N/m
Atm. pr. = 1.013 x 105N/m2
SOLUTION :
2T
Excess pressure Pex 
r
2  72  103
Pex   1440N / m 2
1 103
For air bubble in water as on free surface since the bubble is just below the water surface,
the external pressure on it is equal to the atm. pressure P,
hence the pressure inside the bubble.
P + Pex = 1.013 x 105 + 1440
= 1.0274 x 105 N/m2
7. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the
same capillary tube. If the denisty of mercury is 13.6 g/c.C) and the angles of contact for
mercury and water are 1350 and 00 , respectively, the ratio of surface tension for water and
mercury is
A) 1: 0.15 B) 1: 3 C) 1: 6.5 D) 1.5 :1
108
SOLUTION :
2T cos 
Heoight , h 
r g
2  Tw  cos 00
 For water, hw  r  1 g
2  Tm  cos1350
And, for mercury, hm 
r 13.6  g
hw 2  Tw  1 r  13.6  g  2
 h  r  1 g  2  Tm  1
m
10 T
 cos1350  1/ 2    w  13.6  2
  3.42 Tm
Tw 10 1
 T  3.42  13.6  1.414  6.5
m
 Tw : Tm  1: 6.5
8. Two separate air bubbles (r1=0.02cm, r2 = 0.004cm) formed of same liquid T = 0.07N/m come
together to form a double bubble. Find the radius and sense of curvature of the internal film surface
common to both the bubbles
SOLUTION :
r1r2
r
r2  r1
0.003  0.004
r  0.004m
0.004  0.002
P1 P2
As the excess pressure is always towards convace surface & pressure in smaller bubble is greater
than larger bubble, the common surface is concave towards the centre of the smaller bubble.
9. Two soap bubbles of radii R1 and R2 are kept in vacuum at constant temperature, the ratio of
masses of air inside them, is
SOLUTION :
m
From, PV  RT
M
i.e., m  PV
109
4T 4T 4T 4T
P1  P0  0 ; P2  P0  0
R1 R1 R2 R2
 4T  4 3
   R1
m1 PV  R1  3 m R2
 1 1
  1  12
m2 PV  4T  4 3 m2 R2
  R2
2 2

 R2  3
10. Two soap bubble of raddi R1 and R2 are in atmosphere of pressur eP0 at constant temperature.
Ratio of masses of air inside them is
SOLUTION :
m
PV  RT
M
PV
., m  PV  PV
1 1
2 2
4T 4T
p1  p0  , p2  p0 
R1 R2
 4T 4 3  4T  3
 P0    R1  P0   R1
R1 3 R1
  
 4T 4 3  4T  3
 P0    R2  P0   R2
 R2 3  R2 
11. A capillary of the shape as shown is dipped in a liquiD) Contact angle between the liquid and
the capillary is 00 and mass of liquid inside the meniscus is to be neglecteD) T is surface
tension of the liquid, r is raidus of the meniscus, g is acceleration due to gravity and  is
density of the liquid then height h in equilibrium is
2T 2T 2T
A) Greater than B) Equal to C) less than
r g r g r g
D) of any value depending upon actual values
110
SOLUTION :
. As weight of liquid in capillary is balanced by surface tension,
2
then T  2 r   r h, pg (for uniform r radius tube)
r
2T
h1 
h1
;
rpg
But weight of liquid in tapered tube is more than uniform tube of radius r, then in order to balance is
< h1
r
2T
; h1 
rpg
12. Two soap bubbles are combined isothermally to form a big bubble of radius R. If V is
change in volume, S change in surface area and P1 is atmospheric pressure then show that
3P0  V   4T  S   0
SOLUTION :
PV=constant.
 4T 
After comibining the two bubbles  P0   V =constant;
 R 
4T 4 3
PV
0  . R C
R 3
4T
Before combining the two bubbles P0 V1  V2    S1  S2   C.... 2 
3
According to Boyle’s law PV 1 1  PV
2 2  PV
from equations (1) and (2)

4T 4T
P0 V1  V2    S1  S 2   PV
0  S
3 3
4T
P0 V  S  0;3P0  V   4T  S   0
3
111
13. Soapy water drips from a capillary tube. When the drop breaks away, the diameter of its neck
is D. The mass of the drop is m. Find the surface tension of soapy water?
F
Capillary
mg
mg mg mg mg
1) 2) 3) 4)
 D
2
D 2
D 2 D
SOLUTION :
When the drop breaks away from the capillary, weight of the drop = force of surface tension.
mg
or mg   D  T or T 
D
14. When air bubble comes from bottom to the top of a lake its radius becomes n times. If
temperature remains constant through out the lake the depth of the lake will be,
SOLUTION :
From Boyle’s law PV=constant
 PV
1 1  PV
2 2
 2T  4 3  2T  4
   nR 
3
 P0  h g    R   P0 
 R 3  nR  3
2T 3 2T
h  g  P0  n 3  1  n 
nR R
2T 2
P0  n3  1   n  1
h R
g
15. A capillary tube is immersed vertically in water such that the height of liquid column is found to
be ‘x’ on the surface of the earth. When it is taken to minute the capillary rise is ‘y’ if ‘R’ is the
radius of the earth. Then the depth of mine is
 y  x  y  x  x   y 
1) d  R 2) d  R 3) d  R   4) d  R  
x y  y x  y x
SOLUTION :
g1  g  l  d / R 
112
16. A vertical glass capillary with inside diameter 0.50mm is submerged into water so that the
length of its part emerging outside the water surface is equal to 25 mm.Find the radius of
curvature of the meniscus.Surface tension of water is 73  10 3 N / m .
A) R  0.6m B) R  6mm C) R  0.6mm D) R  0.6 Km
SOLUTION :
2T
In the capillary tube, the water should rise to a height h 
r g
Here T  73  103 N / m
0.50mm
r  0.25 10 3 m
2
2  73 103
h  3
 59  103 m  59mm
0.25 10  10  9.8
3
Now h  h ' ;
i.e., length is outside water surface.
Therefore, radius of meniscus >radius of capillary r.
If R is the radius of meniscus,
then we have
2T
 h'g
R
2T
R
h'g
Here h '  25mm  25  10 3 m
2  73  103
R  3
 0.6 10 3 m
25 10  9.8
17. The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of
water in a beaker. What is the pressure required in the tube in order to blow a hemispherical
bubble at its end in water. [The surface tension of water at temperature of the experiment is
7.30 x 10-2 Nm-1. 1 atmospheric pressure=1.01x105 Pa, density of water=1000kgm-3, g=9.80ms-
2
].
SOLUTION :
The pressure required to blow the bubble is
P=Patm+hdg+2T/r
=1.01784x105+(2x7.3x10-2/10-3)
=(1.01784+0.00146) x105=1.02x105Pa.
18. A glass U-tube is such that the diameter of one limb is 3.0 mm and that of the other is 6.00 mm. The
tube is inverted with the open ends below the surface of water in a beaker. What is the difference
between the heights to which water rises in the two limbs? Surface tension of water is 0.07N/m.
Assume that the angle of contact between water and glass is 00.
113
SOLUTION :
Let PA and PB are the pressure at points A and B respectively.
2T
The pressure at point C. PC  PA  R
1
r1
. Where R1   r1
cos 00
2T
The pressure at point D, PD  PB 
R2
r2
Where, R2   r2
cos 00
A
C
h B
D
If ‘h’ is the difference in levels of liquid in two limbs, then

 2T   2T 
PD  PC  h g   PB     PA    h g
 R2   R1 
As PA  PB and R1  r1  1.5mm
1 1
R2  r2  3.0mm, so 2T     h g
 r1 r2 
 1 1 
0.2  0.07  3
   h  1000  9.8
 1.5  10 3  10 3 
After solving, we get h = 4.76 x 10-3 m
19. A soap bubble is being blown at the end of a very narrow tube of radius b. Air (density  )
moves with a velocity ‘v’ inside the tube and comes to rest inside the bubble. The surface
tension of the soap solution is T. After some time the bubble, having grown to a radius ‘r’,
separates from the tube. Find the value of ‘r’. Assume that r>>b so that you can consider the
air to be falling normally on the bubble’s surface.
r
V
A
4T 4T 2T 4T
1) 2) 3) 4)
v2 v v v2
114
SOLUTION :
The bubble sill separate from the tube when thrust force exerted by the air is equal to the force due to
excess pressure.
 4T 
i.e,  Av  
2
A
 r 
4T
r
 v2
20. Two verical parallel glass plates are particlly submerged in water. The distance between the
plates is d and the length is l . Assume that the water between the plates does not reach the
upper edges of the plates and that the wetting is complete. the water will rise to height (  -
density of water and  = surface tension of water)
2  4 5
A) B) C) D)
 gd 2  gd  gd  gd
SOLUTION :
Total upward force dur to surface tension = 2 l
Weight of lifted liquid = (hlD)  g
2
Equating, we get h 
 gd
21. A capillary tube of radius’ r’ is lowered into water whose surface tension is ' ' and density
‘d’.The liquid rises to a height. Assume that the contact angle is Zero. Choose the correct
statement(s):
4 2
A) Magnitude of work done by force of surface tension is
dg
2 2
B) Magnitude of work done by force ofsurface tension is
dg
2 2
C) Potential energy acquired by the water is
dg
2 2
D) The amount of heat developed is
dg
KEY : A , C, D
SOLUTION :
2
h
dgr
hence work done by force of surface tension is
4 2
Ws    2 r  h 
dg
But centre of mass of liquid in the capillary tube is at a hight h/2.
h 2
2
Mgh
Hence potencial energy gained    r  h d  g 
2

2 2 dg
115
 2 2 
Hence work done by gravity  
 
 dg 
2 2
Amount of heat developed 
dg
22. When a capillary tube is dipped in a liqiud, the liquid rises to a height h in the tube. the free
liquid surface inside the tube is hemispherical in shape. The tube is now pushed down so that
the height of the tube outside the liquid is less than h . Then
A) The liquid will come out of the tube like in a small fountain
B) The liquid will ooze out of the tube slowly
C) The liquid will fill the tube but not come out of its upper end
D) The free liquid surface inside the tube will not be hemispherical
KEY : C, D
SOLUTION :
The nagle of contact at the free liquid surface inside the capillary tube will change such that the certical
component of the surface tension forces just balances the weight of the liquid column.
23. A glass rod of radius r1 is inserted symmetrically into a vertical capillary tube of radius r2 such
that their lower ends are at the same lebel. The arrangement is now dipped in water. The height
to which water will rise into the tube will be  = surface tension of warer ,  = density of water)
2  2 2
A) r  r  g B) r  r  g C) r  r  g D)  r 2  r 2   g
 2 1  2 1  2 1 2 1
SOLUTION :
Total upward force due to surface tension =   2 r1  2 r2  .
This supports the weight of the liquid column of heighrt h.
Weight of liquid column = h   r2   r1   g Equating,
2 2
we get h  r2  r1  r2  r1   g  2  r1  r2 
or h  r2  r1   g  2
2
h
 r2  r1   g
116
::THEORY BITS ::
1. The weight of the body is maximum in
1) air 2) hydrogen 3) water 4) vacuum
KEY:4
2. If a big drop of liquid at 27 0 C is broken into number of small drops,
then the temperature of the droplets is
1)  27 0 C 2)  27 0 C 3)  27 0 C 4)  54 0 C
KEY:3
3 When a body is full immersed in a liquid, the loss of weight of the body is equal to
1) Apparent weight of the body 2) Force of buoyance
3) Half the force of buoyancy 4)Twice the force of buoyancy
KEY:2
4. A boat full of scrap iron is floating on water in a lake. If all the iron is dropped into the water,the
level of water will
1) go up 2) fall down 3) remain the same 4) Can not be decided
KEY:2
5. A large block of ice floats in a liquid.When ice melts the liquid level rises.The density of liquid is
1) Greater than that of water 2) Less than that of water
3) Equal to that of water 4) Half of that of water
KEY:1
6. Identify the correct choice.
A)When a abody floats in a liquid, it displaces the liquid whose weight is equal to its own weight.
B)When a body sinks in a liquid, it displaces the liquid whose volume is equal to its own volume.
1) A is true but B is false 2) A is false but B is true
3) Both A and B are true 4) Both A and B are false
KEY:3
7. A swimmer goes from the surface of water to a depth of 20m, the change in the pressure on his
body is nearly
1) 3 atmospheres 2) 1 atmosphere 3) 2 atmospheres 4) zero
KEY:3
8. A bucket of water contain a wooden block floating in water with(4/5)th of its volume sub merged in the
water.The bucket is placed on the floor of a lift and the lift now starts moving down with uniform
accleration. The block of wood now
1) moves upward 2) moves downward
3) remains at same place 4)moves horizontally
KEY:3
9. Surface tension of water is T1 . When oil spreads on water surface tension becomes T2 , then
T2
1) T1 > T2 2) T1 = T2 3) T1 < T 4) T1 
2
2
KEY:1
10. A wooden block with a coin placed on its top floats in water as shown. After some time the coin
falls into water. Then
117
coin
1) l decreases and h increases 2) l increases and h decreases

3) both l and h increase 4) both l and h decrease
KEY:4
11. Which factor controls the better flow rate of a liquid through the syringe?
1) the pressure exerted by the thumb 2) the length of the needle
3) the nature of the liquid 4) the radius of the syringe bore.
KEY:4
12. In order that a floating object be in a stable equillibrium, its centre of bouyancy should be
1) vertical below its centre of gravity 2) horizontally inline with its centre of gravity
3) vertically above its centre of gravity 4) may be anywhere
KEY:3
13. When there are no external forces, shape of the liquid is determined by
1) Density of liquid 2) Temperature only 3) Surface tension 4) Viscosity
KEY:3
14. A piece of ice floats in a liquid denser than water. The liquid fills the vessel upto the edge.If ice
melts completely then
1) water level remains unchanged 2) water level decreases
3) water overflows 4) data is insufficient.
KEY:3
15. With the increase in temperature the anlge of contact between glass and water
1) decreases 2) increases 3) remains cont.
4) some times increases and some times decreases
KEY:1
16. An object of uniform density is allowed to float in water kept in a beaker. The object has
triangular cross-section as shown in the figure. If the water pressure measured at the three
point A,B and C below the object are PA, PB and PC respectively then:
A B C
1) PA > PB > PC 2) PA > PB< PC 3) PA = PB= PC 4) PA = PC < PB

KEY:3
17. In a gravity free space, shape of a large drop of liquid is
1) Spherical 2) Cylindrical
3) Neither Spherical nor cylindrical 4) May be Spherical or cylindrical
KEY:1
118
18. A jar is filled with two non-mixing liquids 1 and 2 having densities 1 and  2 , respectively. A
solid ball, made of a material of density  3 is dropped in the jar. It come to equilibrium in the
position shown in the figure. Which of the following is true for 1 ,  2 and 3 ? (AIEEE-2008)
Liquid 1 1
3
Liquid 2 2
1) 1   2  3 2) 1  3   2 3) 3  1   2 4) 1  3   2
KEY:2
19. When a capillary tube is immersed into a liquid the liquid neither rises nor falls in the capaillary,
then the angle of contact is
1) 200 2) 900 3) 300 4) 700
KEY:2
Equation of continuity, Bernoulli’s Theorem
20. Stream line motion becomes turbulent motion when the velocity of the liquid is
1) beyond critical velocity 2) critical velocity
3) below critical velocity 4) variable velocity
KEY:1
21. A liquid does not wet the solid surface if the angle of contact is
1) 00 2) =450 3) = 900 4) >900
KEY:4
22. In a laminar flow at a given point the magnitude and direction of the velocity of the fluid
1) both are constant 2) magnitude is only constant
3) direction is only constant. 4) both are not constant.
KEY:1
23. The liquid flow is most stream lined when
1) liquid of high viscosity and high density flowing through a tube of small radius.
2) liquid of high viscosity and low density flowing through a tube of small radius
3) liquid of low viscosity and low density flowing through a tube of large radius
4) liquid of low viscosity and high density flowing through a tube of large radius
KEY:2
24. The rate of flow of the liquid is the product of
1) Area of cross section of the liquid and velocityof the liquid.
2) Length of the tube of the flow and velocity of the liquid.
3) Volume of the tube of the flow and velocity of the liquid.
4) Viscous force acting on the liquid layer and velocity of the liquid
KEY:1
25. The fundemantal quantity which has the same power in the dimensional formula of surface
tension and coefficient of viscosity is
1) Mass 2) Length 3) Time 4) None
KEY:1
119
26. The volume of liquid flowing per second out of an orifice at the bottom of the tank does not
depend upon
1) the density of the liquid 2) acceleration due to gravity
3) the height of the liquid above orifice 4) the area of the orifice
KEY:1
27. Water flows through a horizontal pipe of radius 'r' at a speed V. If the radius of the pipe is
doubled, the speed of flow of water under similar conditions is
1) 2V 2) V/2 3) V/4 4) 4V
KEY:3
28. A liquid is under stream lined motion through a horizontal pipe of nonuniform cross section. If the
volume rate of flow at cross section 'a' is V, the volume rate of flow at cross section a/2 is
1) V/2 2) V 3) V/4 4) V
KEY:4
29 100 kg of iron and cotton are weighed by using a spring balance on the surface of the earth. If
R1 and R2 are the reading shown by the balance,then
1) R1  R2 2) R1  R2 3) R1  R2 4) R1  R2  0
KEY:3
30. Three tubes A, B, C are connected to a horizontal pipe in which liquid is flowing. The radii of
the pipes at the joints of A, B and C are 2 cm, 1 cm and 2 cm respectively. The height of the
A B C
1) in A is maximum 2) in A and C is equal

3) is same in all the three 4) in A and B is same
KEY:2
31. If air is blown with a straw under of the pans of a physical balance present in equilibrium
position, then that pan.
1) rises up 2) remains in the same position
3) lowers down 4) rises or lowers depending upon the velocity of air blown.
KEY:3
32. The liquid meniscus in a capillary tube will be convex, if the angle of contact is
1) greater than 900 2) less than 900 3) equal to 900 4) equal to zero
KEY:1
33. A train goes past a person standing at the edge of a platform at high speed. Then the person
will be
1) attracted towards the train 2) unaffected by the train
3) reflected by the train 4) affected only if its speed is greater than critical velocity.
KEY:1
34. The velocity distribution curve of the stream line flow of a liquid advancing through a capillary
tube is
1) Circular 2) elliptical 3) parabolic 4) a straight line
KEY:3
120
35. Water stands at level A in the arrangement shown in figure . If a jet of air is gently blown into
the horizontal tube in the direction shown in figure, then
horizontal tube
air
A
1) Water will fall below A in the capillary tube

2) Water will rise above A in the capillary tube
3) There will be no effect on the level of water in the capillary tube
4) Air will emerge from end B in the form of bubbles.
KEY:2
36. A car moving on a road when overtaken by a bus
1) is pulled towards the bus 2) is pulled away from the bus
3) is not affected by the bus 4) information is insufficient.
KEY:1
37. The rise of liquid into capillary tube is h1. If the apparatus is taken in a lift moving up with
acceleration, the height is h2, then
1) h1 = h2 2) h1 > h2 3) h2 > h1 4) h2 = 0
KEY:2
38. A capillary tube, made of glass is dipped into mercury. Then
1) mercury rises in the capillary tube 2) mercury descends in capillary tube
3) mercury rises and flows out of capillary tube
4) mercury neither rises nor descends in the capillary tube.
KEY:2
39. A water barrel having water up to depth 'd' is placed on a table of height 'h'. A small hole is made
on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the
ground at a horizontal distance 'R' from the barrel, then the value of 'd' is
4h R2 h
1) 2 2) 4hR 2 3) 4) 4 R 2
R 4h
KEY:3
40. The end of a glass tube becomes round on heating due to
1) friction 2) Viscosity 3) Gravity 4) Surface tension
KEY:4
41. The main cause of viscosity is
1) Force of repulsion between molecules 2) Cohesive forces
3) adhesive forces 4) both cohesive and adhesive forces.
KEY:2
42. Viscosity is the property by virtue of which a liquid
1) occupies minimum surface area
2) offers resistance for the relative motion between its layers.
3) becomes spherical in shape
4) tends to gain its deformed position.
KEY:2
43. Viscosity is most closely related to
121
1) density 2) velocity 3) friction 4) energy
KEY:3
44. Rain drops fall with terminal velocity due to
1) Buoyancy 2) Viscosity 3) Low weight 4) Surface tension
KEY:2
45. Two identical lead shots are dropped at the same time in two glass jars containing water and
glycerine. The lead shot dropped in glycerine descends slowly because.
1) Viscous force is more in water than in glycerine
2) Viscous force is more in glycerine than in water
3) Surface tension is more in water
4) Surface tension is more in glycerine
KEY:2
46. After the storm, the sea water waves subside due to
1) Surface tension of sea-water 2) Disapperance of heavy currents
3) The viscosity of sea water 4) Gravitational pull of the storm
KEY:3
47. When a metallic sphere is dropped in a long column of a liquid, the motion of the sphere is
opposed by the viscous force of the liquid. If the apparent weight of the sphere equals to the
retarding forces on it, the sphere moves down with a velocity called.
1) Critical velocity 2) Terminal velocity 3) Velocity gradient 4) Constant velocity
KEY:2
48. The tangential forces per unit area of the liquid layer required to maintain unit velocity gradient
is known as
1) Coefficient of gravitation of liquid layer 2) Coefficient of friction between layers
3) Coefficient of viscosity of the liquid 4) Temperature coefficient of viscosity
KEY:3
49. The quality of fountain-pen ink depends largely on
1) Surface tension of the liquid 2) Viscosity of ink
3) Impurities in ink 4) Density of ink
KEY:2
50. The tangential force (or) viscous force on any layer of the liquid is directly proportional to
 dv 
velocity gradient   . Then the direction of velocity gradient is :
 dx 
1) Perpendicular to the direction of flow of liquid
2) Parallel to the direction of flow of liquid
3) Opposite to the direction of flow of the liquid
4) Independent of the direction of flow of liquid.
KEY: 1
51. The equation of continuity leads to
1) Law of conservation of moments of liquid flow. 2) law of conservation of energy
3) law of equipartition of energy 4) law of conservation of mass.
KEY:4
52. Viscosity of the fluids is analogous to
1) Random motion of the gas molecules 2) Friction between the solid surfaces
3) integral motion 4) Non uniform motion of solids
KEY:2
53. The surface tension of a liquid __with rise of temperature.
1) Increases 2) Decreases 3) Remains same
122
4) First decreased and then increases
KEY:2
54. The viscous drag is
1) inversely proportional to the velocity gradient
2) directly proportional to the surface area of layers in contact
3) independent of nature of liquid
4) perpendicular to the direction of liquid flow
KEY:2
5 5 . A capillary is dipped in water vessel kept on a freely falling lift,
th e n
1) water will not rise in the tube
2) water will rise to the maximum available height of the tube
3) water will rise to the height observed under normal condition
4) water will rise to the height below that observed under normal condition.
KEY:2
56. Viscosity is exhibited by
1) Solids, liquids, and gases. 2) liquids and gases
3) Solids and gases 4) Solids and liquids
KEY:2
57. A spinning ball is moving in a direction opposite to the direction of the wind. The ball moves in
a curved path as
1) The pressure at the top and the bottom of the ball are equal.
2) The pressure at the top > the pressure at the bottom
3) The pressure at the top < the pressure at the bottom
4) There is no relation between the pressures.
KEY:2
58. A good Lubricant must have
1) high viscosity 2) low viscosity 3) high density 4) low surface tension
KEY:1
59. Coefficient of viscosity of a gas
1) increases with increase of temperature
2) decreases with increase of temperature
3) remains constant with increase of temperature
4) may increase or decrease with increase of temperature
KEY:.1
60. Viscosity of water at constant temperature is
1) more in deep water 2) more in shallow waters
3) less in deep water 4) same in both deep water and shallow waters
KEY:1
61. Hot syrup flows faster because
1) Surface tension increases with temperature 2) Viscosity decreases with temperature
3) Viscosity increases with temperature 4) Surface tension decreases with temperature
KEY:2
62. The pressure at a depth ‘h’ in a liquid of density “  ” is plotted on the Y-axis and the value of
‘h’ on the X-axis , the graph is a straight line. The slope of the straight line is ( g = acceleration
due to gravity )
1)  g 2) 1/  g 3)  /g 4) g/ 
KEY:1
63. If the flow is stream lined then Reynolds number is less than
1) 2000 2) 3000 3) 1000 4) 4000
123
KEY:3
64. A drop of water of radius ‘r’ is falling through the air of coefficient of viscosity ‘  ’ with a
constant velocity of ‘v’. The resultant force on the drop is
1
1) 2) 6 rv 3) 6 rv 4) zero
6 rv
KEY:4
65. The paint -gun works on the principle of
1) Boyle's law 2) Bernoulli's principle
3) Archimedi's principle 4) Newton's laws of motion
KEY:
66. The rate of flow of a liquid through a capillary tube is
1) directly proportional to the length of the tube
2) inversely proportional to the difference of pressure between the ends of the tube.
3) directly proportional to the 4th power of the radius of the tube.
4) independent of the nature of the liquid
KEY:3
67. Poiseuili's equation holds good when
1) the flow is steady and stream line 2) the pressure is constant at every cross section
3) The liquid in contact with the walls is stationary 4) All the above
KEY:4
68. If l is length of the tube and r is the radius of the tube, then the rate of volume flow of a liquid is
maximum for the following measurements, under the same pressure difference.
l r
1) l, r 2) , 2r 3) 2l , 4) 2l , 2r
2 2
KEY:2
69. When a boat in a river enters the sea water, then it
1) sinks a little 2) rises a little 3) remains same 4) will drawn
KEY:2
70. After terminal velocity is reached the acceleration of a body falling through a viscous fluid is
1) zero 2) g 3) less than g 4) greater than g
KEY:1
71. Clouds appear to float in air due to
1) low density 2) Air current 3) Viscosity of air 4) Buoyancy
KEY:4
72. A small ball is dropped in a viscous liquid.
Y
Its fall in the liquid is best described by the figure
Velocity in liquid
A B
C
D
X
distance
1) curve A 2) curve B 3)curve C 4)curve D
KEY:3
73. A solid rubber ball of density 'd' and radius 'R' falls vertically through air. Assume that the air
resistance acting on the ball is F = KRV where K is constant and V is its velocity. Because of
this air resistance the ball attains a constant velocity called terminal velocity VT after some
124
time. Then VT
4 R 2 dg 3K 4  r 3 dg
1) 2) 3) 4)  rdgk
3K 4 R 2 dg 3 K
KEY:1
74. Bernoulli's theorem is applicable in the case of
1) Compressible liquid in stream lined flow 2) Compressible liquid in turbulent flow
3) incompressible liquid in stream lined flow 4) incompressible liquid in turbulent flow.
KEY:3
75. The terminal velocity of a small ball falling in a viscous liquid depends upon
i) its mass m ii) its radius r
iii) the coefficient of viscosity of the liquid  and
iv) acceleration due to gravity. Which of the following relations is dimensionally true for the
terminal velocity.
Kmg Kmgr Kmg Kr
1) V  2) V  3) V  4) V 
r  r mg
KEY:1
76. A ball is dropped into coaltar. Its velocity-time curve will be
y y y y
v v v v
1) 2) 3) 4)
O x O x O x O x
t t t t
KEY:2
77. Two needles are floating on the surface of water. A hot needle when touches water surface
between the needles, then they move
1) Closer 2) Away 3) Out of the liquid 4) Into the liquid
KEY:2
78 Which of the following is a characteristic of turbulen flow?
1) velocity more than critical velocity 2) irregular flow
3) molecules crossing from one layer to the other 4) 1, 2, 3.
KEY:4
79. Liquid drops acquire spherical shape due to
1) gravity 2) surface tension
3) viscosity 4) intermolecular separation
KEY:2
80. Vertical sections of a wing of a fan are shown in the following figures. The maximum up thrust
will be in figure.
1) 2) 3) 4)
KEY:1
81. The height upto which water will rise in a capillary tube will be:
1) maximum when water temperature is 40C 2) minimum when water temperature is 40C
3) minimum when water temperature is 0 C 0
4) same at all temperatures
KEY:2
82. At critical temperature surface tension becomes
1) 0 2) 1 3) Infinite 4) Negative
KEY:1
83. Droplets of a liquid are generally more spherical in shape than large drops of the same liquid
because
1) Force of surface tension is equal and opposite to the force of gravity
125
2) Force of surface tension predominates the force of gravity
3) Force of gravity predominates the surface tension
4) Force of surface tension and force of gravity act in the same direction and are equal.
KEY:2
84. Mercury does not wet glass, wood or iron because
1) cohesive force is less than adhesive force 2) cohesive force is greater than adhesive force
3) angle of contact is less than 90 0 4) cohesive force is equal to adhesive force
KEY:2
85. With the increase of temperature,
1) The viscosity of a liquid increases 2) The viscosity of a liquid decreases
3) The viscosity of a gas decreases 4) The viscosity of a gas remains unchanged.
KEY:2
86. The surface tension of a liquid at its boiling point is
1) Maximum 2 ) Zero
3) Same as at room temperature 4) Minimum but more than zero
KEY:2
87. In turbulent flow, the velocity of the liquid molecules in contact with the walls of the tube.
1) is zero 2) is maximum 3) is equal to critical velocity 4) may have any value
KEY:1
88. The addition of soap changes the surface tension of water to T1 and that of salt solution changes
to T2. Then
1) T1 = T2 2) T1 > T2 3) T1 < T2 4) T1  T2
KEY:3
89. Machine parts are jammed in winter due to
1) increase in viscosity of lubricant 2) decrease in viscosity of lubricant
3) increase in surface tension of lubricant 4) decrease in surface tension of lubricant
KEY:1
90. Two pieces of glass plate one upon the other with a little water between them cannot be separated
easily because of
1) inertia 2) pressure 3) viscosity 4) surface tension
KEY:4
91. When stirring of a liquid is stopped, the liquid comes to rest due to
1) surface tension 2) gravity 3) viscosity 4) buoyancy
KEY:3
92. The quantity on which the rise of liquid in a capillary tube does not depend is
1) density of liquid 2) radius of capillary tube
3) angle of contact 4) atmospheric pressure
KEY:4
93. The force which tends to destroy the relative motion between liquid layers is known as
1) Force due to surface tension 2) Viscous force
3) Gravitational force 4) Force of Cohesion
KEY:2
94. The potential energy of molecule on the surface of a liquid as compared to inside the liquid is
1) zero 2) smaller 3) the same 4) Greater
KEY:4
95. A drop of water breaks into two droplets of equal size. In this process which of the following
statements is correct ?
1) the sum of temperature of the two droplets together is equal to the original temperature of
126
the drop.
2) the sum of masses of the two droplets is equal to the original mass of the drop.
3) the sum of the radii of the two droplets is equal to the radius of the original drop.
4) the sum of the surface areas of the two droplets is equal to the surface area of the original
drop.
1) 1 is correct 2) 2 is correct 3) 3 is correct 4) 4 is correct
KEY:2
96. The dynamic lift of an aeroplane is based on
1) Torricelli theorem 2) Bernoulli's theorem
3) Conservation of angular Momentum 4) Principle of continuity
KEY:2
97. It is difficult to fill a capillary tube with mercury than with water since
1) Angle of contact between glass & mercury is more than 900 and the angle of contact between glass
and water is less than 900.
2) Angle of contact is between glass and mercury is less than 900 and the angle of contact between glass
and water is more than 900.
3) Angle of contact is same for both water and mercury.
4) Mercury is less dense than water.
KEY:1
98. When temperature is increased: (2004M)
a) viscosity of the gas increases
b) viscosity of the gas decreases
c) viscosity of the liquid decreases d) viscosity of the liquid increases
1) a and c are true 2) b and c are true 3) b and d are true 4) a and d are true
KEY:1
99. A water proofing agent changes the angle of contact from
1) Acute to  /2 2)  /2 to obtuse
3) Acute to obtuse value 4) Obtuse to acute value
KEY:3
100. The nature of r-h graph ('r' is radius of capillary tube and 'h' is capillary rise) is
1) Straight Line 2) Parabola
3) Ellipse 4) Rectangular hyperbola
KEY:4
101. If 'L' is the capillary rise or dip and 'A' the cross sectional area of the tube, other conditions
being the same, then
1) L A = Constant 2) L A = Constant
3) L /A = Constant 4) L / A = Constant
KEY:2
102. Water rises in a capillary tube to a height H, when the capillary tube is vertical. If the same
capillary is now inclined to the vertical the length of water column in it will
1) Increase 2) Decrease 3) will not change
4) May increase or decrease depending on the angle of inclination.
KEY:1
103. The excess pressure inside a soap bubble is
1) Inversely proportional to the surface tension 2) Inversely proportional to its radius
3) Directly proportional to square of its radius 4) Directly proportional to its radius
KEY:2
104. The force of buoyancy is equal to
1) Weight of the body 2) Weight of the liquid displaced by the body
127
3) Apparent weight of the body 4) Divorce force
KEY:2
105. Which of the following substances has the greatest viscosity?
1) Mercury 2) Water 3) Kerosine 4) Glycerine
KEY:4
106. If two soap bubbles of different radii are connected by a tube.
1) air flows from the bigger bubbles to the smaller bubble till the sizes become equal.
2) air flows from bigger bubble to the smaller bubble till the sizes are interchanged
3) air flows from the smaller bubble to the bigger.
4) there is no flow of air.
KEY:2
107. When the value of Reynold's number is less, the predominant forces are
1) Viscous forces 2) inertial forces
3) Surface tension forces 4) gravitational forces
KEY:1
108. A gale is on a house. The force on the roof due to the gale is
1) directed downward 2) zero
3) directed upward 4) information insufficient
KEY:3
109. A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water
solution. Which of the following shows the relative nature of the liquid column in the two tubes?
A B A B
1) 2)
A B A B
3) 4)
KEY:1
110 For an ideal fluid, viscosity is
1) zero 2) infinity 3) finite but small 4) unity
KEY:1
11 1 . The water proofing agents:
1) increase the surface tension T and decrease the angle of contact  2 )
increase both T and 
3) decrease both T and  4) decrease T and
increase 
KEY:2
112. Water is flowing in a pipe of uniform cross section under constant pressure. At some place the
128
pipe becomes narrow. The pressure of water at this place.
1) remains same 2) may increase or decrease
3) increases 4) decreases.
KEY:4
113. As the depth of the river increases, the velocity of flow
1) increases 2) decreases
3) remains unchanged 4) may increase or decrease
KEY:2
114. A triangular element of the liquid is shown in the fig., Px, Py and Pz represent the pressures on
the element of the liquid. Then:
Pz Px
Py
1) Px  Py  Pz 2) Px  Py  Pz
3) Px  Py  Pz 4) Px2  Py2  Pz2 = constant
KEY:2
11 5. A vertical glass capillary tube, open at both ends, contains some
water. Which of the following shapes may be taken by the water in the
tube ?
1) 2) 3) 4)
KEY:2
116 .If air is blown through the space between a calendar suspended from a nail on wall and the
wall, then
1) The calendar moves close to the wall. 2) The calendar moves farther from the wall.
3) The position of the calendar does not change.
4) The position of the calendar may or may not change.
KEY: 1
11 7. Which of the following graphs may represent the relation between
capillary rise h and the radius r of the capillary.
h h h h
1) 2) 3) 4)
r r r r
KEY:3
129
: :PRACTICE BITS ::
1. A bucket containing water of depth 15cm is kept in a lift which is moving vertically upward with
an acceleration 2g. Then the pressure on the bottom of the bucket in kgwt/cm2 is
1) 0.45 2) 0.045 3) 0.015 4) 0.15
KEY : 2
HINT :
P  hd  g  a 
2. A bird of mass 1.23kg is able to hover by imparting a down ward velocity of 10m/s uniformly to
air of density ‘  ’ kg/m3 over on effective area 0.1m2. The acceleration due to gravity is 10m/
s2. Then the magnitude of ‘  ’. in kg/m3.
1) 0.34 2) 0.89 3) 1.23 4) 4.8
KEY : 2
HINT :
3. One end of a U-tube of uniform bore (area A) containing mercury is connected to a suction
pump. Because of it the level of liquid of density  falls in one limb. When the pump is
removed, the restoring force in the other limb is:
1) 2x  Ag 2) x  g 3) A  g 4) x  Ag
KEY:1
HINT:
Force due to excess pressure
=Restoring force=   gh  A   g  2 x  A
4. A boat having length 2m and width 1m is floating in a lake. When a man stands on the boat, it
is depressed by 3 cm. The mass of the man is
1) 50kg 2) 55kg 3) 60 kg 4) 70 kg
KEY:3
HINT:
Total weight = Force of buoyancy,
Mg=Vdg
 M  m  g  V  g   M  m   xA
5. A cube of wood supporting 200g mass just floats in water. When the mass is removed, the cube
rises by 1cm, the linear dimension of cube is
1) 10cm 2) 20cm 3) 10 2 cm 4) 5 2 cm
130
KEY:3
HINT:
Let l be the side of the cube.
Volume of the cube outside=volume of water displaced due to mass.
Water displaced is 200 gm and its volume is 200 cm3.
m  A  y   ;1l 2  200  l  10 2
6. A large block of ice 4 m thick has a vertical hole drilled through it and is floating in the middle of
water in a lake. The minimum length of rope required to scoop up a bucket full of water through
the hole is (density of ice = 0.9 CGS unit, density of water = 1 CGS unit)
1) 40 cm 2) 24 cm 3) 20 cm 4)360 cm
KEY:1
HINT:
Mg  Vin d1 g , Ald ice  Alin d1
Length of rope = side of the block - limmersed
7. A hollow metal sphere is found to float in water with the highest point just touching the free
surface of water. If ‘d’ is the density of the metal in cgs units, the fraction that represents the
volume of the hollow in terms of the volume of the sphere is
d  1
1  1 1  
1) 2) 1  
 d 3)   1
d 4)  d 
d
KEY:2
HINT:
Vcavity
VS  Vmetal V d
  1  metal  1  w
VS VS VS d
8. Excess pressure of one soap bubble is four times that of other. Then the ratio of volume of
first bubble to second one is
1) 1:64 2) 64:1 3) 4:1 4) 1:2
KEY : 1
HINT :
4T  4 3 1

 ; V   R , R  V 3 
R  3 
th
 
9. A solid body is found floating in water with   of its volume submerged. The same solid is

th
 
found floating in a liquid with   of its volume above the liquid surface. The specific gravity

of the liquid is
     
1) 2)  3)    4)   

KEY:3
HINT:
 dw  d
 ....1 1   b ....  2 
 dw  dl
131
d 
from 1 and 2 d    
l
10. A solid sphere of radius ‘R’ has a concentric cavity of radius ‘R/2’ inside it. The sphere is found to
just float in water with the highest point of it touching the water surface. The specific gravity of the
material of the sphere is
1) 1 2) 7/8 3) 8/7 4) 8/9
KEY:3
HINT:
Vcavity VS  Vmetal V d 1
  1  metal  1  w  1 
VA VS VS d S .G
11. An inverted u-tube has its two limbs in water and kerosene contained in two beakers. If water
rises to a height of 10cm, to what height does kerosene (density=0.8gm/cc ) rise in the
other limb?
1)10 cm 2) 12.5 cm 3)15 cm 4)20 cm
KEY : 2
HINT :
P1  P2  h11 g  h2  2 g  g11  h2  2
12. A vessel contains oil of density 0.8gm/cc. over mercury of density 13.6gm/c.c. A sphere floats
with half of it’s volume immersed in mercury and the other half in the oil. The density of
material of sphere. (in gm/c.c)
1) 14.4 2) 7.2 3) 3.6 4) 12.2
KEY : 2
HINT :
vdg  v1d1 g  v2 d 2 g
13. An air tight container having a lid with negligible mass and an area of 8cm2 is partially evacuated. If a
48N force is required to pull the lid off the container and the atmospheric pressure is 1.0 x 105 Ps, the
pressure in the container before it is opened must be
1) 0.6atm 2) 0.5 atm 3) 0.4 atm 4) 0.2 atm
KEY : 3
HINT :
F F
P1  P2 
; Patm  Pin 
A A
14. A brass sphere weighs 100 gm.wt in air. It is suspended by a thread in a liquid of specific
gravity =0.8. If the specific gravity of brass is 8, the tension in the thread in the in newtons is
1) 0.882 2)8.82 3)0.882 4)0.00882
KEY : 3
HINT :
 dliquid 
T  mg 1  
 dbody 

15. A cube of side 20cm is floating on a liquid with 5cm of the cube outside the liquid. If the density
of liquid is 0.8 gm/c.c then the mass of the cube is
1) 4.2 kg 2) 4.8 kg 3) 5kg 4) 5.2 kg
KEY : 2
132
HINT :
mg  Vin  g
16. A soap film in formed on a frame of area 4x10–3m2. If the area of the film in reduced t o
half, then the change in the Potential energy of the film is ( surface tension of soap solution
 40  10 3 N / m )
1) 32x10–5J 2)16x10–5J 3)8x10–5J 4) 16x105J
KEY:2
HINT:
PE1=W1=2TA
 A
PE2  W2  2T   .
2
17. If a body floats with (m/n) of its volume above the surface of water, then the relative density
th
of the material of the body is

1) (n-m) / n 2) m/n 3) n/m 4) (n-m) / m
KEY : 1
HINT :
weight of the body W d
Specific gravity  force of buoancy  F  d
B
B w
m nm
Vin  V  Vout  V   V V  
n  n 
Vm d B
   S .G
V dw
18. Water from a top emerges vertically downwards with initial velocity 4 ms-1. The cross-sectional
area of the tap is A. The flow is steady and pressure is constant throughout the stream of
water. The distance h vertically below the tap, where the cross-sectional area of the stream
2
becomes   A is (g=10m/s2) (EAMCET-2010)
3
1) 0.5 m 2) 1 m 3) 1.5 m 4) 2.2 m
KEY:2
HINT:
2
The equation of continuity A1v1  A2v2 , A  4  A  v2 ,
3
1 2 1 2
From Bernoulli’s theorem P   gh1   v1  P   gh2   v2
2 2
1 2
or g  h1  h2    v2  v1 
2
2
1 2
 6    4    h1  h2  h  h  1m
2
gh 
2 
19. A pipe having an internal diameter D is connected to another pipe of same size. Water flows
into the second pipe through ‘n’ holes, each of diameter d. If the water in the first pipe has
speed v, then speed of water leaving the second pipe is (EAMCET-2012)
133
D 2v nD 2v nd 2v d 2v
1) 2) 3) 4)
nd 2 d2 D2 nd 2
KEY : 1
HINT :
2 2
D d D 2v
A1v1  A2v2    v     v '  v ' 
2 2 nd 2
20.. The velocity of the wind over the surface of the wing of an aeroplane is 80 ms-1 and under the wing 60
ms-1. If the area of the wing is 4m2, the dynamic lift experienced by the wing is [ density of air = 1.3 kg.
m-3]
1) 3640 N 2) 7280 N 3)14560N 4)72800 N
KEY : 2
HINT :
d 2
F
2
V2  V12  A
21. A vessel has a small hole at its bottom. If water can be poured into it upto a height of 7 cm without
leakage (g=10 ms-2), the radius of the hole is (surface tension of water is 0.07 Nm-1).
1) 2 mm 2) 0.2 mm 3) 0.1 mm 4) 0.4 mm
KEY : 2
HINT :
rhdg
T
2
22. An aeroplane of mass 5000 kg in flying at an altitude of 3 km. If the area of the wings is 50m 2
and pressure at the lower surface of wings is 0.6x105 pa, the pressure on the upper surface of
wings is (in Pascal) (g=10 ms-2)
1) 59x103 2) 2x104 3) 6x103 4) 59
KEY : 1
HINT :
 p1  p2  A  mg
23. Water flows through a non-uniform tube of area of cross sections A, B and C whose values are
25, 15 and 35 cm2 respectively. The ratio of the velocities of water at the sections A, B and C
is
1) 5 : 3 : 7 2) 7 : 3 : 5 3) 21 : 35 : 15 4) 1 : 1 : 1
KEY :3
HINT :
Av  const, A1v1  A2 v2
24. An incompressible liquid flows through a horizontally tube L M N as shown in the figure. Then
the velocity ‘V’ of the liquid through the tube N is
-1
s
L 4m
2A A/2
-1
4ms
A
N Vm -1
s
1) 1 ms-1 2) 2 ms-1 3) 4.5 ms-1 4) 6 ms-1
KEY : 4
134
HINT :
A1v1  A2v2  A3v3
25. A liquid is kept in a cylindrical jar, which is rotated about the cylindrical axis. The liquid rises at its
sides. The radius of the jar is ‘r’, and speed of rotation is ‘  ’. The difference in height at the center
and the sides of the jar is
r 2 2 r 2 2 g 2g
1) 2) 3) 4)
g 2g r
2 2
r 2 2
KEY : 2
HINT :
Difference in potential energy = Rotational
1 2
Kinetic energy  mg  h1  h2  
I
2
26. At the mouth of the tap, the area of cross-section is 2.0 cm2 and the speed of water is 3m/s.
The area of cross-section of the water column 80cm below the tap is (use g=10m/s 2)
1) 0.6 cm2 2)1.2 cm2 3)1.5 cm2 4) 2.0 cm2
KEY : 2
HINT :
v22  v12  2 gh; A1v1  A2 v2
27. Capillary tubes of diameters 1, 1.5, 2 mm are dipped vertically in the same liquid. The capillary
ascents of the liquid in the tube are in the ratio
1) 2 : 3 : 4 2) 6 : 4 : 3 3) 3 : 4 : 6 4) 4 : 3 : 2
KEY : 2
HINT :
rhdg 1
T ;h 
2 cos  r
28. When two capillary tubes A and B are immersed in water, the heights of water columns are
found to be in the ratio 2 : 3. The ratio of the radii of tubes A and B is
1) 2 : 3 2) 4 : 9 3) 9 : 4 4) 3 : 2
KEY : 4
HINT :
rhdg 1
T ;r 
2 cos  h
29. A cylindrical tank of 1 meter radius rests on a platform 5m high. Initially the tank i filled with
water to a height of 5m. A plug whose of area 104 m2 is removed from an orifice on the side of
the tank at the bottom. The initial speed with which water flows out from the orifice in ms-1 is
(g=10ms-2)
1) 10 2) 5 3) 5. 2 4) 10. 2
KEY : 1
HINT :
v  2 gh
30. In the above problem, the initial speed with which water strikes the ground in ms-1 is
1) 10 2) 5 3) 5 2 4) 10 2
135
KEY : 4
HINT : vx  2 gh1 , v y  2 gh2 ; v  vx2  v y2
31. T her e is a hole at the side-bottom of a big water tank . T he ar ea of the hole is 4mm 2. Through
out it a pipe is connected. The upper surface of water is 5 m above the hole. The rate of flow
of water through the pipe is ( in m3s-1) ( g= 10ms-2)
1) 4 x 10-5 2) 4 x 105 3) 4 x 10-6 4) 28 x 10-5
KEY : 1
HINT :
Q  A 2 gh
32. The flow rate from a tap of diameter 1.25 cm is 3 lit/min. The coefficient of viscosity of water
is 10-3 Pas.The nature of flow is
1) Turbulent 2) Laminar
3) Neither laminar (or) turbulent 4) Data inadequate
KEY : 1
HINT :
 vd  Al Q
R , Q   Av  v  
  t A
33. A force of 10N is required to draw rectangular glass plate on the surface of a liquid with some
velocity. Force needed to draw another glass plate of 3 times length and 2 times width is
1) 5/3N 2) 10N 3) 60N 4) 30N
KEY : 3
HINT :
dv
F A ; F A
dx
34. Water is flowing through a capillary tube at the rate of 20 x 10-6 m3 /s. Another tube of same
radius and double the length is connected in series to the first tube. Now the rate of flow of
water in m3 s 1 is
1) 10 106 2) 3.33 106 3) 6.67 106 4) 20 106
KEY : 3
HINT :
 Pr 4 1
Q ; Q
8 l l
1
35. An artery in a certain person has been widened 1 times the original diameter. If the pressuree
2
difference across the artery is maintained constant, the blood flow through the artery will be
increased
1) 3/2 times 2) 9/4 times 3) no change 4) 81/16 times
KEY : 4
HINT :
 Pr 4
) Q : Q r 4
8 l
36. The potential energy of the liquid of surface tension “T” and density  that rises into the
capillary tube is
136
2 T 2 T 2
1)  T  g
2 2 2
2) 4 T 2
 g
2
3)
g
4)
g
KEY:3
HINT:
P.E  mg  h / 2    r 2 h  gh / 2
37. Water flowing from a hose pipe fills a 15 litre container in one minute. The speed of water from the
free opening of radius 1 cm is ( in ms-1)
 2 .5
1) 2.5 2) 3) 4) 5 
2 .5 
KEY : 3
HINT :
 Pr4
Q ; Q   r 2v
8l
38. A small air bubble of 0.1 mm diameter is formed just below the surface of water. If surface
tension of water is 0.072 Nm-1, the pressure inside the air bubble in kilo pascal is (Atmospheric
pressure = 1.01 x 105 pa)
1) 28.9 2) 0.289 3) 0.0289 4) 103.88
KEY:4
HINT:
2T
Pin  P0 
R
39. If a soap bubble of radius 3 cm coalesce with another soap bubble of radius 4 cm under isothermal
conditions, the radius of the resultant bubble formed is in cm
1) 7 2) 1 3) 5 4) 12
KEY : 4
HINT :
r  r12  r22
40. Two liquids are allowed to flow through two capillary tubes of lengths in the ratio 1 : 2 and radii
in the ratio 2 :3 under the same pressure difference. If the volume rates of flow of the liquids
are in the ratio 8 : 9, the ratio of their coefficients of viscosity is
1) 1 : 3 2) 3 :1 3) 4 : 9 4) 9 : 4
KEY : 3
HINT :
 Pr 4 r4
Q ;
8 l Ql
41. The viscous resistance of a tube to liquid flow is R. Its resistance for a narrow tube of same
length and 1/3 times radius is
1) R/3 2) 3R 3) 27R 4) 81R
KEY : 4
HINT :
137
8 l 1
R ; R 4
r 4
r
42. Two identical tall jars are filled with water to the brim. The first jar has a small hole on the side
wall at a depth h/3 and the second jar has a small hole on the side wall at a depth of 2h/3, where
‘h’ is the height of the jar. The water issuing out from the first jar falls at a distance R1 from the
base and the water issuing out from the second jar falls at a distance R 2 from the base. The
correct relation between R1 and R2 is
1) R1 > R2 2) R1 < R2 3) R2 = 2 x R 1
4)R1=R2
KEY:4
HINT:
R1  v1t1; R2  v2t2
43. There are two holes O1 and O2 in a tank of height H. The water emerging from O1 and O2
strikes the ground at the same points, as shown in fig. Then:
h1 O1
h2
H O2
R
1) H = h1 + h2 2) H = h2 – h1 3) H = h1h2 4) H = h2/h1
KEY:1
HINT:
R  Vefflux  time; R1  R2
44. A tank full of water has a small hole at the bottom. If one-fourth of the tank is emptied in t 1
seconds and the remaining three-fourths of the tank is emptied in t 2 seconds. Then the ratio
t1
t2 is
1 2
1) 3 2) 2 3) 4) 1
2 3
KEY:4
HINT:
t1   
h  3h / 4 ; t2 3h / 4 
45. There are two holes one each along the opposite sides of a wide rectangular tank. The cross
section of each hole is 0.01m2 and the vertical distance between the holes is one meter. The
tank is filled with water. The net force on the tank in Newton when water flows out of the holes
is: (Density of water 1000kg/m3)
1) 100 2) 200 3) 300 4) 400
KEY:2
HINT:
Fnet  F1  F2  V12   aV22   a  2 gh1  2 gh2 
138
 2  ga  h1  h2   2  gah
46. A tank with vertical walls is mounted so that its base is at a height H above the horizontal
ground. The tank is filled with water to a depth 'h' . A hole is punched in the side wall of the
tank at a depth ' x ' below the water surface. To have maximum range of the emerging stream,
the value of x is
H h H h H h 3 H  h
1) 2) 3) 4)
4 2 3 4
KEY:2
HINT:
2  H   h  x
R  2 gx
g
47. A hole is made at the bottom of tank filled with water (density = 103 kg/m3). If the total pressure
at the bottom of the tank is 3 atm (1 atm = 105 N/m2), then the velocity of efflux is
1) 400 m/s 2) 200 m/s 3) 600 m/s 4) 500 m/s
KEY:1
HINT:
Pressure due to water in the tank=3atm-1 atm=2atm=20m of water column
height of the water in teh tank is h=20m
Velocity of efflux= 2 gh  2  10  20  400m / sec
48. Eight spherical rain drops of the same mass and radius are falling down with a terminal speed
of 6 cm-s-1. If they coalesce to form one big drop, what will be the terminal speed of bigger
drop? (Neglect the buoyancy of the air) (EAMCET-2009)
1) 1.5 cms- 1 2) 6 cms-1 3) 24 cms-1 4) 32 cms-1
KEY : 3
HINT :
VB  n 2 / 3VS
49. The velocity of a ball of mass ‘m’ density ‘d1’ when dropped in a container filled with glycerin
of density ‘d2’ becomes constant after some time. The viscous force acting on the ball will be
 d1   d2   d1  d 2   d1  d 2 
1) mg  d  2) mg  1  d  3) mg  d  4) mg  d 
 2  1   1   2 
KEY : 2
HINT :
Viscous force=Apparent weight-Force of buoyancy
 FY  mg  FB
50. One spherical ball of radius R, density d released in a liquid of density d/2 attains a terminal
velocity V. Another ball of radius 2R and density 1.5d, released in the liquid will attain a terminal
velocity
1) 2V 2) 4V 3) 6V 4) 8V
KEY:4
HINT:
139
2 r    0  g v1 r12    0 
2
v   1 
9  v2 r22  2  0 
51. When a solid ball of volume V is falling through a viscous liquid, a viscous force F acts on it. If
another ball of volume 2V of the same material is falling through the same liquid then the viscous
force experienced by it will be (when both fall with terminal velocities).
1) F 2) F/2 3) 2F 4) F/4
KEY:3
HINT:
F  6 rv  v  r 2 , F  r 3 , F  V 
F1 V1
 
F2 V2
52. The level of a liquid in a vessel kept constant at 50cm. It has three identical horizontal tubes
each of length 60cm coming out at heights 5, 10 and 15 cm respectively. If a single tube of the
same radius as that of the three tubes can replace the three tubes when placed horizontally at
the bottom of the vessel length of that tube is
1) 25 cm 2) 40 cm 3) 12.5 cm 4) 50 cm
KEY:1
HINT:
 Pr 4  P1 r 4  P2 r 4  P3r 4
  
8 l 8l1 8 l2 8l3
H H  h1 H  h2 H  h3
  
L l l l
53. A tube of radius R and length L is connected in series with another tube of radius R/2 and
length L/8. If the pressure across the tubes taken together is P, the pressure across the two
tubes seperately are :
1) and 2) and 3) and 4) and

KEY:4
HINT:
54. A capillary tube is attached horizontally at a constant head arrangement. If the radius of the
capillary tube is increased by 10%, the rate of flow of liquid changes by about
1) -40% 2) +40% 3) +21% 4) +46%
KEY:4
HINT:
140
55. Three horizontal capillary tubes of same radii and lengths L1, L2 and L3 are filled side by side a little
above the bottom, to the wall of a tank that filled with water. The length of a single capillary tube of
same radius that can replace the three tubes such that the rate of flow of water through the single tube
equals the combined rate of flow through the three tubes is
1) 2) 3) 4)
KEY:2
HINT:
56. A metallic wire of diameter “d” is lying horizontally on the surface of water. The maximum
length of wire so that is may not sink will be
2T 2Tg 2 d
1) 2) 3) 4)any length
 dg d Tg
KEY:4
HINT:
the expression is independent of length
57. A liquid is filled into a semi elliptical crossection with a as semi major axis and b as semi minor
axis. The ratio of surface tension forces on the curved part and the plane part of the tube in
vertical position will be
1) 2) 3) 4)
KEY:1
HINT:
58. The length of a rubber cord floating on water is 5 cm. The force needed to pull the cord out of
water is ..... N (surface tension of water is 7.2 x 10-2 Nm-1) .
1) 7.2 x 10-3 2) 7.2x10-4 3) 7.2x10- 5
4)7.2x 10-2
KEY : 3
HINT :
59. Calculate the force required to separate the glass plates of area 10 -2 m2 with a film of water
0.05 mm thickness between them (surface tension of water = 70 x 10-3 N/m)
1) 28 N 2) 112 N 3) 5.6 N 4) 11.2 N
KEY : 1
HINT :
60. A thin wire ring of 3 cm radius float on the surface of a liquid. The pull required to raise the ring
141
before the film breaks is 30.14 x 10-3N more than its weight. The surface tension of the liquid
(in Nm–1) is
1) 80x10–3 2) 87x10–3 3) 90x10–3 4) 98x10–3
KEY : 1
HINT :
61. When a ‘U’ shaped slider of negligible mass is dipped in a soap solution and lifted, a thin film
of soap is formed in the frame. It supports a weight of 2.0c10-2N. If the length of the slider is 40
cm, the surface tension of the film of soap is
1) 25 Nm-1 2) 2.5 Nm-1 3) 2.5 x 10-2 Nm-1 4) 2.5 x 10-3 Nm-1
KEY : 3
HINT :
62. A ring of inner and outer radii 8 and 9 cm is pulled out of water surface with a force of [S.T of
water(T)=70 dyne/cm]
1) 26 x 10-2 N 2) 12.6 x 10-2 N 3) 7.48 x 10-2 N 4) 3.08 x 10-2 N
KEY : 3
HINT :
63. In Fig(i) a thin film supports a small weight . The weight suppported by a film of the
same liquid at the same tempertaure in fig.(ii) is ___
1)3.5x10–2N 2)3.5x10–3N 3)3.5x10–1N 4)3.5x10–4 N

KEY : 1
HINT :
Force due to S.T.=Weight
same liquid, same temperature, same length of the film, supports same weight.
64. A film of water is formed between two straight parallel wires of length 10cm each separated by
0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism how
much work will have to be done of water (Surface tension of water )
142
1) 2) 3) 4)
KEY:2
HINT:
Find out
65. The work done is blowing a soap bubble of volume “V” is W. The work done in blowing
a soap bubble of volume “2V” is
2 2
1) W 2) 2 3 W 3) 3 3 W 4) 2 W
KEY:2
HINT:
66. Work of joule is required to be done in increasing the size of a soap film from 10cm x
6 cm to 10cm x 11 cm. The surface tension of the film is(in N/m )
1) 5 x 10–2 2) 6 x 10–2 3) 1.5 x 10–2 4) 1.2 x 10–2
KEY : 2
HINT :
67. The work done in increasing the raduis of a soap bubble from 4 cm to 5 cm is ......Joule(given
surface tension of soap water to be 25 x 10-3 N/m)
1) 0.5657x10-3 2) 5.657x10-3 3) 56.5x10-3 4) 565x10-3
KEY : 1
HINT :
68. The reading of a pressure meter attached with a closed water pipe is 3.5 x 10 5 N m-2. On
opening the valve of the pipe, the reading of pressure meter is reduced to 3
x 10 N m . Calculate the speed of water flowing in the pipe.
5 -2
1)10 cm/s 2)10 m/s 3) 0.1 m/s 4) 0.1 cm/s

KEY : 2
HINT :
69. A mercury drop of radius 1 cm is sprayed into 106 drops of equal size. The energy expended in
joule is (surface tension of mercury is 460 x 10-3N/m)
1) 0.057 2) 5.7 3) 4)
KEY : 1
HINT :
143
70. A wooden cube is found to float in water with ½ cm of its vertical side above the water. On
keeping a weight of 50gm over its top, it is just submerged in the water. The specific gravity of
wood is
1) 0.8 2) 0.9 3) 0.85 4) 0.95
KEY:4
HINT:
weight of the body=weight of displaced liquid
71. 8000 identical water drops combine together to form a big drop. Then the ratio of the final
surface energy to the initial surface energy of all the drops together is.
1) 1 : 10 2) 1 : 15 3) 1 : 20 4) 1 : 25
KEY : 3
HINT :
72. A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density 800 kg m-3 and of
surface tension 3x10-2 Nm-2. The angle of contact of liquid-glass is and given . If
g = 10ms-2 the rise of liquid in the capillary tube is ..... Cm
1) 9 2) 0.9 3) 9 x 10-3 4) 0.09
KEY : 2
HINT :
73. When a clean lengthy capillary tube is dipped vertically in a beaker containing water, the
water rises to a height of 8 cm. What will happen if another capillary tube of length 4 cm and
same radius is dipped vertically in the same beaker containing water. (Angle of contact of
water is 00.)
1) Water will flow out like a fountain.
2) Water will rise to a height of 4 cm only and the angle of contact will be zero.
3) Water will rise to a height of 4 cm only and the angle of contact will be 600.
4) Water will not rise at all
KEY : 3
HINT :
74. A capillary tube is taken from the Earth to the surface of the Moon. The rise of the liquid column on
the Moon (acceleration due to gravity on the Earth is 6 times that of the Moon) is
1) six times that on the Earth surface 2) 1/6 that on the Earth’s surface
3) equal to that on the Earth’s surface 4) zero
KEY : 1
HINT :
75. When a capillary tube is lowered into water, the mass of the water raised above the outside
level is 5 gm. If the radius of the tube is doubled the mass of water that raises in the tube above
144
the outside level is
1)1.25 gm 2) 5 gm 3)10 gm 4) 20 gm
KEY : 3
HINT :
76. When a body lighter than water is completely submerged in water, the buoyant force acting on
it is found to be ‘n’ times its weight. The specific gravity of the material of the body is
1 1 1
1) 2) 3) n 4) n 
1 n n n
KEY : 2
HINT :
77. A 20 cm long capillary tube is dipped in water. The water rises upto 8 cm. If the entire arrangement
is put in a freely falling elevator the length of water coloumn in the capillary tube will be
1) 4 cm 2) 20 cm 3) 8 cm 4) 10 cm
KEY : 2
HINT :
In a freely falling lift capillary height=full length of the capillary tube.
78. When a cylindrical tube is dipped vertically into a liquid the angle of contact is 1400. when the
tube is dipped with an inclination of 400 the angle of contact is
1)1000 2) 1400 3) 1800 4) 600
KEY : 2
HINT :
Angle of contact is independent of tilting angle
79. Water rises in a striaght capillary tube upto a height of 5 cm when held vertical in water. If the
tube is bent as shown in figure then the height of water column in it will be
1) 5 cm 2) less than 5 cm 3) more than 5 cm 4) 5 cos
KEY :1
HINT :
Capillary rise independent of shape of capillary pipe if radius of the pipe does not changes.
80. A tube is mounted so that it’s base is at height ‘h’ above the horizontal ground. The tank is filled
with water to a depth ‘h’. A hole is punched in the side of the tank at depth ‘y’ below water surface.
Then the value of ‘y’ so that the range of emerging stream would be maximum is
1) h 2) h/2 3) h/4 4) 3h/4
KEY:1
HINT:
145
dR
R to be maximum, dy  0 , which gives y=h
81. Twoliquiddropshavetheirdiameters as
1 mm and 2 mm. The ratio of excess pressures in them is
1) 1 : 2 2) 2 : 1 3) 4 : 1 4) 1 : 4
KEY : 2
HINT :
2T
P 
r
82. The pressure inside soap bubble is 1.01 and 1.02 atmosphere respectively the ratio of their
volume
1) 102:101 2) (102)3:(101)3 3) 8 : 1 4) 2 : 1
KEY : 3
HINT :
4T  4 1

P  ;  V   R3 , R  V 3 
R  3 
83. A spherical soap bubble of radius 1 cm is formed inside another of radius 3 cm. The radius of
single soap bubble which maintains the same pressure difference as inside the smaller and
outside the larger soap bubble is ........cm.
1) 1 2) 0.8 3) 0.5 4) 0.25
KEY:2
HINT:
r1r2
r
r1  r2
84. If the shearing stress between the horizontal layers of water in a river is 1.5 milli newton/ m 2
and  water  1  10  3 pa.s , The velocity gradient is ... s-1
1) 1.5 2) 3 3) 0.7 4) 1
KEY : 1
HINT :
dv F

dx A
85. The depth of water at which air bubble of radius 0.4mm remains in equilibrium is
Twater  72  103 N / m 
1) 3.67cm 2)3.67 m 3)6.37 cm 4)5.32 cm
KEY:1
HINT:
2T
hdg 
r
146
86. Two separate air bubbles having radii ( r1  .002cm, r2  .004cm ) formed of same liquid
T  0.07 N/m come together to form a double bubble. Find the radius of curvature of the
internal film surface common to both the bubbles.
1) 0.001cm 2) 0.002 cm 3) 0.004 cm 4) 0.003 cm
KEY:3
HINT:
r1r2
r
r2  r1
as the excess pressure is always towards concave surface & pressure in smaller bubble is greater than
larger bubble, the common surface is concave towards the centre of the small bubble.
87. The excess pressure inside a spherical soap bubble of radius 1 cm is balanced by a column of
oil (Specific gravity = 0.8), 2 mm high, the surface tension of the bubble is (EAM-10)
1) 3.92 N/m 2) 0.0392 N/m 3) 0.392 N/m 4) 0.00392 N/m
KEY:2
HINT:
The excess pressure of soap bubble
4T 4T Rh  g
p  , h g  ;T   0.0392 N / m
R R 4
88. In a car lift compressed air experts a force F1 on a small piston having a radius of 5cm. This
pressure is transmitted to a second piston of radius 15cm. If the mass of the car to be lifted is
1350 kg. What is F1?
1) 14.7 x103 N 2)1.47x103 N 3) 2.47x103 N 4) 24.7x103 N
KEY : 2
HINT :
F1 F2

A1 A2
89. The lower end of a capillary tube of radius r is placed vertically in water then with the rise of water in the
capillary heat evolved is
 r 2 h 2 dg  r 2 h 2 dg  r 2 h 2 dg  r 2 h 2 dg
1)  2)  3)  4) 
2J J 2J J
KEY:1
HINT:
mgh
Work done = heat evolved; W  JQ; Q 
2J
90. Four identical capillary tubes a, b, c and d are dipped in four beakers containing water with
tube ‘a’ vertically, tube ‘b’ at 300, tube‘c’ at 450 and tube ‘d’ at 600 inclination with the vertical.
Arrange the lengths of water column in the tubes in descending order.
1) d, c, b, a 2) d, a, b, c 3) a,c,d,b 4) a,b,c,d
KEY:1
HINT:
h  angle of inclination
91. A vessel whose bottom has round holes with diameter of 1mm is filled with water Assuming
that surface tension acts only at holes, then the maximum height to which the water can be
147
filled in vessel without leakage is (Given surface tension of water is 75
x 10 -3 N/m and g = 10m / s2)
1) 3 cm 2) 0.3 cm 3) 3 mm 4) 3 m
KEY:1
HINT:
Gauge pressure=excess pressure above the meniscus
2T 2T
 hdg  h
r rdg
92. Water rises to a height h1 in a capillary tube in a stationary lift. If the lift moves up with
uniform acceleration it rises to a height h2 , then acceleration of the lift is
 h2  h1   h2  h1   h1  h2   h1  h2 
1)  h  g 2)  g 3)  g 4)  h  g
 2  h
 1   h1   2 
KEY:4
HINT:
h2  g  a   h1 g
93. A liquid drop of diameter D breaks up into 27 drops. Find the resultant change in energy.
 TD 2
1) 2 TD 2 2)  TD 2 3) 4) 4 TD 2
2
KEY:1
HINT:
W  T  4 R 2  n1/ 3  1
94. The radii of the two columns in a ‘U’ tube are ‘r1’ & ‘r2’, when a liquid of density ( angle of
contact is 0o } is filled in it, the level difference of the liquid in the two arms in ‘h’. The surface
tension of the liquid is {g = acceleration due to gravity } (2004-M)
 ghr1r2  gh  r2  r1  2  r1  r2  2  r1  r2 
1)
2  r2  r1  2)
2r2 r1
3)
 ghr2 r1
4)
 gh
KEY:1
HINT:
2T 1 1 
h   
dg  r1 r2 
148
149
Previous JEE Mains Questions and Solutions
VISCOSITY
Pressure , Density , Pascal’s Law and Archimedis Principle :
1. A hollow spherical shell at outer radius 𝑹 floats just submerged under the water surface. The
𝟐𝟕
inner radius of the shell is 𝒓. Ifthe specific gravity ofthe shell material is 𝐰. 𝐫. 𝐭 water, the
𝟖
value of 𝒓 is: [5 Sep. 2020 (I)]
𝟖 𝟒 𝟐 𝟏
(a) 𝟗 𝑹 (b) 𝟗 𝑹 (c) 𝟑 𝑹 (d) 𝟑 𝑹
SOLUTION : (a)
In equilibrium, 𝒎𝒈 = 𝑭𝐁
𝑭𝑩 = 𝑽𝐩𝟎 𝒈 and mass =volume × density
𝟒 𝟒
𝝅 𝑹𝟑 − 𝒓𝟑 𝐩𝟎 𝒈 = 𝝅𝑹𝟑 𝐩𝒘 𝒈
𝟑 𝟑
𝐩 𝟐𝟕
Given, relative density, 𝐩 𝟎 =
𝒘 𝟖
𝒓 𝟑 𝟐𝟕
⇒ 𝟏− 𝐩 = 𝐩𝒘
𝑹 𝟖 𝒘
𝒓𝟑 𝟗 𝟏 𝒓𝟑 𝟐 𝒓𝟑
⇒𝟏− = ⇒ 𝟏 − = ⇒ =
𝑹𝟑 𝟐𝟕 𝟑 𝑹𝟑 𝟑 𝑹𝟑
𝟏/𝟑
𝒓 𝟐 𝒓𝟑 𝟖 𝒓𝟑 𝟖 𝟏𝟗
⇒ = ⇒𝟏− 𝟑 = ⇒ 𝟑 =𝟏− =
𝑹 𝟑 𝑹 𝟐𝟕 𝑹 𝟐𝟕 𝟐𝟕
𝟖
𝒓 = 𝟎. 𝟖𝟗𝑹 = 𝑹.
𝟗
2. An air bubble of radius 1 cm in water has an upward acceleration 𝟗. 𝟖 cm 𝐬−𝟐 . The density of
water is 1 gm 𝐜𝐦−𝟑 and water offers negligible drag force on the bubble. Themass ofthe bubble
is 𝐠 = 𝟗𝟖𝟎𝐜𝐦/𝐬𝟐 [4 Sep. 2020 (I)]
(a) 𝟒. 𝟓𝐥𝐠𝐦 (b) 𝟑. 𝟏𝟓𝒚𝐧 𝐜 𝟒. 𝟏𝟓𝕹 (d) 𝟏. 𝟓𝟐𝕹
SOLUTION : (c)
Given: Radius of air bubble = 𝟏 cm,
Upward acceleration ofbubble, 𝒂 = 𝟗. 𝟖𝐜𝐦/𝐬𝟐 ,
𝐩𝐰𝐚𝐭𝐞𝐫 = 𝟏𝐠𝐜𝐦−𝟑
𝟒𝝅 𝟒𝝅
Volume 𝑽 = 𝒓𝟑 = × 𝟏 𝟑
= 𝟒. 𝟏𝟗𝐜𝐦𝟑
𝟑 𝟑
↑𝒂
𝑭𝐛𝐮𝐨𝐲𝐚𝐧𝐭
Fbuoyant −𝒎𝒈 = 𝒎𝒂 ⇒ 𝒎 = 𝒈+𝒂
𝑽𝐩𝟎𝐉 𝒈 𝑽𝐩𝟎𝐉 𝟒. 𝟏𝟗 .× 𝟏 𝟒. . 𝟏𝟗
𝒎= = 𝒂 = = = 𝟒. 𝟏𝟓𝐠
𝒈+𝒂 𝟏+𝒈 𝟗𝟖 𝟏𝟎𝟏
𝟏+
𝟗𝟖𝟎
3. Two identical cylindrical vessels are kept on the ground and each contain the same liquid
ofdensity 𝒅. The area of the base ofboth vessels is 𝑺 but the height ofliquid in one vessel is 𝒙𝟏
and in the other, 𝒙𝟐 . When both cylinders are connected through a pipe ofnegligible volume
very close to the bottom, the liquid flows from one vessel to the other until it comes to
equilibrium at a new height. The change in energy ofthe system in the process is:
[4 Sep. 2020 (II)]

𝟑 𝟏
(a) 𝒈𝒅𝑺 𝒙𝟐𝟐 + 𝒙𝟐𝟏 (b) 𝒈𝒅𝑺 𝒙𝟐 + 𝒙𝟏 𝟐
(c) 𝟒 𝒈𝒅𝑺 𝒙𝟐 − 𝒙𝟏 𝟐
(d) 𝟒 𝒈𝒅𝑺 𝒙𝟐 − 𝒙𝟏 𝟐
SOLUTION : (d)
𝒙𝟏 𝒙𝟐
Initial potential energy, 𝑼𝟏 = 𝐩𝑺𝒙𝟏 𝒈 ⋅ + 𝐩𝑺𝒙𝟐 𝒈 ⋅
𝟐 𝟐
𝒙𝒇
Final potential energy, 𝑼𝒇 = 𝐩𝑺𝒙𝒇 𝒈 ⋅ ×𝟐
𝟐
By volume conservation, 𝑺𝒙𝟏 + 𝑺𝒙𝟐 = 𝑺 𝟐𝒙𝒇
𝒙𝟏 + 𝒙𝟐
𝒙𝒇 =
𝟐
When valve is opened loss in potentail energy occur till water level become same.
𝜟𝑼 = 𝑼𝒊 − 𝑼𝒇
𝒙𝟐𝟏 𝒙𝟐𝟐
𝜟𝑼 = 𝝆𝑺𝒈 + − 𝒙𝟐𝒇
𝟐 𝟐
𝒙𝟐𝟏 𝒙𝟐𝟐 𝒙𝟏 + 𝒙𝟐 𝟐
= 𝐩𝑺𝒈 + −
𝟐 𝟐 𝟐
𝐩𝑺𝒈 𝒙𝟐𝟏 𝒙𝟐𝟐 𝐩𝑺𝒈 𝟐

= + − 𝒙𝟏 𝒙𝟐 = 𝒙𝟏 − 𝒙𝟐
𝟐 𝟐 𝟐 𝟒
4. A leak proofcylinder oflengLh 1 , made ofa metal which has very low coefficient ofexpansion
is floating vertically in water at 𝟎∘ 𝐂 such that its height above the water surface is 20 cm. When
the temperature ofwater is increased to 𝟒∘ 𝐂, the height of the cylinder above the water surface
becomes 21 cm. The density ofwater at 𝑻 = 𝟒∘ 𝐂, relative to the density at 𝑻 = 𝟎∘ 𝐂 is close to:
[𝟖𝐉𝐚𝐧𝟐𝟎𝟐𝟎(𝐃]
(a) 𝟏. 𝟐𝟔 (b) 1. (c) 𝟏. 𝟎𝟏 (d) 𝟏. 𝟎𝟑
SOLUTION : (c)
When cylinder is floating in water at 𝟎∘ 𝐂
Net thrust = 𝑨 𝒉𝟐 − 𝒉𝟏 𝐩𝟎∘ 𝒄 𝒈 = 𝑨 𝟏𝟎𝟎 − 𝟖𝟎 𝐩𝟎∘ 𝒄 𝒈
𝟎𝐨
When cylinder is floating in water at 𝟒∘ 𝐂
Net thrust = 𝑨 𝒉𝟐 − 𝒉𝟏 𝐩𝟒∘ 𝒄 𝒈 = 𝑨 𝟏𝟎𝟎 − 𝟐𝟏 𝐩𝟒∘ 𝒄 𝒈
𝟒∘
𝐩𝟒∘ 𝒄 𝟖𝟎
= = 𝟏. 𝟎𝟏
𝐩𝟎∘ 𝒄 𝟕𝟗
𝒓𝟐
5. Consider a solid sphere of radius 𝑹 and mass density 𝒓 = 𝐩𝟎 𝟏 − 𝑹𝟐 , 𝟎 < 𝒓 ≤ 𝑹. The
minimum density ofa liquid in which it will float is: [8 Jan 2020 (𝐃]
𝐩𝟎 𝐩𝟎 𝟐𝐩𝟎 𝟐𝐩𝟎
(a) (b) (c) (d)
𝟑 𝟓 𝟓 𝟑
SOLUTION : (c)
For minimum density ofliquid, solid sphere has to float (completely immersed) in the liquid.
𝒎𝒈 = 𝑭𝑩 (also 𝑽𝐢𝐫𝐞𝐫𝐬𝐞𝐝 = 𝑽𝐭𝐨𝐭𝐚𝟏 )
𝟒
𝐨𝐫 𝐩 𝒅𝑽 = 𝝅𝑹𝟑 𝝆𝓵
𝟑
𝒓𝟐
[𝐩 𝒓 = 𝐩𝟎 𝟏 − 𝑹𝟐 𝟎 < 𝒓 ≤ 𝑹 given]
𝑹 𝒓𝟐 𝟒
⇒ 𝐩 𝟒𝝅
𝟎 𝟎
𝟏 − 𝑹𝟐 . 𝒓𝟐 𝒅𝒓 = 𝟑 𝝅𝑹𝟑 𝐩𝒍
𝑹
𝒓𝟑 𝒓𝟓 𝟒
⇒ 𝟒𝝅𝐩𝟎 − = 𝝅𝑹𝟑 𝐩𝓵
𝟑 𝟓𝑹𝟐 𝟎
𝟑
𝟒𝝅𝐩𝟎 𝑹𝟑 𝟐 𝟒
× = 𝝅𝑹𝟑 𝐩𝒑
𝟑 𝟓 𝟑
𝟐𝐩𝟎
𝐩𝓵 =
𝟓
=𝟎
6. Two liquids of densities 𝐩𝟏 and 𝐩𝟐 𝐩𝟐 = 𝟐𝐩𝟏 are filled up behind a square wall of side 10 as
shown in figure. Each liquid has a height of5 . The ratio ofthe forces due to these liquids
exerted on upper part MNto that at the lower part NO is(Assume that the liquids are not
mixing): [8 Jan 2020 (II)]
(a) 1/3 (b) 2/3 (c) ½ (d) ¼
SOLUTION : (𝐝 )
Let 𝑷𝟏 , 𝑷𝟐 and 𝑷𝟑 be the pressure at points 𝑴, 𝑵 and 𝑶 respectively.
Pressure is given by 𝑷 = 𝐩𝒈𝒉
Now, 𝑷𝟏 = 𝟎 𝒉 = 𝟎
𝑷𝟐 = 𝐩𝒈 𝟓
𝑷𝟑 = 𝐩𝒈 𝟏𝟓
= 𝟏𝟓𝐩𝒈
𝑷𝟏 +𝑷𝟐
Force on upper part, 𝑭𝟏 = 𝑨
𝟐
𝑷𝟐 +𝑷𝟑
Force on lower part, 𝑭𝟐 = 𝑨
𝟐
𝑭𝟏 𝟓𝐩𝒈 𝟓 𝟏
= = =
𝑭𝟐 𝟐𝟎𝐩𝒈 𝟐𝟎 𝟒
7. A cubical block ofside 𝟎. 𝟓𝐦 floats on water with 30% of its volume under water. What is the
maximum weight that can be put on the block without fully submerging it under water? [Take,
density ofwater = 𝟏𝟎𝟑 𝐤𝐠/𝐦𝟑 ] [10 April 2019 (II)]
(a) 𝟒𝟔. 𝟑 kg (b) 𝟖𝟕. 𝟓 kg (c) 𝟔𝟓. 𝟒𝐤𝐠 (d) 30. 𝐥𝐤𝐠
SOLUTION : (b)
𝟑𝟎
When a 𝜾𝐗) 𝐝𝐲 floats then the weight ofthe body = upthrust (50)3 × 𝟏𝟎𝟎 × 𝟏 × 𝒈 = 𝑴𝐜𝐮𝐛𝐞 𝒈 (i)
Let 𝐦 mass should be placed, then (50) 𝟑 × 𝟏 × 𝒈 = 𝑴𝐜𝐮𝐛𝐞 + 𝒎 𝒈 (ii)
Subtracting equation (i) 𝐟𝐢𝐢 𝐨𝐦 equation (ii), we get

𝟑
⇒ 𝐦𝐠 = 𝟓𝟎 × 𝐠 𝟏 − 𝟎. 𝟑 = 𝟏𝟐𝟓 × 𝟎. 𝟕 × 𝟏𝟎𝟑 𝐠
⇒ 𝐦 = 𝟖𝟕. 𝟓𝐤𝐠
8. A submarine experiences apressure 𝐨𝐟 𝟓. 𝟎𝟓 × 𝟏𝟎𝟔 Pa at depth of 𝒅𝟏 in a sea. When it goes

further to a depth of 𝒅𝟐 , it experiences a pressure of 𝟖. 𝟎𝟖 × 𝟏𝟎𝟔 Pa. Then𝒅𝟏 − 𝒅𝟏 is
approximately (density ofwater= 𝟏𝟎𝟑 𝐤𝒚𝐦𝟑 and acceleration due to gravity = 𝟏𝟎 ms- 2):
[10 April 2019 (II)]
(a) 300 m (b) 400 m (c) 𝟔𝟎𝟎𝐦 (d) 500 m
SOLUTION : (a)
𝐏𝟏 = 𝐏𝟎 + 𝐩𝐠𝐝𝟏
𝐏𝟐 = 𝐏𝟎 + 𝐩𝐠𝐝𝟐
𝜟𝐏 = 𝐏𝟐 − 𝐏𝟏 = 𝐩𝐠𝜟𝐝
𝟑. 𝟎𝟑 × 𝟏𝟎𝟔 = 𝟏𝟎𝟑 × 𝟏𝟎 × 𝜟𝐝
⇒ 𝜟𝐝 = 𝟑𝟎𝟎𝐦
𝟒
9. A wooden block floating in a bucket ofwater has 𝟓 ofits volume submerged. When certain
amount ofan oil poured into the bucket, it is found that the block is just under the oil
surface with halfofits volume under water and halfin oil. The density ofoil relative to that
ofwater is: [9 April 2019 (II)]
(a) 𝟎. 𝟓 (b) 𝟎. 𝟖 (c) 𝟎. 𝟔 (c) 𝟎. 𝟕

SOLUTION : (c)
𝟒𝑽
𝑴𝒈 = 𝒑𝝎𝒈
𝟓
𝑴 𝟒𝒑𝝎 𝟒𝐩𝐨)
or = or 𝐩 =
𝑽 𝟓 𝟓
When block floats fully in water and oil, then
𝑴𝒈 = 𝑭𝒃𝐉 + 𝑭𝒃𝟐
𝑽 𝑽
𝒑𝑽 𝒈 = 𝝆𝐨𝐢𝟏 𝒈 + 𝝆𝝎𝒈
𝟐 𝟐
𝟑
or 𝐩𝐨𝐢𝟏 = 𝟓 𝐩𝖈𝟎 = 𝟎. 𝟔𝝆𝖈𝟎
10. A load of mass kg is suspended 𝐟𝐢: 𝐨𝐦 a steel wire of length 2 and radius 𝟏. 𝟎 mm in
Searle’s apparatus experiment. The increase in length produced in the wire is 𝟒. 𝟎 mm. Now
the load is fully immersed in a liquid ofrelative density2. The relative density ofthe
material ofload is 8. The new value of increase in length ofthe steel wire is:
[12 Jan. 2019 (II)]
(a) 𝟑. 𝟎mm (b) 𝟒. 𝟎mm (c) 𝟓. 𝟎mm (d) Zero
SOLUTION : (a)
𝜞 𝜟𝓵
Using 𝐀 = 𝐘 ⋅ 𝓵
⇒ 𝜟𝓵 ∝ 𝜞 …….. (i)
𝐏𝒑 𝟐
𝐓 = 𝐌𝐠 − 𝐟𝐁 = 𝐌𝐠 − 𝐏𝐛 = 𝟏 − 𝐏𝐛 𝐌𝐠 = 𝟏 − 𝟖 𝐌𝐠
𝟑
𝐓= 𝐌𝐠
𝟒
From eqn (i)
𝜟𝓵′ 𝐓′ 𝟑 𝟑 𝟑
= = 𝟒 [Given: 𝜟𝓵 = 𝟒 mm] 𝜟𝓵′ = 𝟒 ⋅ 𝜟𝓵 = 𝟒 × 𝟒 = 𝟑 mm
𝜟𝓵 𝐓
11. A soap bubble, blown by a mechanical pump at the mouth ofa tube, increases in volume,
with time, at a constant rate. The graph that correctly depicts the time dependence of
pressure inside the bubble is given by: [12 Jan. 2019 (II)]
SOLUTION : (d)
𝟒
𝐕 = 𝐜𝐭𝐨𝐫, 𝝅𝐫 𝟑 = 𝐜𝐭
𝟑
𝟏
⇒ 𝐫 = 𝐤𝐭 𝟑
𝟒𝐓
𝐏 = 𝐏𝟎+
𝐤𝐭 𝟏/𝟑
𝟏
𝐏 = 𝐏𝟎 + 𝐜
𝐭 𝟏,𝟑
12. Aliquid ofdensity is coming out ofa hose pipe ofradius a with horizontal speed vand hits a
mesh. 50% ofthe liquid passes through the mesh unaffected. 25% looses all ofits
momentum and 25% comes back with the same speed. The resultant pressure on the mesh
will be: [11 Jan. 2019 (I)]
𝟏 𝟑 𝟏
(a) 𝟒 𝐩𝐯 𝟐 (b) 𝟒 𝐩𝐯 𝟐 (c) 𝟐 𝐩𝐯 𝟐 (d) 𝐩𝐯 𝟐
SOLUTION : (b)
Mass per unit time ofthe liquid = 𝐩𝐚𝐯
Momentum per second carried by liquid = 𝐩𝐚𝐯 × 𝐯

𝟏
Net force due to bounced back liquid,𝜞𝟏 = 𝟐 × 𝐩𝐚𝐯 𝟐
𝟒
𝟏
Net force due to stopped liquid, 𝜞𝟐 = 𝟒 𝐩𝐚𝐯𝟐
𝟏 𝟏 𝟑
Total force, 𝜞 = 𝜞𝟏 + 𝜞𝟐 = 𝟐 𝐩𝐚𝐯 𝟐 + 𝟒 𝐩𝐚𝐯 𝟐 = 𝟒 𝐩𝐚𝐯𝟐
𝟑
Net pressure = 𝟒 𝐩𝐯 𝟐
13. A thin uniform tube is bent into a circle ofradius 𝒓 in the verticalplane. Equal volumes oftwo
immiscible liquids, whose densities are 𝐩𝟏 and 𝐩𝟏 (𝐩𝟏 > 𝐩𝟐 ) fill half the circle. The angle 𝜽
between the radius vector passing through the common interface and the vertical is

𝝅 𝐩𝟏 −𝝆𝟐 𝝅 𝐩𝟏 −𝐩𝟐
(a) 𝜽 = 𝐭𝐚𝐧−𝟏 (b) 𝜽 = 𝐭𝐚𝐧−𝟏 𝟐
𝟐 𝐩𝟏 +𝐩𝟐 𝐩𝟏 +𝐩𝟐
𝐩𝟏
(c) 𝜽 = 𝐭𝐚𝐧−𝟏 𝝅 (d) None of above
𝐩𝟐
SOLUTION : (d)
Pressure at interface A must be same from both the sides to be in equilibrium.
𝐑 𝐜𝐨𝐬 𝜽 + 𝐑 𝐬𝐢𝐧 𝜽 𝐩𝟐 𝐠 = 𝐑 𝐜𝐨𝐬 𝜽 − 𝐑 𝐬𝐢𝐧 𝜽 𝐩𝐥 𝐠

𝐝 𝐜𝐨𝐬 𝜽+ 𝐬𝐢𝐧 𝜽 𝟏+ 𝐭𝐚𝐧 𝜽
⇒ 𝐝𝟏 = = 𝟏− 𝐭𝐚𝐧 𝜽
𝟐 𝐜𝐨𝐬 𝜽− 𝐬𝐢𝐧 𝜽
⇒ 𝐩𝟏 − 𝐩𝟏 𝐭𝐚𝐧 𝜽 = 𝐩𝟐 + 𝐩𝟐 𝐭𝐚𝐧 𝜽
⇒ 𝐩𝟏 + 𝐩𝟐 𝐭𝐚𝐧 𝜽 = 𝐩𝟏 − 𝐩𝟐
𝐩𝟏 − 𝐩𝟐
𝜽 = 𝐭𝐚𝐧−𝟏
𝐩𝟏 + 𝐩𝟐
14. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities
𝐝𝟏 and 𝐝𝟐 are filled in the tube. Each liquid subtends 𝟗𝟎𝐨 angle at centre. Radius joining
𝐝
their interface makes an angle 𝜶 with vertical. Ratio 𝐝𝟏 is: [2014]
𝟐
𝐝𝟏
𝟏+ 𝐬𝐢𝐧 𝜶 𝟏+ 𝐜𝐨𝐬 𝜶 𝟏+ 𝐭𝐚𝐧 𝜶 𝟏+ 𝐬𝐢𝐧 𝜶
(a) 𝟏− 𝐬𝐢𝐧 𝜶 (b) 𝟏− 𝐜𝐨𝐬 𝜶 (c) 𝟏− 𝐭𝐚𝐧 𝜶 (d) 𝟏− 𝐜𝐨𝐬 𝜶
SOLUTION : (c)
Pressure at interface A must be same 𝐟𝐢𝐢𝐎 𝐦 both the sides to be in equilibrium.
𝑹 𝐜𝐨𝐬 𝜶 + 𝑹 𝐬𝐢𝐧 𝜶 𝒅𝟐 𝒈 = 𝑹 𝐜𝐨𝐬 𝜶 − 𝑹 𝐬𝐢𝐧 𝜶 𝒅𝟏 𝒈
𝒅𝟏 𝐜𝐨𝐬 𝜶 + 𝐬𝐢𝐧 𝜶 𝟏 + 𝐭𝐚𝐧 𝜶

⇒ = =
𝒅𝟐 𝐜𝐨𝐬 𝜶 − 𝐬𝐢𝐧 𝜶 𝟏 − 𝐭𝐚𝐧 𝜶
15. A uniform cylinder oflength and mass having crosssectional areaA is suspended, with
its length vertical, from a fixed point by a massless spring such that it is half submerged in a
liquid ofdensity at equilibrium position. The extension 𝐱 𝟎 of the spring when it is in
equilibrium is: [2013]
𝐌𝐠 𝐌𝐠 𝐋𝐀𝒐 𝐌𝐠 𝐋𝐀𝒐 𝐌𝐠 𝐋𝐀𝒐
(a) (b) 𝟏− (c) 𝟏− (d) 𝟏+
𝐤 𝐤 𝐌 𝐤 𝟐𝐌 𝐤 𝐌
SOLUTION : (c)
From figure, 𝐤𝐱 𝟎 + 𝜞𝐁 = 𝐌𝐠
𝐋
𝐤𝐱 𝟎 + 𝒐 𝟐 𝐀𝐠 = 𝐌𝐠 [mass =density × 𝐯𝐨𝐥𝐮𝐦𝐞]
𝐋
⇒ 𝐤𝐱 𝟎 = 𝐌𝐠 − 𝟎 𝐀𝐠
𝟐
𝒐𝐋𝐀𝐠
𝐌𝐠 −
⇒ 𝐱𝟎 = 𝟐 = 𝐌𝐠 𝟏 − 𝐋𝐀𝒐
𝐤 𝐤 𝟐𝐌
Hence, extension ofthe spring when it is in equilibrium is,
𝐌𝐠 𝐋𝐀𝒐
𝐱𝟎 = 𝟏−
𝐤 𝟐𝐌
16. Aball is made ofa material ofdensity pwhere 𝐩𝐨𝐢𝟏 < 𝒑 < 𝐩𝐰𝐚𝐭𝐞𝐫 with 𝐩𝐨𝐢𝟏 and 𝐩𝐰𝐚𝐭𝐞𝐫
represe‐nting the densities of oil and water, respectively. The oil and water are immiscible.
Ifthe above ball is in equilibrium in a mixture ofthis oil and water, which ofthe following
pictures represents its equilibrium position? [2010]
SOLUTION : (b)
Oil will float on water so, (b) or (d) is the correct option. But density ofball is more than that
of oil,, hence it will sink in oil.
17. Two identical charged spheres are suspended by strings of equal lengths. The strings make
an angle of 𝟑𝟎∘ with each other. When suspended in a liquid of density 𝟎. 𝟖𝐠𝐜𝐦−𝟑 , the angle
remains the same. If density of the material ofthe sphere is . 𝟔𝐠𝐜𝐦−𝟑 , the dielectric constant
of the liquid is [2010]
(a) 4 (b) 3 (c) 2 (d) 1
SOLUTION : (c)
𝑭𝒆 = 𝑻 𝐬𝐢𝐧 𝟑𝟎∘ and 𝒎𝒈 = 𝑻 𝐜𝐨𝐬 𝟑𝟎∘

𝐅
⇒ 𝐭𝐚𝐧 𝟑𝟎∘ = 𝒎𝒈
𝐞
…………. (1)
In liquid, 𝑭𝒆 ‘ = 𝑻 ‘ 𝐬𝐢𝐧 𝟑𝟎∘ (A)
𝒎𝒈 = 𝑭𝑩 + 𝑻′ 𝐜𝐨𝐬 𝟑𝟎∘
But 𝑭𝑩 = Buoyant force

mg
= 𝑽 𝒅 − 𝐩 𝒈 = 𝑽 𝟏. 𝟔 − 𝟎. 𝟖 𝒈 = 𝟎. 𝟖𝑽𝐠
𝒎 𝟎. 𝟖. 𝒎𝒈 𝒎𝒈
= 𝟎. 𝟖 𝒈= =
𝒅 𝟏𝟔 𝟐
𝒎𝒈
𝒎𝒈 = + 𝑻 ‘ 𝐜𝐨𝐬 𝟑𝟎∘
𝟐
𝒎𝒈
⇒ = 𝑻′ 𝐜𝐨𝐬 𝟑𝟎∘ (B)
𝟐
𝟐𝑭𝒆
From (A) and (B), 𝐭𝐚𝐧 𝟑𝟎∘ = 𝒎𝒈
From (1) and (2)
⇒ 𝑭𝒆 = 𝟐𝑭𝒆
If 𝑲 be the dielectric constant, then
𝑭𝒆
𝑭𝒆 =
𝑲
𝟐𝑭𝒆
𝑭𝒆 = ⇒𝑲=𝟐
𝑲
18. Ajar is filled with two non‐mixing liquids 1 and 2 having densities 𝐩𝟏 and, 𝐩𝟐 respectively. A
solid ball, made of a material of density 𝐩𝟑 , is dropped in the jar. It comes to equilibrium in
the position shown in the figure.Which of the following is true for 𝐩𝟏 , 𝐩𝟐 𝐚𝐧𝐝𝐩𝟑 ? [2008]
(a) 𝐩𝟑 < 𝐩𝟏 < 𝐩𝟐 (b) 𝐩𝟏 > 𝐩𝟑 > 𝐩𝟐 (c) 𝐩𝟏 < 𝐩𝟐 < 𝐩𝟑 (d) 𝐩𝟏 < 𝐩𝟑 < 𝐩𝟐
SOLUTION : (d)
As liquid 1 floats over liquid 2. The lighter liquid floats over heavier liquid. So, 𝐩𝟏 < 𝐩𝟐
Also 𝐩𝟑 < 𝐩𝟐 because the ball of density 𝐩𝟑 does not sink to the bottom ofthejar.
Also 𝐩𝟑 > 𝐩𝟏 otherwise the ball would have floated in liquid 1.
we conclude that 𝐩𝟏 < 𝐩𝟑 < 𝐩𝟐 .

Fluid Flow , Reynold’s Number and Bernouli’s Principle :
19. Afluid is flowing through ahorizontal pipe ofvarying crosssection, with speed 𝐯 ms 1 at a
𝑷
point where the pressure is 𝑷 Pascal. At another point where pressure is 𝟐 Pascal its speed
is V ms 1. Ifthe density ofthe fluid is 𝐩 kg 𝐦𝟑 and the flow is streamline, then V is equal to:
[6 Sep. 2020 (𝐈𝐃]
𝑷 𝟐𝑷 𝑷 𝑷
(a) +𝐯 (b) + 𝐯𝟐 (c) + 𝐯𝟐 (d) + 𝐯𝟐
𝐩 𝐩 𝟐𝐩 𝐩
SOLUTION : (d)
Using Bernoulli’s equation
𝟏 𝟏
𝑷𝟏 + 𝐩𝐮𝟐𝟏 + 𝝆𝒈𝒉𝟏 = 𝑷𝟐 + 𝐩𝐮𝟐𝟐 + 𝐩𝒈𝒉𝟐
𝟐 𝟐
𝑷
For horizontal pipe, 𝒉𝟏 = 𝟎 and 𝒉𝟐 = 𝟎 and taking 𝑷𝟏 = 𝑷, 𝑷𝟐 = 𝟐 , we get
𝟏 𝑷 𝟏
⇒ 𝑷 + 𝐩𝜾)𝟐 = + 𝐩𝑽𝟐
𝟐 𝟐 𝟐
𝑷 𝟏 𝟏
⇒ + 𝐩𝜾)𝟐 = 𝐩𝑽𝟐
𝟐 𝟐 𝟐
𝑷
⇒𝑽= 𝟎𝟐 +
𝐩
20. Water flows in a horizontal tube (see figure). The pressure ofwater changes by 700 Nm 2
between A and 𝐁 where the area of cross section are 40 𝐜𝐦𝟐 and20 𝐜𝐦𝟐 , respectively. Find
the rate of flow ofwater through the tube. (density ofwater = 𝟏𝟎𝟎𝟎 kgm 3) [9 Jan. 2020 (𝐃]
(a) 3020 𝐜𝐦𝟑 /𝐬 (b) 2720 𝐜𝐦𝟑 /𝐬 (c) 2420 𝐜𝐦𝟑 /𝐬 (d) 1810 𝐜𝐦𝟑 /𝐬
SOLUTION : (b)
According to question, area of cross‐section at 𝑨, 𝒂𝑨 = 𝟒𝟎𝐜𝐦𝟐 and at 𝑩, 𝒂𝑩 = 𝟐𝟎𝐜𝐦𝟐
Let velocity ofliquid flow at 𝑨, = 𝑽𝑨 and at 𝑩, = 𝑽𝑩
Using equation of continuity 𝒂𝑨 𝑽𝑨 = 𝒂𝑩 𝑽𝑩
𝟒𝟎𝑽𝑨 = 𝟐𝟎𝑽𝑩 ⇒ 𝟐𝑽𝑨 = 𝑽𝑩

Now, using Bernoulli’s equation
𝟏 𝟏 𝟏
𝑷𝑨 + 𝐩𝑽𝟐𝑨 = 𝑷𝑩 + 𝐩𝑽𝟐𝑩 ⇒ 𝑷𝑨 − 𝑷𝑩 = 𝐩 𝑽𝟐𝑩 − 𝑽𝟐𝑨
𝟐 𝟐 𝟐
𝟏 𝑽𝟐𝑩 𝟑𝑽𝟐𝑩
⇒ 𝜟𝑷 = 𝟏𝟎𝟎𝟎 𝑽𝟐𝑩 − ⇒ 𝜟𝑷 = 𝟓𝟎𝟎 ×
𝟐 𝟒 𝟒
𝜟𝑷 ×𝟒 𝟕𝟎𝟎 ×𝟒
⇒ 𝑽𝑩 = = 𝐦/𝐬 = 𝟏. 𝟑𝟕 × 𝟏𝟎𝟐 𝐜𝐦/𝐬 Volume flow rate 𝐐 = 𝒂𝑩 × 𝒗𝑩
𝟏𝟓𝟎𝟎 𝟏𝟓𝟎𝟎
= 𝟐𝟎 × 𝟏𝟎𝟎 × 𝑽𝑩 = 𝟐𝟕𝟑𝟐𝐜𝐦𝟑 /𝐬 ≈ 𝟐𝟕𝟐𝟎𝐜𝐦𝟑 /𝐬
21. An ideal fluid flows (laminar flow) through apipe ofnonuniform diameter. The maximum
and minimum diameters of the pipes are 𝟔. 𝟒 cm and 𝟒. 𝟖 cm, respectively. The ratio of the
minimum and the maximum velocities offluid in this pipe is: [7 Jan. 2020 𝟎𝐃]
𝟗 𝟑 𝟑 𝟖𝟏
(a) 𝟏𝟔 (b) (c) 𝟒 (d) 𝟐𝟓𝟔
𝟐
SOLUTION : (a)
From the equation ofcontinuity 𝑨𝟏 𝒗𝟏 = 𝑨𝟐 𝒗𝟐
Here, 𝒗𝟏 and 𝒗𝟐 are the velocities at two ends ofpipe. 𝐀 𝟏 and 𝐀 𝟐 are the area ofpipe at two ends
𝟐
𝒗𝟏 𝑨𝟐 𝝅 𝟒. 𝟖 𝟗
⇒ = = 𝟐
=
𝒗𝟐 𝑨𝟏 𝝅 𝟔. 𝟒 𝟏𝟔
22. Water from a tap emerges vertically downwards with an initial speed of 𝟏. 𝟎 ms 1. The
cross‐sectional area of the tap is 𝟏𝟎⊲ 𝐦𝟐 . Assume that the pressure is constant throughout
the stream ofwater and that the flow is streamlined. The cross‐sectional area ofthe stream,
𝟎. 𝟏𝟓𝐦 below the tap would be: [Take 𝐠 = 𝟏𝟎 ms 2) [10 April 2019 (II)]
(a) 𝟐 × 𝟏𝟎𝟓 𝐦𝟐 (b) 𝟓 × 𝟏𝟎𝟓 𝐦𝟐 (c) 𝟓 × 𝟏𝟎⊲ 𝐦𝟐 (d) 𝟏 × 𝟏𝟎𝟓 𝐦𝟐
SOLUTION : (b)
𝟏
Using Bernoullie’s equation 𝑷 + 𝟐 𝒗𝟐𝟏 − 𝒗𝟐𝟐 + 𝝆𝒈𝒉 = 𝑷
⇒ 𝒗𝟐𝟐 = 𝒗𝟐𝟏 + 𝟐𝒈𝒉 ⇒ 𝒗𝟐 = 𝒗𝟐𝟏 + 𝟐𝒈𝒉
Equation ofcontinuity 𝐀 𝟏 𝐯𝟏 = 𝐀 𝟐 𝐯𝟐
𝟏𝟓
𝟏𝐜𝐦𝟐 𝟏𝐦/𝐬 = 𝐀 𝟐 𝟏 𝟐 + 𝟐 × 𝟏𝟎 ×
𝟏𝟎𝟎
𝟏𝟎−𝟒
𝟏𝟎⊲ × 𝟏 = 𝐀 𝟐 × 𝟐 ; 𝐀 𝟐 = = 𝟓 × 𝟏𝟎𝟓 𝐦𝟐
𝟐
23. Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5
cm, the Reynolds number for the flow is ofthe order of: (density ofwater = 1000 𝐤𝐠/𝐦𝟑 ,
coefficient ofviscosity ofwater = 𝟏𝐦𝐏𝐚𝒔) [𝟖April 2019 I]
(a) 𝟏𝟎𝟑 (b) 𝟏𝟎𝟒 (c) 𝟏𝟎𝟐 (d) 𝟏𝟎𝟔
SOLUTION : (b)
𝟏𝟎𝟎×𝟏𝟎−𝟑 𝟓
Rate offlow ofwater(V) = 𝟏𝟎𝟎𝐥𝐢𝐭/ 𝐦𝐢𝐧 = × 𝟑 × 𝟏𝟎−𝟑 𝐦𝟑
𝟔𝟎
𝑽 𝟓×𝟏𝟎−𝟑 𝟏𝟎 𝟐
Velocity of flow ofwater 𝒗 = 𝑨 = 𝟐 = 𝟏𝟓𝝅 = 𝟑𝝅 𝐦/𝐬 = 𝟎. 𝟐𝐦/𝐬
𝟑×× 𝟓×𝟏𝟎−𝟐
𝑫𝒗𝐩 𝟑𝝅
Reynold number 𝐍𝐑 = = 𝟏𝟎 × 𝟏𝟎−𝟐 × 𝟐 × 𝟏𝟎𝟎𝟎 = = 𝟐 × 𝟏𝟎𝟒
𝜼 𝟏
Order 𝐨𝐟𝐍𝐑 = 𝟏𝟎𝟒
24. Water flows into a large tank with flat bottom at the rate of 1𝟎−𝟒 𝐦𝟑 𝐬−( 𝟏) Water is also
leaking out ofa hole of area 1 𝐜𝐦𝟐 at its bottom. If the height ofthe water in the tank
remains steady, then this height is: [10 Jan. 2019 I]
(a) 𝟓. 𝟏 cm (b) 7 cm (c) 4 cm (d) 9 cm
SOLUTION : . (a)
−𝝉𝐐𝐨𝐮𝐭
Since height ofwater column is constant therefore, water inflowrate 𝐐𝐦
= water outflow rate
𝐐𝐢𝐧 = 𝟏𝟎→↓ 𝐦𝟑 𝐬𝟏
𝐐𝐨𝐮𝐭 = 𝐀𝐮 = 𝟏𝟎→𝒍 × 𝟐𝐠𝐡
𝟏𝟎⊲ = 𝟏𝟎⊲ × 𝟐𝟎 × 𝐡
𝟏
𝐡 = 𝟐𝟎 𝐦 = 𝟓 cm
25. The top of a water tank is open to air and its water lavel is maintained. It is giving out
𝟎. 𝟕𝟒𝐦𝟑 water per minute through a circular opening of2 cm radius in its wall. The depth
ofthe centre ofthe opening from the level ofwater in the tank is close to: [𝟗𝐉𝐚𝐧. 2019 (II)]
(a) 𝟔. 𝟎𝐦 (b) 𝟒. 𝟖𝐦 (c) 𝟗. 𝟔𝐦 (d) 𝟐. 𝟗𝐦
SOLUTION : . (b)
𝟎.𝟕𝟒
Here, volume tric flow rate = = 𝝅𝐫 𝟐 𝐯 = 𝝅 × 𝟒 × 𝟏𝟎→𝖙 × 𝟐𝐠𝐡
𝟔𝟎
𝟕𝟒 × 𝟏𝟎𝟎 𝟕𝟒𝟎
⇒ 𝟐𝐠𝐡 = ⇒ 𝟐𝐠𝐡 =
𝟐𝟒𝟎𝝅 𝟐𝟒𝝅
𝟕𝟒𝟎 × 𝟕𝟒𝟎
⇒ 𝟐𝐠𝐡 = ⋅.⋅ 𝝅𝟐 = 𝟏𝟎
𝟐𝟒 × 𝟐𝟒 × 𝟏𝟎
𝟕𝟒 × 𝟕𝟒
⇒𝐡= ≈ 𝟒. 𝟖𝐦
𝟐 × 𝟐𝟒 × 𝟐𝟒
i. e., The depth ofthe centre ofthe opening from the level ofwater in the tank is close to 𝟒. 𝟖𝐦
26. When an air bubble ofradius 𝐫 rises from the bottom to thesurface of a lake, its radius
𝟓𝐫
becomes Taking the atmospheric pressure to be equal to 𝟏𝟎𝐦 height of water column,
𝟒
the depth of the lake would approximately be (ignore the surface tension and the effect
oftemperature): [Online Apri115, 2018]
(a) 𝟏𝟎. 𝟓𝐦 (b) 𝟖. 𝟕𝐦 (c) 𝟏𝟏𝟐𝐦 (d) 𝟗. 𝟓𝐦
SOLUTION : . (d)
Using 𝐏𝟏 𝐕𝟏 = 𝐏𝟐 𝐕𝟐
𝟒 𝟑 𝟒 𝟏𝟐𝟓𝒓𝟑
𝑷𝟏 𝝅𝒓 = 𝑷𝟐 𝝅
𝟑 𝟑 𝟔𝟒
𝐩𝒈 𝟏𝟎 + 𝐩𝒈𝒉 𝟏𝟐𝟓
=
𝐩𝒈 𝟏𝟎 𝟔𝟒
640+64 𝒉 = 𝟏𝟐𝟓𝟎
On solving we get 𝒉 = 𝟗. 𝟓𝐦
27. Two tubes ofradii 𝒓𝟏 and 𝒓𝟐 , and lengths 𝒍𝟏 and 𝒍𝟐 , respectively, are connected in series and
a liquid flows through each ofthem in streamline conditions. 𝐏𝟏 and 𝐏𝟐 are pressure
𝒍
differences across the two tubes. If 𝐏𝟐 is 𝟒𝐏𝟏 and 𝒍𝟐 is 𝟒𝟏 , then the radius 𝐫𝟐 will oe equal to:

𝒓𝟏
(a) 𝒓𝟏 (b) 𝟐𝒓𝟏 (c) 𝟒𝒓𝟏 (d) 𝟐
SOLUTION : (d)
The volume ofliquid flowing through both the tubes i. e., rate offlow ofliquid is same.
𝝅𝐏𝟏 𝐫𝟏𝟒 𝝅𝐏𝟐 𝐫𝟐𝟒 𝐏𝟏 𝐫𝟏𝟒 𝐏𝟐 𝐫𝟐𝟒

Therefore, 𝐕 = 𝐕𝟏 = 𝐕𝟐 i.e., = or =
𝟖𝜼𝒍𝟏 𝟖𝜼𝒍𝟐 𝒍𝟏 𝒍𝟐
𝐏𝟐 = 𝟒𝐏𝟏 and 𝒍𝟐 = 𝒍𝟏 /𝟒
𝐏𝟏 𝐫𝟏𝟒 𝟒𝐏𝟏 𝐫𝟐𝟒 𝟒

𝐫𝟏𝟒
= ⇒ 𝐫𝟐 =
𝒍𝟏 𝒍𝟏 /𝟒 𝟏𝟔
𝐫𝟐 = 𝐫𝟏 /𝟐
28. Consider a waterjar ofradius 𝐑 that has water filled up to height 𝐇 and is kept on a stand
ofheight 𝐡 (see figure). Through a hole ofradius 𝐫(𝐫 << 𝑹) at its bottom, the water leaks out
and the stream of water coming down towards the ground has a shape like a funnel as
shown in the figure. Ifthe radius ofthe cross‐section ofwater stream when it hits the ground
is 𝐱. Then : [Online April 9, 2016]
→ 𝟐𝐱 ←
𝟏 𝟏
𝐇 𝐇 𝐇 𝟐 𝐇
𝟒 𝟐
(a) 𝐱 = 𝐫 (b) 𝐱 = 𝐫 (c) 𝐱 = 𝐫 (d) 𝐱 = 𝐫
𝐇+𝐡 𝐇+𝐡 𝐇+𝐡 𝐇+𝐡
SOLUTION : (a)
𝟏 𝟏
According to Bernoulli’s Principle, 𝝆𝐯𝟏𝟐 + 𝝆𝐠𝐡 = 𝟐 𝝆𝐯𝟐𝟐
𝟐
𝐯𝟏𝟐 + 𝟐𝐠𝐡 = 𝐯𝟐𝟐
𝟐𝐠𝐇 + 𝟐𝐠𝐡 = 𝐯𝟐𝟐 (i)
𝐚𝟏 𝐯𝟏 = 𝐚𝟐 𝐯𝟐
𝐫𝟐
𝝅𝐫 𝟐 𝟐𝐠𝐡 = 𝝅𝐱 𝟐 𝐯𝟐 ; 𝟐𝐠𝐡 = 𝐯𝟐
𝐱𝟐
𝐫𝟒
Substituting the value 𝐨𝐟𝐯𝟐 in equation (i) 𝟐𝐠𝐇 + 𝟐𝐠𝐡 = 𝐱𝟒 𝟐𝐠𝐡
𝟏
𝐇
or, 𝐱 = 𝐫| 𝐇+𝐡 |𝟒
𝟐
29. Ifit takes 5 minutes to fill a 15 litre bucket 𝐟𝐢𝐢 𝐨𝐦 a water tap of diameter cm then the
𝝅
Reynolds number for the flow is (density of water = 𝟏𝟎𝟑 𝐤𝐠/𝐦𝟑 ) and viscosity of water
= 𝟏𝟎−𝟑 Pa. s) close to : [Online April 10, 2015]
(a) 1100 (b) 11,000 (c) 550 (d) 5500
SOLUTION : . (d)
𝟐 𝟏
Given: Diameter of water tap = cm Radius, 𝒓 = × 𝟏𝟎−𝟐 𝐦
𝝅 𝝅
𝒅𝒎
= 𝐩𝐀𝐕
𝒅𝒕
𝟏𝟓 𝟏 𝟐
= 𝟏𝟎𝟑 × 𝝅 × 𝟏𝟎−𝟒 𝐕 ⇒ 𝐕 = 𝟎. 𝟎𝟓𝐦/𝐬
𝟓×𝟔𝟎 𝝅
𝐩𝑽𝒓
Reynold’s number, 𝑹𝒆 = 𝒏
𝟐
𝟏𝟎𝟑 × 𝟎. 𝟓 × 𝟏𝟎−𝟐
𝝅
= ≅ 𝟓𝟓𝟎𝟎
𝟏𝟎−𝟑
30. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends
above the mercury level. The open end ofthe tube is then closed and sealed and the tube is
raised vertically up by additional 46 cm. What will be length ofthe air colunm above
mercury in the tube now? (Atmospheric pressure = 𝟕𝟔 cm ofHg) [2014]
(a) 16 cm (b) 22 cm (c) 38 cm (d) 6 cm
SOLUTION : (a)
Length ofthe air column above mercury in the tube is, 𝑷 + 𝒙 = 𝑷𝟎
⇒ 𝑷 = 𝟕𝟔 − 𝒙
⇒ 𝟖 × 𝑨 × 𝟕𝟔 = 𝟕𝟔 − 𝒙 × 𝐀 × 𝟓𝟒 − 𝒙
𝒙 = 𝟑𝟖
Thus, length ofair column = 𝟓𝟒 − 𝟑𝟖 = 𝟏𝟔 cm.

31. In the diagram shown, the difference in the two tubes of the manometer is 5 cm, the cross
section ofthe tube at 𝐀 and 𝐁 is 6 𝐦𝐦𝟐 and 10 𝐦𝐦𝟐 respectively. The rate at which water
flows through the tube is 𝐠 = 𝟏𝟎𝐦𝐬−𝟐 [Online Apri119, 2014]
𝟓𝐜𝐜 𝟎𝐜𝐜
(a) 𝟕. (b) 𝟖. 𝐜 𝟏𝟎. 𝟎𝐜𝐜/𝐬 (d) 𝟏𝟐. 𝟓𝐜𝐜/𝐬
𝐬 𝐬
SOLUTION : (a)
𝟏 𝟏
According to Bernoulli’s theorem, 𝐏𝟏 + 𝟐 𝐩𝐯𝟏𝟐 = 𝐏𝟐 + 𝟐 𝝆𝐯𝟐𝟐
𝐯𝟐𝟐 − 𝐯𝟏𝟐 = 𝟐𝐠𝐡 …….. (1)
According to the equation of continuity 𝐀 𝟏 𝐯𝟏 = 𝐀 𝟐 𝐯𝟐 ………. (2)
𝐀𝟏 𝟔𝐦𝐦𝟐
=
𝐀 𝟐 𝟏𝟎𝐦𝐦𝟐
𝐀 𝐯 𝟔
From equation (2), 𝐀𝟏 = 𝐯𝟐 = 𝟏𝟎
𝟐 𝟏
𝟔
or, 𝐯𝟐 = 𝟏𝟎 𝐯𝟏
Putting this value 𝐨𝐟𝐯𝟐 in equation (1)

𝟐
𝟔 𝟐
𝐯 − 𝐯𝟏 = 𝟐 × 𝟏𝟎𝟑 × 𝟓
𝟏𝟎 𝟏
𝐠 = 𝟏𝟎𝐦/𝐬𝟐 = 𝟏𝟎𝟑 𝐜𝐦/𝐬𝟐

𝟏𝟎
Solving we get 𝐯𝟏 = 𝟖
𝟔×𝟏𝟎
Therefore the rate at which water flows through the tube = 𝐀 𝟏 𝐯𝟏 = 𝐀 𝟐 𝐯𝟐 = = 𝟕. 𝟓𝐜𝐜/𝐬
𝟖
32. A cylindrical vessel ofcross‐section A contains water to a height 𝐡. There is a hole in the
bottom ofradius ‘a’. The time in which it will be emptied is: [Online Apri112, 2014]
𝟐𝐀 𝐡 𝟐𝐀 𝐡 𝟐 𝟐𝐀 𝐡 𝐀 𝐡
(a) 𝝅𝐚𝟐 (b) 𝝅𝐚𝟐 (c) (d)
𝐠 𝐠 𝝅𝐚𝟐 𝐠 𝟐𝝅𝐚𝟐 𝐠
SOLUTION : (b)
𝒅𝒉
Let the rate offalling water level be − 𝒅𝒕
Integrating both sides

𝒕 𝟎
𝑨
𝒅𝒕 = − 𝒉−𝟏/𝟐 𝒅𝒉
𝟎 𝟐𝒈𝝅𝒂𝟐 𝒉
𝟎
𝒕
𝑨 𝒉𝟏/𝟐
𝒕 𝟎 =− .
𝟐𝒈𝝅𝒂𝟐 𝟏/𝟐 𝒉
𝟐𝑨 𝒉
𝒕=
𝝅𝒂𝟐 𝒈
33. Water is flowing at a speed of 𝟏. 𝟓𝐦𝐬−𝟏 through horizontal tube ofcross‐sectional area
𝟏𝟎−𝟐 𝐦𝟐 and you are trying to stop the flow by your palm. Assuming that the water stops
immediately after hitting the palm, the minimum force that you must exert should be
(density ofwater = 𝟏𝟎𝟑 𝐤𝐠𝐦−𝟑 ) [Online April 9, 2014]
(a) 𝟏𝟓𝐍 (b) 𝟐𝟐. 𝟓𝐍 (c) 𝟑𝟑. 𝟕𝐍 (d) 𝟒𝟓𝐍
SOLUTION : (a)
For 1 𝐦 length ofhorizontal tube
Mass ofwater 𝐌 =density × volume = 𝟏𝟎𝟑 ×area × lengLh = 𝟏𝟎𝟑 × 𝟏𝟎−𝟐 × 𝟏 = 𝟏𝟎 kg

𝜟𝒑
Therefore minimum force = (rate of change of momentum) = 𝟏𝟎 × 𝟏. 𝟓 = 𝟏𝟓𝑵
𝜟𝒕
34. Air ofdensity 𝟏. 𝟐 kg 𝐦−𝟑 is blowing across the horizontal wings ofan aeroplane in such a
way that its speeds above and below the wings are 150 𝐦𝐬−𝟏 and 100 𝐦𝐬 −𝟏 , respectively.
The pressure difference between the upper and lower sides ofthe wings, is:
[OnlineApril 2𝟐, 𝟐𝟎𝟏𝟑]
(a) 𝟔𝟎𝐍𝐦−𝟐 (b) 𝟏𝟖𝟎𝐍𝐦−𝟐 (c) 𝟕𝟓𝟎𝟎𝐍𝐦−𝟐 (d) 𝟏𝟐𝟓𝟎𝟎𝐍𝐦−𝟐

SOLUTION : (c)
𝟏 𝟏
Pressure difference𝐏𝟐 − 𝐏𝟏 = 𝟐 𝐩 𝐯𝟐𝟐 − 𝐯𝟏𝟐 = 𝟐 × 𝟏. 𝟐 𝟏𝟓𝟎 𝟐
− 𝟏𝟎𝟎 𝟐
𝟏
= 𝟐 × 𝟏. 𝟐 𝟐𝟐𝟓𝟎𝟎 − 𝟏𝟎𝟎𝟎𝟎 = 𝟕𝟓𝟎𝟎𝐍𝐦−𝟐
35. In a cylindrical water tank, there are two small holes 𝑨 and 𝑩 on the wall at a depth 𝐨𝐟𝒉𝟏 ,
from the surface ofwater and at a height 𝐨𝐟𝒉𝟐 𝐟𝐢𝐢 𝐨𝐦 the bottom ofwater tank. Surface
ofwater is at heigh 𝑯 from the bottom ofwater tank. Water coming out from both holes
strikes the ground at the same point 𝑺. Find the ratio os h1 and h2
(a) Depends on 𝑯 (b) 1 : 1 (c) 2: 2 (d) 1: 2
SOLUTION : . (b)
i.e. 𝑹𝟏 = 𝑹𝟐 = 𝑹 or, 𝒗𝟏 𝒕𝟏 = 𝒗𝟐 𝒕𝟐 (i)
Where 𝒗𝟏 = velocity of efflux at 𝑨 = 𝟐𝒈𝒉𝟏 and
𝒗𝟐 = velocity of efflux at 𝑩 = (𝟐𝒈 𝐇 − 𝒉𝟐
𝟐(𝐇−𝒉𝟏
𝒕𝟏 = time of fall water stream through 𝑨 = 𝒈
𝟐𝒉𝟐
𝒕𝟐 = time of fall ofthe water stream through 𝑩 = Putting these values is eqn (i) we get
𝒈
𝑯 − 𝒉𝟏 𝒉𝟏 = 𝑯 − 𝒉𝟐 𝒉𝟐 or 𝑯 − 𝒉𝟏 + 𝒉𝟐 𝒉𝟏 − 𝒉𝟐 = 𝟎
Here, 𝑯 = 𝒉𝟏 + 𝒉𝟐 is irrelevant because the holes are at 𝒍𝜼 = 𝟏
two different heights. Hence 𝒉𝟏 = 𝒉𝟐

Viscosity and Terminal Velocity :
36. Water is flowing through a horizontal tube having crosssectional areas ofits two ends being
𝑨 and 𝑨′ such that the ratio AlA’ is 5. Ifthe pressure difference ofwater between the two
ends is 𝟑 × 𝟏𝟎𝟓 𝐍𝐦−𝟐 , the velocity of water with which it enters the tube will be (neglect
gravity effects) [Online May 12, 2012]
(a) 𝟓𝐦𝐬 −𝟏 (b) 𝟏𝟎𝐦𝐬−𝟏 (c) 25 𝐦𝐬 −𝟏 (d) 𝟓𝟎 𝟏𝟎𝐦𝐬−𝟏
SOLUTION : (a)
𝟏 𝟏
According to Bernoulli’ 𝐬 theorem 𝑷𝟏 + 𝟐 𝐩𝒗𝟐𝟏 = 𝑷𝟐 + 𝟐 𝐩𝒗𝟐𝟐 (i) From question,
𝑨𝟏
𝑷𝟏 − 𝑷𝟐 = 𝟑 × 𝟏𝟎𝟓 , =𝟓
𝑨𝟐
According to equation of continuity 𝑨𝟏 𝒗𝟏 = 𝑨𝟐 𝒗𝟐 ⇒ 𝒗𝟐 = 𝟓𝒗𝟏

𝟏
From equation (i) 𝑷𝟏 − 𝑷𝟐 = 𝟐 𝐩 𝒗𝟐𝟐 − 𝒗𝟐𝟏
𝟏
or 𝟑 × 𝟏𝟎𝟓 = 𝟐 × 𝟏𝟎𝟎𝟎 𝟓𝒗𝟐𝟏 − 𝒗𝟐𝟏
⇒ 𝟔𝟎𝟎 = 𝟔𝒗𝟏 × 𝟒𝒗𝟏
⇒ 𝒗𝟐𝟏 = 𝟐𝟓
𝒗𝟏 = 𝟓𝐦/𝐬
37. A square hole of side length 𝓵 is made at a depth ofh and a circular hole ofradius 𝐫 is made
at a depth 𝐨𝐟𝟒𝐡𝐟𝐢𝐢 𝐨𝐦 the surface ofwater in a water tank kept on a horizontal surface. If
𝓵 << 𝒉, 𝒓 << 𝒉 and the rate ofwater flow from the holes is the same, then 𝐫 is equal to
[May7, 2012]
𝓵 𝓵 𝓵 𝓵
(a) (b) (c) 𝟑𝝅 (d) 𝟐𝝅
𝟐𝝅 𝟑𝝅
SOLUTION : . (a)
As 𝑨𝟏 𝒗𝟏 = 𝑨𝟐 𝒗𝟐 (Principle of continuity)
or, 𝓵𝟐 𝟐𝒈𝒉 = 𝝅𝒓𝟐 𝟐𝒈 × 𝟒𝒉
(Efflux velocity = 𝟐𝐠𝐡)
𝓵𝟐 𝓵𝟐 𝓵
𝒓𝟐 = 𝟐𝝅 or 𝒓 = =
𝟐𝝅 𝟐𝝅
38. Water is flowing continuouslyfiom atap having an internal diameter 𝟖 × 𝟏𝟎−𝟑 𝐦. The water
velocity as it leaves the tap is 𝟎. 𝟒𝐦𝐬 −𝟏 . The diameter ofthe water stream at a distance 2
× 𝟏𝟎−𝟏 𝐦 below the tap is close to: [2011]
(a) 𝟕. 𝟓 × 𝟏𝟎−𝟑 𝐦 (b) 𝟗. 𝟔 × 𝟏𝟎−𝟑 𝐦 (c) 𝟑. 𝟔 × 𝟏𝟎−𝟑 𝐦 (d) 𝟓. 𝟎 × 𝟏𝟎−𝟑 𝐦
SOLUTION : (c)
𝟏 𝟏
Using Bernoulli’s theorem, for horizontal flow 𝑷𝟎 + 𝟐 𝐩𝒗𝟐𝟏 + 𝐩𝒈𝒉 = 𝑷𝟎 + 𝟐 𝐩𝒗𝟐𝟐 + 𝟎
𝒗𝟐 = 𝒗𝟐𝟏 + 𝟐𝒈𝒉 = 𝟎𝟏𝟔 + 𝟐 × 𝟏𝟎 × 𝟎. 𝟐 = 𝟐. 𝟎𝟑𝐧 𝐬
According to equation of continuity 𝑨𝟐 𝒗𝟐 = 𝑨𝟏 𝒗𝟏
𝑫𝟐𝟐 𝑫𝟐𝟏
𝝅 × 𝒗𝟐 = 𝝅 𝒗
𝟒 𝟒 𝟏
𝒗𝟏
⇒ 𝑫𝟐 = 𝑫𝟏 = 𝟑. 𝟓𝟓 × 𝟏𝟎−𝟑 𝐦
𝒗𝟐
39. A cylinder ofheight 20 𝐦 is completely filled with water. The velocity ofefflux ofwater (in
𝐦𝐬 −𝟏 ) through a small hole on the side wall ofthe cylinder near its bottom is [2002]
(a) 10 (b) 20 (c) 𝟐𝟓. 𝟓 (d) 5
SOLUTION : (b)
Given, Height ofcylinder, 𝐡 = 𝟐𝟎 cm Acceleration due to gravity, 𝐠 = 𝟏𝟎𝐦𝐬 −𝟐
Velocity ofefflux 𝒗 = 𝟐𝒈𝒉
Where 𝒉 is the height ofthe free surface ofliquid from th 𝐞 hole
⇒ 𝒗 = 𝟐 × 𝟏𝟎 × 𝟐𝟎 = 𝟐𝟎𝐦/𝐬
40. In an experiment to 𝐕𝐞𝐫𝐢𝖍𝒓𝐬tokeS l𝐚𝐖, asmall spherical ball of radius 𝒓 and density 𝐩 falls
under gravity through a distance 𝒉 in air before entering a tank ofwater. Ifthe terminal
velocityof the ball inside water is same as its velociLyjust before enterin 𝐠 the water
surface, then the value ofh is proportional to: (ignore viscosity of air) [5 Sep. 2020 (II)]
(a) 𝒓𝟒 (b) 𝒓 (c) 𝒓𝟑 𝐝 𝒓𝟐
SOLUTION : (a)
Using, 𝒗𝟐 − 𝒖𝟐 = 𝟐𝒈𝒉 ⇒ 𝒗𝟐 − 𝟎𝟐 = 𝟐𝒈𝒉 ⇒ 𝒗 = 𝟐𝒈𝒉
𝟐 𝒓𝟐 𝐩−𝟎 𝒈
Terminal velocity, 𝑽𝑻 = 𝟗 𝜼
After falling through 𝒉 the velocity should be equal to terminal velocity
𝟐 𝒓𝟐 𝐩 − 𝟎 𝒈 𝟒𝒓𝟒 𝒈𝟐 𝐩 − 𝟎 𝟐
𝟐𝒈𝒉 = ⇒ 𝟐𝒈𝒉 =
𝟗 𝜼 𝟖𝟏𝜼𝟐
𝟐𝒓𝟒 𝒈 𝐩 − 𝟎 𝟐
⇒𝒉= ⇒ 𝒉𝜶𝒓𝟒
𝟖𝟏𝜼𝟐
41. A solid sphere, of radius 𝐑 acquires a terminal velocity 𝒗𝟏 when falling (due to gravity)
through a viscous fluid having a coefficient ofviscosity 𝜼. The sphere is broken into 27
identical solid spheres. If each of these spheres acquires a terminal velocity, 𝒗𝟐 , when
falling through the Same fluid, the ratio 𝒗𝟏 𝒍𝒗𝟐 equals: [12 April 2019 (II)]
(a) 9 (b) 1/27 (c) 1/9 (d) 27
SOLUTION : (a)
𝟒 𝟒 𝑹
𝟐𝟕 × 𝟑 𝝅𝒓𝟑 = 𝟑 𝝅𝒓𝟑 or 𝒓 = 𝟑
Terminal velocity, 𝒗 ∝ 𝒓𝟐
𝒗𝟏 𝒓𝟐𝟏
=
𝒗𝟐 𝒓𝟐𝟐
𝒗𝟏
𝒓𝟐𝟐
or 𝒗𝟐 = 𝒓𝟐𝟏
𝟏
𝒗𝟏 =
𝟗
42. Which of the following option correctly describes the variation ofthe speed 𝐯 and
acceleration’a’ ofa point mass falling vertically in a viscous medium that applies a force
𝜞 = −𝐤𝐯, where ↑ 𝐤 ′ is a constant, on the 𝐭 𝐗) 𝐝𝐲? (Graphs are schematic and not drawn to
scale) [Online April 9, 2016]
SOLUTION : (c)
When a point mass is falling vertically in a viscous medium, the medium or viscous fluid
exerts drag force on the body to oppose its motion and at one stage body falling with
constant terminal velocity.
43. The velocity ofwater in a river is 18 𝐤𝐦/𝐡𝐫 near the surface. Ifthe river is 5 𝐦 deep, find the
shearing stress between the horizontal layers ofwater. The co‐efficient ofviscosity of water
= 𝟏𝟎−𝟐 poise. [Online April 19, 2014]
(a) 𝟏𝟎−𝟏 𝐍/𝐦𝟐 (b) 𝟏𝟎−𝟐 𝐍/𝐦𝟐 (c) 𝟏𝟎−𝟑 𝐍/𝐦𝟐 (d) 𝟏𝟎−𝟒 𝐍/𝐦𝟐
SOLUTION : (b)
𝜼 = 𝟏𝟎−𝟐 poise
𝟏𝟖𝟎𝟎𝟎
𝒗 = 𝟏𝟖𝐤𝐦/𝐡 = = 𝟓𝐦/𝐬 𝒍 = 𝟓𝐦
𝟑𝟔𝟎𝟎
𝒗
Strain rate = 𝒍
𝐬𝐡𝐞𝐚𝐫𝐢𝐧𝐠𝐬𝐭𝐫𝐞𝐬𝐬
Coefficient of viscosity, 𝜼 = 𝐬𝐭𝐫𝐚𝐢𝐧𝐫𝐚𝐭𝐞
𝟓
Shearing stress = 𝜼 × strain rate = 𝟏𝟎−𝟐 × = 𝟏𝟎−𝟐 𝐍𝐦−𝟐
𝟓
44. The average mass ofrain drops is 𝟑. 𝟎 × 𝟏𝟎−𝟓 kg and their avarage terminal velocity is 9
𝐦/𝐬. Calculate the energy transferred by rain to each square metre ofthe surface at a place
which receives 100 cm ofrain in a year. [Online April 11, 2014]
(a) 𝟑. 𝟓 × 𝟏𝟎𝟓 𝐉 (b) 𝟒. 𝟎𝟓 × 𝟏𝟎𝟒 𝐉 (c) 𝟑. 𝟎 × 𝟏𝟎𝟓 𝐉 (d) 𝟗. 𝟎 × 𝟏𝟎𝟒 𝐉
SOLUTION : (b)
Total volume ofrain drops, received 100 cm in a year byarea 1 𝐦𝟐
𝟏𝟎𝟎
= 𝟏𝐦𝟐 × 𝐦 = 𝟏𝐦𝟑
𝟏𝟎𝟎
As we know, density ofwater, 𝒅 = 𝟏𝟎𝟑 𝐤𝐠/𝐦𝟑
Therefore, mass ofthis volume ofwater 𝑴 = 𝒅 × 𝒗 = 𝟏𝟎𝟑 × 𝟏 = 𝟏𝟎𝟑 kg
Average terminal velocity ofrain drop 𝒗 = 𝟗𝐦/𝐬 (given)

𝟏 𝟏
Therefore, energy transferred by rain,𝑬 = 𝟐 𝒎𝒗𝟐 = 𝟐 × 𝟏𝟎𝟑 × 𝟗 𝟐
𝟏
= × 𝟏𝟎𝟑 × 𝟖𝟏 = 𝟒. 𝟎𝟓 × 𝟏𝟎𝟒 𝐉
𝟐
45. A tank with a small hole at the bottom has been filled with water and kerosene (specific
gravity 𝟎. 𝟖). The height of water is 𝟑𝐦 and that of kerosene 𝟐𝐦. When the hole is opened
the velocity of fluid coming out from it is nearly: (take 𝐠 = 𝟏𝟎𝐦𝐬 −𝟐 and density ofwater
= 𝟏𝟎𝟑 kg 𝐦−𝟑 ) [Online April 11, 2014]
(a) 𝟏𝟎. 𝟕𝐦𝐬 −𝟏 (b) 𝟗. 𝟔𝐦𝐬 −𝟏 (c) 𝟖. 𝟓𝐦𝐬−𝟏 (d) 𝟕. 𝟔𝐦𝐬−𝟏
SOLUTION : (b)
According to Toricelli’s theorem, Velocity ofefflex,
𝑽𝐞𝐟𝐟 = 𝟐𝒈𝒉 = 𝟐 × 𝟗𝟖 × 𝟓 ≅ 𝟗. 𝟖𝐦𝐬−𝟏
46. In an experiment, a small steel ball falls through a liquid at a constant speed of10 𝐜𝐦/𝐬.
Ifthe steel ball is pulled upward with a force equal to twice its effective weight, how fast will
it move upward? [Online April 25, 2013]
(a) 5 𝐜𝐦/𝐬 (b) Zero (c) 10 𝐜𝐦/𝐬 (d) 20 𝐜𝐦/𝐬

SOLUTION : ( c)
𝟒
Weight of the body𝐖 = 𝐦𝐠 = 𝟑 𝝅𝐫 𝟑 𝐩𝐠
𝟒
𝐓 = 𝟑 𝝅𝐫 𝟑 𝒐𝐠 and 𝜞 = 𝟔𝝅𝜼𝐯𝐫
When the body attains terminal velocity net force acting on the body is zero.
i. e., W‐T −𝜞 = 𝟎
𝟐 𝐫 𝟐 𝐩−𝟎 𝐠
And terminal velocity 𝐯 = 𝟗 𝜼
As in case of upward motion upward force is twice its effective weight, therefore,
it will move with same speed 10 𝐜𝐦/𝐬
47. The terminal velocity of a small sphere of radius 𝐚 in a viscous liquid is proportional to
(a) 𝐚𝟐 (b) 𝐚𝟑 (c) 𝐚 (d) 𝐚−𝟏
SOLUTION : . (a)
Terminal velocity in a viscous medium is given by:
𝟐𝒂𝟐 𝐩 − 𝟎 𝒈
𝑽𝑻 =
𝟗𝜼
𝑽𝑻 ∝ 𝒂𝟐
48. Ifa ball of steel (density 𝐩 = 𝟕. 𝟖𝐠𝐜𝐦−𝟑 ) attains a terminal velocity of 10 cm 𝐬 −𝟏 when falling
in water (Coefficient of viscosity 𝜼𝐰𝐚𝐭𝐞𝐫 = 𝟖. 𝟓 × 𝟏𝟎−𝟒 Pa.s), then, its terminal velocity in
glycerine (𝐩 = 𝟏. 𝟐𝐠𝐜𝐦−𝟑 , 𝜼 = 𝟏𝟑. 𝟐 Pa. s) would be, nearly [2011 𝐑𝐒]
(a) 𝟔. 𝟐𝟓 × 𝟏𝟎−𝟒 cm 𝐬 −𝟏 (b) 𝟔. 𝟒𝟓 × 𝟏𝟎−𝟒 cm 𝐬−𝟏
(c) 𝟏. 𝟓 × 𝟏𝟎−𝟓 cm 𝐬−𝟏 (d) 𝟏. 𝟔 × 𝟏𝟎−𝟓 cm 𝐬−𝟏
SOLUTION : (a)
When the ball attains terminal velocity
Weight ofthe ball = viscous force + buoyant force

𝑽𝐩𝒈 = 𝟔𝝅𝜼𝒓𝒗 + 𝑽𝐩𝒑 𝒈
⇒ 𝑽𝒈 𝐩 − 𝐩𝓵 = 𝟔𝝅𝜼𝒓𝒗
Also 𝑽𝒈 𝐩 − 𝐩′𝒊 = 𝟔𝝅𝜼′ 𝒓𝒗′
𝐩 − 𝐩𝒍
𝒗′ 𝜼′ = × 𝒗𝜼
𝐩 − 𝐩𝒍
𝐩 − 𝐩𝒑 ′ 𝒗𝜼
⇒ 𝒗′ = ×
𝐩 − 𝐩𝒑 𝜼’
𝟕. 𝟖 − 𝟏. 𝟐 𝟏𝟎 × 𝟖. 𝟓 × 𝟏𝟎−𝟒
= ×
𝟕. 𝟖 − 𝟏 𝟏𝟑. 𝟐
𝒗′ = 𝟔. 𝟐𝟓 × 𝟏𝟎−𝟒 𝐜𝐦/𝐬
49. A spherical solid ball ofvolume 𝑽 is made of a material of density 𝐩𝟏 . It is falling through a
liquid of density 𝐩𝟏 (𝐩𝟐 < 𝐩𝟏 ) . Assume that the liquid applies a viscous force on the ball that
is proportional to the square of its speed 𝐯, i.e., 𝜞𝐯𝐢𝐬𝐜𝐨𝐮𝐬 = −𝒌𝒚𝟐 (𝒌 > 𝟎) . The terminal speed
ofthe ball is [2008]
𝑽𝒈 𝐩𝟏 −𝐩𝟐 𝑽𝒈𝐩𝟏 𝑽𝒈𝐩𝟏 𝑽𝒈 𝐩𝟏 −𝐩𝟐

(a) (b) (c) (d)
𝒌 𝒌 𝒌 𝒌
SOLUTION : (a)
When the ball attains terminal velocity
Weight ofthe ball = Buoyant force + Viscous force
𝑽𝐩𝟐 𝒈
𝑾 = 𝑽𝐏𝟏 𝒈
𝑽𝐩𝟏 𝒈 = 𝑽𝐩𝟐 𝒈 + 𝒌𝒗𝟐𝒕 ⇒ 𝒚𝒈 𝐩𝟏 −𝐩𝟐 𝒈=𝒌𝒗𝟐𝒕
𝑽𝒈 𝝆𝟏 − 𝝆𝟐
⇒ 𝒗𝒕 =
𝒌
50. If the terminal speed of a sphere of gold (density = 𝟏𝟗. 𝟓𝐤𝐠/𝐦𝟑 ) is 𝟎. 𝟐𝐦/𝐬 in a viscous
liquid (density = 𝟏. 𝟓𝐤𝐠/𝐦𝟑 ) , fmd the terminal speed of a sphere of silver (density
= 𝟏𝟎. 𝟓𝐤𝐠/𝐦𝟑 ) ofthe same size in the same liquid [2006]
(a) 𝟎. 𝟒 𝐦/𝐬 (b) 𝟎. 𝟏𝟑𝟑 𝐦/𝐬 (c) 𝟎. 𝟏 𝐦/𝐬 (d) 𝟎. 𝟐 𝐦/𝐬
SOLUTION : (c)
Given,
Density ofgold, 𝐩𝐆 = 𝟏𝟗. 𝟓𝐤𝐠/𝐦𝟑
Density of silver, 𝐩𝟓 = 𝟏𝟎. 𝟓𝐤𝐠/𝐦𝟑
Density ofliquid, 𝟎 = 𝟏. 𝟓𝐤𝐠/𝐦𝟑
𝟐𝒓𝟐 𝐩−𝟎 𝒈
Terminal velocity, 𝒗𝑻 = 𝟗𝜼
𝒗𝑻𝟐 𝟏𝟎. 𝟓 − 𝟏. 𝟓 𝟗
= ⇒ 𝒗𝑻𝟐 = 𝟎. 𝟐 ×
𝟎. 𝟐 𝟏𝟗. 𝟓 − 𝟏. 𝟓 𝟏𝟖
𝒗𝑻𝟐 = 𝟎. 𝟏𝐦/𝐬
51. Spherical balls ofradius ‘R’ are falling in a viscous fluid ofviscosity ’ with a velocity 𝐯’. The
retarding viscous force acting on the spherical ball is [2004]
(a) inversely proportional to both radius ‘R’ and velocity‘v’
(b) directly proportional to both radius ‘R’ and velocity‘v’
(c) directlyproportional to ‘R’ but inverselyproportional to ‘v’
(d) inverselyproportional to ‘R’ but directly proportional to velocity‘v’
SOLUTION : (b)
From Stoke’s law, force of viscosity acting on a spherical body is 𝑭 = 𝟔𝝅𝜼𝒓𝒗
hence 𝑭 is directly proportional to radius & velocity.

SURFACETENSION
Surface Tension , Surface Energy and Capillarity :
1. When a long glass capillary tube of radius 𝟎. 𝟎𝟏𝟓 cm is dipped in a liquid, the liquid rises to
a height of 15 cm within it. Ifthe contact angle between the liquid and glass to close to 𝟎∘ ,
the surface tension of the liquid, in milliNewton 𝐦−𝟏 , is 𝐩 𝟏𝐢𝐪𝐮𝐢𝐝 = 𝟗𝟎𝟎𝐤𝐠𝐦−𝟑 , 𝒈 = 𝟏𝟎𝐦𝐬−𝟐
(Give answer in closest integer) [NA 3 Sep. 2020 (I)]
SOLUTION : (101)
Given: Radius ofcapillary tube, 𝒓 = 𝟎. 𝟎𝟏𝟓 cm = 𝟏𝟓 × 𝟏𝟎−𝟓 mm 𝒉 = 𝟏𝟓 cm = 𝟏𝟓 × 𝟏𝟎−𝟐 mm
Using, 𝒉 = 𝟐𝑻 𝐜𝐨𝐬 𝜽 𝐜𝐨𝐬 𝜽 = 𝐜𝐨𝐬 𝟎∘ = 𝟏
𝐩𝒈𝒓
𝒓𝒉𝐩𝒈 𝟏𝟓×𝟏𝟎−𝟓 ×𝟏𝟓×𝟏𝟎−𝟐 ×𝟗𝟎𝟎×𝟏𝟎

Surface tension, 𝑻= = = 𝟏𝟎𝟏 milli newton 𝐦−𝟏
𝟐 𝟐
2. Pressure inside two soap bubbles are 𝟏. 𝟎𝟏 and 𝟏. 𝟎𝟐 atmosphere, respectively. The ratio of
their volumes is: [3 Sep. 2020 (I)]
(a) 4: 1 (b) 𝟎. 𝟖: 1 (c) 8: 1 (d) 2: 1
SOLUTION : (c)
According to question, pressure inside, 1st soap bubble,

𝟒𝑻
𝜟𝑷𝟏 = 𝑷𝟏 − 𝑷𝟎 = 𝟎. 𝟎𝟏 = 𝑹 (i)
𝟏
And 𝜟𝑷𝟐 = 𝑷𝟐 − 𝑷𝟎 = 𝟎. 𝟎𝟐 = 𝟒𝑻 (ii)
𝑹𝟐
Dividing, equation(ii) by(i),
𝟏 𝑹𝟐
= ⇒ 𝑹𝟏 = 𝟐𝑹𝟐
𝟐 𝑹𝟏
𝟒
Volume 𝑽 = 𝟑 𝝅𝑹𝟑
𝑽𝟏 𝑹𝟑𝟏 𝟖𝑹𝟑𝟐 𝟖
= = 𝟑 =
𝑽𝟐 𝑹𝟑𝟐 𝑹𝟐 𝟏
3. A capillarytube made ofglass ofradius 𝟎. 𝟏𝟓 mm is dipped vertically in a beaker filled with

methylene iodide (surface tension = 𝟎. 𝟎𝟓𝐍𝐦−𝟏 , density = 𝟔𝟔𝟕 kg 𝐦−𝟑 ) which rises to
height 𝒉 in the tube. It is observed that the two tangents drawn 𝐟𝐢𝐢 𝐨𝐦 liquid‐glass interfaces
(from opp. sides of the capillary) make an angle 𝐨𝐟𝟔𝟎∘ with one another. Then 𝒉 is close to
𝒈 = 𝟏𝟎𝐦𝐬−𝟐 . [2 Sep. 2020 (II)]
(a) 𝟎. 𝟎𝟒𝟗 𝐦 (b) 𝟎. 𝟎𝟖𝟕 𝐦 (c) 𝟎. 𝟏𝟑𝟕 𝐦 (d) 𝟎. 𝟏𝟕𝟐 𝐦
SOLUTION : (b)
Given,Angle ofcontact 𝜽 = 𝟑𝟎∘
Surface tension, 𝑻 = 𝟎. 𝟎𝟓𝐍𝐦−𝟏
Radius ofcapillarytube, 𝒓 = 𝟎. 𝟏𝟓 mm = 𝟎. 𝟏𝟓 × 𝟏𝟎−𝟑 𝐦
Density ofmethylene iodide, 𝐩 = 𝟔𝟔𝟕 kg 𝐦−𝟑
𝟑
𝟐𝑻 𝐜𝐨𝐬 𝜽 𝟐×𝟎.𝟎𝟓×
Capillaryrise, 𝒉 = 𝟐
= 𝟔𝟔𝟕×𝟏𝟎×𝟎.𝟏𝟓×𝟏𝟎−𝟑 = 𝟎. 𝟎𝟖𝟕𝐦
𝐩𝒈𝒓
4. A small spherical droplet ofdensity 𝒅 is floating exactly halfimmersed in a liquid ofdensity 𝐩

and surface tension T. The radius of the droplet is (take note that the surface tension
applies an upward force on the droplet): [9 Jan. 2020 (II)]
𝟐𝐓 𝐓 𝐓 𝟑𝐓
(a) 𝒓 = (b) 𝒓 = (c) 𝒓 = (d) 𝒓 =
𝟑 𝒅+𝐩 𝒈 𝒅−𝐩 𝒈 𝒅+𝐩 𝒈 𝟐𝒅−𝐩 𝒈
SOLUTION : (d)
For the drops to be in equilibrium upward force on drop = downward force on drop
𝟒 𝟐
𝑻. 𝟐𝝅𝑹 = 𝝅𝑹𝟑 𝒅𝒈 − 𝝅𝑹𝟑 𝝆𝒈
𝟑 𝟑
𝟐
⇒ 𝑻 𝟐𝝅𝑹 = 𝝅𝑹𝟑 𝟐𝒅 − 𝐩 𝒈
𝟑
𝑹𝟐 𝟑𝑻
⇒𝑻= 𝟐𝒅 − 𝐩 𝒈 ⇒ 𝑹 =
𝟑 𝟐𝒅−𝐩 𝒈
5. The ratio of surface tensions of mercury and water is given to be 𝟕. 𝟓 while the ratio oftheir
densities is 𝟏𝟑. 𝟔. Their contact angles, with glass, are close to 𝟏𝟑𝟓𝐨 and 𝟎𝐨 , respectively. It is
observed that mercury gets depressed by an amount 𝐡 in a capillary tube ofradius 𝐫𝟏 , while
water rises by the same amount 𝐡 in a capillary tube of radius 𝐫𝟐 . The ratio, 𝐫𝟏 /𝐫𝟐 , is then
close to: [10 April 2019 (I)]
(a) 4/5 (b) 2/5 (c) 3/5 (d) 2/3
SOLUTION : (b)
As we know that
𝟐𝐓 𝐜𝐨𝐬 𝜽 = 𝐑 𝐡
𝐫𝐩𝐠
𝐓𝐇𝐠
= 𝟕. 𝟓
𝐓𝐖𝐚𝐭𝐞𝐫
𝐩𝐇𝐠 𝐜𝐨𝐬 𝜽𝐇𝐠 𝐜𝐨𝐬 𝟏𝟑𝟓𝐨 𝟏
= 𝟏𝟑. 𝟔& = =
𝐏𝐰 𝐜𝐨𝐬 𝜽𝐖 𝐜𝐨𝐬 𝟎𝐨 𝟐
𝐑 𝐇𝐠
=
𝐑 𝐖𝐚𝐭𝐞𝐫
𝟏 𝟏 𝟐
= 𝟕. 𝟓 × × = 𝟎. 𝟒 =
𝟏𝟑. 𝟔 𝟐 𝟓
6. If 𝐌’ is the mass ofwater that rises in a capillary tube of radius 𝐫’, then mass ofwater which
will rise in a capillary tube of radius 𝟐𝐫 ’ is: [9 April 2019 I]
𝐌
(a) 𝐌 (b) (c) 4 𝐌 (d) 2 𝐌
𝟐
SOLUTION : (d)
𝟐𝑻 𝐜𝐨𝐬 𝜽
We have, 𝐡 = 𝒓𝐩𝒈
Mass of the water in the capillary
𝟐𝑻 𝐜𝐨𝐬 𝜽
𝐦 = 𝐩𝐕 = 𝐩 × 𝝅𝐫 𝟐 𝐡 = 𝐩 × 𝝅𝒓𝟐 ×
𝒓𝐩𝒈
⇒ 𝐦𝜶𝐫
7. If two glass plates have water between them and are separated by very small distance (see
figure), it is very difficult to pull them apart. It is because the water in between forms
cylindrical surface on the side that gives rise to lower pressure in the water in comparison
to atmosphere. If the radius of the cylindrical surface is 𝐑 and surface tension ofwater is 𝐓
then the pressure in water between the plates is lower by: [Online Apri110, 2015]
𝟐𝐓 𝟒𝐓 𝐓 𝐓
(a) (b) (c) 𝟒𝐑 (d) 𝐑
𝐑 𝐑
SOLUTION : (d)
8. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take
the bubbles to be spheres ofradius 𝐑 and making a circular contact ofradius 𝐫 with the
bottom ofthe vessel. 𝐈𝐟𝐫 ≪ 𝐑 and the surface tension of water is 𝐓, value of 𝐫 just before
bubbles detach is: (density of water is 𝐩𝐰 ) [2014]
↔𝟐𝐫
𝟐𝐩𝐰 𝐠 𝐩𝐰 𝐠 𝐩𝐰 𝐠 𝟑𝐩𝐰 𝐠
(a) 𝐑𝟐 (b) 𝐑𝟐 (c) 𝐑𝟐 (d) 𝐑𝟐
𝟑𝐓 𝟔𝐓 𝐓 𝐓
SOLUTION : (a)
When the bubble gets detached, Buoyant force = force due to surface tension
𝟐𝐓
Force due to excess pressure = upthrust Access pressure in air bubble = 𝐑
𝟐𝑻 𝟒𝝅𝑹𝟑
𝝅𝒓𝟐 = 𝝆 𝒈
𝑹 𝟑𝑻 𝒘
𝟐𝑹𝟒 𝝆𝒘 𝒈 𝟐𝝆𝒘 𝒈
⇒ 𝒓𝟐 = ⇒ 𝒓 = 𝑹𝟐
𝟑𝑻 𝟑𝑻
9. A large number ofliquid drops each ofradius 𝐫 coalesce to from a single drop ofradius R.
The energyreleased in the process is converted into kinetic energy of the big drop so
formed. The speed of the big drop is (given, surface tension ofliquid 𝐓, density p)
[Online Apri119, 2014, 2012]
𝐓 𝟏 𝟏 𝟐𝐓 𝟏 𝟏 𝟒𝐓 𝟏 𝟏 𝟔𝐓 𝟏 𝟏
(a) −𝐑 (b) −𝐑 (c) −𝐑 (d) −𝐑
𝐩 𝐫 𝐩 𝐫 𝐩 𝐫 𝐩 𝐫
SOLUTION : (d)
When drops combine to form a single drop ofradius R.

𝟏 𝟏
Then energy released, 𝐄 = 𝟒𝝅𝐓𝐑𝟑 −𝐑
𝐫
If this energy is converted into kinetic energy then

𝟏 𝟏 𝟏
𝐦𝐯 𝟐 = 𝟒𝝅𝐑𝟑 𝐓 −
𝟐 𝐫 𝐑
𝟏 𝟒 𝟏 𝟏
× 𝝅𝐑𝟑 𝐩 𝐯 𝟐 = 𝟒𝝅𝐑𝟑 𝐓 −
𝟐 𝟑 𝐫 𝐑
𝟔𝐓 𝟏 𝟏
𝐯𝟐 = −
𝐩 𝐫 𝐑
𝟔𝐓 𝟏 𝟏
𝐯= −
𝐩 𝐫 𝐑
10. Two soap bubbles coalesce to form a single bubble. 𝐈𝐟𝐕 is the subsequent change in volume
ofcontained air and 𝐒 change in total surface area, 𝐓 is the surface tension and 𝐏
atmospheric pressure, then which ofthe following relation is correct?
(a) 𝟒𝐏𝐕 + 𝟑𝐒𝐓 = 𝟎 (b) 𝟑𝐏𝐕 + 𝟒𝐒𝐓 = 𝟎
(c) 𝟐𝐏𝐕 + 𝟑𝐒𝐓 = 𝟎 (d) 𝟑𝐏𝐕 + 𝟐𝐒𝐓 = 𝟎
SOLUTION : (b)
11. An air bubble ofradius 𝟎. 𝟏 cm is in a liquid having surface tension 𝟎. 𝟎𝟔𝐍/𝐦 and density
𝟏𝟎𝟑 𝐤𝐠/𝐦𝟑 . The pressure inside the bubble is 1100 𝐍𝐦−𝟐 greater than the
atmospheric pressure. At what depth is the bubble below the surface of the liquid?
𝐠 = 𝟗. 𝟖𝐦𝐬 −𝟐 [Online April 11, 2014]
(a) 𝟎. 𝟏𝐦 (b) 𝟎. 𝟏𝟓𝐦 (c) 𝟎. 𝟐𝟎𝐦 (d) 𝟎. 𝟐𝟓𝐦
SOLUTION : (a)
Given: Radius of air bubble, 𝒓 = 𝟎. 𝟏 cm = 𝟏𝟎−𝟑 𝐦
Surface tension of liquid, 𝑺 = 𝟎. 𝟎𝟔𝐍/𝐦 = 𝟔 × 𝟏𝟎−𝟐 𝐍/𝐦
Density ofliquid, 𝐩 = 𝟏𝟎𝟑 𝐤𝐠/𝐦𝟑
Excess pressure inside the bubble, 𝐩𝐞𝐱𝐞 = 𝟏𝟏𝟎𝟎𝐍𝐦−𝟐
Depth of bubble below the liquid surface, 𝒉 =?

𝟐𝒔
As we know, pExcess = 𝒉𝐩𝒈 + 𝒓
𝟑
𝟐 × 𝟔 × 𝟏𝟎−𝟐
⇒ 𝟏𝟏𝟎𝟎 = 𝒉 × 𝟏𝟎 × 𝟗. 𝟖 +
𝟏𝟎−𝟑
⇒ 𝟏𝟏𝟎𝟎 = 𝟗𝟖𝟎𝟎𝒉 + 𝟏𝟐𝟎
⇒ 𝟗𝟖𝟎𝟎𝒉 = 𝟏𝟏𝟎𝟎 − 𝟏𝟐𝟎

𝟗𝟖𝟎
⇒𝒉= = 𝟎. 𝟏𝐦
𝟗𝟖𝟎𝟎
12. Acapillarytube is immersed verticallyin water and the height ofthe water column is 𝐱.
When this arrangement is taken into a mine ofdepth 𝐝, the height ofthe water colunm is
𝐱
𝐲. 𝐈𝐟𝐑 is the radius ofearth, the ratio 𝐲 is: [Online April 9, 2014]
𝐝 𝟐𝐝 𝐑−𝐝 𝐑+𝐝
(a) 𝟏 − 𝐑 (b) 𝟏 − (c) (d)
𝐑 𝐑+𝐝 𝐑−𝐝
SOLUTION : (a)
𝐝
Acceleration due to gravity changes with the depth, 𝐠 ′ = 𝐠 𝟏 − 𝐑
Pressure, 𝐏 = 𝐩𝐠𝐡
𝐱 𝐝
Hence ratio, 𝐲 is 𝟏 − 𝐑
13. Wax is coated on the inner wall ofa capillary tube and the tube is then dipped in water.
Then, compared to the unwaxed capillary, the angle ofcontact 𝜽 and the height 𝒉 upto
which water rises change. These changes are: [Online April 23, 2013]
(a) 𝜽 increases and 𝒉 also increases (b) 𝜽 decreases and 𝒉 also decreases
(c) 𝜽 increases and 𝒉 decreases (d) 𝜽 decreases and 𝒉 increases
SOLUTION : (c)
𝐓𝐒𝐀 −𝐓𝐒𝐋
Angle of contact 𝜽 𝐜𝐨𝐬 𝜽 = 𝐓𝐋𝐀
when water is on a waxy or oily surface
𝐓𝐒𝐀 < 𝐓𝐒𝐋 𝐜𝐨𝐬 𝜽 is negative i. e.,
𝟗𝟎∘ < 𝜽 < 𝟏𝟖𝟎∘
i.e., angle of contact 𝜽 increases
And for 𝜽 > 𝟗𝟎∘ liquid level in capillary tube fall.
i.e., 𝐡 decreases
14. A thin tube sealed at both ends is 100 cm long. It lies horizontally, the middle 20 cm
containing mercury and two equal ends containing air at standard atmospheric pressure. If
the tube is now turned to a vertical position, bywhat amount will the mercurybe displaced?
(Given : cross‐section of the tube can be assumed to be unifoim) [Online April 23, 2013]
(a) 𝟐. 𝟗𝟓 cm (b) 𝟓. 𝟏𝟖 cm (c) 𝟖. 𝟔𝟓 cm (d) 𝟎. 𝟎 cm
SOLUTION : (b)
15. This question has Statement‐l and Statement‐2. Ofthe four choices given after the
Statements, choose the one that best describes the two Statetnents.
Statement‐l: A capillary is dipped in a liquid and liquid rises to a height 𝒉 in it. As the
temperature ofthe liquid is raised, the height 𝒉 increases (ifthe density of the liquid and the
angle ofcontact remain the same).
Statement‐2: Surface tension of a liquid decreases with the rise in its temperature.
(a) Statement‐l is true, Statement‐2 is true; Statement‐2 is not the correct explanation for
Statement‐l.
(b) Statement‐ 1 is false, Statement‐2 is true.
(c) Statement‐ 1 is true, Statement‐2 is false.
(d) Statement‐l is true, Statement‐2 is true; Statement‐2 is the correct explanation for
Statement‐l.
SOLUTION : (b )
Surface tension of a liquid decreases with the rise in temperture. At the boiling point ofliquid,
surface tension is zero.
𝟐𝐓 𝐜𝐨𝐬 𝜽
Capillary rise 𝐡 = 𝐫𝐝𝐠
As surface tension 𝐓 decreases with rise in temperature hence capillary rise also decreases.
16. A thin liquid film formed between a 𝐔‐shaped wire and a light slider supports a weight of
𝟏. 𝟓 × 𝟏𝟎−𝟐 𝐍 (see figure). The length ofthe slider is 30 cm and its weight is negligible. The
surface tension ofthe liquid film is [2012]
(a) 𝟎. 𝟎𝟏𝟐𝟓 𝐍𝐦−𝟏 (b) 𝟎. 𝟏 𝐍𝐦−𝟏 (c) 𝟎. 𝟎𝟓 𝐍𝐦−𝟏 (d) 𝟎. 𝟎𝟐𝟓 𝐍𝐦−𝟏
SOLUTION : (d)
Let 𝑻 is the force due to surface tension per unit length, then 𝜞 = 𝟐𝒍𝐓
𝒍 = length of the slider.
At equilibrium, 𝑭 = 𝑾
𝟐𝑻𝒍 = 𝒎𝒈
𝒎𝒈 𝟏.𝟓×𝟏𝟎−𝟐 𝟏.𝟓
⇒𝑻= = = = 𝟎. 𝟎𝟐𝟓𝐍𝐦−𝟏
𝟐𝒍 𝟐×𝟑𝟎×𝟏𝟎−𝟐 𝟔𝟎
17. Work done in increasing the size ofa soap bubble 𝐟𝐢𝐢 𝐨𝐦 a radius of 3 cm to 5 cm is nearly
(Surface tension of soap solution = 𝟎. 𝟎𝟑𝐍𝐦−𝟏 ) [2011]
(a) 𝟎. 𝟐𝝅𝐦𝐉 (b) 𝟐𝝅𝐦𝐉 (c) 𝟎. 𝟒𝝅𝐦𝐉 (d) 𝟒𝝅𝐦𝐉
SOLUTION : (c)
Work done = increase in surface area × surface tension
⇒ 𝑾 = 𝟐𝑻𝟒𝝅 𝟓𝟐 − 𝟑 𝟐
× 𝟏𝟎−𝟒
= 𝟐 × 𝟎. 𝟎𝟑 × 𝟒𝝅 𝟐𝟓 − 𝟗 × 𝟏𝟎−𝟒 𝐉
= 𝟎. 𝟒𝝅 × 𝟏𝟎−𝟑 𝐉 = 𝟎. 𝟒𝝅𝐦𝐉
18. Two mercury drops (each of radius ‘r’) merge to form bigger drop. The surface energy ofthe
bigger drop, if 𝑻 is the surface tension, is : [2011 𝐑𝐒]
(a) 𝟒𝝅𝒓𝟐 𝑻 (b) 𝟐𝝅𝒓𝟐 𝑻 (c) 𝟐𝟖/𝟑 𝝅𝒓𝟐 𝑻 (d) 𝟐𝟓/𝟑 𝝅𝒓𝟐 𝑻
SOLUTION : (c)
As volume remains constant
Sum of volumes of 2 smaller drops = Volume of the bigger drop

𝟒 𝟒
2. 𝟑 𝝅𝒓𝟑 = 𝟑 𝝅𝑹𝟑 ⇒ 𝑹 = 𝟐𝟏/𝟑 𝒓
Surface energy = Surface tension × Surface area = 𝑻. 𝟒𝝅𝑹𝟐
= 𝑻𝟒𝝅𝟐𝟐/𝟑 𝒓𝟐 = 𝑻. 𝟐𝟖/𝟑 𝝅𝒓𝟐
19. Acapillary tube(A) is dipped in water. Another identical tube (B) is dipped in a soap‐water
solution. Wmich of the following shows the relative nature ofthe liquid columns in the two
tubes? [2008]
SOLUTION : (c)
In case of water, the meniscus shape is concave upwards.

𝟐𝒐 𝐜𝐨𝐬 𝜽
From ascent formula 𝒉 = 𝒓𝐩𝒈
The surface tension (𝝇) of soap solution is less than water.
Therefore height ofcapillaryrise for soap solution should be less as compared to water.
As in the case ofwater, the meniscus shape of soap solution is also concave upwards.
20. A20 cm long capillary tube is dipped in water. The water rises up to 8 cm. Ifthe entire
arrangement is put in a freely falling elevator the length ofwater column in the capillary
tube will be [2005]
(a) 10 cm (b) 8 cm (c) 20 cm (d) 4 cm
SOLUTION : (c)
Water fills the tube entirely in gravityless condition i.e., 20 cm.
21. If two soap bubbles of different radii are connected by a tube [2004]
(a) air flows 𝐟𝐢𝐢 𝐨𝐦 the smaller bubble to the bigger
(b) air flows 𝐟𝐢𝐢𝐎 𝐦 bigger bubble to the smaller bubble till the sizes are interchanged
(c) air flows 𝐟𝐢𝐢 𝐨𝐦 the bigger bubble to the smaller bubble till the sizes become equal
(d) there is no flow ofair.
SOLUTION : (a)
Let pressure outside be 𝑷𝟎 and 𝒓 and 𝑹 be the radius of smaller bubble and bigger bubble
respectively.
𝟐𝑻
Pressure 𝑷𝟏 For smaller bubble = 𝑷𝟎 + 𝒓
𝟐𝑻
𝑷𝟐 For bigger bubble = 𝑷𝟎 + (𝑹 > 𝒓)
𝑹
𝑷𝟏 > 𝑷𝟐
hence air moves from smaller bubble to bigger bubble.

THERMOMETRY
Heat
The form of energy which is exchanged among various bodies or system on account of temperature difference
isdefinedasheat.
We can change the temperature of a body by giving heat (temperature rises) or by removing heat
(temperaturefalls)frombody.
The amount of heat (Q) is given to a body depends upon it’s mass (m), change in it’s temperature
( °=)andnatureofmateriali.e. Q  m .c.  ;wherec=specificheatofmaterial.
Heatisascalarquantity.It’sunitsarejoule,erg,cal,kcaletc.
The calorie (cal) is defined as the amount of heat required to raise the temperature of 1 gm of water
from 14.5°C to 15.5°C.
Also 1 kcal = 1000 cal = 4186 J and 1 cal = 4.18 J
British Thermal Unit (BTU) : One BTU is the quantity of heat required to raise the temperature of
one pound (1 lb) of water from 63°F to 64°F
1 BTU = 778 ft. lb = 252 cal = 1055 J
Insolidsthermal energyispresentintheformofkineticenergy,inliquids,intheformoftranslatory
energy of molecules. In gas it is due to the random motion of molecules.
Heat always flowsfroma bodyof highertemperatureto lower temperature till their temperature becomes
equal(Thermalequilibrium).
The heat required for a given temperature increase depends only on how many atoms the sample contains,
not on themassof an individualatom.
Temperature
Temperatureisdefinedasthe degreeofhotness orcoldnessof abody.The natural flowof heatisfrom higher

temperature to lower temperature.
Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the
other. That is when both the bodies are at the same temperature.
Temperatureisone ofthesevenfundamentalquantities withdimension[ ].Itisascalarphysical

quantitywithS.I.unitkelvin.
When heat is given to a body and its state does not change, the temperature of the body rises and if
heatistakenfromabodyitstemperaturefallsi.e.temperaturecanberegardedastheeffectof
cause“heat”.
According to kinetic theory of gases, temperature (macroscopic physical quantity) is a measure of
averagetranslationalkineticenergyofamolecule(microscopicphysicalquantity).
Although the temperature of a body can to be raised without limit, it cannot be lowered without limit
and theoretically limiting low temperature is taken to be zero of the kelvin scale.
Highest possible temperature achieved in laboratory is about 108K while lowest possible temperature
attained is 10–8 K.
Temperature of the core of the sun is 107 K while that of its surface is 6000 K.
Normal temperature of human body is 310. 15 K (37°C = 98.6°F).
NTP or STP implies 273.15K (0°C = 32°F)
Scales of Temperature
Boiling 212°F 100°C 373 K

water
Freezing 32°F 0°C 273 K

water
Fahrenheit Celsius Kelvin

Fig. 12.1
The centigrade (°C), Farenheite (°F), Kelvin (K), Reaumer (R), Rankine (Ra) are commonly used
temperature scales.
Toconstructascaleoftemperature,twofixedpointsaretaken.Firstfixedpointisthefreezingpoint(ice
point)ofwater,itiscalledlowerfixedpoint(LFP).Thesecondfixedpointistheboilingpoint(steam
point)ofwater,itiscalledupperfixedpoint(UFP).
Celsius scale : In this scale LFP (ice point) is taken 0° and UFP (steam point) is taken 100°. The
temperature measured on this scale all in degree Celsius (°C).
Farenheite scale : This scale of temperature has LFP as 32°F and UFP as 212°F. The change in
temperature of 1°F corresponds to a change of less than 1° on Celsius scale.
Kelvin scale : The Kelvin temperature scale is also known as thermodynamic scale. The triple point
of water is also selected to be the zero of scale of temperature. The temperature measured on this
scale are in Kelvin (K).
The triple point of water is that point on a P-Tdiagramwherethethreephasesofwater,thesolid,theliquid

andthegas,cancoexistinequilibrium.
Different measuring scales
Scale Symbol LFP UFP Number of divisions

Celsius °C 0°C 100°C 100
Fahrenheit °F 32°F 212°F 180
Reaumer °R 0°R 80°R 80
Rankine °Ra 460 Ra 672 Ra 212
Kelvin K 273.15 K 373.15 K 100
 Temperature on one scale can be converted into other scale by using the following identity.
Reading on any scale  LFP

UFP  LFP = Constant for all scales
 All these temperatures are related to each other by the following relationship
C 0 F  32 K  273 . 15 R 0 Ra  460
   
100 212  32 373 . 15  273 . 15 80  0 672  460
C F  32 K  273 R Ra  460
or    
5 9 5 4 10 . 6
 The Celsius and Kelvin scales have different zero points but the same size degrees. Therefore any temperature differenceis the
same on the Celsius and Kelvin scales (T2 – T1)°C = (T2 – T1) K.
Thermometry
A branch of science which deals with the measurement of temperature of a substance is known as thermometry.
 The linear variation in some physical properties of a substance with change of temperature is the basic
principle of thermometry and these properties are defined as thermometric property (x) of the substance.
Thermometric properties (x) may be as follows
 Length of liquid in capillary
 Pressure of gas at constant volume.
 Volume of gas at constant pressure.
 Resistance of a given platinum wire.
In old thermometry, freezing point (0°C) and steam point (100°C) are taken to define the temperature scale.
So if the thermometric property at temperature 0°C, 100°C and t°C are x0, x100 and x respectively then
t0 x  x0

100  0 x 100  x 0
x  x0
tC  x 100  x 0
 100 C
 General equationused to measure temperature t. X t  X 0 1   t 
 In modern thermometry instead of two fixed points only one reference point is chosen (triple point of
water 273.16 K) the other is itself 0 K where the value of thermometric property is assumed to
be zero.
So if the value of thermometric property at 0 K, 273.16 K and TK are 0, xTr and x respectively then
T x

273 .16 x Tr
 x 
T  273 .16  x K
 Tr 
Faulty Thermometer:
 If the reading on a faulty thermometer is ‘x’ and its lower and upper fixed points are L and U respectively
then
 correct reading on Celsius scale is
C XL

100 UL
F 3 2 X L
 Correct reading on Fahrenheit scale is =
180 U L
K  273 X  L
 Correct reading on Kelvin sclae is 
100 UL
 Error in measurement by faulty thermometer = measures value true value
Correction = Error
Types of Thermometers:
Types of Thermometric
thermometer property Advantages Disadvantages Where is it used
and its range
Mercuryin Length of (i) Quick and (i) Fragile Laboratory use
glass column of easy to use (ii) Small size where high
mercury in (direct reading) limits precision accuracy is not
390C to 450C capillary tube (ii) Easily portable (iii) Limited range required
Constant Pressure of a (i) Very acurate (i) Very large volume (a) Standard when
volume gas fixed mass of gas (ii) Very sensitive of bulb compared to other
at constant (iii)Wide range (ii) Slow to use and thermometer
thermometer volume (iv) Easily convenient
270C to15000C reproducible
Platinum Electrical (i) Accurate Not suitable for Best thermometer
resistance resistance of a (ii) Wide range varying temperature for small steady
1800C to platinum coil (i.e, is slow to respond temperature
to changes difference
11500C
Emf produced
Thermocouple (i) Fast response Accuracy is lost if Best thermometer
between junctions because of low heat emf is measured using for varying
2500C to
of dissimilar capacity a moving coil temperature
11500C metals at (ii) Wide range voltmeter (as may be
different (iii) Can be necessary for rapid
temperature for employed for changes when
measurement remote readings potentiometer is
of emfs unsuitable)
Radiation Colour of Does not come into (i) cumbersome Only
pyrometer radiation emitted contact when (ii) some direct thermometer
by a hot body temperature is reading (needs a possible for very
Above 30000C high thermometers
measured trained observer)
PROBLEMS
1. On the Celsius scale the absolute zero of temperature is at
(a) 0°C (b) – 32°C (c) 100°C (d) – 273.15°C
SOLUTION:
T  273 . 15  t C
 0  273 .15  tC
 t  273 .15 C
2. Oxygen boils at – 183°C. This temperature is approximately
(a) 215°F (b) – 297°F (c) 329°F (d) 361°F

SOLUTION:
C F  32

5 9
183 F  32
 
5 9
 F  297  F
3. The temperature of a body on Kelvin scale is found to be x K. When it is measured by Fahrenheit

thermometer, it is found to be x°F, then the value of x is
(a) 40 (b) 313 (c) 574.25 (d) 301.25
SOLUTION:
F  32 K  273

9 5
x  32 x  273
 
9 5
 x  574 .25
4. A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is
lowered until the Fahrenheit thermometer registers 140°. What is the fall in temperature as registered
by the Centigrade thermometer
(a) 30° (b) 40° (c) 60° (d) 80°
SOLUTION:
C F  32

5 9
C (140  32 )
 
5 9
 C  60 
5. At what temperature the centigrade (Celsius) and Fahrenheit, readings are the same
(a) – 40° (b) + 40° (c) 36.6° (d) – 37°
SOLUTION:
C F  32

5 9
t t  32
 
5 9
 t   40 
6. If a thermometer reads freezing point of water as 20°C and boiling point as 150°C, how much thermometer
read when the actual temperature is 60°C
(a) 98°C (b) 110°C (c) 40°C (d) 60°C
SOLUTION:
x  LFP
Temperature on any scale can be converted into other scale by UFP  LFP = Constant
x  20 60
forallscales 150  20 
100
 x  98 C
7. The graph between two temperature scales A and B is shown in Fig. Between upper fixed point
and lower fixed point there are 150 equal divisions on scale A and 100 on scale B. The relation
between the temperatures in two scales is given by__
Temperature  A
0
1800
t A  1500
300
tB 1000
O 1000 Temperature  0 B 
0 t A  30 t B  0
SOLUTION: When t B  0, t A  30  
150 100
8. Graph shows the relation between Centigrade and Fahrenheit scales of temperature. Find slope
in each graph?
F
C
2120 F
5
9 Slope 
Slope=
5
9
320 F
O 320 F F
17.78 C O
0 0
100 C C 0
17.78 C
SOLUTION:
Case (i): A plot of Fahrenheit temperature (F) versus Celsius temperature (C)
9
F  C  32  y  mx  c 
5
Slope of the graph m=9/5
C F  32 5 160
Case (ii):  ;C  F   y  mx  c 
5 9 9 9
Slope of the graph, m=5/9
A plot of Celsius temperature (C) versus Fahrenheit temperature (F)
9. An accurate Celsius thermometer and a faulty Fahrenheit thermometer register 600 and 1410
respectively when placed in the same constant temperature enclosure. What is the error in the
Fahrenheit thermometer?
SOLUTION:
C F32 60 F32
From =  =  F=140 0 F
5 9 5 9
Error=141140=10 F; C orrection=10 F
10. Two absolute scales X and Y have triple points of water defined to be 300 X and 450 Y. How are
TX and TY related to each other?
SOLUTION:
Here, temperature 300 on absolute scale X=273.16K (Triple point of water)
 Value of temperature T
TX on absolute
273.16
scale X  TX
300
Similarly, value of temperature TY on absolute scale
273.16
Y TY
450
Since both these values are equal,
273.16 273.16 2
TX  TY  TX  TY
300 450 3
11. The readings corresponding to the ice point and steam point for a constant pressure gas thermometer
are 500cc, and 545 cc. If the reading corresponding to room temperature be 510 cc, find the room
temperature?
SOLUTION:
Given: V0  500cc; V100  545cc. and Vt  510cc.
 V V   510  500 
Using, t   V t  V0   100   545  500  100
 100 0   
 22.220 C
12.. The resistance of a platinum wire is 15 at 200C. This wire is put in a hot furnace and the resistance
of the wire is found to be 40 . Find the temperature of the hot furnace if temperature coefficient
of resistance of platinum is 3.6  10 3 0C 1
SOLUTION:
R2 1   t2 
Rt  R0 1   t   
R1 (1   t1 )
40 1   t2 
  40  15   15t2  40t1 
15 1   t1 
25
15t2   40  20  7745
3.6  103
7745
 t2   5160 C
15
13. The resistance of a platinum resistance thermometer is found to be 11.0 ohm when dipped in a
triple point cell. When it is dipped in a bath, resistance is found to be 28.887 ohm. Find the
temperature of the bath in 0C on platinum scale.
SOLUTION: In terms of triple point of water,
 R 
TK   273.16 K
 RTr 
28.887
so TK  273.16   717.32 K
11.0
Now as TC  717.32  273.15  444.17 0 C

14. What is the temperature for which the reading on Kelvin and Fahrenheit scales are same?
SOLUTION:
On the Kelvin and Fahrenheit scales
K 2 7 3 .1 5 F 3 2
= (if X = K = F )
100 180
X 2 7 3 .1 5 X 3 2
=
100 180
9
X =  2 5 5 .3 8  = 5 7 4 .6
4
 5 7 4 .6 K = 5 7 4 .6 0 F .
15. At what temperature is the Fahrenheit scale reading equal to half that on the Celsius scale?
SOLUTION:
9 1
Ast F  tc  32 and t F  tc ,
5 2
1 9
tc  tc  32
2 5
320
 24.60 C
or tc  
13
16. A constant volume gas thermometer shows pressure readings of 50 cm and 90 cm of mercury at
00C and 1000C respectively. What is the temperature on gas scale when the pressure reading is
60 cm of mercury?
SOLUTION:
Given that P0  50cm of Hg,
P100  90cm of Hg
Pt  60cm of Hg
Pt  P0 60  50
t  100   100  250 C
P100  P0 90  50
Thermal Expansion:
When matter is heated without any change in it’s state, it usually expands. According to atomic theory of
matter,asymmetryinpotentialenergycurveisresponsibleforthermalexpansion.Aswithriseintemperaturethe
amplitudeofvibrationandhenceenergyofatomsincreases,hencetheaveragedistancebetweentheatomsincreases.
So the matter as a whole expands.
Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular force
is maximum in solids but minimum in gases.

Solids can expand in one dimension (linear expansion), two dimension (superficial expansion) and three
dimension(volume expansion)whileliquids and gases usually sufferschangeinvolume only.
Linear expansion :
Whenasolidisheatedandit’slengthincreases,thentheexpansioniscalledlinearexpansion.
L0 L0 + DL = L
(A) Before heating (B) After heating
 Change in length  L = L0T

(L0 = Original length, T = Temperature change)
 Final length L = L0 (1 + T)
L
 Co-efficientoflinearexpansion   L0 T
 Unit of  is C 1 or K 1 . It’s dimension is [ 1 ]

Superficial (areal) expansion :
When the temperature of a 2D object is changed, it’s area changes, then the expansion is called
superficialexpansion.
L0
DL
L0
DL
(i) Change in area is A = A0T
(A0 = Original area, T = Temperature change)
(ii) Final area A = A0(1 + T)
A
(iii)Co-efficientofsuperficialexpansion  
A0 T
(iv) Unit of  is °C–1 or K–1.

Volume or cubical expansion :
When a solid isheated and it’s volume increases, then the expansion is called volumeor cubical
expansion.
L0 L0
L0 V
(i) Change in volume is V  V0T

(V0 = Original volume, T = change in temperature)
(ii)Finalvolume V  V0 (1  T )
V
(iii)Volumeco-efficientofexpansion  
V0 T
iv) Unit of  is °C–1 or K–1.
Contraction on heating : Some rubber like substances contract with rising temperature, because transverse
vibrationofatomsofsubstancedominateoverlongitudinalvibrationwhichisresponsiblefor
expansion.
Coefficients of Linear Expansion
The ratio of increase in length of a solid per degree rise in temperature to its original length is called coefficient
of linear expansion  
l 2  l1
  /0 C
l1   t 2  t 1 
1  dl  0
 in differential form    / C
l0  d t 
length of the solid after heating
 2   1 1    t2  t1  
The coefficient of linear expansion of a solid depends on the nature of the material and scale of temperature.
(it is independent on dimension of material)
 The linear expansion of a solid l2 l1 e l1 t2 t1
 It depends on three factors.
a) Its original length  l1 
b) The nature of the material  
c) Change in temperature  t2  t1 
 Increase in length l  lt
l
 Fractional change in length  t
l
l
Percentage change in length     t  100
l
 For anisotropic solids, if  x ,  y and  z are coefficients of linear expansions along x, y and z directions
x y z
respectively then the average coefficient of linear expansion is  
3
 Numerical value of coefficient of linear expansion of a solid is  C when the temperature is measured in
Celsius scale and its value is  F when the temperature is measured in Fahrenheit scale then
5 9
a)  F   C (or)  C    F
9  5
b)  F   C
 A composite rod is made by joining two rods of different materials and of same cross section. If l1 , l2 are
their initial lengths at t1 0C , then
(a) the increase in length of composite rod at t2 0C is given by l   l11  l2 2  t2  t1 
l11  l2 2
b) The effective coefficient of linear expansion of the composite rod is given by   l1  l2
 If two metal rods of coefficients of linear expansions 1 and  2 have same length at
t1 0C and t2 0C respectively, then the common temperature at which they have again the same length is
1t1   2t2
t
1   2
 If two rods of same length l having different coefficients of linear expansion 1 and  2 (1   2 ) are at the
same temperature t1 0C then a) difference in their lengths at higher temperature t2 0C is given by
l  x  1   2  l  t2  t1 
 The diameter of a metal ring is ‘D’ and the coefficient of linear expansion is  . If the temperature of the ring
is increased by t then the increase in circumference of the ring
  D
 C  C   t  2 r  t  2  2    t   D  t
 
Increase in circumference =  D   t
Coefficient of Areal (or) Superficial expansion:

The ratio of increase in its area per degree rise in temperature to its original area is called coefficient of areal
expansion    .
A2  A1
 /0 C
A1   t2  t1 
Final area A2  A1 1    t2  t1  
 Change in area A  At .
A
 Fractional change in area   t
A
A
Percentage change in area     t  
A
 The diameter of a metal ring is ‘D’ and the coefficient of areal expansion is  . If the temperature of the ring
is increased by t then
a) The increase in area of the ring
 D 2  t
A  At   R 2 t 
4
 D 2
 t    2 
2
Coefficient of volume expansion:
 The ration of increase in its volume per degree rise in temperature to its original volume is called coefficient
of volume expansion  .
V2  V1 0
 / C; V2  V1 1    t2  t1  
V1  t2  t1 
V2  V1
 Final temperature t 2   t1
V1
 Change in volume V   V1  t2  t1  .
V
 Fractional change in volume  t
V
 Percentage change in volume
V
 100   t  100
V
 Volume expansion of a body is independent of its cavities.
Relation among α, β, γ :
   2a ,   3a
  :  :    : 2 : 3  1: 2 : 3
  
  
1 2 3
Variation of density of substance with temperature
 When a solid is heated its volume increases and hence its density decreases, as mass remains constant.
If 1 and  2 are densities of a solid at t1 0C and t2 0C , and as m1  m2 ; 1V1   2V2
1V1   2V1 1    t2  t1  
1   2 1    t2  t1  
If 1 and  0 are densities at t 0C and 00 C .
0 1
1   o r   1   0 1   t 
1   t 
1  0 1   t 
 For anisotropic materials  is the sum of linear coefficients in three mutually perpendicular directions.
  x  y z
For isotropic solids   3
Applications:
Same Expansion In Different Rods:
If two rods of different materials have the same difference between their lengths at all temperatures
only when their linear expansions are equal.
l1  l2 ; l11t  l2 2 t
l1  2
Then l11  l2 2 , l  
2 1
if the constant difference in their lengths is x then
x 2 x1
l1  , l2  , x  l2  l1
1  2 1  2
Bimetallic Strip:
l
2
1
r1 t
t
r2
R

Where t is thickness of each strip
 Bimetallic strip works on the principle that different metals expand differently for the same rise in temperature.
 If a bimetallic strip made of brass and iron is heated brass bends on convex side  b  i 
 If it is cooled brass bends on concave side.
 Radius of curvature of a bimetallic strip.
1 dl l2  l1 l l  2  1  T
  R  or    dr  r  r ; R  2t
2 1
2t
R 
 2  1  T  2  1 
t = thickness of each metal strip used.
 Bimetallic strip can be used as temperature sensor in thermometers and fire alarms.
 As an automatic switch or circuit breaker in electric iron, refrigerators, incubators, thermostats, flash lights
etc.
 As a balance wheel in wrist watches.
Pendulum Clocks:
Variation of Time Period of Pendulum Clocks:
l0
T0  2
g
If temperature is increased by t ,
l o 1    t 
T  2
g
(by using Binomial expansion)
  
T  T0 1  t 
 2 

 T  T  T0  T0 t
2
T = increase in time period.
 Pendulum clocks looses time in summer and gains time in winter
1
The loss or gain per day  t  86400 Sec.
2
Compensated pendulum length is always constant at all temperatures, so it shows correct time at all
temperatures.
Grid Iron Pendulum:
The total expansion of brass rods should be equal to that of steel rods. l1  l2
n1 l1 1  n 2 l 2  2
Measuring Tapes:
Expansion of a Measuring Scale
0 1 2 3 4 5
OBJECT
A scale made of material having coefficient of linear expansion  s  is caliberated at t10C. The scale gives
correct measurement only at calibrated temperature (t10C). If the length of an object is measured at higher
temperature t20C(t20>t10), there will be an error in the measurement at that temperature (t20C) due to expansion
of sacle i.e. due to increase in length of each division. The error in measurement of scale is L  L s t
Where L measured length of object at t20C and t is raise in temperature
a) When temperature increases the length of object oberved by the scale is less than correct length. Hence
true value at measured temperature t20C is LTV  L  L s T ; LTV  L 1   s T 
b) When temperature decrease below the caliberation temperature, length of object observed by the scale
is more than correct length. Hence true value at measured temperature t20C is LTV  L  L s T ;
LTV  L 1   s T 
c) At room temperature (t0C) the length of a material rod is measured using a metal centimeter scale. The
measured length is lcm. If the scale is calibrated to read accurately at temperature )0C, then actual length of
metal rod at 00C is .....
The length of the metal rod at t0C is lTV 1   M t 
The one cm on the metal scale at t0C is 11   st 
The length of the metal rod at t0C
lTV 1   M t 
 lTV 1   M t 1   s t 
1   s t 
l  lTV 1   M   s  t 
lTV  l 1   M   s  t 
when lTV = correct length at calibration temperature.
 M = coefficient of linear expansion of metal rod.
 s = coefficient of linear expansion of metal scale.
Note (1) : If  M   s lTV  1
(2) If  M   s lTV  1
Thermal Stress:
 It is developed due to prevention of expansion of a solid when it is heated.
 A rod of length l0 clamped between two fixed walls.
For t Change in temperature
Young’s modulus
F / A Fl0 F
Y    l  l0t 
l / l0 Al At
F
or  Y t
A
 Thermal force F  YA  t .
Thermal force is independent of length of rod.
 Thermal stress= Y t
 For same thermal stress in two different rods heated through the same rise in temperature,
Y1 1  Y2 2
 Two rods of different metals having the same area of cross section A are placed between the two massive
walls as shown in the fig. The first rod has a length l1, coefficient of linear expansion 1 and Young’s modulus
Y1 . The corresponding quantities for second rod are l2 ,  2 & Y2 . The temperature of both rods is now
raised by t 0C .

Y11 Y2 2
Total length prevented from expansion
At  l11    l2 2  
F  l1 F  l2 Thermal force  F
l1  l2    l1 l2 
Y1  A Y2  A   
 Y1 Y2 
 Fl 
 l  l t  
 YA 
F t   l1 1    l 2 2  
Thermal stress  
A  l1 l 2 
  
 Y1 Y2 
 Lengths of individual rods due to thermal stress:
Length of the first rod=Original length + increase in length due to rise in temperaturedecrease in length due
to thermal force
 Fl 
l11  l1  l11t   1 
 AY1 
Length of the second rod
 Fl 
l21  l2  l2 2t   2 
 AY2 
 Junction displacement=difference in lengths of any one of the rods after heating and before heating.
Fl1 F l2
 x  l1 1t   x  l 2 2 t 
(or) A Y2
AY1
Effect of Moment of inertia of a Rigid body due to thermal expansion

Moment of inertia of a rigid body about an axis of rotation is given by I=MK2.......(1)
where M is mass of the body and K is its radius of gyration. On heating, due to thermal expansion radius of
gyration of the about same axis increases and hence its moment of inertia increases.
As mass remains constant on heating I  K 2
Differentiating and then dividing the same equation we have
I K
2 .......(2)
I K
K
Here, is fractional change in radius of gyration
K
If  is coefficient of linear expansion of the material
Then K  Kt
Sunstituting in equation (2)
I
fractional change in moment of inertia  2t
I
Change in moment of inertia I  2t
Effect of angular velocity of a rotating rigid body due to thermal expansion
On heating since no external torque acts on the body its angular momentum remains constant i.e, when I
increases  decreases.
Angular momentum of an object is L  I
Where I is moment of inertia and  angular velocity. On heating I and  changes but L remains constant,
since no external torque is acitng on object, then I  = constant.
I11  I 22 I   11  2t  2

2    1  2t 
1  2t

Then   2t

 If a cube of coefficient of cubical expansion  is heated, then the pressure to be applied on it to prevent its
expansion is P then V  V t
P P
K 
V t  P  K t
V
 P  3K  t2  t1 
where K is bulk modulus
Barometer With Brass Scale:

 Relation between faulty and actual barometric heights is given by
h2  h1 1  ( Hg   s )  t2  t1  
h1 =height of barometer at t10C where the scale is marked.

h2 =height of barometer at t20C where the measurement is made.
 Hg =real coefficient of expansion of mercury
 s =Coefficient of linear expansion of scale
EXAMPLES:
Between the rails a gap is left to allow for their expansion in summer. If l is the length of the rail and t is
the change in temperature then the gap is given by l = lt
A wire of length l is bent in the form of a ring with a small gap of length x1 at t1 0C . On heating the ring
x2  x1
to t2 0C the gap increases to x2 in length. The  
x1  t2  t1 
Gap behaves like the material for all thermal expansions.
Telephone wires are loosely connected between the poles in summer, to allow for their contraction in
winter.
Concrete roads are laid in sections and gaps are provided between them to allow for expansion.
Pipes used to convey steam from boiler must have loops to prevent cracking of pipes due to thermal
expansion.
Huge iron girders used in the construction of bridges and buildings are allowed to rest on rollers on either
side providing scope for expansion.
Hence the damage to the structure can be avoided.
When a drop of water falls on a hot glass chimney, the portion of the spot where the water falls, contracts
and the remaining portion expands. So, the glass chimney breaks (brittle nature of the glass also)
Pyrex glass is used to prepare test tubes for heating purpose because its linear expansion coefficient is
small.   3  1 0  6 0C 1

Silica glass (quartz) is used for making bulbs of thermometer because of low linear expansion
coefficient.   0.5  106 0C 1 
Invar is an alloy of Iron, Nickel and Carbon. Invar has very low linear expansion coefficient, so used in
wrist watches, pendulum clocks and standard scales.
A hole is drilled at the centre of a metallic plate. When plate is heated, the diameter of hole increases.
When two holes are drilled on a metal plate and heated the distance between the holes increases.
When a solid and hollow sphere with same outer radius made up of same metal are heated to same
temperature then both expand equally.
Platinum is used to seal glass because their coefficients of expansion are almost same.
PROBLEMS
1. A bimetallic strip is made of aluminium and steel   Al  steel  . On heating, the strip will
1) remain straight 2) get twisted
3) will bend with aluminium on concave side 4) will bend with steel on concave side
SOLUTION:
Due to unequal expansions the one having more  bend towards convexside.
2. What length of brass and iron at 00 C must be used if the difference between their lengths is
always 0.2m? The value of  for brass and iron are 18 106 / 0 C and 12  10 6 / 0 C respectively.
(2014E) (2013 M).
SOLUTION:
l2 2
l11  l2 2 and l2  l1  x; l1 
1
l2 2    2 
l2  l1  x  l2   l2  1 
1  1 
x1 0.2  12  106

l2    0.40m
1   2 18  12   106
x 2 0.2  18  106
l1    0.60m
1   2 18  12   106
3. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is
heated uniformly to raise its temperature slightly
1) its speed of rotation increases. 2) its speed of rotation decreases.
3) its speed of rotation remains same. 4) its speed increases because its moment of inertia increases.
SOLUTION:
If the rod heated its length increases and moment of inertia increases, so that its angular velocity decreases.
4. In the given figure, when temperature is increased then which of the following increases ?
R1
R2
1) R1 2) R2 3) R2 R1 4) all

SOLUTION:
All of the above
represents expanded Boundary
_____ represents original Boundary
As the intermolecular distance between atoms increases on heating nence the inner and outer perimeter
increases. Also if teh atomic arrangement in radial direction is observed then we can say that it also increases
hence all A,B,C are true.
5. As the temperature is increased, the time period of a pendulum
1) increases as its effective length increases even though its centre of mass still remains at the centre of the bob.
2) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.
3) increases as its effective length increases due to shifting of centre of mass below the centre of the bob.
4) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.
SOLUTION:
If the temperature of a pendulum increases length of the pendulum increases T  l  .
So that time period increases. V  V t
6. The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion
of the metal is  . The sphere is heated a little by a temperature  T so that its new temperature
is T +  T . The increase in the volume of the sphere is approximately
1) 2RT 2) R 2 T 3) 4R 3T / 3 4) 4R 3T
SOLUTION:
4 3
The increase in the volume of the sphere   r  3 .t
3
V  4 r 3t
7. A rail track made of steel having length 10m is clamped on a railway line at its two ends (figure).
On a summer day due to r ise in temper atur e by 200C. It is deformed as shown in figure. Find x
(displacement of the centre) if  steel=1.2 x 10-5 /0C-1
L
x
1/2 LL
1) 5cm 2) 20cm 3) 15cm 4) 11cm

SOLUTION:
Diagram show the deformmation of a railway track due to rise in temperature
l
900 900
1/ 2  L  L 
Applying Pythagoras theorem in right angled triangle ,
2 2
2 L  L   L 
x    
 2  2
2 2
 L  L   L 
x    
 2  2
2 2 2
 L  2LL  L   L 
x       
2 4  2  2
1 1

2
 L  L  2LL   L
2 2 2

2
 L  2LL 
2
2
As increase in length L is very small, therefore neglecting  L  ,
2LL
we get x  ....(i)
2
But L  Lt ...(ii)
According to the problem, L = 10m
 =1.2 x 105 /0C1 ,
0
 T=20 C
Substituting value  L in Eq. (i) from Eq (ii)
1 1
x 2L  Lt  L 2t
2 2
= 11cm
8. A pendulum clock loses 12s a day if the temperature is 400C and gains 4s day if the temperature
is 200C. The temperature at which the clock will show correct time and the coefficient of linear
expansion    of the metal of the pendulum shaft are respectively (JEE Mains-2016)
1) 250C; =1.85 x 10-5/0C 2) 600C; =1.85 x 10-4/0C
3) 300C; =1.85 x 10-3/0C 4) 550C; =1.85 x 10-2/0C
SOLUTION:
Let at temperature , clock gives correct time
1 
T     T, T = 1 day = 86400s
2 
1
12    40    T ...(i)
2
1
4     20  T...(ii)
2
i / ii  = 250C
substituting  inequation(ii),weget
1
4   25  20   86400
2
1
 1.85x105 / 0 C

5  86400
9. The variation of lengths of two metal rods A and B with change in temperature are shown in figure.
The coefficient of linear expansion  A for the metal A and the temperature T will be:(Given
 B  9  10 6 / 0 C)
1006 A
B
Length(mm)
1004
1002
1000
998
996
994
O Temperature ( 0C ) T
1)  A  3  106 / 0 C ,5000 C 2)  A  3  106 / 0 C , 222.220 C

3)  A  27 10 6 / 0 C ,5000 C 4)  A  27  10 6 / 0 C , 222.220 C
SOLUTION:
1006  1000 L
Slopeoftheline A 
T T
6
 1000mm  A    (1)
T
similarlyforBline
2
 1002mm  B    (2)
T
From (1) & (2)  A  3 B
10. Two bar s ar e unst r essed and have l engt hs of 25 cmand 30 cmat 200C as shown in Figure. Bar (1)
is of aluminium and bar (2) is of steel. The cross-sectional area of bars are 20 cm2for aluminium
and 10cm2 for steel. Assuming that the top and bottom supports are rigid, stress in Al steel bars in
N (Ya  0.70 105 N / mm2 .Ys  2.1105 N / mm2 .
0
when the temperature is 70 C. (Nearly)
mm 2  a  24 106 / 0C and  a  12 106 / 0C )
Al 25cm
St 30cm
1) 75, 150 2) 25, 50 3) 50, 100 4) 100, 200

SOLUTION:
Contraction of the two bars due to compressive stress = Elongation of the two bars due to rise of temperature
 Sl   Sl 
       Lt  Al   Lt  St
 Y  Al  Y  St
Forceinsteel=forinaluminium
S Al  AAl  S St  ASt
11. A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period
2s at T=00C. If the temperature of the wire is increased and the corresponding change in its time
period is plotted against it temperature, the resulting graph is a line of slope S. If the coefficient of
linear expansion of metal is  then the value of S is (JEE Mains-2016 online)
t
S

T
1)  2) /2 3) 2 4) 1/
SOLUTION:
l0
t 0  2  2
g
l0 1  T  1/ 2
t 0  2  2 1  T   2  T
g
t  t 0  T  t  T
12. A wire of length L0 is supplied heat to raise its temperature by T. if  is the coefficient of volume
expansion of the wire and Y is Young’s modulus of the wire then the energy density stored in the
wireis
1 2 2 1 2 2 3 1  2T 2 1 2 2
1)  T Y 2)  T Y 3) 4)  T Y
2 3 18 Y 18
SOLUTION:
Elasticpotentialenergyperunitvolume
1 1 2
E   Stress  Strain   Y   Strain 
2 2
2
1  L  1 2 2
E Y      Y    T
2  L  2
13. The coefficient of linear expansion of an in homogeneous rod changes linearly form 1to  2 from
one end to the other end of the rod. The effective coefficient of linear expansion of rod is
1   2
1) 1   2 2) 3) 1 2 4) 1   2
2
SOLUTION:
L
  2  1 
 n  1    x; L    n dxt
 l  0
   2   1   2 
L 1  LT ;  eff   
 2   2 
14. An equilateral triangle ABC is formed by joining three rods of equal length and D is the mid-point
of AB. The coefficient of linear expansion for AB is 1 and for AC and BC is  2 . The relation
between 1 and  2, if distance DC remains constant for small changes in temperature is (2010 E)
A D 1 B
2 2
1
1) 1   2 2) 1  4 2 3)  2  41 4) 1   2
2
SOLUTION: Before andafter changing the temperature,
l12 2 1 2
l22 
 l2  l   t    l1 1   t  
4 4
and l1  l2
15. When composite rod is free, composite length increases to 2.002m from temperature 200C to
1200C. When composite rod is fixed between the support, there is no change in component length.
Find Y and  of steel if Ycu  1.5 1013 N / m 2  cu  1.6  105 / 0 C
Steel Copper
0.5m
2m
SOLUTION:
l  ls s T  lc c T
0.002  1.5 s  0.5  1.6 10 5  100
1.2 105
s   8 106 0C
1.5
If there is no change in composite length, thermal force of steel and copper rod should be equal
Fst  Fcu ; Ys A s t  Ycu A Cu t
Ys  c  1.5 1013 1.6 105

 ; Ys  Yc  c  Ys  3  1013 N / m 2
Yc  s s 106
16. A blacksmith fixes iron ring on the rim of the wooden wheel of bullock cart. The diameter of the
rim and the iron ring are 5.243 m and 5.231 m respectively at 27 0 C. The temperature to which the
ring should be heated so as to fit the rim on the wheel  α iron = 1.20×10-5 / 0 C  ?
SOLUTION:
Given t1  27 0 C ; l1  5.231m; l2  5.243m
l2  l1 1  1  t2  t1  
5.243=5.231 1  1.20  10 5  t2  27  

or t2  2180 C
17. An aluminium sphere of 20 cm diameter is heated from 00 C to 1000 C . Its volume changes by
(given that the coefficient of linear expansion for aluminium  Al  23 106 / 0 C  (AIEEE 2011)
SOLUTION:
Given d=20cm
V  V0 1   t   V0 1  3 t  since   3 
change in volume = V  V0  3V0 t
3
4 d
 3      23  106 100
3 2
3
4  0.2  6
 3     23  10  100
3  2 
= 28.9cc 1cc  106 m3 
18. An iron rod of length 50 cm is joined at an end to copper rod of length 100 cm at 200C. Find the length
of the system at1000Cand average coefficient of linear expansion of the system.
α iron = 12×10-6 / 0 C and αcopper = 17×10-6 / 0 C. 
SOLUTION:
Increase in length of composite rod is
l  l1  l2  1l1   2l2  t
 12  106  50  17  106  100   100  20  =0.192cm

Length of the composite rod at 1000 C is l  l  150.192cm
Average linear expansion coefficient
l 0.192
 avg    16  10 6 / 0 C
l t 150  80
19. Density of gold is 19.30 g / cm3 at 200 C . Compute the density of gold at 900 C by adding steam
to it.  α = 14.2×10 -6
/ 0 C
SOLUTION:
2 V1 V1 1
   1  3T 
1 V2 V1 1  3T 

or   1  3T  or
2
 2  1 1  3T 
 19.30  1  3 14.2 10   70    19.24 g / cm

6 3
20: A steel bar of cross sectional area 1 cm 2 and 50 cm long at 300 C fits into the space between two
fixed supports. If the bars now heated to 2800 C , what force will it exert against the supports?
(α for steel = 11×10-6 / 0 C and

Young's modulus for steel = 2×1011 N / m2 )
SOLUTION:
Force exerted on the supports
 Stress × Area of cross section  Y  A  t2  t1  .
 2  1011  11 106  104  250  55000N
21. Uniform pressure P is exerted on all sides of a solid cube of bulk modulus, B and volume coefficient of
expansion  , at temperature t0C. By what amount should the temperature of cube be raised in order to
bring its volume back to the value it had before the pressure was applied?
SOLUTION:
P VP
As B  , V  ......(i )
V / V B
If T is the required increase in temperature,
V   V T .....(ii )
From eqns. (i) and (ii),
VP P
 V T  or T 
B B
1
22. The balance wheel of a mechanical wrist watch has a frequency of oscillation given by f = C/I,
2π
where I is the moment of inertia of the wheel and C is the torsional rigidity of its spring. The wrist watch
keeps accurate time at 250C. How many seconds would it gain a day at-250C if the balance wheel made
ofAluminium?
(Given,  Al  25.5 10 6 / 0 C )
SOLUTION:
1 C 1 C
f    I  M K 2 
2 1 2 k M
1 1 df dT dk
f ; f   
T k f T k
dk dT
As   dt    dt
k T
Number of seconds gained/day
dT   8.64 10 4   dT   110.2 s / day
23. An aluminium measuring rod, which is correct at 50C measures the length of a line as 80 cm at
450C. If thermal coefficient of linear expansion of aluminium is 2.50 10 5 / 0 C . The correct length
of the line is
SOLUTION:
L2  L1  L1t
L2  80   2.50  105   80  40   80.08cm
24. A mass of 2kg is suspended from a fixed pointed by a wire of length 3m and diameter 0.5 mm.
Initially the wire is just unstretched, the mass resting on a fixed support. By how much must the
temperature fall if the mass is to be entirely supported by the wire (Given Y for wire = 206 G Pa,
  11 10  6 / 0 C )
SOLUTION:
Contraction due to cooling is equal to the stretching produced by the weight ‘mg’.
m gL 2     3
 L   2
AY   0 .2 5   1 0  6  2 06  1 0 9
Now the contraction due to cooling
 Lt  3  11 10 6  t
solving t  440 C
25. A metallic rod of length 1 cm and cross-sectional area A cm 2 is heated through t 0C . After expansion
if a mechanical force is applied normal to its length on both sides of the rod and restore its original
length, what is the value of force? The young’s modulus of elasticity of the metal is E and mean
coefficient of linear expansion is  per degree Celsius.
SOLUTION:
Change in the length  l  l t
Length of rod at t 0C is l  l t
Decrease in length due to stress= l
But length of rod remains constant lt  l  0
stress F l   lt
  l   lt  l t  E   
strain A  lt
EA  lt  EAl t EA t
F   
l   lt l  l t 1   t 
Here, negative sign indicates that forces are compressive in nature.
26. Two metal rods are fixed end to end between two rigid supports as shown in figure. Each rod is
length l and area of cross-section is A. When the system is heated up, determine the condition when
the junction between rods does not shift? ( Y1 and Y2 are Young’s modulus of materials of rods,
1 and  2 are coefficients of linear expansion)
L L
1Y1  2Y2
SOLUTION:
Since, each rod is prevented from expansion so, they are under compression and mechanical strain.
The strain in each rod is zero.
e1 F e2 F e1 e2
l  1T  AY ; l   2 T  AY ; l  l  0
1 2
F F
1T   0 and  2 T  0
AY1 AY2
F F
1T  AY .........(1) and  2 T  AY ........(2)
1 2
Dividing (1) by (2), we get 1Y1   2Y2
27. A bimetallic strip of thickness 2 cm consists of zinc and silver rivetted together. The approximate
radius of curvature of the strip when heated through 500 C will be: (linear expansivity of zinc and
silver are 32 106 / 0 C and 19 106 / 0 C respectively)
SOLUTION:
2t
Radius of curvature R 
 2  1  T
2 1
R  30.77m
 32  19  106  50
28. A clock with a metallic pendulum is 5 seconds fast each day at a temperature of 150Cand 10
seconds slow each day at a temperature of 300C. Find coefficient of linear expansion for the metal.
SOLUTION:
The time lost or gained per day is
1
t  T  86400  as 1 day  86400 s.
2
If graduation temperature of clock is T0 then gain in time at 150 C is
1
5  T0  15  86400 .........(i)
2
At 300 C clock is loosing time thus
1
10    30  T0  86400 .......(ii )
2
Dividing equation (ii) by (i), we get
2 T0  15    30  T0  or T0  200 C
Thus from equation (i)
1
5  2 0  1 5  8 6 4 0 0
2
  2.31105 / 0 C
29. A steel rail 30 m long is firmly attached to the road bed only at its ends. The sun raises the
temperature of the rail by 500 C , causing the rail to buckle. Assuming that the buckled rail consists
of two straight parts meeting in the centre, calculate how much centre of the rail rise?
Given,  steel  12  106 / 0 C.
x+ x y x+ x
2x
SOLUTION:
Let the initial length be 2x and the final total length be 2  x  x  as shown.
Let y be the height of the centre of the buckled rail.
Clearly, x   x T and
2
y  x  x   x 2  2 x  x   2 x 2t
y  x 2t  neglecting  x 2 
 
 
Thus, y  15 2 12 10  50  cm  0.52m
6
 
30. A metal rod of Young’s modulus F and coefficient of thermal expansion  is held at its two ends
such that its length remains invariant. If its temperature is raised by t 0C , then the linear stress
developed in it is (AIE-2011)
SOLUTION:
FL F
L   L T   Stress   Y T  Y  t
AY A
31. A wooden wheel of radius R is made of two semi circular parts (see figure). The two parts are held
together by a ring made of a metal strip of cross-sectional area S and length L. L is slightly less
than 2 R . To fit the ring on the wheel, it is heated so that its temperature rises by T and it just
steps over the wheel. As it cools down to surrounding temperature. It presses the semi-circular
parts together. If the coefficient of linear expansion of the metal is  and its Young’s modulus is
Y, then the force that one part of the wheel applies on the other part is (AIEEE 2012)
SOLUTION:
T T
Increase in length
L
L   L T    T
L
the thermal stress developed is
T L
Y  Y T ; T  SY T
S L
From the FBD of one part of the wheel, F = 2T
Where F is the force applied by one part of the wheel on other part, F=2SY T
32. Two rods of different materials and identical cross sectional area, are joined face to face at one
end and their free ends are fixed to the rigid walls. If the temperature of the surroundings is
increased by 300C, the magnitude of the displacement of the joint of the rod is (length of rods
l1  l2  1 unit , ratio of their young’s moduli, Y1 / Y2  2, coefficients of linear expansion are
1 and  2)
1) 5( 2  1 ) 2) 10(1   2 ) 3) 10( 2  21 ) 4) 5(21   2 )
SOLUTION:
Y1  Strain 1  Y2  Strain 2
  l T  x    l T  x 
Y1  1 1   Y21  2 2 
 l   l
Displacement of the rod
(Y2 2  Y11 )
x  l  T
Y1  Y2
33. Calculate the compressional force required to prevent the metallic rod of length lcm and cross-
sectional area A cm2 when heated through t0C, from expanding along length wise. The Young’s
modulus of elasticity of the metal is E and mean coefficient of linear expansion is  per degree
Celsius
EA t EA t
1) EA t 2) 3) 4) El t
1 t 1t
SOLUTION:
Natural length of the rod at toC=l  l t
decreaseinnaturallengthduetostress=l
 F  l  l   EA t
l1  l  0; E  ;F 
Al 1 t
34. The coefficient of linear expansion for a certain metal varies with temperature as  T . If L0 is
the initial length of the metal and the temperature of metal changed from T0 to T T0  T  then
T T
 T 
1) L  L0 T  T dT 2)L  L0 1  T0  T dT  3) L  L0 1    (T )dT  4) L  L0
0  T0 
dL
SOLUTION:   T  dt ; Integrate the equation
L0
35. A solid sphere of radius r and mass m is spinning about a diameter as axis with a speed 0. The
temperature of the sphere increases by 1000C without any other disturbance. If the coefficient of
linear expansion of material of sphere is 2 10 4 / 0 C , the ratio of angular speed at 1000C and
0is
1)1:1 2)1:1.04 3)1.04:1 4)1:1.02
SOLUTION:
I11  I 22 ; R121  R222

2
2  R2  1
  
1  R1  1  t  2
Thermal Expansion in Liquids:
(1) Liquids do not have linear and superficial expansion but these only have volume expansion.
Sinceliquidsare alwaystobeheatedalongwith avesselwhich containsthem soinitiallyon heating the
system(liquid+vessel),thelevelofliquidinvesselfalls(asvesselexpandsmoresinceitabsorbsheatand
liquidexpandsless)butlateron,itstartsrisingduetofasterexpansionoftheliquid.
R
P
Q
Fig. 12.11
PQ  represents expansion of vessel

QR  representsthe realexpansionofliquid
PR  Represent the apparent expansion of liquid
The actual increase in the volume of the liquid = The apparent increase in the volume of liquid + the
increaseinthevolumeofthevessel.
Liquids have two coefficients of volume expansion.
Co-efficient of apparent expansion (a) : It is due to apparent (that appears to be, but is not) increase in the
volumeofliquidifexpansionofvesselcontainingtheliquidisnottakenintoaccount.
Apparent expansion in volume (V )a

a  
Initial vo lume   V  
Co-efficientofrealexpansion(r):It isdue tothe actual increaseinvolumeofliquiddue toheating.
Real increase in volume (V )r

r  
Initial vo lume   V  
(V )Vessel
 Alsocoefficientof expansionofflask  Vessel 
V  
  Real   Apparent   Vessel
Change(apparentchange)involumeinliquidrelativetovesselis
Vapp  V  app  = V ( Real   Vessel )  V ( r  3 )
 =Coefficientoflinearexpansionofthevessel.
Anomalous expansion of water :
Generally matter expands on heating and contracts on cooling. In case of water, it expands on
heating if its temperature is greater than 4°C. In the range 0°C to 4°C, water contracts on heating and expands on
cooling, i.e. g is negative. This behaviour of water in the range from 0°C to 4°C is called anomalous expansion.
This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter
approaches, the water temperature increases initially at the surface. The water there sinks because of its increased
density. Consequently, the surface reaches 0°C first and the lake becomes covered with ice. Aquatic life is able to
survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C.
A t 4°C, density of water is maximum while its specific volume is minimum.
max
min Ano Ano
vol malo De
/ us nsi
ma be ty
0°C 4°C Temperature 0°C 4° Temperature
(A) (B)
Variation of Density with Temperature

Most substances (solid and liquid) expand when they are heated,
 1
i.e., volume of a given mass of a substance increases on heating, so the density should decrease  as   V .
 
1  V V V 1
For a given mass      V   V  V  V  V  1   
V

    1      (1    )1 =  (1    )
CALORIMETRY
Specific Heat
When a body is heated it’s temperature rises (except during a change in phase).
Gram specific heat : The amount of heat energy required to raise the temperature of unit mass of a body through 1°C (or K) is called
specific heat of the material of the body.
Q
If Q heat changes the temperature of mass m by  then specific heat c 
m 
(i) Units : Calorie/gm °C (practical), J/kg K (S.I.)
Dimension : [L2 T 2 1 ]
(ii) For an infinitesimal temperature change d and corresponding quantity of heat dQ.
1 dQ
pecific heat c  .
m d
Molar specific heat : Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram
mole of the substance through a unit degree it is represented by (capital) C.
Molar specific heat (C)  M  Gram specific heat (c)
(M = Molecular mass of substance)
Q 1 Q  m
CM   where, Number of moles   
m     M
Units : calorie/mole  °C (practical); J/mole  kelvin (S.I.)
Dimension : [ML2 T 2 1 ]
Specific Heat of Solids
Y 3R
CV
X
T Debye
When a solid is heated through a small range of temperature, its volume remains more or less constant. Therefore
specific heat of a solid may be called its specific heat at constant volume CV.
(1) From the graph it is clear that at T = 0, CV tends to zero
(2) With rise in temperature, CV increases and at a particular temperature (called Debey’s temperature) it becomes constant = 3R = 6 cal/
mole  kelvin = 25 J/mole  kelvin
(3) For most of the solids, Debye temperature is close to room temperature.
Dulong and Petit law :

Average molar specific heat of all metals at room temperature is constant, being nearly equal to
3R = 6 cal. mole–1 K–1 = 25 J mole–1 K–1, where R is gas constant for one mole of the gas. This statement is known as Dulong
and Petit law.
Debey’s law :
It was observed that at very low temperature molar specific heat  T 3 (exception are Sn, Pb and Pt)
Specific heat of ice :
cal
In C.G.S. cice  0 .5
gm  C
cal Joule
InS.I.cice  500  2100 .
kg  C kg  C
Specific Heat of Liquid (Water)
(1) Among all known solids and liquids specific heat of water is maximum i.e. water takes more time to heat and more time to cool w.r.t.
other solids and liquids.
(2) It is observed that by increasing temperature, initially specific heat of water goes on decreasing, becomes minimum at 37°C and then
itstartincreasing.
1 cal cal J
Specific heat of water is –  1000  4200
gm  C kg  C kg  C
(This value is obtained between the temperature 14.5°C to 15.5°C)
Sp.
1.008
heat
cal/ 1.004
g C°
1.000
0.996
20 40 60 80 100
Temp. in °C
(3) The variation of specific heat with temperature for water is shown in the figure. Usually this temperature dependence of
specificheatis neglected.
(4) As specific heat of water is very large; by absorbing or releasing large amount of heat its temperature changes by small amount.
This is why, it is used in hot water bottles or as coolant in radiators.
Specific Heat of Gases
(1) In case of gases, heat energy supplied to a gas is spent not only in raising the temperature of the gas but also in expansion of
gas against atmospheric pressure.
(2) Hence specific heat of a gas, which is the amount of heat energy required to raise the temperature of one gram of gas through a
unit degree shall not have a single or unique value.
(3) If the gas is compressed suddenly and no heat is supplied from outside i.e. Q = 0, but the temperature of the gas raises on
the account of compression.
Q 0
 c  0
m ( ) m 
(4) If the gas is heated and allowed to expand at such a rate that rise in temperature due to heat supplied is exactly equal to fall
in temperature due to expansion of the gas. i.e.  = 0
Q Q
 c  
m ( ) 0
(5) If rate of expansion of the gas were slow, the fall in temperature of the gas due to expansion would be smaller than the rise
in temperature of the gas due to heat supplied. Therefore, there will be some net rise in temperature of the gas i.e. T
will bepositive.
Q
c  Positive
m ( )
(6) If the gas were to expand very fast, fall of temperature of gas due to expansion would be greater than rise in temperature due
to heat supplied. Therefore, there will be some net fall in temperature of the gas i.e.  will be negative.
Q
 c  Negative
m (  )
Hence the specific heat of gas can have any positive value ranging from zero to infinity. Further it can even be
negative. The exact value depends upon the mode of heating the gas. Out of many values of specific heat of a gas, two are of
special significance, namely CP and CV , in the chapter “Kinetic theory of gases” we will discussed this topic in detail.
Specific heat of steam : c steam  0 . 47 cal / gm  C
Phase Change and Latent Heat
(1) Phase : We use the term phase to describe a specific state of matter, such as solid, liquid or gas. A transition from one phase to another
is called a phase change.
(i) For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of
heat and a change of volume and density.
(ii) In phase change ice at 0°C melts into water at 0°C. Water at 100°C boils to form steam at 100°C.
Heat taken (– D Q)
Solid
at 0°C
Liquid at 0°C
Heat given (+ D Q)
(A)
Heat taken (– D Q)
Liquid at 100°C Vapours at 100°C
Heat given (+ D Q)
(B)
(iii) In solids, the forces between the molecules are large and the molecules are almost fixed in their positions inside the solid. In
a liquid, the forces between the molecules are weaker and the molecules may move freely inside the volume of the
liquid. However, they are not able to come out of the surface. In vapours or gases, the intermolecular forces are
almost negligible and the molecules may move freely anywhere in the container. When a solid melts, its molecules
move apart against the strong molecular attraction. This needs energy which must be supplied from outside. Thus,
the internal energy of a given body is larger in liquid phase than in solid phase. Similarly, the internal energy of a given
body in vapour phase is larger than that in liquid phase.
(iv) In case of change of state if the molecules come closer, energy is released and if the molecules move apart, energy is
absorbed.
Latent heat : The amount of heat required to change the state of the mass m of the substance is written as : Q = mL, where L is the
latent heat. Latent heat is also called as Heat of Transformation. It’s unit is cal/gm or J/kg and Dimension: [L2 T 2 ]
(i) Latentheatof fusion: The latent heat of fusion is the heat energy required to change 1 kg of the material in its solid state at its
melting point to 1 kg of the material in its liquid state. It is also the amount of heat energy released when at melting point
1 kg of liquid changes to 1 kg of solid. For water at its normal freezing temperature or melting point (0°C), the latent heat
of fusion(or latent heat ofice) is
LF  Lice  80 cal / gm  60 kJ / mol  336 kilo joule / kg

Latent heat of vaporisation : The latent heat of vaporisation is the heat energy required to change 1 kg of the material in its liquid
state at its boiling point to 1 kg of the material in its gaseous state. It is also the amount of heat energy released when 1
kg of vapour changes into 1 kg of liquid. For water at its normal boiling point or condensation temperature (100°C), the
latent heat of vaporisation (latent heat of steam) is
LV  Lsteam  540 cal / gm  40 . 8 kJ / mol  2260 kilo joule / kg
(iii) Latent heat of vaporisation is more than the latent heat of fusion. This is because when a substance gets converted from liquid to
vapour, there is a large increase in volume. Hence more amount of heat is required. But when a solid gets converted to a
liquid, then the increase in volume is negligible. Hence very less amount of heat is required. So, latent heat of vaporisation
is more than the latent heat of fusion.
Thermal Capacity and Water Equivalent
Thermal capacity : It is defined as the amount of heat required to raise the temperature of the whole body (mass m) through 0°C or 1K.
Q
Thermal capacity  mc  C 

The value of thermal capacity of a body depends upon the nature of the body and its mass.
Dimension : [ML2 T 2 1 ],
Unit:cal/°C(practical) Joule/k(S.I.)
Water Equivalent : Water equivalent of a body is defined as the mass of water which would absorb or evolve the same amount of heat as
is done by the body in rising or falling through the same range of temperature. It is represented by W.
If m = Mass of the body, c = Specific heat of body,  = Rise in temperature.
Then heat given to body Q  mc  …..(i)
If same amount of heat is given to W gm of water and its temperature also rises by .
Thenheat given to water Q  W  1   … (ii) [As c water  1 ]
From equation (i) and (ii) Q  mc   W  1  
 Water equivalent (W) = mc gm
(i)Unit:Kg(S.I.) Dimension : [ML0 T 0 ]
(ii) Unit of thermal capacity is J/kg while unit of water equivalent is kg.
(iii) Thermal capacity of the body and its water equivalent are numerically equal.
(iv) If thermal capacity of a body is expressed in terms of mass of water it is called water-equivalent of the body.
Some Important Terms
Evaporation : Vaporisation occurring from the free surface of a liquid is called evaporation. Evaporation is the escape of molecules from
the surface of a liquid. This process takes place at all temperatures and increases with the increase of temperature. Evaporation
leads to cooling because the faster molecules escape and, therefore, the average kinetic energy of the molecules of the liquid (and
hence the temperature) decreases.
Melting (or fusion)/freezing (or solidification) : The phase change of solid to liquid is called melting or fusion. The reverse
phenomenon is called freezing or solidification.
When pressure is applied on ice, it melts. As soon as the pressure is removed, it freezes again. This phenomenon is called regelation.
Vaporisation/liquefication (condensation) : The phase change from liquid to vapour is called vaporisation. The reverse transition is called
liquefication or condensation.
Sublimation : Sublimation is the conversion of a solid directly into vapours. Sublimation takes place when boiling point is less than the
melting point. A block of ice sublimates into vapours on the surface of moon because of very very low pressure on its surface. Heat
required to change unit mass of solid directly into vapours at a given temperature is called heat of sublimation at that temperature.
Hoar frost : Direct conversion of vapours into solid is called hoar frost. This process is just reverse of the process of sublimation, e.g.,
formation of snow by freezing of clouds.
Vapour pressure : When the space above a liquid is closed, it soon becomes saturated with vapour and a dynamic equilibrium is
established. The pressure exerted by this vapour is called Saturated Vapour Pressure (S.V.P.) whose value depends only on the
temperature – it is independent of any external pressure. If the volume of the space is reduced, some vapour liquefies, but the
pressure is unchanged.
A saturated vapour does not obey the gas law whereas the unsaturated vapour obeys them fairly well. However, a vapour
differs from a gas in that the former can be liquefied by pressure alone, whereas the latter cannot be liquefied unless it is first cooled.
Boiling : As the temperature of a liquid is increased, the rate of evaporation also increases. A stage is reached when bubbles of vapour
start forming in the body of the liquid which rise to the surface and escape. A liquid boils at a temperature at which the S.V.P. is equal
to the external pressure.
It isa fast process. Theboiling point changes on mixing impurities.
Dew point : It is that temperature at which the mass of water vapour present in a given volume of air is just sufficient to saturate it,
i.e. the temperature at which the actual vapour pressure becomes equal to the saturated vapuor pressure.
Humidity : Atmospheric air always contains some water vapour. The mass of water vapour per unit volume is called absolute
humidity.
The ratio of the mass of water vapour (m) actually present in a given volume of air to the mass of water vapour (M)
required to saturate the same volume at the same temperature is called the relative humidity (R.H.). Generally, it is expressed as
a percentage,
m
i.e., R.H.(%)   100 (%)
M
R.H. May also be defined as the ratio of the actual vapour pressure (p) of water at the same temperature,
p
i.e. R.H.(%)   100 (%)
P
Thus R.H. may also be defined as
S.V.P. at dew point
R.H.(%)   100
S.V. P. at given temperatu re
Variation of melting point with pressure : For those substances with contract on melting (e.g. water and rubber), the melting point
decreases with pressure. The reason is the pressure helps shrinking and hence melting. Most substances expand on melting.
(e.g.max,sulpheretc.)
An increase of pressure opposes the melting of such substances and their melting point is raised.
Variation of latent heat with temperature and pressure : The latent heat of vapourization of a substance varies with
temperature and hence pressure because the boiling point depends on pressure. It increases as the temperature is decreased.
For example, water at 1 atm boils at 100°C and has latent heat 2259 Jg–1 but at 0.5 atm it boils at 82°C and has latent heat
2310 Jg–1
P P Critical point
Critical point Fu
L si L
Vaporization Vaporization
S
S Triple point
Triple point
V
V
T T
The latent heat of fusion shows similar but less pronounced pressure dependence.
The figures show the P-T graphs for (a) a substance (e.g., water) which contracts on melting an (b) a substance (e.g. wax) which expands
on melting. The P-T graph consists of three curves.
(i) Sublimation curve which connects points at which vapour (V) and solid (S) exist in equilibrium.
(ii) Vapourization curve which shows vapour and liquid (L) existing in equilibrium.
(iii) Fusion curve which shows liquid and solid existing in equilibrium.
The three curves meet at a single point which is called the triple point. It is that unique temperature-pressure point for a substance at
which all the three phases exist in equilibrium.
Freezing mixture : If salt is added to ice, then the temperature of mixture drops down to less than 0°C. This is so because, some ice
melts down to cool the salt to 0°C. As a result, salt gets dissolved in the water formed and saturated solution of salt is obtained; but the ice
point (freeing point) of the solution formed is always less than that of pure water. So, ice cannot be in the solid state with the salt solution
at 0°C. The ice which is in contact with the solution, starts melting and it absorbs the required latent heat from the mixture, so the temperature
of mixture falls down.
Principle of Calorimetry
Calorimetry means ‘measuring heat’.
When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from
body at higher temperature to a body at lower temperature till both acquire same temperature. The body at higher temperature releases heat
while body at lower temperature absorbs it, so that
Heat lost = Heat gained
i.e. principle of calorimetry represents the law of conservation of heat energy.
(1) Temperature of mixture (mix) is always e” lower temperature (L) and d” higher temperature (H), i.e.,  L   mix   H .
It means the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of
cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Furthermore usually
rise in temperature of one body is not equal to the fall in temperature of the other body though heat gained by one body is equal to the heat
lost by the other.
(2) Mixing of two substances when temperature changes only : It means no phase change. Suppose two substances having masses
m 1 and m 2 , gram specific heat c 1 and c 2 , temperatures  1 and  2 ( 1   2 ) are mixed together such that temperature of mixture at
equilibrium is mix
Hence, Heat lost = Heat gained
m 1 c1 1  m 2 c 2 2
 m 1 c1 ( 1   mix )  m 2 c 2 ( mix   2 )   mix  m 1 c1  m 2 c 2
Mixing of two substances when temperature and phase both changes or only phase changes: A very common example for this
category is ice-water mixing.
Suppose water at temperature W°C is mixed with ice at 0i°C, first ice will melt and then it’s temperature rises to attain
thermal equilibrium. Hence; Heat given = Heat taken
 m W CW ( W   mix )  m i Li  m iCW ( mix  0 )
m i Li
m W W 
CW
  mix 
mW  m i
Li
W 
(i) If m W  m i then  CW
mix 
2
(ii) By using this formulae if  mix   i then take  mix  0 C
Heating Curve:
C Boiling
T2
Te Boil D point
mp.
T1 B Melting
A Melt point
O Time
t1 t2 t3 t4
If to a given mass (m) of a solid, heat is supplied at constant rate P and a graph is plotted between temperature and
time, the graph is as shown in figure and is called heating curve. From this curve it is clear that
(1) In the region OA temperature of solid is changing with time so, Q  mc S T  P t  mc S T [as Q = Pt]
But as (T/t) is the slope of temperature-time curve
1
cs 
Slope of line OA
i.e. specific heat (or thermal capacity) is inversely proportional to the slope of temperature-time curve.
(2) In the region AB temperature is constant, so it represents change of state, i.e., melting of solid with melting point T1. At A
melting starts and at B all solid is converted into liquid. So between A and B substance is partly solid and partly
P(t 2  t1 )
liquid. If LF is the latent heat of fusion. Q  mL F or L F  [as Q  P(t 2  t1 )]
m
or LF  length of line AB
1
i.e. Latent heat of fusion is proportional to the length of line of zero slope. [In this region specific heat   ]
tan 0
(3) In the region BC temperature of liquid increases so specific heat (or thermal capacity) of liquid will be inversely
proportional to the slope of line BC
1
i.e., cL 
Slope of line B C
(4) In the region CD temperature is constant, so it represents the change of state, i.e., boiling with boiling point T2. At C all
substance is in liquid state while at D in vapour state and between C and D partly liquid and partly gas. The length of
line CD is proportional to latent heat of vaporisation
i.e., LV  Length of line CD [In this region specific heat  ]
(5) The line DE represents gaseous state of substance with its temperature increasing linearly with time. The reciprocal of
slope of line will be proportional to specific heat or thermal capacity of substance in vapour state.
Note:
The temperature of mixture can never be lesser than lower temperature and can never be greater than higher
temperature
 L  mix   H
If ‘m’ g of steam at 1000 C is mixed with ‘m’ g of ice at 00 C then
a) Resultant temperature of mixture is 1000 C
m
b) Mass of steam condensed  g
3
2m
c) Mass of steam left uncondensed  g
3
4m 2m
d) The final mixture contains g of water and g of steam both at 1000 C
3 3
Super cooling :
 Most liquids, if cooled in a pure state in a perfectly clean vessel, with least disturbance, can be
lowered to a temperature much below the normal freezing point, without solidifying. This is
known as super cooling or super fusion.
 In super cooling, water can be cooled upto 10ºC without becoming solid.
Saturated and Unsaturated Vapours :
(a)When the pressure exerted by a vapour is maximum it is called saturated vapour, when pressure exerted is
not maximum, it is called unsaturated vapour.
(b)Saturated vapours do not obey the gas laws and saturated vapour pressure of liquid is independent of
volume occupied. But unsaturated vapour obey the gas laws.
(c)At boiling point of a liquid saturated vapour pressure is equal to atmospheric pressure at that place.
Units
Physical
Quantity SI CGS
( Practical )
Heat Joule Calories
Specific Heat Joule/KgK Cal/g0C
Molar specific
Joule/molK Cal/mol0C
Heat
NOTE: Thermal
Joule/Kg Cal/0C
capacity
Water
Kg g
Equivalent
PROBLEMS
1. Find the water equivalent of copper block of mass 200g. The specific heat of copper is 0.09 cal / g 0C .
SOLUTION:
Water equivalent w= mS  200  0.09  18g
2. A lead piece of mass 25g gives out 1200 calories of heat when it is cooled from 900 C to 100 C .
What is its (i) specific heat
(ii) thermal capacity
(iii) water equivalent.
SOLUTION:
Mass of lead piece (m) = 25 g = 0.025 kg
Heat energy given out  dQ   1200  4.2 J
1 dQ 1 1200  4.2
(i) specific heat S     2520JKg 1K 1
m d 0.025 80
(ii) Thermal capacity = mS = 0.025  2520 = 63 J/K
63
(iii) Water equivalent Kg  0.015 Kg
4200
3. Two spheres of radii in the ratio 1:2, have specific heats in the ratio 2:3. The densities are in the
ratio 3:4. Find the ratio of their thermal capacities.
SOLUTION:
Thermal capacity of a body = mS.
The ratio of thermal capacities
4 3
m1 S1 V1 1 S1
 r1 1 S1  r  3     S 
  3   1   1  1 
m 2 S 2 V2  2 S 2 4  r 3  S
2 2 2
 r2    2   S 2  .
3
r1 1 S1 2 1 3
Here, r  2 ; S  3 ;   4
2 2 2
3
 1   3  2  1
The ratio of thermal capacities =      
 2   4   3  16
4. A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling
water, so that the sphere is at 1000 C . It is then immediately transferred to 0.14 kg copper
calorimeter containing 0.25 kg of water at 200 C . The temperature of water rises and attains a
steady state at 230 C . Calculate the specific heat capacity of aluminium.
( Scu  386 J / Kg  K , S w  4180 J / Kg  K )

SOLUTION:
Heat lost by aluminium sphere =
(heat gained by water) + (heat gained by calorimeter)
0.047  S Al  1000  230   0.25  4180  230  200 
0.14  386  230  200 
 S Al  911J / Kg  K
5: The temperature of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC
respectively. The common temperature when A and B are mixed is 16ºC and when B and C are
mixed is 23ºC.What should be the common temperature when A and C are mixed?
SOLUTION:
Given  A  12º C , B  19º C and C  28º C . Let S A , S B and S C are the specific heats of r e
spective liquids.
When liquid A and B are mixed
Heat gain = Heat lost
mS A 16  12   mS B 19  16 
4
or S B  S A ........(i )
3
When liquid B and C are mixed
mS B  23  19   mSC  28  23  or
5
S B  S C .....(ii )
4
15
From (i) and (ii), we get S A  SC
16
When A and C are mixed, let equilibrium temperature of mixture is  , then
mSA  12  mSC  28     20.26º C
6: A piece of ice of mass 100 g and at temperature 00 C is put in 200 g of water at 250 C . How much
ice will melt as the temperature of the water reaches 00 C ? ( specific heat capacity of water
 4200J kg 1 K 1 and latent heat of fusion of ice  3.4 105 J Kg 1 )
SOLUTION:
The heat released as the water cools down from 250 C to 00 C is
Q  mS    0.2  4200  25   21000 J

The amount of ice melted by this heat is
Q 21000
m    62 g
L 3.4  10 5
7: 10 litres of hot water at 70º C is mixed with an equal volume of cold water at 20º C . Find the
resultant temperature of the water. (Specific heat of water = 4200 J/kg -K)
SOLUTION:
m1S11  m2S22
Resultant temperature,   m S  m S
1 1 2 2
Here, m1  m2  10kg ,
(since mass of 1 litre of water is 1 kg).
1  70º C;2  20º C
and S1  S2  4200 J / kg  K
10  4200  70  10  4200  20
  45º C
10  4200  10  4200
8: The following graph represents change of state of 1 gram of ice at 200 C . Find the net heat required
to convert ice into steam at 1000 C
0
T( C)
100 d
e
b c
0
a Q(cal)
20 Q1 Q2 Q3 Q4
SOLUTION:
Sice  0.53cal / g  0 C
In the figure :
a to b: Temperature of ice increases until it reaches its melting point 0 0 C .
Q1  mSice  0   20    1 0.53 20   10.6cal

b to c: Temperature remains constant until all the ice has melted
Q2  mL f  1 80   80cal
c to d : Temperature of water again rises until it reaches its boiling point 1000 C
Q3  mSwater 100  0  11.0 100   100cal

d to e : Temperature is again constant until all the water is transformed into the vapour phase
Q4  mLv  1 539   539cal
Thus, the net heat required to convert 1g of ice at 200 C into steam at 1000 C is
Q  Q1  Q2  Q3  Q4  729.6cal
9 : A calorimeter of water equivalent 83.72 Kg contains 0.48 Kg of water at 35ºC. How much mass of ice at
0ºC should be added to decrease the temperature of the calorimeter to 20ºC. (SW= 4186J / Kg-K and
Lice  335000 J / Kg )
SOLUTION:
Heat capacity of the calorimeter = 83.72J K1
From law of method of mixtures,
Heat lost by calorimeter 

   Heat gained by the ice
Heat lost by water 

83.7215 0.48418615  m 335000 83720

 m  0.07498 Kg
10. Steam is passed into a calorimeter with water having total thermal capacity 110 cal/gm and
initial temperature 30ºC. If the resultant temperature is 90ºC, the increase in the mass of the
water is
1) 12 gm 2) 1.2 gm 3) 5 gm 4) 12.4 gm
SOLUTION:
From principle of calorimetry
 mS water 1  m  Lsteam  S 2 
(mS)water= 110, Sw = 1, Lice = 540 ,
1  60 &  2  10
11: A steam at 100ºC is passed into 1 kg of water contained in a calorimeter of water equivalent 0.2
kg at 9ºC till the temperature of the calorimeter and water in it is increased to 90ºC. Find the mass
of steam condensed in kg ( SW=1 cal/g ºC, & Lsteam = 540 cal/g)(EAM-14E)
SOLUTION:
Let, m be the mass of the steam condensed. mass of the steam passed into calorimeter,
m2 = 1kg = 1000 g.
Water equivalent of calorimeter,
m1Sl = 0.2 kg = 200g
1 = temperature of the steam = 100ºC
2 = temperature of the water = 9ºC
3 = resultant temperature = 90ºC
From law of method of mixtures,
Heat lost = heat gained (calorimeter + water)
m  Lsteam  SW 1  3     m1S1  m2 SW  3  2 
m 540  1100  90     200  1000 1  90  9 

 m  176 g  0.176kg  0.18kg
12: 1g steam at 100ºC is passed in an insulating vessel having 1g ice at 0ºC. Find the equilibrium
composition of the mixture. (Neglecting heat capacity of the vessel).
SOLUTION:
Available heat from steam mL  1 540  540 cal
Heat required for melting of ice and to rise its temperature to 100º C  mice Lice  mwater S water 
 1 80  11 100  0   180cal
Let m be the mass of steam condensed, then
180 1
m 540  180  m  540  3 g
1 4
Final contents : Water = 1   g ,
3 3
1 2
steam = 1   g
3 3
13: 20g of steam at 100ºC is passed into 100g of ice at 0ºC. Find the resultant temperature if latent
heat of steam is 540 cal/g, latent heat of ice is 80 cal/ g and specific heat of water is 1 cal/gºC.
SOLUTION:
For steam
Heat lost by the steam in condensation
Q1  ms Ls  20  540  10800cal .........(1)
For ice
Heat gained by the ice in melting and to rise its temperature from 0o C to 1000 C is Q2  mice Lice  mice S w t
 100  80  100 1100  18000cal ......(2)
From eq. (1) and (2) ; Q2  Q1
Let  = resultant temperature of the mixture
According to law of method of mixtures
Heat lost by steam = Heat gained by ice
mL s water 100 miceLice miceSwater  0
s s mS
 20540 201100  10080 1001 

   40º C
14. A thermally insulated vessel contains some water at 00C. The vessel is connected to a vacuum
pump to pump out water vapour. This results in some water getting frozen. The maximum percentage
amount of water that will be solidified in this manner will be ( Lsteam  21105 J / kg and
Lice  3.36105 J / kg ).
1) 86.2% 2) 33.6% 3) 21% 4) 24.36%
SOLUTION:
Let m1 mass is vaporised and m2 mass gets solidified
Then heat taken in vaporisation = heat given during
or m1  2110    m   3.36 10 

5
2
5
m2  6.25m1;
m2
%  100
m1  m2
15 : 6 gm of steam at 1000 C is mixed with 6 gm of ice at 00 C . Find the mass of steam left uncondensed
( L f  80cal / g , Lv  540cal / g ,
SWater  1cal / g  0 C )
SOLUTION:
For steam
Heat lost by the steam in condensation
Q1  ms Ls  6  540  3240cal .........(1)
For ice
Heat gained by the ice in melting and to rise its temperature from 0o C to 1000 C is
Q2  mice Lice  mice S w t
 6  80  6 1100  1080cal ......(2)
From eq (1) and (2) Q1  Q2
i.e , the total steam should not condensed in to water.
Let ‘m’ gm of steam is condensed into water by giving 1080cal. of heat .
1080
mLs  1080 ; m   2 gm
540
 mass of the steam left uncondensed  6  2  4g
16. 2kg of water contained in a vessel of negligible heat capacity is heated with a coil of 1kw. The loss
of energy form the vessel is at the rate of 160J/s. How much time temperature will raise from
1) 8 min 20 sec 2) 6 min 2 sec 3) 7 min 4) 14 min
SOLUTION:
Q
Net   100  160
t
msT
 840
t
17. 30 gms of water at 30ºC is in a beaker. Which of the following, when added to water, will have
greatest cooling effect? (Specific heat of copper = 0.1 cal/gmºC)
1) 100gm of water at10ºC 2) 15gm of water at 0ºC ]
3) 3gm of ice at 0ºC 4) 18gm of copper at 0ºC
SOLUTION:
(i) mSw  30  1   m1Sw 1  10 
(ii) mSw  30  2   m2 Sw 2  0 
(iii) mSw  30  3   m3 Lice  m3 Sw 3  0 
(iv) mSw  30  4   m4 Scu  4  0 
here m1  100 g , m2  15 g , m3  3g , m4  18 g
above calculations will be show that 1 is least.
18: A piece of ice(heat capacity =2100J/Kg 0C and latent heat  3.36 105 J / Kg ) of mass m grams is at
5º C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-
water mixture is in equilibrium, it is found that 1gm of ice has melted. Assuming there is no other heat
exchange in the process. Find the value of m. (JEE-2010)
SOLUTION:
Here, heat given is used to increase the temperature of the ice to 0º C and to melt 1gm of ice.
Given m is mass of ice in gm.
 
 420  m  2100  5  1 3.36 105  10 3
 m  8 gm .
19. In an industrial process 10 kg of water per hour is to be heated from 200C to 800C. To do this
steam at 1500C is passed from a boiler into a copper coil immersed in water. The steam condenses
in the coil and is returned to the boiler as water at 900C . How many kilograms of steam is required
per hour (specific heat of steam = 1 cal/gm , latent heat of vapourisation = 540 cal/gm)?
1) 1gm 2) 1 kg 3) 10 gm 4) 10 kg
SOLUTION:
Let m kg of steam is required for this process
m w  S w  8 0  2 0 
 mSsteam 150100  mLsteam  mSw 100 
20.When a small ice crystal is placed into super cooled water, it begins to freeze instantaneously. What
amount of ice is formed from 1kg of water super cooled to 8º C .
SOLUTION: mL  m1S ; m 80  1000 1 8 ; m = 100g
 
21 : The specific heat of a substance varies as 3 2   103 cal /g º C. What is the amount of heat
required to raise the temperature of 1kg of substance from 10ºC to 20ºC?
SOLUTION:
For small change in temperature d  , heat required, dQ  mSd .
2
 Q   mSd
1
20 20
2
 2

Q   1000 3   10 d    3 3
2
10 10
 3 202   3 102 
 20 
   10    8200 1050  7150cal
 2   2 
22: The melting point of ice is 0ºC at 1 atm. At what pressure will it be -1ºC?
 1  3 3
(Given, V2  V1  1    10 m )
 0.9 
SOLUTION:
 1  3 3
Here T   1  0   1, T  273  0  273K and V2  V1  1    10 m (given)
 0.9 
L  80cal / g
P L
we have, T  T V  V
 2 1
P 80  4.2 103

 1 
273 1 
1  3
 10
 0.9 
P  110.8 105 N / m 2  110.8 atm
P2  P1  110.8 atm  P2 110.8  P1 111.8 atm
23. 30g of ice 00c and 20 g of steam at 1000c are mixed. The composition of the resultant mixture is
1) 40g of water and 10g steam at 1000c 2) 10g of ice and 40g of water at 00c
3) 50g of water at 1000c 4) 35g of water and 15g of steam at 1000c
SOLUTION:
msLs = (mice Lice  mice s )
Where ms=mass of steam condensed to rise temperature of ice to 1000c water.
24. Ice at 0ºC is added to 200gm of water initially at 70ºC in a vacuum flask. When 50gm of ice has
been added and has all melted, the temperature of flask and contents is 40ºC. When a further
80gm of ice is added and has all melted, the temperature of whole become 10ºC. Neglecting heat
lost to surroundings the latent heat of fusion of ice is
1) 80 cal/gm 2) 90 cal/gm 3) 70 cal/gm 4) 540 cal/gm
SOLUTION:
Accordingtoprincipleofcalorimetry
ML f  MS    mS   water   mS   flask
5Lf  400 3w....(i), here w   mS  flask
Now the system contains (200+50)gm of water at 40ºC,
so when further 80gm of ice is added
8 L f  670  3w .....( ii )
from(i)&(ii)weget L f .
25. ‘n’ number of liquids of masses m,2m,3m,4m, ....... having specific heats S, 2S,3S, 4S, .... are at
temperatures t, 2t, 3t, 4t . . . . are mixed. The resultant temperature of mixture is
3n 2n  n  1 3n  n  1 3n  n  1
1) t 2) 3  2n  1 t 3) 2  2n  1 t 4)  2n  1 t
2n  1
SOLUTION:
m1S11  m2 S22  ......

m1S1  m2 S2  .....

1  2
3 3
 33  ........  n3 t
1  2
2 2 2
 3  .........  n 2

3n  n  1
= 2  2n  1 t
26. 2 k g of ice at -200C is mixed with 5 kg of water at 200C in an insulating vessel having a negligible
heat capacity. The final mass of water in the vessel. ( The specific heat of water and ice are 1k cal/
kg0C and 0.5 k cal/kg/0C respectively and the latent heat of fusion of ice is 80 k cal/kg) is
1) 7 kg 2) 6 kg 3) 4 kg 4) 2 kg
SOLUTION:
Let m be mass of ice melted into water
mice  Sice  20  m Lice  mwater  Sw  20
final mass of water in vessel = m + 5kg.

27. The specific heat of a substance varies with temperature as s=0.20+0.14  +0.023  2 (cal/gmºC)
.Heat required to raise the temperature of 2 gm of the substance from 50 C to 150 C is (  is in º C )
1) 24 cal 2) 56 cal 3) 82 cal 4) 100 cal
SOLUTION:
2
Q   m  S  d
115
 
Q  2  0.2  0.14  0.023 2 d
5
28. A heater melts 0ºC ice in a bucket completely into water in 6 minutes and then evaporates all that
water into steam in 47 minutes 30 sec. If latent heat of fusion of ice is 80 cal/gram, latent heat of
steam will be (specific heat of water is 1 cal /gam-ºC)
1) 536 Cal/gram 2) 533.3 Cal/gram 3) 540 Cal/gram 4) 2.268  106 J/Kg
SOLUTION:
Let m be the mass of ice in bucket
Heat given out by heater in 6min is 80m
Heat given out in 47.5min is 100m+mLv
m  80 -- 6 minutes
(m1100 )+ (m  Lv)47.5 minutes
 8047.5 = 6(100+Lv)
Transmission of Heat
Heat energy transfers from a body at higher temperature to a body at lower temperature. The transfer of
heat from one body to another may take place by one of the following modes.
Conduction
Conduction :
The process of transmission of heat energy in which the heat is transferred from one particle to other particle
without dislocation of the particle from their equilibrium position is called conduction.
(1) Heat flows from hot end to cold end. Particles of the medium simply oscillate but do not leave their place.
(2) Medium is necessary for conduction
(3) It is a slow process
(4) The temperature of the medium increases through which heat flows
(5) Conduction is a process which is possible in all states of matter.
(6) When liquid and gases are heated from the top, they conduct heat from top to bottom.
(7) In solids only conduction takes place
(8) In nonmetallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore
they are poor conductors.
(9) In metallic solids free electrons carry the heat energy, therefore they are good conductor of heat.
Conduction in Metallic Rod :
When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end.
Radiat ion
A
Conduct ion
Hot end Cold end ’
Temperature gradient (T.G.) :
The rate of change of temperature with distance between two isothermal surfaces is called temperature
gradient.
Hence
 
(i) Temperature gradient = x
(ii) The negative sign show that temperature  decreases as the distance x increases in the direction of
heat flow.
1   2 
(iii) For uniform temperature fall l

x
(iv) Unit : K/m or °C/m (S.I.) and Dimensions [L1 ]
q q – Dq
Heat Heat
A
q 1 q 2
Dx
l
Law of thermal conductivity :
Consider a rod of length l and area of crosssection A whose faces are maintained at temperature q1
and q2 respectively. The curved surface of rod is kept insulated from surrounding to avoid leakage of heat
q 1 q 2
Q Q
A Q
l
(i) In steady state the amount of heat flowing from one face to the other face in time t is given by
KA ( 1   2 ) t
Q
l
where K is coefficient of thermal conductivity of material of rod.
1 2 Q KA (   )
(ii) Rate of flow of heat i.e. heat current t  H  l
(iii) In case of nonsteady state or variable crosssection, a more general equation can be used to solve
problems.
dQ d
  KA
dt dx
More about K :
It is the measure of the ability of a substance to conduct heat through it.
(i) Units : Cal/cmsec oC (in C.G.S.), kcal/msecK (in M.K.S.) and W/m K (in S.I.).
Dimension : [MLT 3 1 ]
(ii) The magnitude of K depends only on nature of the material.
(iii) Substances in which heat flows quickly and easily are known as good conductor of heat. They
possesses large thermal conductivity due to large number of free electrons
e.g. Silver, brass etc. For perfect conductors, K   .
(iv) Substances which do not permit easy flow of heat are called bad conductors. They possess low
thermal conductivity due to very few free electrons e.g. Glass, wood etc. and for perfect
insulators, K  0 .
(v) The thermal conductivity of pure metals decreases with rise in temperature but for alloys thermal
conductivity increases with increase of temperature.
(vi) Human body is a bad conductor of heat (but it is a good conductor of electricity).
Decreasing order of conductivity :

For some special cases it is as follows
(a) K Ag  KCu  K Al
(b) KSolid  K Liquid  KGas
(c) K Metals  K Non metals
Relation between temperature gradient and thermal conductivity :
dQ d
In steady state, rate of flow of heat dt
  KA
dx
= – KA ´ (T.G.)
1 dQ
(T.G.) 
K
( dt
= constant)
Temperature difference between the hot end and the cold end in steady state is inversely
proportional to K,
i.e. in case of good conductors temperature of the cold end will be very near to hot end.
In ideal conductor where K = ¥, temperature difference in steady state will be zero.
Thermal resistance (R) :
The thermal resistance of a body is a measure of its opposition to the flow of heat through it.
It is defined as the ratio of temperature difference to the heat current (= Rate of flow of heat)
1   2 1   2 l
(i) Hence R  H

KA(1   2 ) / l

KA
(ii) Unit : o C  sec / cal or K  sec / kcal and
Dimension : [M 1 L2 T 3 ]
Wiedmann-Franz law :
At a given temperature T, the ratio of thermal conductivity to electrical conductivity is constant
i.e., (K / T ) = constant,
i.e., a substance which is a good conductor of heat (e.g., silver) is also a good conductor of electricity.
Mica is an exception to above law.
Thermometric conductivity or diffusivity :
It is a measure of rate of change of temperature (with time) when the body is not in steady state (i.e., in
variable state)
It is defined as the ratio of the coefficient of thermal conductivity to the thermal capacity per unit volume
mc
of the material. Thermal capacity per unit volume = V
=  c
K
(  = density of substance) Þ Diffusivity (D) =  c
Unit : m2/sec
Combination of Metallic Rods
Series combination :
Let n slabs each of cross-sectional area A, lengths l1 , l2 , l3 ,...... ln and conductivities K1 , K 2 , K3 ...... Kn respectively be
connected in the series
q1 q2 q3 q n – 1 qn
K1 K2 Kn
l1 l2 ln
Heat current :
Heat current is the same in all the conductors.
Q
i.e.,  H 1  H 2  H 3 .........  H n
t
K1 A(1   2 ) K 2 A( 2   3 ) Kn A( n 1   n )
 
l1 l2 ln
Equivalent thermal resistance :
R  R1  R 2  ..... Rn
Equivalent thermal conductivity :
Itcanbecalculatedasfollows
From RS  R1  R2  R3  ...
l1  l2  ...ln l l l
 1  2  ....  n
Ks K1 A K 2 A Kn A
l1  l2  ...... ln
Ks 
l1 l l
 2  ........ n
K1 K 2 Kn
n
Ks 
1 1 1 1
(a) For n slabs of equal length    .....
K1 K2 K3 Kn
2 K1 K 2
(b) For two slabs of equal length, K s  K  K
1 2
Temperature of interface of composite bar :

Let the two bars are arranged in series as shown in the figure.
q1 q q2
K1 K2
l1 l2
Then heat current is same in the two conductors.
Q K1 A(1   ) K 2 A(   2 )
i.e.,  
t l1 l2
K1 K2
1  2
l l2
  1
By solving we get K1

K2
l1 l2
K11  K 2 2
(a) If l1 = l2 then  
K1  K 2
1   2
(b) If K1 = K2 and l1 = l2 then  
2
‘Parallel Combination :
Let n slabs each of length l, areas A1 , A2 , A3 ,..... An and thermal conductivities K1, K 2 , K 3 ,..... Kn are
connected in parallel then
q1 l q2
A1
K1
A2
K2
K3 A3
A3
Kn
Equivalent resistance :
1 1 1 1 1
    .....
Rs R1 R2 R3 Rn
R1 R2
For two slabs Rs  R  R
1 2
Temperature gradient :
Same across each slab.
Heat current :
in each slab will be different. Net heat current will be the sum of heat currents through individual slabs.
i.e., H  H1  H 2  H 3  .... H n
K ( A1  A2  .....  An ) ( 1   2 ) K1 A1 ( 1   2 ) K 2 A2 ( 1   2 ) K A (   2 )
=   ...  n n 1
l l l l
K1 A1  K 2 A2  K 3 A3  ..... K n An
K
A1  A2  A3  ..... An
K1  K 2  K 3  ..... K n
(a) For n slabs of equal area K 
n
K1  K 2
(b) For two slabs of equal area K  .
2
Growth of Ice on Lake

(1) Water in a lake starts freezing if the atmospheric temperature drops below 0 o C . Let y be the thickness of ice layer in the
lake at any instant t and atmospheric temperature is   o C .
(2) The temperature of water in contact with lower surface of ice will be zero.
KA [0  ( )] dt
(3) If A is the area of lake, heat escaping through ice in time dt is dQ 1 
y
(4) Suppose the thickness of ice layer increases by dy in time dt, due to escaping of above heat.
Then dQ 2  mL  (dy A) L
(5) As dQ 1  dQ 2 ,
hence, rate of growth of ice will be
(dy / dt )  (K  / Ly )
So, the time taken by ice to grow to a thickness y is
L y L 2
t
K  0
y dy 
2 K
y
(6) If the thickness is increased from y 1 to y 2 then time taken

 L y2 L 2
t  ydy  (y 2  y12 )
K 1 y 2 K
(7) Take care and do not apply a negative sign for putting values of temperature in formula and also do not convert it to
absolutescale.
(8) Ice is a poor conductor of heat, therefore the rate of increase of thicknessof ice on ponds decreases with time.
(9) It follows from the above equation that time taken to double and triple the thickness, will be in the ratio of
t 1 : t 2 : t 3 :: 1 2 : 2 2 : 3 2 , i.e., t1 : t 2 : t 3 :: 1 : 4 : 9
(10) The time intervals to change the thickness from 0 to y, from y to 2y and so on will be in the ratio
t 1 : t 2 : t 3 :: (1 2  0 2 ) : (2 2  1 2 ) : (3 2 : 2 2 )
 t1 : t 2 : t 3 :: 1 : 3 : 5
Convection :
Mode of transfer of heat by means of migration of material particles of medium is called convection. It is of two types.
(1) Natural convection :
Convection
current
This arise due to difference of densities at two places and is a consequence of gravity because on account of gravity
the hot light particles rise up and cold heavy particles try setting down. It mostly occurs on heating a liquid/fluid.
Forced convection :
Fig. 15.17
If a fluid is forced to move to take up heat from a hot body then the convection process is called forced convection.
In this case Newton’s law of cooling holds good. According to which rate of loss of heat from a hot body due to moving fluid is
directly proportional to the surface area of body and excess temperature of body over its surroundings
Q
i.e .  A(T  T0 )
t
Q
 h A(T  T0 )
t
where h = Constant of proportionality called convection coefficient, T = Temperature of body and T0 = Temperature of surrounding
Convection coefficient (h) depends on properties of fluid such as density, viscosity, specific heat and thermal conductivity.
 Natural convection takes place from bottom to top while forced convection in any direction.
 In case of natural convection, convection currents move warm air upwards and cool air downwards. That is why heating is done
from base, while cooling from the top.
 Natural convection plays an important role in ventilation, in changing climate and weather and in forming land and sea breezes and
trade winds.
 Natural convection is not possible in a gravity free region such as a free falling lift or an orbiting satellite.
 The force of blood in our body by heart helps in keeping the temperature of body constant.
 If liquids and gases are heated from the top (so that convection is not possible) they transfer heat (from top to bottom) by
conduction.
 Mercury though a liquid is heated by conduction and not by convection.
Radiation
 The process of the transfer of heat from one place to another place without heating the
intervening medium is called radiation.
 Precisely it is electromagnetic energy transfer in the form of electromagnetic wave through any
medium. It is possible even in vacuum e.g. the heat from the sun reaches the earth through
radiation.
 The wavelength of thermal radiations ranges from 7 .8  10 7 m to 4  10 4 m . They belong to
infra-red region of the electromagnetic spectrum. That is why thermal radiations are also called
infra-red radiations.
 Medium is not required for the propagation of these radiations.
 They produce sensation of warmth in us but we can’t see them.
 Every body whose temperature is above zero Kelvin emits thermal radiation.
 Their speed is equal to that of light i.e. ( 3  10 8 m / s) .
 Their intensity is inversely proportional to the square of distance of point of observation from
the source (i.e. I  1 / d 2 ).
 Just as light waves, they follow laws of reflection, refraction, interference, diffraction and
polarisation.
 When these radiations fall on a surface then exert pressure on that surface which is known as
radiation pressure.
 While travelling these radiations travel just like photons of other electromagnetic waves. They
manifest themselves as heat only when they are absorbed by a substance.
 Spectrum of these radiations can not be obtained with the help of glass prism because it absorbs heat
radiations. It is obtained by quartz or rock salt prism because these materials do not have free
electrons and interatomic vibrational frequency is greater than the radiation frequency, hence they
do not absorb heat radiations.
Colour of Heated Object
When a body is heated, all radiations having wavelengths from zero to infinity are emitted.
 Radiations of longer wavelengths are predominant at lower temperature.
 The wavelength corresponding to maximum emission of radiations shifts from longer wavelength to shorter
wavelength as the temperature increases. Due to this the colour of a body appears to be changing.
 A blue flame is at a higher temperature than a yellow flame
Interaction of Radiation with Matter :
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly
transmitted.
Q Qr
Qa
Qt
(1) Q  Qa  Qr  Qt
Qa Qr Qt
(2)    art 1
Q Q Q
Qa
(3) a  Q
= Absorptance or absorbing power
Qr
r  Q
= Reflectance or reflecting power
Qt
t  Q
= Transmittance or transmitting power
(4) r, a and t all are the pure ratios so they have no unit and dimension.
(5) Different bodies
(i) If a = t = 0 and r = 1 ® body is perfect reflector
(ii) If r = t = 0 and a = 1 ® body is perfectly black body
(iii) If, a = r = 0 and t = 1 ® body is perfect transmitter
(iv) If t = 0 Þ r  a  1 or a  1  r i.e. good reflectors are bad absorbers.
Emissive Power, Absorptive Power and Emissivity
If temperature of a body is more than it’s surrounding then body emits thermal radiation
Monochromatic Emittance or Spectral emissive power (el) :
For a given surface it is defined as the radiant energy emitted per sec per unit area of the surface
with in a unit wavelength around l
 1  1
i.e. lying between    2  to    2  .
Energy
Spectral emissive power (e  )  Area  time  wavelength
Joule
Unit : 2
m  sec  Å
Dimension : [ML1 T 3 ]
Total emittance or total emissive power (e) :
It is defined as the total amount of thermal energy emitted per unit time, per unit area of the
body for all possible wavelengths.

e   0 ed
Joule Watt
Unit : 2 or
m  sec m2
Dimension : [MT 3 ]
Monochromatic absorptance or spectral absorptive power (al) :

It is defined as the ratio of the amount of the energy absorbed in a certain time to the total heat
energy incident upon it in the same time, both in the unit wavelength interval. It is dimensionless and unit
less quantity. It is represented by al.
Total absorptance or total absorpting power (a):
It is defined as the total amount of thermal energy absorbed per unit time, per unit area of the
body for all possible wavelengths.

a 0
a d
Emissivity (e) :
Emissivity of a body at a given temperature is defined as the ratio of the total emissive power of
the body (e) to the total emissive power of a perfect black body (E) at that temperature
e
i.e.  
E
(e ® read as epsilon)
 For perfectly black body e = 1
 For highly polished body e = 0
 But for practical bodies emissivity (e) lies between zero and one (0 < e < 1).
Perfectly Black Body
(1) A perfectly black body is that which absorbs completely the radiations of all wavelengths incident on it.
(2) As a perfectly black body neither reflects nor transmits any radiation, therefore the absorptance of a
perfectly black body is unity i.e. t = 0 and r = 0 Þ a = 1.
(3) We know that the colour of an opaque body is the colour (wavelength) of radiation reflected by it. As a
black body reflects no wavelength so, it appears black, whatever be the colour of radiations incident on it.
(4) When perfectly black body is heated to a suitable high temperature, it emits radiation of all possible
wavelengths. For example, temperature of the sun is very high (6000 K approx.) it emits all possible
radiation so it is an example of black body.
Ferry’s black body :
A perfectly black body can’t be realised in practice. The nearest example of an ideal black
body is the Ferry’s black body. It is a doubled walled evacuated spherical cavity whose inner wall is
blackened. The space between the wall is evacuated to prevent the loss of heat by conduction and
radiation. There is a fine hole in it. All the radiations incident upon this hole are absorbed by this black
body. If this black body is heated to high temperature then it emits radiations of all wavelengths. It is the
hole which is to be regarded as a black body and not the total enclosure
P
O
A perfectly black body can not be realised in practice but materials like Platinum black or Lamp
black come close to being ideal black bodies. Such materials absorbs 96% to 85% of the incident radiations.
Prevost Theory of Heat Exchange :
(1) Every body emits heat radiations at all finite temperature (Except 0 K) as well as it absorbs radiations from
the surroundings.
(2) Exchange of energy along various bodies takes place via radiation.
(3) The process of heat exchange among various bodies is a continuous phenomenon.
(4) At absolute zero temperature (0 K or – 273°C) this law is not applicable because at this temperature the
heat exchange among various bodies ceases.
 If Qemission > Qabsorbed ® temperature of body decreases and consequently the body appears colder.
 If Qemission < Qabsorbed ® temperature of body increases and it appears hotter.
 If Qemission = Qabsorbed ® temperature of body remains constant (thermal equilibrium)
Kirchoff’s Law:
According to this law the ratio of emissive power to absorptive power is same for all surfaces at
the same temperature and is equal to the emissive power of a perfectly black body at that temperature.
e1 e2 E
Hence a   ...  
1 a2  A  Perfectly black bod y
But for perfectly black body A = 1
e
i.e. a  E
If emissive and absorptive powers are considered for a particular wavelength l,
 e 
 a  ( E )black

 
Now since (El)black is constant at a given temperature, according to this law if a surface is a good
absorber of a particular wavelength it is also a good emitter of that wavelength.
This in turn implies that a good absorber is a good emitter (or radiator)
Stefan’s Law:
According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive
power of black body) is directly proportional to the fourth power of its absolute temperature,
i.e . E  T 4
E = sT4
where s is a constant called Stefan’s constant having dimension [MT 3 4 ]

value 5.67  10 8 W / m 2 K 4 .
(i) For ordinary body : e = eE  T 4
(ii) Radiant energy : If Q is the total energy radiated by the ordinary body then
Q
e  T 4
At
Q  A  T 4 t
(iii) Radiant power (P) : It is defined as energy radiated per unit area
Q
i.e. P   A T 4 .
t
(iv) If an ordinary body at temperature T is surrounded by a body at temperature T0, then Stefan’s law
may be put as
e    (T 4  T04 )
Rate of Loss of Heat (RH) and Rate of Cooling (RC):
Rate of loss of heat (or initial rate of loss of heat) :
If an ordinary body at temperature T is placed in an environment of temperature T0 (T0 < T) then
heatlossbyradiationisgivenby Q  Qemission  Qabsorption  A   (T 4  T04 )
dQ
Rate of loss of heat (R H )   A   (T 4  T04 )
dt
If two bodies aremadeof same material, have same surfacefinish and are at thesame initial
temperaturethen
 dQ 
 
 dt 1 A
dQ  1
 A  dQ  A2
dt  
 dt  2
Initial rate of fall in temperature (Rate of cooling):
Ifmisthebodyandcisthespecificheatthen
dQ dT d
 mc .  mc ( Q  mc T and dT  d )
dt dt dt
d  (dQ / dt )
Rate of cooling (Rc )  
dt mc
A  4
 (T  T04 )
mc
A  4
 (T  T04 )
V c
where m = density (r) ´ volume (V)
for two bodies of the same material under identical environments,
(R c )1 A V
the ratio of their rate of cooling is (R  1. 2
c )2 A 2 V1
Dependence of rate of cooling :
When a body cools by radiation the rate of cooling depends on
 Nature of radiating surface i.e. greater the emissivity, faster will be the cooling.
 Area of radiating surface, i.e. greater the area of radiating surface, faster will be the cooling.
 Mass of radiating body i.e. greater the mass of radiating body slower will be the cooling.
 Specific heat of radiating body i.e. greater the specific heat of radiating body slower will be cooling.
 Temperature of radiating body i.e. greater the temperature of body faster will be cooling.
 Temperature of surrounding i.e. greater the temperature of surrounding slower will be cooling.
Newton’s Law of Cooling:
When the temperature difference between the body and its surrounding is not very large
i.e. T – T0 = DT
then T 4  T04 may be approximated as 4 T03 T
dT A  4
By Stefan’s law, dt  [T  T04 ]
mc
dT A 
Hence dt  4 T03 T
mc
dT
dt
 T
d
  0
dt
i.e., if the temperature of body is not very different from surrounding,
rate of cooling is proportional to temperature difference between the body and its surrounding.
This law is called Newton’s law of cooling.
 Greater the temperature difference between body and its surrounding greater will be the rate of cooling.
d
 If    0 , dt  0 i.e. a body can never be cooled to a temperature lesser than its surrounding by
radiation.
 If a body cools by radiation from  1o C to  2o C in time t,
d  1   2 1   2
then dt

t
and    av 
2
.
 1   2     2 
The Newton’s law of cooling becomes   K 1  0  .
t   2 
This form of law helps in solving numericals.
 Practical examples
 Hot water loses heat in smaller duration as compared to moderate warm water.
 Adding milk in hot tea reduces the rate of cooling.
Cooling Curves
Curve between log( – 0) and time :
loge
(q –
q 0)
O t
d d
As  (   0 )    Kdt
dt (   0 )
Integrating log e (   0 )   Kt  C
log e (   0 )   Kt  log e A
This is a straight line with negative slope
Curve between temperature of body and time:
q-0
0 Time t
As log e (   0 )   Kt  log e A
  0
 log e   Kt
A
    0  Ae kt
which indicates temperaturedecreases exponentially withincreasing time.
Curve between the rate of cooling:
R
Kq 0 q
(R) and body temperature (). R  K(   0 )  K  K 0
This is a straight line intercept R-axisat  K 0

Curve between rate of cooling (R):
O
(q – q0 )
R and temperature difference betweenbody () and surrounding (0)

R  (   0 ).
This is a straight line passing through origin.’
Wien’s Displacement Law:

According to Wien’s law the product of wavelength corresponding to maximum intensity of
radiation and temperature of body (in Kelvin) is constant,
i.e. m T  b  constant
where b is Wien’s constant and has value 2 .89  10 3 m K .
As the temperature of the body increases, the wavelength at which the spectral intensity (El) is
maximum shifts towards left. Therefore it is also called Wien’s displacement law.
This law is of great importance in ‘Astrophysics’ as through the analysis of radiations coming
from a distant star, by finding m the temperature of the star T ( b / m ) is determined.
Law of Distribution of Energy (Plank’s Hypothesis):
(1) The theoretical explanation of black body radiation was done by Planck.
(2) According to Plank’s atoms of the walls of a uniform temperature enclosure behave as oscillators, each
with a characteristic frequency of oscillation.
(3) These oscillations emits electromagnetic radiations in the form of photons (The radiation coming out from
a small hole in the enclosure are called black body radiation). The energy of each photon is hn. Where n
is the frequency of oscillator and h is the Plank’s constant. Thus emitted energies may be hn, 2hn, 3hn ...
nhn but not in between.
8hc 1
According to Planck’s law E  d  5 [e hc / KT  1]
d
where c = speed of light
k = Boltzmann’s constant.
This equation is known as Plank’s radiation law. It is correct and complete law of radiation
(4) This law is valid for radiations of all wavelengths ranging from zero to infinite.
 hc 
(5) For radiations of short wavelength    
KT  Planck’s law reduces to Wien’s energy distribution
A
law E  d  e  B / T d 
5
 hc 
(6) For radiations of long wavelength    
KT  Planck’s law reduces to Rayleigh-Jeans energy
8KT
distribution law E  d  4 d
Temperature of the Sun and Solar Constant:
R Earth
Sun
If R is the radius of the sun and T its temperature, then the energy emitted by the sun per sec
through radiation in accordance with Stefan’s law will be given by
4
P  AT  4  R 2 T 4
In reaching earth this energy will spread over a sphere of radius r (= average distance between
sun and earth); so the intensity of solar radiation at the surface of earth (called solar constant S) will be
given by
P 4R 2 T 4
S  2

4r 4 r 2
1/4 1/4
 1 .5  10 8 2 
 r  2 S   3
T
i.e.  R   
     1 .4  10  ~ 5800 K
 7  10 5  5 . 67  10 8 
    
As r  1 .5  10 8 km, R  7  10 5 km,
cal kW
S 2  1 .4 2
cm 2 min m
W
  5 . 67  10 8
m2K4
This result is in good agreement with the experimental value of temperature of sun, i.e., 6000 K.
PROBLEMS
1. The heat is flowing through two cylindrical rods of same material. The diameters of the rods are in the ratio 1 : 2 and their lengths are
in the ratio 2 : 1. If the temperature difference between their ends is the same, the ratio of rate of flow of heat through them will be
(a) 1:1 (b) 2:1 (c) 1:4 (d) 1:8
‘
Q KA 

t l
Q A d2
  (d = Diameter of rod)
t l l
2 2
(Q / t )1  d 1  l 1 1 1
    2      
(Q / t)2  d 2  l1  2   2  8
2.. Two identical square rods of metal are welded end to end as shown in figure (i), 20 calories of heat flows through it in 4 minutes. If the
rods are welded as shown in figure (ii), the same amount of heat will flow through the rods in
o o
0C 100 C
o o
0C 100 C
(i) (ii)
(a) 1 minute (b) 2 minutes (c) 4 minutes (d) 16 minutes
SOLUTION:
Q KA   
   (R = Thermal resistance)
t l (l / KA) R
t  R ( Q and  are same)
tP R R/2 1
 P  
tS RS 2R 4
Series resistance RS  R1  R 2
R1 R2
parallelresistance R P  R  R )
1 2
tS 4
tP    1 min .
4 4
3. Two rods A and B are of equal lengths. Their ends are kept between the same temperature and their
area of crosssections are A1 and A2 and thermal conductivities K1 and K 2 . The rate of heat
transmission in the two rods will be equal, if
(a) K1 A2  K 2 A1 (b) K1 A1  K 2 A2 (c) K1  K 2 (d) K1 A12  K 2 A22
SOLUTION:
Q K A (   2 )
   1 1 1
 t 1 l
Q K A (   2 )
   2 2 1
t
 2 l
Q Q
given  t    t 
 1   2
K 1 A1  K 2 A 2
4. Two rectangular blocks A and B of different metals have same length and same area of crosssection. They
are kept in such a way that their crosssectional area touch each other. The temperature at one end of A is
100°C and that of B at the other end is 0°C. If the ratio of their thermal conductivity is 1 : 3, then under steady
state, the temperature of the junction in contact will be
(a) 25°C (b) 50°C (c) 75°C (d) 100°C
SOLUTION:
Junction tem
perature q
Q
A B
Q
100°C 0°C
K1 1
It is given that K  3
2
K1  K then K 2  3 K
K 1 1  K 2 2
the temperature of the junction in contact   K1  K 2
1  100  3  0 100
  = 25°C
13 4
5. Two walls of thicknesses d1 and d2 and thermal conductivities k1 and k2 are in contact. In the steady
state, if the temperatures at the outer surfaces are T1 and T2 , the temperature at the common wall is
k 1 T1 d 2  k 2 T2 d 1 k 1 T1  k 2 d 2  k 1 d1  k 2 d 2  k 1 d 1 T1  k 2 d 2 T2
(a) k1d 2  k 2 d1 (b) d1  d 2 (c)  T1 T2
 (d)
 T1  T2  k 1 d1  k 2 d 2
SOLUTION:
T1 
T2
K1 K2
d1 d2
In series both walls have same rate of heat flow. Therefore

dQ K1 A(T1   ) K 2 A(  T2 )
  ’
dt d1 d2
 K1d 2 (T1   )  K 2 d1 (  T2 )
K1d 2 T1  K 2 d1T2
 
K1d 2  K 2 d1
6. If two metallic plates of equal thicknesses and thermal conductivities K1 and K 2 are put together face to
face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be
K1 K2
K1 K 2 2 K1 K 2 (K 12  K 22 )3 / 2 (K 12  K 22 )3 / 2
(a) K (b) K (c) (d)
1  K2 1  K2 K1 K 2 2 K1 K 2
SOLUTION:
In series Req  R1  R 2
2l l l
 K A  K AK A
eq 1 2
2 1 1 2 K1 K 2
 
K eq K1 K 2  K 1  K 2
7. Two spheres of different materials one with double the radius and onefourth wall thickness of the other, are
filled with ice. If the time taken for complete melting ice in the large radius one is 25 minutes and that for
smaller one is 16 minutes, the ratio of thermal conductivities of the materials of larger sphere to the smaller
sphere is
(a) 4 : 5 (b) 5 : 4 (c) 25 : 1 (d) 1 : 25
SOLUTION:
KA ( )t
Q
l
 Q and  are same for both spheres hence
l l
K  2
At r t
2
Klarger ll  rs  ts
     .
K smaller ls  rl  tl
1
It is given that rl  2rs , ll  ls and t1  25 min, t s  16 min.
4
2
Klarger  1  1  16 1
      
K smaller 4
   2 25 25
8. The ratio of the diameters of two metallic rods of the same material is 2 : 1 and their lengths are in the ratio 1 : 4. If
the temperature difference between their ends are equal, the rate of flow of heat in them will be in the ratio
(a) 2 : 1 (b) 4 : 1 (c) 8 :1 (d) 16 : 1
SOLUTION:
Q KA( )
 Q A r2
  
t l t l l
2
Q / t 1
r
 1
 l 2
  2   2    4   16

(Q / t)2  r2 
 l1  1   1  1
9. Two cylinders P and Q have the same length and diameter and are made of different materials having thermal
conductivities in the ratio 2 : 3. These two cylinders are combined to make a cylinder. One end of P is kept
at 100°C and another end of Q at 0°C. The temperature at the interface of P and Q is
(a) 30 o C (b) 40 o C ‘ (c) 50 o C (d) 60 o C
SOLUTION:
K1 1  K 2 2
Temperature of interface   K1  K 2
 K1 2 
 
where K1 = 2K and K2 = 3K  K  3
 2 
2 K  100  3 K  0
  2K  3K
200 K
  40 C
5K
10. Two bars of thermal conductivities K and 3K and lengths 1cm and 2cm respectively have equal cross
sectional area, they are joined lengths wise as shown in the figure. If the temperature at the ends of this
composite bar is 0 o C and 100 o C respectively (see figure), then the temperature  of the interface is

o o
0C K 3K 100 C
1 cm 2 cm
100 o 200 o
(a) 50 o C (b) 3
C (c) 60 o C (d) 3
C
SOLUTION:
Temperatureofinterface
K1 1 l 2  K 2 2 l1

K 1 l 2  K 2 l1
K  0  2  3 K  100  1

K  2  3K 1
300 K
 = 60°C
5K
11. A metal rod of length 2m has cross sectional areas 2A and A as shown in figure. The ends are maintained at
temperatures 100°C and 70°C. The temperature at middle point C is
100°C C
70°C
2A A
1m 1m
(a)80°C (b)85°C (c)90°C (d)95°C

‘
Let  be temperature middle point C and in series rate of heat flow is same
 K(2 A)(100   )  KA(  70 )
 200  2    70
 3  270
   90 C
12. A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are
different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature
difference between the two ends is 36 o C . Then the difference of temperature at the two surfaces of A
will be
(a) 6o C (b) 12 o C (c) 18 o C (d) 24 o C
SOLUTION:
A B
2K K
1 x  x 2
Q Q
Suppose thickness of each wall is x then  t   
 combinatio n  t  A
K S A(1   2 ) 2 KA(1   )
 
2x x
2  2K  K 4
 K S  (2 K  K )  3 K
(1   2 )  36 
4
KA  36
 3 2 KA(1   )

2x x
Hence temperature difference across wall A is (1   )  12 o C

13. Two rods (one semicircular and other straight) of same material and of same crosssectional area are
joined as shown in the figure. The points A and B are maintained at different temperature. The
ratio of the heat transferred through a crosssection of a semicircular rod to the heat transferred through
a cross section of the straight rod in a given time is
A Straight rod B
(a) 2 : p (b) 1 : 2 (c) p : 2 (d) 3 : 2

‘
dQ KA 
 ,
dt l
dQ 1
For both rods K, A and  are same  
dt l
(dQ / dt )semi circular lstraight 2r 2
   .
(dQ / dt )straight lsemicircular r 
14. A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell

of inner radius R and outer radius 2R made of material of thermal conductivity K 2 . The two ends of the
combined system are maintained at two different temperatures. There is no loss of heat across the
cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is
K1 K 2 K1  3 K 2 3 K1  K 2
(a) K 1  K 2 (b) K (c) (d)
1  K2 4 4
SOLUTION:
Both the cylinders are in parallel, for the heat flow from one end as shown.
K2
2R
R
K1
K1 A1  K 2 A2
Hence Keq 
A1  A2 ;
where A1 = Area of cross-section of inner cylinder = R2
A2 Areaofcross-sectionofcylindricalshell   {(2 R)2  (R)2 }  3R 2
K1 (R 2 )  K 2 (3R 2 ) K1  3 K 2
 Keq  
R 2  3R 2 4
15. Three rods made of the same material and having the same cross section have been joined as shown in
the figure. Each rod is of the same length. The left and right ends are kept at 0 o C and 90 o C
respectively. The temperature of the junction of the three rods will be
o
90 C
B
o
0C
A
C
o
90 C
(a) 45 o C (b) 60 o C (c) 30 o C (d) 20 o C
SOLUTION:
Let the temperature of junction be . Since roads B and C are parallel to each other (because both having the same temperature
difference).
Hence given figure can be redrawn as follows
90°
R
R  R
P
2 0°  90°
Q
R R/2
R C B A
0°  l l
R
90°
Q ( 1   2 )
 
t R
Q Q
   
  AB  t  BC
t
(90   ) (  0 )
 
R/2 R
 180  2  
  60 C
16. A composite metal bar of uniform section is made up of length 25 cm of copper, 10 cm of nickel and
15 cm of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end
of the composite rod is maintained at 100 o C and the aluminium end at 0 o C . The whole rod is covered
with belt so that there is no heat loss occurs at the sides. If KCu  2 K Al and K Al  3 K Ni , then what will be
the temperatures of
Cu Ni Al
100°C 0°C
(a) 23 .33 o C and 78 .8 o C (b) 83 .33 o C and 20 o C (c) 50 o C and 30 o C (d) 30 o C and 50 o C
SOLUTION:
If suppose K Ni  K  K Al  3 K and KCu  6 K .
Since all metal bars are connected in series
Q Q Q Q
So  t       
 Combinatio n  Cu   Al  t  Ni
t t
3 1 1 1 1 1 1 9
and K  K  K  K  6 K  3 K  K  6 K
eq Cu Al Ni
25 cm 10 cm 15 cm
Q Cu Ni Al Q
100°C 0°C
1 2
Q Q
Hence, if  t   
 Combinatio n  t Cu
K eq A(100  0 ) K Cu A(100   1 )
 
lCombinatio n lCu
2 K A (100  0) 6 K A (100   1 )
    1  83 . 33 C
(25  10  15 ) 25
Q Q
Similarif  t   
 Combinatio n  t  Al
2 K A(100  0 ) 3 K A( 2  0 )
 
50 15
  2  20 o C
17. Three rods of identical area of crosssection and made from the same metal form the sides of an
isosceles triangle ABC , right angled at B . The points A and are maintained at temperatures and
respectively. In the steady state the temperature of the point C is . Assuming that only heat conduction
takes place, is equal to
1 3 1 1
(a) ( 2  1)
(b) ( 2  1)
(c) 2( 2  1)
(d) 3 ( 2  1)
SOLUTION:
 TB  TA
 Heat will flow B to A via two paths (i) B to A (ii) and along BCA as shown.
Rate of flow of heat in path BCA will be same
(T)A
a 2
a
2T B a C(Tc)
Q Q k ( 2 T  TC ) A k (TC  T ) A

i.e.  t     
t
  BC   CA a 2a
TC 3
 
T 1 2
18. Three rods of the same dimension have thermal conductivities 3K, 2K and K. They are arranged as shown in
fig. Given below, with their ends at 100oC, 50oC and 20oC. The temperature of their junction is
o
50 C
2K
o
100 C
3K
K
o
20 C
(a) 60 o C (b) 70 o C (c) 50o C (d) 35o C

SOLUTION:
Let the temperature of junction be  then according to following figure.
50°
2K
H1
H
100° 3K  H2
K
20°
H = H1 + H2
3 K  A  (100   ) 2 KA(  50 ) KA (  20 )
  
l l l
 300 – 3 = 3 – 120   = 70°C
19. Two identical conducting rods are first connected independently to two vessels, one containing water at
100o C and the other containing ice at 0oC. In the second case, the rods are joined end to end and
connected to the same vessels. Let q1 and q2 g / s be the rate of melting of ice in two cases respectively.
The ratio of q1 / q 2 is
1 2 4 1
(a) 2
‘ (b) 1
(c) 1
(d) 4
SOLUTION:
R
RP 
R 2
100°C R 0°C
l
100°C 0°C
Initially the rods are placed in vessels as shown below
Q mL (100  0)
Q (1   2 )     q1 L 
   t 1 t R ...(i)
t R
2
Finally when rods are joined end to end as shown
Q mL (100  0)
  t   t  q2 L  2R ...(ii)
 2
q1 4
From equation (i) and (ii), q  1
2
20. A solid cube and a solid sphere of the same material have equal surface area. Both are at the same
temperature 120 o C , then
(a) Both the cube and the sphere cool down at the same rate
(b) The cube cools down faster than the sphere
(c) The sphere cools down faster than the cube
(d) Whichever is having more mass will cool down faster
SOLUTION:
 A  (T 4  T04 )
Rate of cooling of a body R  
t mc
A Area
 R 
m Volume
1
 For the same surface area. R 
Volume
 Volume of cube < Volume of sphere
 RCube  RSphere
i.e.cube,coolsdown withfasterrate.
21. Three rods of same dimensions are arranged as shown in figure they have thermal conductivities K 1 , K 2
and K 3 The points P and Q are maintained at different temperatures for the heat to flow at the same rate
along PRQ and PQ then which of the following option is correct
R
K1 K2
P K3 Q
1 K1 K 2
(a) K 3 
2
(K 1  K 2 ) (b) K 3  K1  K 2 (c) K 3 
K1  K 2 (d) K 3  2(K 1  K 2 )
SOLUTION:
2 K1 K 2
K
K1  K 2
H1 H1 K1 K2
l l
K3
H H2 H
The given arrangement of rods can be redrawn as follows

It is given that H1 = H2
KA(1   2 ) K 3 A(1   2 )
 
2l l
K K1 K 2
 K3  2  K  K
1 2
22. Four identical rods of same material are joined end to end to form a square. If the temperature
difference between the ends of a diagonal is 100 o C , then the temperature difference between the ends
of other diagonal will be
100 o 100 o
(a) 0 o C (b) l
C ; where l is the length of each rod (c) 2l
C (d) 100 o C
SOLUTION:
C
H/2 H/2
A B
H H
H/2 H/2
D
Suppose temperature difference between A and B is 100°C and A > B

Heat current will flow from A to B via path ACB and ADB. Since all the rod are identical so ()AC =()AD

(Because heat current H  ; here R = same for all.)
R
  A C   A  D
 C   D
i.e. temperature difference between C and D will be zero.
23. A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1gm of
ice per second. If the rod is replaced by another with half the length and double the radius of the first
and if the thermal conductivity of material of second rod is that of first, the rate at which ice melts in will
be ‘
(a) 3.2 (b) 1.6’ (c) 0.2 ‘ (d) 0.1
SOLUTION:
Q KA  mL K(r 2 )
  
t l t l
 m  Kr 2
 Rate of melting of ice   
 t  l
1
Since for second rod K becomes th r becomes double and length becomes half, so rate of melting will be twice
4
m  m 
i.e.  t   2  t   2  0 .1  0 .2 gm / sec .
 2  1
24. The temperature of the two outer surfaces of a composite slab, consisting of two materials having
coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2 >
 A(T2  T1 )K 
T1). The rate of heat transfer through the slab, in a steady state is  f , with ¦ which equal to
x 
T2 K 2K T1
x 4x
1 2 1
(a) 1 (b) 2
(c) 3 (d) 3
SOLUTION:
l1  l2 x  4x 5
Keq    K
l1 l2 x 4x 3 .
Equation of thermal conductivity of the given combination  
K1 K 2 K 2K
5
Q Keq . A(T2  T1 ) 3 K A (T2  T1 )
Hence rate of flow of heat through the given combination is  
t (x  4 x ) 5x
1
K A (T2  T1 )
=3
x
1
On comparing it with given equation we get f 
3
25. The figure shows a system of two concentric spheres of radii r1 and r2 and kept at temperatures T1 and T2,
respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional
to
r1
T1
r2 T2
r1 r2  r2 
(a) (r (b) (r2  r1 ) ’ (c) (r2  r1 )(r1 r2 ) (d)In  r 

1  r2 )  1
SOLUTION:
dr
r
r1
r2
Consider a concentric spherical shell of radius r and thickness dr as shown in fig.
The radial rate of flow of heat through this shell in steady state will be
dQ dT dT
H   KA   K (4 r 2 )
dt dr dr
4 K T1
r2 dr
 
r1 r 2

H T1
dT 
Which on integration and simplification gives
dQ 4Kr1r2 (T1  T2 )
H 
dt r2  r1
dQ r1r2
 dt  (r  r )
2 1
26. Ice starts forming in lake with water at 0 o C and when the atmospheric temperature is  10 o C . If the time

taken for 1 cm of ice be 7 hours, then the time taken for the thickness of ice to change from 1 cm to 2
cm is
(a) 7 hours (b) 14 hours (c)Less than 7 hours (d)More than 7 hours
SOLUTION:
L
t (x 22  x12 )
2 K
 t  (x 22  x 12 )
t (x 22  x 12 ) 9 (1 2  0 2 )
 t'  2 2  
( x '2  x '1 ) t ' (2 2  12 )
 t'  21 hours
27. There is formation of layer of snow x cm thick on water, when the temperature of air is   o C (less than

freezing point). The thickness of layer increases from x to y in the time t , then the value of is given by
(x  y )( x  y )L (x  y )L (x  y )( x  y )L (x  y )Lk
(a) 2k 
(b) 2k 
(c) k
(d) 2
SOLUTION:
L 2
Since t  ( x 2  x 12 )
2k 
L 2 2 L( x  y )(x  y )
 t  2k  ( x  y )  2 K
28. A 5cm thick ice block is there on the surface of water in a lake. The temperature of air is –10°C; how much
time it will take to double the thickness of the block(L = 80 cal/g, Kicc = 0.004 Erg/s-k, dice = 0.92 g cm–3)
(a) 1 hour (b) 191 hours (c) 19.1 hours (d) 1.91 hours

SOLUTION:
Ql mLl VLl
t  
KA(1   2 ) KA(1   2 ) KA(1   2 )
5  10
5  A  0 . 92  80 
 2  19 .1 hours.
0 . 004  A  10  3600
29. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of
the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength
 B corresponding to maximum spectral radiancy in the radiation from is shifted from the wavelength
corresponding to maximum spectral radiancy in the radiation from , by . If the temperature of is
(a)The temperature of is (b)  B  1 .5 m
(c)The temperature of is (d)The temperature of is
SOLUTION:
(a,b) According to Stefan’s law
E  eA T 4  E1  e1 A T14
E2  e 2 A T24
 E1  E 2
4 4
 e 1 T1  e 2 T2
1 
 e1 4  4  1 4
 T2   T1     (5802 )4 
 e 2   81 
And, from Wein’s law  A  T A   B  TB
 A TB
 
B TA
 B   A T A  TB
 
B TA
1 5802  1934 3968

      B  1 .5 m
B 5802 5802
30. A black body is at a temperature of . The energy of radiation emitted by this object with wavelength
between and is , between and is and between and is . The Wein’s constant . Then
(a) U1  0 (b) U3  0 (c) U1  U 2 (d) U 2  U1
SOLUTION:’
E
U2
U1
U3
1499
1500
1000
499
500
900
 (nm)
. Wein’s displacement law is m T  b
b 2 . 88  10 6
 m    1000 nm .
T 2880
Energy distribution with wavelength will be as follows
From the graph it is clear that U2 > U1.

31. A black metal foil is warmed by radiation from a small sphere at temperature T and at a distance . It is
found that the power received by the foil is ‘P’. If both the temperature and the distance are doubled,
the power received by the foil will be
(a) 16P (b) 4P
(c) 2P (d) P
SOLUTION:
Energy received per second i.e., power P  (T 4  T04 )
 P  T 4 ( T0  T )
1
Also energy received per sec (p)
d2
(inverse square law)
4 2
T4 P1  T1   d 2 
 P 2       
d P2  T2   d 1 
2 2
P  T   2d  1
       P2  4 P.
P2  2 T   d  4
32. Two metallic spheres S 1 and S 2 are made of the same material and have identical surface finish. The

mass of is three times that of . Both the spheres are heated to the same high temperature and placed in
the same room having lower temperature but are thermally insulated from each other. The ratio of the
initial rate of cooling of to that of is
(a) 1 / 3 (b) (1 / 3)1 / 3 (c) 1 / 3 (d) 3 /1
SOLUTION:
 A   (T 4  T04 )
‘ Rate of cooling (R)  
t mc
A Area r2 1
 R   3 
m volume r r
1 1  4 3 1/3 
 Rate (R )  
r m1 / 3  m    3 r  r  m 
 
1/3 1/3
R1  m 2  1
    
R 2  m 1 
 3
33. Three discs A, B and C having radii 2m, 4m, and 6m respectively are coated with carbon black on their other

surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively.
The power radiated by them are Qa, Qb, and Qc respectively
(a) Qa is maximum (b) Qb is maximum (c) Qc is maximum (d) Qa = Qb = Qc
SOLUTION:
Radiated power P  A T 4  P  AT 4
1
From Wein’s law, m T  constant  T  
m
A r2
 P 4

(m ) (m )4
22 42 62
 Q A : Q B : QC  : :
(300 ) (400 ) (500 )4
4 4
 Q B will be maximum
34. The total energy radiated from a black body source is collected for one minute and is used to heat a
quantity of water. The temperature of water is found to increase form 20 o C to 20 .5 o C . If the absolute
temperature of the black body is doubled and the experiment is repeated with the same quantity of
water at , the temperature of water will be
(a) 21 o C (b) 22 o C (c) 24 o C (d) 28 o C
SOLUTION:
The total energy radiated from a black body per minute.
4
Q2  2 T 
QT 4     16
Q1  T 
 Q 2  16 Q1
If m be mass of water taken and S be its specific heat capacity,
then Q1  ms (20 .5  20)
Q 2  ms (  20 )
 C  Final temperature of water
Q2   20 16   20
 Q  0 .5     28 C
1 1 0 .5
35. A solid sphere and a hollow sphere of the same material and size are heated to the same temperature
and allowed to cool in the same surroundings. If the temperature difference between each sphere and its
surroundings is , then
(a) The hollow sphere will cool at a faster rate for all values of
(b) The solid sphere will cool at a faster rate for all values of
(c) Both spheres will cool at the same rate for all values of
(d) Both spheres will cool at the same rate only for small values of
SOLUTION:
 A  (T 4  T04 )
Rate of cooling 
t mc
As surface area, material and temperature difference are same, so rate of loss of heat is same in both the spheres.
Now in this case rate of cooling depends on mass.
 1
 Rate of cooling 
t m
 m solid  m hollow . Hence hollow sphere will cool fast.
36. A solid copper cube of edges is suspended in an evacuated enclosure. Its temperature is found to fall
from to in . Another solid copper cube of edges , with similar surface nature, is suspended in a similar
manner. The time required for this cube to cool from to will be approximately
(a) 25 s (b) 50 s (c) 200 s (d) 400 s
SOLUTION:
 A  (T 4  T04 )
Rate of cooling 
t mc
m
 t [  , t,  , (T 4  T04 ) are constant ]
A
m Volume a3
 t   2
A Area a
t1 a1
t  a  t  a
2 2
100 1

 t2 2  t2  200 sec .
37. A body initially at 80o C cools to 64o C in 5 minutes and to 52o C in 10 minutes. The temperature of the

body after 15 minutes will be
(a) 42.7 o C (b) 35 o C (c) 47 o C (d) 40 o C
SOLUTION:
5 min
80°C 64°C
1
10 min
52°C
2
15 min
=?
3
According to Newton law of cooling

1   2    2 
K 1  0 
t  2 
(80  64 )  80  64 
For first process :  K   0  ...(i)’
5  2 
(80  52 )  80  52 
For second process :  K  0  ...(ii)
10  2 
(80   )  80   
For third process :  K  0  …(iii)
15  2 
1
On solving equation (i) and (ii) we get K  and  0  24 C .
15
Putting these values in equation (iii) we get   42 .7 C
38. A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1gm of
ice per second. If the rod is replaced by another with half the length and double the radius of the first
1
and if the thermal conductivity of material of second rod is 4 that of first, the rate at which ice melts in
gm / sec will be
(a) 3.2 (b) 1.6 (c) 0.2 (d) 0.1

SOLUTION:
Q KA  mL K(r 2 )
  
t l t l
 m  Kr 2
 Rate of melting of ice   
 t  l
1
Since for second rod K becomes th r becomes double and length becomes half, so rate of melting will be twice
4
m  m 
i.e.  t   2  t   2  0 .1  0 .2 gm / sec .’
 2  1
39. One end of a copper rod of length 1 .0 m and area of crosssection 10 3 is immersed in boiling water

and the other end in ice. If the coefficient of thermal conductivity of copper is 92 cal / ms o C and the
latent heat of ice is 8  10 4 cal / kg , then the amount of ice which will melt in one minute is
(a) 9 .2  10 3 kg (b) 8  10 3 kg (c) 6 .9  10 3 kg (d) 5 .4  10 3 kg
SOLUTION:
Heat transferred in one minute is utilised in melting the ice
KA ( 1   2 )t
so, mL
l
10 3  92  (100  0)  60
m   6 .9  10  3 kg
1  8  10 4
40. An ice box used for keeping eatable cold has a total wall area of 1 metre 2 and a wall thickness of 5 .0 cm .

The thermal conductivity of the ice box is K  0 .01 joule / metre  o C . It is filled with ice at 0 o C along with
eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is 334  10 3 joules / kg .
The amount of ice melted in one day is( 1day  86 ,400 sec onds )
(a) 776 gms (b) 7760 gms (c) 11520 gms (d) 1552 gms
SOLUTION:
dQ KA 0 .01  1
 d   30 = 6J/sec
dt l 0 .05
Heat transferred in on day (86400 sec)
  6  86400  518400 J
Now Q  mL
Q 518400
 m 
L 334  10 3
= 1.552 kg = 1552g.
41. A solid copper sphere (density and specific heat capacity c) of radius r at an initial temperature 200K is
suspended inside a chamber whose walls are at almost 0K. The time required (in s) for the temperature
of the sphere to drop to 100 K is
72 rc 7 rc 27 rc 7 rc

(a) 7 
(b) 72 
(c) 7 
(d) 27 
SOLUTION:
dT  A
 (T 4  T04 )
dt mcJ
In the given problem fall in temperature of body dT  (200  100 )  100 K ,
temp. of surrounding T0 = 0K,
Initial temperature of body T  200 K ].
100  4r 2
 (200 4  0 4 )
dt 4 3
r  c J
3
r c J r c 4. 2
 dt   10 6 s  .  10  6
48   48
7 r c 7 r c
  s ~– s [As J = 4.2]
80  72 
42. A sphere and a cube of same material and same volume are heated upto same temperature and allowed to
cool in the same surroundings. The ratio of the amounts of radiations emitted will be
1/3 2/3
4   1  4 
(a)1 : 1 (b) :1 (c)  6  :1 (d) 2   :1
3    3 
SOLUTION:
Q =  A t (T4 – T04)
If T, T0,  and t are same for both bodies
Qsphere Asphere 4 r 2
then   …..(i)
Qcube Acube 6a2
But according to problem, volume of sphere = Volume of cube
4
 r 3  a 3
3
1/3
4 
 a   r
3 
Substituting the value of a in equation (i) we get
Qsphere 4 r 2 4 r 2
 2
 2
Qcube 6a  4 1 / 3 
6    r 
 3  
1/3
4 r 2  
 2/3
  :1
4  2 6
6   r
3 
43. The graph. Shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and
y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between
the emissivity (e) and absorptivity (a) of the two bodies
T
y
x
(a) e x  e y & a x  ay (b) e x  e y & a x  ay (c) e x  e y & a x  ay (d) e x  e y & a x  ay
SOLUTION:
 dT 
Rate of cooling   dt   emissivity (e)
 
 dT   dT 
From graph,   dt     dt 
 x  y
 e x  ey
Further emissivity (e)  Absorptive power (a)
 a x  ay
( good absorbers are good emitters).
44. The plots of intensity versus wavelength for three black bodies at temperatures T1, T2 and T3 respectively are
as shown. Their temperature are such that
T3
I T2
T1
(a)T1 >T2 > T3 (b)T1 >T3 > T2 (c) T2 >T3 > T1 (d) T3 >T2 > T1

SOLUTION:
1
According to Wien’s law m 
T
from the figure (m )1  (m )3  (m )2
therefore T1 > T3 > T2.
45. The adjoining diagram shows the spectral energy density distribution E  of a black body at two different

temperatures. If the areas under the curves are in the ratio 16 : 1, the value of temperature T is
TK
E
2000 K

(a)32,000 K (b)16,000 K (c)8,000 K (d)4,000 K
SOLUTION:
AT 16
 (given)
A2000 1
Area under e    curve represents the emissive power of body

emissive power  T 4
(Hence area under curve)
4
AT  T 
  
A2000  2000 
4
16  T 
  
1  2000 
 T  4000 K .
46. A body cools in a surrounding which is at a constant temperature of  0 . Assume that it obeys Newton’ss

law of cooling. Its temperature  is plotted against time t. Tangents are drawn to the curve at the points
P(   1 ) and Q(   2 ) . These tangents meet the time axis at angles of  2 and 1 , as shown

2 P
1 Q
0 2 1
t
tan  2 1   0 tan  2  2  0 tan 1 1 tan 1 2

(a) tan   (b) tan   (c) tan   (d) tan  
1  2  0 1 1   0 2 2 2 1
SOLUTION:
d
For -t plot, rate of cooling  slope of the curve.
dt
d
At P,  tan  2  k ( 2   0 ),
dt
where k = constant.
d
At Q  tan 1  k (1   0 )
dt
tan  2  2   0
 
tan 1 1   0
47. Five identical rods are joined as shown in figure. Point A and C are maintained at temperature 120°C and
20°C respectively. The temperature of junction B will be
120°C
C
A B
20°C
(a)100°C (b)80°C (c)70°C (d) 0°C

SOLUTION:
If thermal resistance of each rod is considered R then, the given combination can be redrawn as follows
R 2R
R
120°C R
C
A B
20°C
R R
2R
R  R
A C
120°C B 20°C
(Heat current)AC = (Heat current)AB
(120  20 ) (120   )
    70 C
R R
48. On a clear sunny day, an object at temperature T is placed on the top of a high mountain. An identical object
at the same temperature is placed at the foot of mountain. If both the objects are exposed to sunrays for two
hours in an identical manner, the object at the top of the mountain will register a temperature
(a) Higher than the object at the foot (b)Lower than the object at the foot
(c) Equal to the object at the foot (d)None of the above
SOLUTION:
According to Newton’s law of cooling
1   2    2 
 K 1  0 
t  2 
(60  50 )  60  50 
In thefirstcase, K  0 
10  2 
1  K (55   )….(i)
(50  42 )  50  42 
In the second case, K  0 
10  2 
0 . 8  K (46   0 ) ….(ii)
1 55   0
Dividing (i) by (ii), we get 0 . 8  46  
0
or 46   0  44  0 . 8 0
  0  10 o C
THEORY BITS
1. Temperature of gas is a measure of
1) The average translational kinetic energy of the gas molecules
2) The average potential energy of the gas molecules
3) The average distance of the gas molecules
4) The size of the molecules of the gas
KEY: 1
2. The coefficient of linear expansion for a cetain metal varies with temperature as  T  . If L0 is the
intial length of the metal and the temperature of the metal is changed from T0 to T (T0>T).Then
T T
1) L  L0 T0  T dt 2) L  L0 1  T0  T dT 
T
3) L  L0  L  T0  T dT  4) L  L0
KEY:2
3. On the Celsius scale the absolute zero of temperature is at
1) 00 C 2) 320 C 3) 1000C 4)273.150C
KEY:3
4. The correct value of 00 C on the Kelvin scale is
1) 273.150 C 2) 273.160 C 3) 2730 C 4) 273.20 C
KEY:1
5. Which of the following is the largest rise in temperature?
1) 1o F 2) 1o R 3) 1K 4) 1o C
KEY:2
6. Melting and Boiling point of water on Fahrenheit scale of temperature respectively
1) 2120 F ,320 F 2) 320 F , 2120 F 3) 00 F ,1000 F 4) 320 F ,1320 F
KEY:2
7. The substance which has negative coefficient of linear expansion is
1) Lead 2) Aluminium 3) Iron 4) Invar steel
KEY:1
8. Mercury boils at 3560 C . However, mercury thermometers are made such that they can measuree
temperatures upto 5000 C . This is done by
1) maintaining vacuum above the mercury column
2) filling Nitrogen gas at high pressure above the mercury column
3) filling Nitrogen gas at low pressure above the mercury column
4) filling oxygen gas at high pressure above the mercury column
KEY:2
9. For measuring temperature near absolute zero, the thermometer used is
1) thermoelectric thermometer 2) radiation thermometer
3) magnetic thermometer 4) resistance thermometer
KEY:3
10. Which of the following is the smallest rise in temperature?
1) 10 F 2) 10 R 3) 1K 4) 10 C
KEY:1
11. For measurements of very high temperature say around 50000 C (of sun), one can use:
1) Gas thermometer 2) Platinum resistance thermometer
3) Vapour pressure thermometer 4) Pyrometer (Radiation thermometer)
KEY:4
12. The temperature at which two bodies appear equally hot or cold when touched by a person is
1) 00 C 2) 37 0 C 3) 250 C 4) 4 0 C
KEY:2
13. Celsius is the Unit of
1) Temperature 2) Heat 3) Specific heat 4) Latent heat
KEY:1
14. The range of clinical thermometer is
1) 370 C to 420 C 2) 950 F to 1100 F 3) 900 F to 1120 F 4) 950 C to 1040 C
KEY:2
15. Solids expand on heating because
1) the K.E. of the atoms increases 2) the P.E. of the atoms increases
3) total energy of the atoms increases 4) the K.E. of the atoms decrease
KEY:1
16. A ring shaped piece of a metal is heated, If the material expands, the hole will
1) Contract 2) Expand 3) Remain same
4) Expand or Contract depending on the width
KEY:2
17. Expansion during heating
1) occurs only in solids 2) decreases the density of the material
3) occurs at same rate for all liquids and gases 4) increases the weight of the material
KEY:2
18. When a metal bar is heated, the increase in length is greater, if
1) the bar has large diameter 2) the bar is long
3) the temperature rise small 4) small diameter
KEY: 2
19. Which of the following scales of temperature has only positive degrees of temperature?
1) Centigrade 2) Fahrenheit scale 3) Reaumur scale 4) Kelvin scale
KEY:4
20. A solid ball of metal has a spherical cavity inside it. The ball is cooled.The Volume of the cavity
will
1) decrease 2) increase 3) remain same 4) have its shape changed
KEY:1
21. The standard scale of temperature is
1) The mercury scale 2) The gas scale 3) The platinum resistance scale 4) liquid scale
KEY: 2
22. Two spheres of same size are made of same material but one is hollow and the other is solid. They
are heated to same temperature, then
1) Both spheres will expand equally. 2) Hollow sphere will expand more than solid one.
3) Solid sphere will expand more than hollow one.4) Hollow sphere will expand double that of solid one
KEY:1
23. When a metal bar is cooled, then which one of these statements is correct.
1) Length, density and mass remain same
2) Length decreases, density increases but mass remains same
3) Length and mass decrease but density remains the same
4) Length and density decrease but mass remains the same
KEY:2
24. If temperature of two spheres of same size but made of different materials changes by T then
1) Both expands equally
2) Sphere with greater  expands or contracts more than other..
3) Sphere with greater  expands or contracts less than other..
4) Both contracts equally.
KEY:2
25. The coefficient of linear expansion of a solid depends upon
1) the unit of pressure 2) the nature of the material only
3) the nature of the material and temperature 4) unit of mass
KEY:2
26. When hot water is poured on a glass plate, it breaks because of
1) unequal expansion of glass 2) equal contraction of glass
3) unequal contraction of glass 4) glass is delicate
KEY:1
27. If  c and  k denote the numerical values of coefficient of linear expansions of the solid, expressed
per 0C and per Kelvin respectively, then.
1)  c >  k 2)  c <  k 3)  c =  k 4)  c = 2  k
KEY:3
28. If  c and  f denote the numerical values of coefficient of linear expansion of a solid, expressed per
0
C and per 0F respectively, then
1)  c >  f 2)  f >  c 3)  f =  c 4)  f +  c = 0
KEY:1
29. The coefficient of linear expansion of a metal rod is 12x10-6 / 0C, its value in per 0C, its value in per
0
F
20 6 0 15 6 0
1)  10 / F 2)  10 / F 3) 21.6  10 6 / 0 F 4) 12  10 6 / 0 F
3 4
KEY:1
30. A brass disc fits into a hole in an iron plate. To remove the disc.
1) the system must be cooled 2) the system must be heated
3) the plate may be heated (or) cooled 4) the disc must be heated
KEY:1
31. The coefficient of volume expansion is
1) equal to the coefficient of linear expansion.
2) Twice the coefficient of linear expansion
3)Equal to the sum of coefficients of linear and superficial expansions.
4) Twice the coefficient of areal expansion.
KEY:3
32. Two metal strips that constitute a bimetallic strip must necessarily differ in their.
1) length 2) mass
3) coefficient of linear expansion 4) resistivity
KEY:3
33. Thermostat is based on the principle of
1) equal expansion of two rods of different lengths.
2) Different expansion of two rods of different
lengths.
3) Different expansion of two rods of same length
4) Equal expansion of two rods of same length.
KEY:3
34. A pendulum clock shows correct time at 00 C. At a higher temperature the clock.
1) loses time 2) gains time
3) neither loses nor gains time 4) will not operate
KEY:1
35. To keep the correct time modern day watches are fitted with balance wheel made of
1) Steel 2) Platinum 3) Invar 4) tungsten
KEY:3
36. When the temperature of a body increases

1) density and moment of inertia increase
2) density and moment of inertia decrease
3) density decreases and moment of inertia increases.
4) density increases and moment of inertia
decreases.
KEY:3
37. The coefficient of linear expansion of crystal in one direction is 1 and that in every direction
perpendicular to it  2 . The coefficient of cubical expansion is
1) 1   2 2) 21   2 3) 1  2 2 4) None
KEY:3
38. In balance wheel of watch, the factors that make its oscillations uniform are
1) tension in string 2) moment of inertia of balance wheel
3) temperature 4) pressure
KEY:2
39. Always platinum is fused into glass, because
1) platinum is good conductor of heat 2) melting point of platinum is very high
3) they have equal specific heats 4) their coefficients of linear expansion are equal
KEY:4
40. A cube of ice is placed on a bimetallic strip at room temperature as shown in the figure. What will
happen if the upper strip of iron and the lower strip is of copper?
ice
Fe
Cu
1) Ice moves downward 2) Ice moves upward 3) Ice remains in rest 4) None of the above
KEY:1
41. To withstand the shapes of concave mirrors against temperature variations used in high resolution
telescope, they are made of
1) Quartz 2) Flint glass 3) Crown glass 4 ) Combination of Flint and Silica
KEY:1
42. The holes through which the fish plates are fitted to join the rails are oval in shape because
1) Bolts are in oval shape
2) To allow the movement of rails in the direction
of length due to change in temperature.
3) To make the fitting easy and tight 4) Only oval shape holes are possible
KEY:2
43. The diameter of a metal ring is D and the coefficient of linear expansion is  . If the temperature
of the ring is increased by 10 C, the circumference and the area of the ring will increase by
 D  D 2
1)  D , 2 D 2) 2 D ,  D 
2 3)  D , 4)  D ,
2 2
KEY:4
44. When a metal ring is heated
1) The inner radius decreases and outer radius
increases
2) The outer radius decreases and inner radius increases
3) Both inner and outer radii increase 4) Both inner and outer radii decrease
KEY:3
45. The linear expansion of a solid depends on
1) Its original mass
2) Nature of the material and temperature
difference.
3) The nature of the material only
4) pressures
KEY:2
46. The moment of inertia of a uniform thin rod about its perpendicular bisector is I. If the temperature
of the rod is increased by  t, the moment of inertia about perpendicular bisector increases by
(coefficient of linear expansion of material of the rod is  ).
1) Zero 2) I  t 3) 2 I  t 4) 3 I  t
KEY:3
47. A bimetal made of copper and iron strips welded together is straight at room temperature. It is
held vertically so that the iron strip is towards the left hand and copper strip is towards right hand.
The bimetal strip is then heated. The bimetal strip will
1) remain straight 2) bend towards right 3) bend towards left 4) have no change
KEY:3
48. If L1 and L2 are the lengths of two rods of coefficients of linear expansion  1 and  2 respectively the
condition for the difference in lengths to be constant at all temperatures is
1) L1  1 = L2  2 2) L1  2 = L2  1 3) L1  12 = L2  22 4) L1  22 = L2  12
KEY:1
49. A semicircular metal ring subtends an angle of 1800 at the center of the circle. When it is heated,
this angle
1) remains constant 2) increases slightly 3) decreases slightly 4) becomes 3600
KEY:1
50. The coefficients of linear expansion of P and Q are 1 and  2 respectively. If the coefficient of
cubical expansion of ‘Q’ is three times the coefficient of superficial expansion of P, then which of the
following is true ?
1)  2 = 2 1 2) 1 = 2  2 3)  2 = 3 1 4) 1 = 3  2
KEY:1
51. When a rod is heated, its linear expansion depends on
a) initial length b) area of cross section c) mass d) temperature rise
1) only a is correct 2) a & d are correct 3) b & c are correct 4) a & c are correct
KEY:2
52. PQR is a right angled triangle made of brass rod bent as shown. If it is heated to a high temperature
the angle PQR.
1) increases
2) decreases
3) remains same
4) becomes 1350
KEY:3
53. A brass scale gives correct length at 00 C . If the temperature be 250 C and the length read by the
scale is 10 cm. Then the actual length will be
1) more than 10 cm 2) less than 10 cm 3) equal to 10 cm 4) we can not say
KEY:1
54. The coefficient of volume expansion is
1) twice the coefficient of linear expansion. 2) twice the coefficient of real expansion.
3) thrice the coefficient of real expansion. 4) thrice the coefficient of linear expansion
KEY:4
55. For a constant volume gas thermometer, one should fill the gas at
1) Low temperature and low pressure 2) Low temperature and high pressure
3) High temperature and low pressure 4) High temperature and high pressure
KEY:3
56. When a metal sphere is heated maximum percentage increase occurs in its
1) Density 2) Surface area 3) Radius 4) Volume
KEY:4
57. When a copper ball is cooled the largest percentage increase will occur in its
1) diameter 2) area 3) volume 4) density
KEY:4
58. A solid sphere and a hollow sphere of same material have same mass. When they are heated by 500C,
increase in volume of solid sphere is 5 c.c. The expansion of hollow sphere is
1) 5 c.c. 2) more than 5 c.c. 3) Less than 5 c.c. 4) None
KEY:2
59. The numerical value of coefficient of linear expansion is independent of units of
a) length b) temperature
c) area d) mass
1) only (a) is correct 2) (a) & (b) are correct
3) (a), (b) & (c) are correct 4) (a), (b) & (c) are correct
KEY:4
60. Expansion during heating
a) occurs in solids only b) causes decrease in weight
c) is due to increase of interatomic spacing
1) only (a) is wrong 2) (a), (b) & (c) are wrong
3) (a) & (b) are wrong 4) (a), (b) & (c) are correct
KEY:3
61. Due to thermal expansion with rise in temperature

a) Metallic scale reading becomes lesser than true value
b) Pendulum clock goes fast
c) A floating body sinks a little more
d) The weight of a body in a liquid increases
1) only (a) is correct 2) (a) & (b) are correct
3) (a), (b) & (d) are correct 4) (a), (c) & (d) are correct
KEY:4
62. Which of the following statements are true
a) Rubber contracts on heating
b) Water expands on freezing
c) Water contracts on heating from 00C to 40C
d) Water expands on heating from 40C to 400C
1) only (a) is correct 2) (b) & (c) are correct
3) (c) & (d) are correct 4) all are correct
KEY:4
63. The substance which contracts on heating is
1) Silica glass 2) Iron 3) Invar steel 4) Aluminum
KEY:1
64. When a metal ring having some gap is heated
a) length of gap increases
b) radius of the ring decreases
c) the angle subtended by the gap at the centre remains same
d) length of gap decreases
1) only (d) is correct 2) (a) & (b) are correct
3) (a) & (c) are correct 4) all are correct
KEY:3
65. At what temperature, does the celsius and fahrenheit scales show the same reading but with opposite
sign ?
1) 44 2) 20 3) 40 4) 10
KEY:3
66. Gas thermometers are more sensitive than liquid thermometers because
1) gases expand more than liquids 2) gases do not easily changed their state
3) gases are much lighter 4) gases are easy to obtain
KEY:1
67. When a copper solid sphere is heated, its
a) moment of inertia increases
b) Elasticity decreases
c) density decreases
d) mass increases
1) only (b) is true 2) (a) & (b) are true
3) (a), (b) & (c) are true 4) all are true
KEY:3
68. A platinum resistance thermometer is constructed which reads 00C at ice point and 1000C at steam
point. The resistance of platinum coil varies with t as Rt  R0 1   t   t 2  . If tp denotes temperature
on the platinum resistance thermometer and t denotes temperature on mercury scale. Then
resistance as the function of tp will be
1) R0 1   t p    (100)t p  2) R0 1   t p    (100)t p 
3) R0 1   t p   t p  4) R0   t p   t p 

KEY:1
69 Which of the following statements is correct for a thermometer ?
1) The bulb of the thermometer is made of a good conducting material
2) The bulb of the thermometer is made of a poor conducting material
3) Sole purpose of making the walls of the bulb thin is to proveide maximum possible space for liquid
4) None of these
KEY:1
70. Of the following thermometers the one which is most useful for the measurement of a rapidly
varying temperature is a
1) platinum resistance thermometer 2) gas thermometer
3) thermoelectric thermometer 4) saturation vapour pressure thermometer
KEY:3
71. Two thermometers are constructed in the same way except that one has a spherical bulb and the
other a cylindrical bulb; which one will respond quickly to temperature changes ?
1) Spherical bulb thermometer 2) Cylindrical bulb thermometer
3) Both equally 4) either
KEY:2
72. Which of the following statements is true for a thermometer ?
1) Coefficient of cubical expansion of liquid must be greater than that of bulb material
2) Coefficient of cubical expansion of liquid may be equal to that to bulb material
3) Coefficient of cubical expansion of liquid must be less than that of bulb material
4) None of the above
KEY:1
73. The coefficient of linear expansion of an in homogeneous rod changes linearly from 1 to  2 from
one end to the other end of the rod. The effective coefficient of linear expansion of the rod is
l11  l2 2
 eq 
l1  l2
1
1) 1   2 2)  1   2  3) 1 2 4) 1   2
2
KEY:2
74. By what temperature a rod should be heated so that its length becomes 2l0 of its initial length l0.
Assume that  (coefficient of linear expansion) in that range of temperature remains constant
and melting point is very high.
2 1 1n 2 1n 1
1) 2) 3) 4)
   
KEY:3
75. Column I gives some devices and Column II gives some processes on which the functioning of
these devices depend. Match the devices in Column I with the processes in Column II.
Column I Coumn II
A. Bimetallic strip p. Radiation from a hot
body
B. Steam engine q. Energy conversion
C. Incandescent lamp r. Melting
D. Electric fuse s. Thermal expansion of solids
KEY:As, Bq, Cp,q, Dq,r
76. List-I List-II
a) Isotorpic solids e) Expands on melting
b) Ice f) Equal expansion in all directions
c) Aniostropic solids g) Contracts on heating
d) Copper h) unequal expansion in different directions
KEY:af, bg, ch, de
77. List-I List-II
a) Thermal expansion e) Pendulum clock
b) , ,  f) Depends on dimension, material, temperature
c) Bimetallic strip g) Depends on nature of material only
d) Invar steel h) Balance wheel of a watch
KEY:af, bg, ch, de
78. List-I List-II
a) Bimetal thermostat e) Pendulum clock
b) Compensated f) Invar steel pendulum
c) Metal tape g) Differential expansion of metals
d) Alvinar h) Hair spring
KEY:ag, be, cf, dh
79. List-I List-II
a) Thermal stress e) 3t100
d
b) Loss in time of a f)    t
 2 1
pendulum clock per sec
c) Percentage increase g) Yt
in volume of a solid
d) Radius of circular arc h) (1/ 2)t
of bimetallic strip
KEY:ag, bh, ce, df
80. Cooking is difficult on mountains because
1) water boils at low temperature
2) water boils at high temperature
3) water does not boil
4) it is cool there
KEY:1
81. Heat capacity of a substance is infinite. It means
1) heat is given out 2) heat is taken in
3) no change in temperature whether heat is taken in (or) given out
4) all of the above
KEY:3
82. Heat required to raise the temperature of one gram of water through 10C is
1) 0.001 Kcal 2) 0.01 Kcal
3) 0.1 Kcal 4) 1.0 Kcal
KEY:1
83. A large block of ice is placed on a table when the surroundings are at 00C
1) ice melts at the sides 2) ice melts at the top
3) ice melts at the bottom 4) ice does not melt at all
KEY:3
84. In defining the specific heat, temperature is represented in 0F instead of 0C. Then the value of
specific heat will
1) decreases 2) increases
3) remain constant 4) be converted to heat capacity
KEY:1
85. Why the specific heat at a constant pressure is more than that at constant volume
1) there is greater inter molecular attraction at constant pressure
2) at constant pressure molecular oscillation are more violent
3) external work need to be done for allowing expansion of gas at constant pressure
4) due to more reasons other than those mentioned in the above
KEY:3
86. The ratio [Cp / Cv] of the specific heats at a constant pressure and at a constant volume of any
perfect gas
1) can’t be greater than 5/4 2) can’t be greater than 3/2
3) can’t be greater than 5/3 4) can have any value
KEY:3
87. During melting process, the heat given to a solid is used in (generally)
1) Increasing the temperature 2) Increasing the density of material
3) Increasing the average distance between the molecules 4) Increasing the average K.E. of the molecules
KEY:3
88. When two blocks of ice are pressed against each other then they stick together (coalesce) because
1) cooling is produced 2) heat is produced
3) increase in pressure, increase in melting point 4) increase in pressure, decrease in melting point
KEY:4
89. Ice is found to be slippery when a man walks on it This is so because
1) increase in pressure causes ice to melt faster
2) increase in pressure causes ice to melt slower
3) its surface is smooth and cold
4) ice is colder
KEY:1
90. A piece of ice at 00C is dropped into water at 00C. Then ice will
1) melt 2) be converted to water 3) not melt 4) partially melt
KEY:3
91. Paraffin wax expands on melting. The melting point of wax with increasing pressure is
1) increases 2) decreases 3) remains same 4)we can’t say
KEY:1
92. In a pressure cooker cooking is done quickly because
1) the cooker does not absorb any heat 2) it has a safety valve
3) boiling point of water rises due to increased pressure
4) it is a prestige to cook in a cooker
KEY:3
93. Which of the following at 1000C produces most severe burns ?
1) hot air 2) water 3) steam 4) oil
KEY:3
94. The latent heat of vaporisation of a substance is always
1) greater than its latent of fusion 2) greater than its latent heat of sublimation
3) equal to its latent heat of sublimation 4) less than its latent heat of fusion
KEY:1
95. The latent heat of vaporisation of water is more than latent heat of fusion of ice, why
1) on vaporisation much larger increase in volume takes place
2) increase in kinetic energy is much larger on boiling
3) kinetic energy decreases on boiling
4) volume decreases when the ice melts
KEY:1
96. The heat capacity of material depends upon
1) the structure of a matter 2) temperature of matter
3) density of matter 4) specific heat of matter
KEY:4
97. Which of the following states of matter have two specific heats ?
1) solid 2) gas 3) liquid 4) vapour
KEY:2
98. The specific heat of a gas in an isothermal process is
1) infinity 2) zero
3) negative 4) remains constant
PRACTICE BITS
1. Specific heat of aluminium is 0.25 cal/g-0c. The water equivalent of an aluminium vessel of mass
one kilogram is
1) 40 cal/0c 2) 250 g 3) 250 cal/0c 4) 40 g
KEY :2
SOLUTION :
water equivalent = mS gram.
2. A metal block absorbs 4500 cal of heat when heated from 300C to 800C. Its thermal capacity is
1) 90 gm 2) 90 cal / 0 C 3) 9 gm 4) 9 cal / 0 C
KEY :2
SOLUTION :
Q
H 

3. Two beakers A and B contain liquids of masses 300 g and 420 g respectively and specific heats
0.8cal/g-0C and 0.6cal/g-0C. The amount of heat on them is equal. If they are joined by a metal rod
1) heat flows from the beaker B to A 2) heat flows from A to B
3) no heat flows 4) heat flows neither from A to B nor B to A
KEY :2
SOLUTION :
Quantity of heat on A = Quantity of heat on B
m A  S A   1  m B  S B   2  1   2
4. The ratio of densities of two substances is 2:3 and that of specific heats is 1 : 2. The ratio of
thermal capacities per unit volume is
1) 1 : 2 2) 2 : 1 3) 1 : 3 4) 3 : 1
KEY :3
SOLUTION :
H1  1  S1 
H  mS  VS  H     S 
2  2  2 
5. Two spheres of copper of diameters 10cm and 20 cm will have thermal capacities in the ratio
1 1 1 1
1) 2) 3) 4)
8 2 4 6
KEY :1
SOLUTION :
3
4 H r 
H  mS    r 3 S  1   1 
3 H 2  r2 
6. Two liquids A and B of equal volumes have their specific heats in the ratio 2 : 3. If they have same
thermal capacity, then the ratio of their densities is
1) 1 : 1 2) 2 : 3 3) 3 : 2 4) 5 : 6
KEY :3
SOLUTION :
1 S2
m1S1 = m2S2  v11S1 = v2 2 S2 ;  = S
2 1
7. Three liquids A,B and C of masses 400gm, 600 gm and 800 gm are at 300c, 400c and 500c respectively.
When A and B are mixed resultant temperature is 360c when B and C are mixed resultant
temperature is 440c Then ratio of their specific heats are
1) 2:1:1 2) 3:2:1 3) 2:2:1 4) 1:4:9
KEY :3
SOLUTION :
When A, B are mixed
mA S A    A  mB SB    B ...... (i)
When B, C are mixed
m B S B     B  mC S C    C ...... (ii)
From (i) and (ii) we get relation between SA and SC.
When A and C are mixed
m AS A  A  m C SC  C
8. 50g of copper is heated to increase its temperature by 100C. If the same quantity of heat is given
to 10 g of water, the rise in its temperature is
(Scu= 420J/kg/0C and Sw= 4200J/kg/0C )
1) 50C 2) 60C 3) 70C 4) 80C
KEY :1
SOLUTION :
Q1  Q2  mSc 1  mSw 2
9. 1gm of ice at 00c is converted to steam at 1000c the amount of heat required will be (Lsteam=536 cal/
g)
1) 756 cal 2)12000 cal 3)716 cal 4)450 cal
KEY :3
SOLUTION :
Q = mLice+mSw(1000)+mLs
10. Boiling water at 1000C and cold water at t0C are mixed in the ratio 1:3 and the resultant maximum
temperature was 370C. Assuming no heat losses, the value of ‘t’ is
1) 40C 2) 90C 3) 120C 4) 160C
KEY :4
SOLUTION :
Heat lost by hot water = Heat gained by cold water.
 m1 1 
m1S1  1  m2 S 2 2  Given,  
 m2 3 
11. The fraction of ice that melts by mixing equal masses of ice at -10°C and water at 60°C is
1)6/11 2) 11/16 3) 5/16 4) 11/15
KEY :2
SOLUTION :
Here a part of ice is melted because heat given by water when it comes to 0ºC is less than the heat required
for ice to melt completely.
Let m| is the mass of the ice melted.
mSice 10   mLice  mwater Sw  60 
12. Power of a man who can chew 0.3 kg ice in one minute is ( in cal/s)
1) 400 2) 4 3) 24 4) 240
KEY :1
SOLUTION :
mL f
P
t
13. A beaker contains 200g of water. The heat capacity of the beaker is equal to that of 20g water.
The initial temperature of water in the beaker is 200c. If 440g of hot water at 920c is poured in it,
the final temperature (neglecting radiation loss) will be nearly
1) 580c 2) 680c 3) 730c 4) 780c
KEY :2
SOLUTION :
mwater  Sw    20   mS     20
 mhot water  Sw  92  
14. If 10g of the ice at 00c is mixed with 10g of water at 1000c, then the final temperature of the
mixture will be
1) 50c 2) 100c 3) 100 K 4) 00c
KEY :2
SOLUTION :
 m  L   m  S 1   m  S 2
15. The final temperature, when 10 g of steam at 1000 C is passed into an ice block of mass 100g
L steam  540 cal / g , Lice  80 cal / g ; S water  1 cal / g 0C  is
1) 21.80C 2)15.70C 3) 16.90C 4) 20.40C
KEY :1
SOLUTION :
Heat lost by steam = Heat gained by ice
msteam  Lv  msteamSw 100    miceLf  miceSw  0
0 0 0
16. The quantity of heat which can rise the temperature of x gm of a substance through t1°C can rise
the temperature of y gm of water through t2°C is same. The ratio of specific heats of the substances
is
1) yt1/xt2 2) xt2/yt1 3) yt2/xt1 4) xt1/yt2
KEY :3
SOLUTION : Q1  Q2  m1S11  m2 S2 2
17. Two liquids A and B are at 300c and 200c respectively. When they are mixed in equal masses the
temperature of the mixture is found to be 260c. The ratio of specific heats is
1) 4 : 3 2) 3 : 4 3) 2 : 3 4) 3 : 2
KEY :4
S A    B
SOLUTION : Heat lost by A = Heat gain by B mS A    A  mS B    B  
SB    A
18. M g of ice at 00c is mixed with M g of water at 100c. The final temperature is
1) 80c 2) 60c 3) 40c 4) 00c
KEY :4
SOLUTION :
 M  80    M 10   Final Temp. is 00 c
19. 10 grams of steam at 1000 C is mixed with 50 gm of ice at 00 C then final temperature is
1) 200 C 2) 500 C 3) 400 C 4) 1000 C
KEY :3
SOLUTION :
mice Lice  mice Sw   
msteam Lsteam  msteam S w 100   

20. The heat energy required to vapourise 5kg of water at 373 K is
1) 2700 K.cal 2) 1000 K.cal 3) 27 K.cal 4) 270 K.cal
KEY :1
SOLUTION : Q = m x Lsteam
21. Two liquids A and B are at temperatures of 750c and 1500c respectively. Their masses are in the
ratio of 2 : 3 and specific heats are in the ratio 3 : 4. The resultant temperature of the mixture,
when the above liquids, are mixed (Neglect the water equivalent of container ) is
1) 1250c 2) 1000c 3) 500c 4) 1500c
KEY :1
SOLUTION :
mASA (  75) = mBSB (150  )
22. 1g of ice at 00 C is mixed 1g of steam at 1000 C . The mass of water formed is
1) 1.33g 2) 13.3 g 3) 0.133 g 4) 13.3g
KEY :1
SOLUTION :
Here the resultant temperature is 1000 C
m| is mass of the steam condensed
m| Lv  mice L f  mice S water  
 water formed = 1g + m|
23. A liquid of mass ‘m’ and specific heat ‘S’ is at a temperature ‘2t’. If another liquid of thermal
capacity 1.5 times, at a temperature of t/3 is added to it, the resultant temperature will be
4 t 2
1) t 2) t 3) 4) t
3 2 3
KEY :2
SOLUTION :
m1S11  m2 S22
From principle of calorimetry  m1S1  m2 S2
 G iv e n , m 2 S 2  1 .5  m 1 S 1 
24. Three liquids with masses m1 , m2 , m3 are thoroughly mixed. If their specific heats are S1 , S 2 , S3 and
their temperatures 1 ,2 ,3 respectively, the temperature of the mixture is
S11  S22  S33 m1S11  m2 S22  m3S33
1) m S  m S  m S 2) m1S1  m2 S2  m3S3
1 1 2 2 3 3
m1S11  m2 S22  m3 S33 m11  m2 2  m3 3

3) m11  m22  m33 4) S   S   S 
1 1 2 2 3 3
KEY :2
SOLUTION :
Let 1   2  3 and  = resultant temperature.
From principle of colorimetry m3 S3 3     m1S1   1   m2 S2   2 

25. A piece of metal of mass 112g is heated to 1000 C and dropped into a copper calorimeter of mass
40g containing 200g of water at 160 C . Neglecting heat loss, the specific heat of the metal is
nearly, if the equilibrium temperature reached is 240 C
( Scu  0.1cal / g º C )
1) 0.292cal / gm 0 C 2) 0.392cal / gm 0 C 3) 0.192cal / gm 0 C 4) 0.492cal / gm 0 C
KEY :3
SOLUTION :
Heat lost by metal = heat gained by calorimeter and water
mmetal Smetal 100  24  
 mcu Scu  mwater Sw   24 16

PREVIOUS JEE MAINS QUESTIONS AND SOLUTOINS
1. Two different wires having lengths and , and respective temperature coefficient
oflinear expansion and ’ are joined end‐to‐end. Then the effective temperature
coefficient of linear expansion is : [Sep. 05, 2020 (II)]
(a) (b) 2√ (c) (d) 4 ( )

∘
Sol : (a) Let ′
and ′
be the lengths of the wire when temperature is changed by C.
At ∘ C,
= +
At + ∘ C
′
= ′ + ′
1+ = (1 + )+ (1 + )
↑
= (1 + )
⇒( ⊲ + ) 1+ = + + +
+
⇒ =
+
2. A bakelite beaker has volume capacity of500 cc at 30∘ C. When it is partially filled with
volume (at 30∘ C) of mercury, it is found that the unfilled volume ofthe beaker remains
∘ )
constant as temperature is varied. If ( ) = 6 × 10 C and i = 1.5 ×
10 ∘ C , Coefficient of , then (in cc) is close to where is the

[NA Sep. 03, 2020 (D]
Sol : Volume capacity ofbeaker, = 500 cc
= +
When beaker is partially filled with volume ofmercury,
= +
Unfilled volume ( − )=( − )
⇒ beaker =
=
.× ×
or, = × ⊲
= 20 cc.
3. When the temperature of a metal wire is increased from 0∘ C to 10∘ C, its length increased by
0.02%. The percentage change in its mass density will be closest to:[Sep. 02, 2020 (II)]
(a) 0.06 (b) 2.3 (c) 0.008 (d) 0.8
Sol : . (a) Change in length of the metal wire ( ) when its temperature is changed by is
given by
=
Here, = Coefficient oflinear expansion
Here, 1= 0.02%, = 10 C
0.02
= =
100 × 10
⇒ = 2 × 10
Volume coefficient of expansion, \[= 3 = 6 × 10
p=
× 100 = = (6 × 10 × 10 × 100) = 6 × 10
Volume increase by 0.06% therefore density decrease by 0.06%.
4. A non‐isotropic solid metal cube has coefficients of linear expansion as: 5 × 10 /∘ C along
the ‐axis and 5 × 10 /∘ C along the and the ‐axis. If the coefficient of volume expansion
of the solid is C × 10 /∘ C then the value of C is [NA7 Jan. 2020 I]
Sol : (60.00) Volume, = ℎ
ℓ ℎ
= = + +
ℓ ℎ
( = coefficient ofvolume expansion)
⇒ = 5 × 10 + 5 × 10⊲ + 5 × 10⊲
= 60 × 10⊲ /∘
Value of = 60.00
5. At 40 C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung
from the free end ofthe wire. When the wire is cooled down from 40 C to 20 C it regains
its original length of 0.2m. The value of M is close to: [12 Apri12019 I]
(Coefficient of linear expansion and Young’s modulus of brass are 10 /oC and 10 N/m ,
respectively; = 10ms )
(a) 9 kg (b) 0.5 kg (c) 1.5kg (d) 0.9kg
Sol : 5. (Bonus) temp = 1oad and = = (10 ) = × 10 ⊲
× .
or 0.2 × 10 × 20 = ( )×
×
= 20 .⋅. = = 2 = 6.28 kg
6. Two rods A and B of identical dimensions are at temperature 30∘ C. IfA is heated upto 180∘ C
and B upto T C, then the new lengths are the same. If the ratio of the coefficients of linear
expansion ofA and B is 4: 3, then the value ofT is: [11 Jan. 2019 II]
(a) 230∘ C (b) 270∘ C (c) 200∘ C (d) 250∘ C
Sol : (a) Change in length in both rods are same i. e.
ℓ = ℓ
ℓ =ℓ
4
= ⋅.⋅ =
3
4 − 30
=
3 180 − 30
7. Athermometer graduated according to alinear scale reads a value x when in contact with
boiling water, and x /3 when in contact with ice. What is the temperature of an object in oC,
ifthis thermometer in the contact with the object reads x /2? [11 Jan. 2019 II]
(a) 25 (b) 60 (c) 40 (d) 35
Sol: a) Let required temperature = T∘ C
M.P. B.P.
⇒ T ∘ C = −= x x x
236
x
& x − = (100 − 0∘ C)
3
2x 300
⇒ = 100 ⇒ x =
3 2
x 150
⇒ T∘C = = = 25∘ C
6 6
8. A rod, of length L at room temperature and uniform area of cross section A, is made of a
metal having coefficient of linear expansion /∘ C. It is observed that an external
compressive force , is applied on each of its ends, prevents any change in the length of the
rod, when its temperature rises by T K. Young’s modulus, Y, for this
metal is: [9 Jan. 2019 I]
(a) (b) ( )
(c) (d)
/
Sol : (a) Young’s modulus Y = = ( ℓ/ℓ)
Using, coefficient oflinear expansion,
ℓ ℓ
= ⇒ = T
ℓ T ℓ
Y=
A( T)
9. An external pressure P is applied on a cube at 0 C so that it is equally compressed from all

sides. K is the bulk modulus of the material of the cube and is its coefficient oflinear
expansion. Suppose we want to bring the cube to its original size by heating. The
temperature should be raised by: [2017]
(a) (b) 3 (c) (d)
Sol : (c) As we know, Bulk modulus

= ⇒ =
= (1 + )
= ⇒ = =
3
10. A steel rail of length 5 m and area of cross‐section 40 cm is prevented from expanding
along its length while the temperature rises by 10∘ C. If coefficient of linear expansion and
Young’s modulus ofsteel are 1.2 × 10 K and 2 × 10 Nm respectively, the force
developed in the rail is approximately: [Online April 9, 2017]
(a) 2 × 10 N (b) 1 × 10 N (c) 2 × 10 N (d) 3 × 10 N
/
Sol : (b) Young’s modulus = =
/
L
Y= Q =
. . L
Force developed in the rail = YA t

= 2 × 10 × 40 × 10 × 1.2 × 10 × 10
= 9.6 × 10 = 1 × 10 N
11. A compressive force, is applied at the two ends of a long thin steel rod. It is heated,
simultaneously, such that its temperature increases by T. The net change in its length is
zero. Let be the length ofthe rod, A its area of cross‐section, Y its Young’s modulus, and
its coefficient oflinear expansion. Then, is equal to:[Online April 8, 2017]
(a) Y T (b) AY T (c) AY T (d)
Sol: (c) Due to thermal exp., change in length ( ) = T (i)

Norma1stress
Young’s modulus (Y) =
Longitudina1strain
/A
Y= ⇒ =
/ AY
1=
AY
From eq (i), = T
= AY T
12. The ratio of the coefficient of volume expansion of a glass container to that of a viscous
liquid kept inside the container is 1 : 4. What fraction of the inner volume of the container
should the liquid occupy so that the volume of the remaining vacant space will be same at
all temperatures? [Online April 23, 2013]
(a) 2:5 (b) 1:4 (c) 1:64 (d) 1:8
Sol: (b) When there is no change in liquid level in vessel then ′rea1 = ′
vessei
Change in volume in liquid relative to vessel VP=V = V( )
13. On a linear temperature scale Y, water :eezes at − 160∘ Y and ) ilsat − 50∘ Y. On this Y
scale, a temperature of340 K wouldbe read as: (water fieezes at273 K andboils at 373 K)
(a) −73. 7∘ Y (b) −233. 7∘ Y (c) −86. 3∘ Y (d) −106. 3∘ Y
Sol: (c) = constant for all scales
340 − 273 ∘ Y − (−160)
=
373 − 273 −50 − (−160)
67 y + 160
⇒ =
100 110
Y = −86. 3∘ Y
14. A wooden wheel ofradius is made oftwo semicircular part (see figure). The two parts are
held together by a ring made ofa metal strip ofcross sectional area and length L. is
slightly less than 2 . To fit the ring on the wheel, it is heated so that its temperature rises
by and it just steps over the wheel. As it cools down to surrounding temperature, it
presses the semicircular parts together. If the coefficient of linear expansion of the metal is
, and its Young’s modulus is , the force that one part ofthe wheel applies on the other part
is: r2012]
(a) 2 (b) (c) (d) 2 ⋅

Sol: (d) The Young modulus is given as
stress /S
Y= =
strain L/L
Here, L = 2 RL = 2 R
Y= ×2 R
S2 R
⇒Y= … (i)
.
The coefficient oflinear expansion =
⇒ = . ⇒ = ... (ii)
From equation (i) and(ii)
= ⇒ = . .
.
The ring is pressing the wheel from both sides, Thus
net = 2 = 2Y
15. Three rods of identical cross‐section and lengths are made of three different materials of
thermal conductivity K , K and K , respectively. They arejoined together at their ends to
make a long rod(see figure). One end ofthe long rod is maintained at 100 C and the other at
0 C (see figure). If the joints of the rod are at 70 C and 20 C in steady state and there is no
loss of energy fi om the surface of the rod, the correct relationship between K , K and K is:
[Sep. 06, 2020 (II)]
(a) K : K = 2 : 3, K < K = 2 : 5 (b) K < K < K

(c) K : K = 5: 2, K : K = 3: 5 (d) K > K > K
Sol: (a) As the rods are identical, so theyhave same length (f) and area ofcross‐section (A) . They
are connected in series. So, heat current will be same for all rods.
Heat current = = =
(100 − 70) (70 − 20) (20 − 0)

⇒ = =
⇒ (100 − 70) = (70 − 20) = (20 − 0)

⇒ (30) = (50) = (20)
⇒ = =
10 6 15
⇒ : : = 10: 6: 15
⇒ : = 2: 3.
16. A bullet of mass 5 g, travelling with a speed of210 m/s, strikes a fixed wooden target. One
halfofits kinetics energy is converted into heat in the bullet while the other half is converted
into heat in the wood. The rise of temperature of the bullet if the specific heat of its material
is 0.030 cal/(g–°C) (1 cal = 4.2 × 10 ergs) close to: [Sep. 05, 2020 (D]
(a) 87. 5 C (b) 83. 3 (c) 119. 2 C (d) 38. 4 C
Sol: (a) According to question, one half ofits kinetic energy is converted into heat in the wood.
1 1
× =
2 2
210 × 210
⇒ = = = 87. 5∘ C
4× 4 × 4.2 × 0.3 × 1000
17. The specific heat ofwater = 4200Jkg K and the latent heat of ice
= 3.4 × 10 Jkg . 100 grams of ice at 0∘ C is placed in 200 g ofwater at 25∘ C. The amount
Of ice that will melt as the temperature ofwater reaches 0∘ C is close to (in grams):
[Sep. 04, 2020 (I)]
(a) 61.7 (b) 63.8 (c) 69.3 (d) 64.6
Sol: (a) Here ice melts due to water.
Let the amount ofice melts =
= ice 1ce
ice =
ice
0.2 × .4200 × 25
= = 0.0617kg = 61.7g
34 × 10
18. A calorimter ofwater equivalent 20 g contains 180 g of water at 25∘ C. ′ grams of steam at
100∘ C is mixed in it till the temperature ofthe mixture is 31∘ C. The value of’m’ is close to
(Latent heat of water = 540 cal g , specific heat of water = 1calg ∘ C )
[Sep. 03, 2020 (II)]
(a) 2 (b) 4 (c) 32 (d) 2.6
Sol: (a) Heat given by water = ( − )
= 200 × 1 × (31 − 25)
Heat taken by steam = + ( − )
= m × 540 + m(1) × (100 − 31)
= m × 540 + m(1) × (69)
From the principal ofcalorimeter,
(200)(31 − 25) = × 540 + (1)(69)
⇒ 1200 = (609) ⇒ ≈ 2.
19. Three containers , and have water at different temperatures. The table below shows
the final temperature Twhen different amounts ofwater (given in liters) are taken : om each
container and mixed (assume no loss of heat during the process) [8 Jan. 2020 II]
C
1 2 60∘ C
1 2 30∘ C
2 1 60∘ C
1 1 1
The value of (inC to the nearest integer) is_______ .
Sol: (50.00)
Let 1
, 2
, 3
be the temperatures ofcontainer 1, 2 and C3 respectively.
Using principle ofcalorimetry in container 1, we have
( 1 − 60) = 2ms(60 − )
⇒ − 60 = 120 − 2
⇒ = 180 − 2 (i)
For container 2
ms ( 2 − 30) = 2ms(30 − )
⇒ = 90 − 2 3 (ii)
For container C3
2ms( − 60) = ms(60 − )
⇒2 − 120 = 60 −
⇒2 + = 180 (i )
Also, 1 + 2 + 3 = 3 (iv)
Adding (i), (ii) and(iii)
3 +3 +3 = 450
⇒ 1 + 2 + 3 = 150
⇒ 3 = 150 ⇒ = 50 C
20. M grams ofsteam at 100∘ C is mixed with 200 g ofice at its melting point in a thermally
insulated container. If it produces liquid water at 40∘ C [heat ofvaporization ofwater is 540
cal/g and heat offusion ofice is 80 cal/g], the value of M is_____ [NA7 Jan. 2020 II]
Sol: (40) Using the principal ofcalorimetry
ice + ice (40 − 0)
= + (100 − 40)
⇒ (540) + × 1 × (100 − 40)
= 200 × 80 + 200 × 1 × 40
⇒ 600 = 24000
⇒ = 40
21. When M gram ofice at − 10 C(Specific heat = 0.5 cal C ) is added to M gram of
water at 50 C, finally no ice is left and the water is at 0 C. The value oflatent heat of
ice, in cal is: [12 April 2019 I]
(a) −5 (b) − 50 (c) (d) −5

Sol: (a) 1 ice × (10) + 1 = 2 (50)
or 1 × C (= 0.5) × 10 + 1 = 2 × 1 × 50
50
⇒L= −5
22. A massless spring (K = 800N/m) , attached with a mass (500g) is completely immersed in
lkg of water. The spring is stretched by2cm and released so that it starts vibrating. What
would be the order of magnitude of the change in the temperature of water when the
vibrations stop completely? (Assume that the water container and spring receive negligible
heat and specific heat ofmass = 400 Jlkg K, specific heat ofwater = 4184J/kgK) [9 April
2019 II]
(a) 10⊲ K (b) 10 5K (c) 10 1K (d) 10 3K
Sol: (b) . = ( )+
or × 800 × 0.02 = 0.5 × 400 × + 1 × 4184 ×
= 1 × 10
3
23. Two materials having coefficients ofthermal conductivity ‘ 3 ’ and ‘K’ and thickness ‘d’
and ‘ 3 ’, respectively, are joined to form a slab as shown in the figure. The temperatures of
the outer surfaces are ’ and ’ respectively, ( > ) . The temperature at the interface
is: [9 April 2019 II]
(a) + (b) (c) + (d) +
Sol: a) 1 = 2 23
3
2− − 1
or (3 ) =
3
or =
24. A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius
2R. The thermal conductivity ofthe material ofthe inner cylinder is K and that of the outer
cylinder is K . Assuming no loss of heat, the effective thermal conductivity ofthe system for
heat flowing along the length ofthe cylinder is: [12 Jan. 2019 I]
(a) (b) K + K (c) (d)
Sol: (d) Effective thermal conductivity of system

=4 R =4
25. Ice at −20∘ C is added to 50 g ofwater at 40∘ C, When the temperature ofthe mixture reaches
0∘ C, it is found that 20 g ofice is still unmelted. The amount of ice added to the water was
close to [11 Jan. 2019 I] (Specific heat ofwater = 4.2J/g/∘ C, Specific heat ofIce = 2.1J ∘ C
Heat offusion ofwater at 0∘ C = 334J/g)
(a) 50g (b) 100 g (c) 60 g (d) 40 g
Sol: (d) Let m gram ofice is added.
From principal ofcalorimeter
heat gained (by ice) = heat lost (by water)
20 × 2.1 × m + (m − 20) × 334
= 50 × 4.2 × 40
376m = 8400 + 6680
m = 40.1
26. When 100 g of a liquid A at 100∘ C is added to 50 g of a liquid B at temperature 75∘ C, the
temperature ofthe mixture becomes 90∘ C. The temperature ofthe mixture, if100 g of liquid A
at 100∘ C is added to 50 g ofliquid B at 50∘ C, will be: [11 Jan. 2019 II]
(a) 85∘ C (b) 60∘ C (c) 80∘ C (d) 70∘ C
Sol: (c) Heat loss = Heat gain = mS
So, mA SA A = mB SB B
⇒ 100 × S × (100 − 90) = 50 × S × (90 − 75)
3
2S = 1.5S ⇒ S = S
4
Now, 100 × S × (100 − ) = 50 × S × ( − 50)
3
2× × (100 − ) = ( − 50)
4
300 − 3 = 2 − 100
400 = 5 ⇒ = 80∘ C
27. A metal ball of mass 0.1 kg is heated upto 500∘ C and dropped into a vessel of heat capacity
800 JK and containing 0.5 kg water. The initial temperature ofwater and vessel is 30∘ C.
What is the approximate percentage increment in the temperature ofthe water? [Specific
Heat Capacities ofwater and metal are, respectively, 4200 Jkg 1K − 1 and400 Jkg K ]
[11 Jan. 2019 II]
(a) 15% (b) 30% (c) 25% (d) 20%
Sol: (d) Assume final temperature = T ∘ C
Heat lass = Heat gain = ms T
⇒m s T =m s T
0.1 × 400 × (500 − T)
= 0.5 × 4200 × (T − 30) + 800 (T‐30)
⇒ 40(500 − T) = (T − 30) (2100+800)
⇒ 20000 − 40T = 2900T − 30 × 2900
⇒ 20000 + 30 × 2900 = T(2940)
T = 30. 4∘ C
T 6.4
× 100 = × 100 = 21%,
T 30
so the closest answer is 20%.
Temp. of Temp. of ] }roir
28. A heat source at T = 10 K is connected to another heat reservoir at T = 10 K by a copper

slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK m , the
energy flux through it in the steady state is: [10 Jan. 2019 I]
(a) 90 Wm 2 (b) 120Wm (c) 65 Wm 2 (d) 200Wm
Sol: (a)
\
dQ kA T
=
dt ℓ
Energy flux, =
ℓ
(0.1)(900)
= = 90W/m
1
29. An unknown metal of mass 192 g heated to a temperature of 100∘ C was immersed into a
brass calorimeter of mass 128 g containing 240 g ofwater at a temperature of8. 4∘ C.
Calculate the specific heat ofthe unknown metal ifwater temperature stablizes at 21. 5∘ C.
(Specific heat ofbrass is 394 Jkg K ) [10 Jan. 2019 II]
(a) 458 Jkg K (b) 1232 Jkg K (c) 916Jkg K (d) 654Jkg K
Sol: (c) Let specific heat ofunknown metal be s’ According to principle of calorimetry, Heat lost
= Heat gain m × s =m s ( +m s + )
⇒ 192 × S × (100 − 21.5)
= 128 × 394 × (21.5 − 8.4)
Solving we get, +240 × 4200 × (21.5 − 8.4)
S = 916Jkg k
∘
30. Temperature difference of 120 C is maintained between two ends ofa uniform rod AB
oflength 2L. Another bent rod PQ, of same cross‐section as AB and length , is connected
across AB (See figure). In steady state, temperature difference between P and Q will be
close to: [9 Jan. 2019 I]
2
∘ ∘
(a) 45 C (b) 75 C (c) 60∘ C (d) 35∘ C
×
Sol: 30. (a) = =
A R/2 P Q R/2 B
In steady state temperature difference between P and
Q,
120 × 5 3 360
T = × R= = 45∘ C
8R 5 8
31. A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of
mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the
temperature of the system is found to be 75∘ C. T is given by (Given: room temperature
= 30∘ C, specific heat of copper = 0.1ca1/gm C [2017]
(a) 1250∘ C (b)825∘ C (c) 800∘ C (d) 885∘ C
Sol: (d) According to principle ofcalorimetry,
Heat lost = Heat gain
100 × 0.1(T − 75) = 100 × 0.1 × 45 + 170 × 1 × 45
10 − 750 = 450 + 7650 = 8100
⇒ − 75 = 810
= 885∘
32. In an experiment a sphere ofaluminium ofmass 0.20 kg is heated upto 150∘ C. Immediately,
it is put into water of volume 150 cc at 27∘ C kept in a calorimeter of water equivalent to
0.025 kg. Final temperature ofthe system is 40∘ C. The specific heat of aluminium is:
(take 4.2 Joule = 1 calorie) [Online April 8, 2017]
(a) 378 J/kg −∘ c (b) 315 J/kg −∘ c
(c) 476 J/kg −∘ c (d) 434 J/kg −∘ c
Sol: (d) According to principle ofcalorimetry,
Qgiven = Qused
0.2 × S × (150 − 40) = 150 × 1 × (40 − 27) + 25 × (40 − 27)
0.2 × S × 110 = 150 × 13 + 25 × 13
Specific heat ofaluminium
13 × 25 × 7
S= = 434J/kg −∘ C
0.2 × 110
33. An experiment takes 10 minutes to raise the temperature of water in a container from 0 C to
100 C and another 55 minutes to convert it totally into steam by a heater supplying heat at
a uniform rate. Neglecting the specific heat of the container and taking specific heat of water
to be 1 cal/g C, the heat of vapourization according to this experiment will come out to be :
(a) 560 cal/g (b) 550 cal/g (c) 540 cal/g (d) 530 cal/g
Sol: (b) As =
So, × 10 × 60 = 100 … (i)
and × 55 × 60 = mL (ii) Dividing equation (i) by (ii) we get
10 × 100
=
55
= 550 cal. /g.
34. Three rods of Copper, Brass and Steel are welded together to form a Y shaped structure.
Area of cross‐ section of each rod = 4cm . End ofcopper rod is maintained at 100 C where
as ends ofbrass and steel are kept at 0 C. Lengths ofthe copper, brass and steel rods are 46,
13 and 12 cms respectively. The rods are thermally insulated from surroundings excepts at
ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units
respectively. Rate ofheat flow through copper rod is:[2014]
(a) 1.2cal/s (b) 2.4cal/s (c) 4.8cal/s (d) 6.0cal/s
Sol: (c) Rate ofheat flow is given by,
KA( − )
Q=
Where, K = coefficient ofthermal conductivity

= length ofrod and A = area ofcross‐section ofrod
s
C
Ifthe junction temperature is T, then
QCopper = Q +Q
0.92 × 4(100 − )
46
0.26 × 4 × ( − 0) 0.12 × 4 × ( − 0)
= +
13 12
⇒ 200 − 2T = 2T + T
⇒ T = 40∘ C
0.92 × 4 × 60
Coppe
= = 4.8cal/s
46
35. A black coloured solid sphere ofradius R and mass M is inside a cavitywith vacuum inside.
The walls of the cavity are maintained at temperature T . The initial temperature ofthe
sphere is 3T . Ifthe specific heat ofthe material of the sphere varies as T per unit mass
with the temperature T ofthe sphere, where is a constant, then the time taken for the
sphere to cool down to temperature 2T will be (0 is Stefan Boltzmann constant) [Online
Apri119, 2014]
(a) In (b) In (c) In (d) In
Sol: (c) In the given problem, fall in temperature ofsphere,

dT = (3T − 2T ) = T
Temperature ofsurrounding, Tsu = T0
Initial temperature ofsphere, Tinitia1 = 3T0
Specific heat ofthe material ofthe sphere varies as,
c = T per unit mass ( = a constant)
Applying formula,
dT A
= (T − T )
dt McJ
T 04 R
⇒ = [(3T ) − (T ) ]
dt M (3T ) J
M 27T J
⇒ dt =
04 R × 80T
Solving we get,
Time taken for the sphere to cool down temperature 2T ,
M 16
t= ln
16 R 3
36. Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is
heated, the container loses energy at a rate of 160 J/s. In how much time will the
temperature ofwater rise fi 27∘ C to 77∘ C? (Specific heat ofwater is 4.2kJ/kg and that
ofthe container is negligible). [Online April 9, 2014]
(a) 8 min 20s (b) 6 min 2s (c) 7 min (d) 14 min
Sol: (a) From question,
In 1 sec heat gained by water
= 1 KW‐160 J/s
= 1000J/s − 160J/s
= 840J/s
Total heat required to raise the temperature ofwater (volume 2L)fi om27∘ c to 77∘ c
=m × sp. ht ×
= 2 × 10 × 4.2 × 50 [mass =density × volume]
And, 840 × t = 2 × 10 × 4.2 × 50
× × . ×
or, t =
= 500s = 8 min 20s
37. Assume that a drop ofliquid evaporates by decrease in its surface energy, so that its
temperature remains unchanged.What should be the minimum radius of the drop for this to
be possible? The surface tension is T, density of liquid is p and L is its latent heat of
vaporization. [2013]
(a) pL/T (b) (c)T/pL (d) 2T/pL
Sol: (d) When radius is decrease by R,

4 R RpL = 4 T[R − (R − R) ]
⇒ pR RL = T[R − R + 2R R − R ]
⇒ pR RL = T2R R [ R is very small]
2
⇒ =
p
38. A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes
2 minutes to cool fi om30∘ C to 25∘ C. Amass 00g ofanother liquid in an identical vessel
with identical surroundings takes the same time to cool from 30∘ C to 25∘ C. The specific
heat of the liquid is: (The water equivalent ofthe vessel is 30g.) [Online April 25, 2013]
(a) 2.0kcal/kg (b) 7 kcal/kg (c) 3 kcal/kg (d) 0.5 kcal/kg
Sol: d) As the surrounding is identical, vessel is identical time taken to cool both water and
liquid(from 30∘ Cto25∘ C) is same 2 minutes, therefore
dQ
water = liquid
dt
39. 500 g ofwater and 100 g ofice at 0∘ C are in a calorimeter whose water equivalent is 40 g.
10g ofsteam at 100∘ C is added to it. Then water in the calorimeter is: (Latent heat of ice
= 80cal/g, Latent heat ofsteam = 540cal/g) [Online April 23, 2013]
(a) 580 g (b) 590 g (c) 600 g (d) 610 g
Sol: (b ) As lg ofsteam at 100∘ C melts 8g ofice at 0∘ C. 10 g of steam will melt 8 × 10g ofice at 0∘ C
Water in calorimeter = 500 + 80 + 10g = 590g
40. Given that 1 g ofwater in liquid phase has volume 1 cm and in vapour phase 1671 cm at
atmospheric pressure and the latent heat ofvaporization ofwater is 2256 J/g; the change in
the internal energy injoules for 1 g ofwater at 373 K when it changes fi om liquid phase to
vapour phase at the same temperature is : [Online Apri122, 2013]
(a) 2256 (b) 167 (c) 2089 (d) 1
Sol : c
41. A large cylindrical rod oflength is made byjoining two identical rods of copper and steel of
length each. The rods are completely insulated fi m the surroundings. If the fi ee end of
copper rod is maintained at 100∘ C and that of steel at 0∘ C then the temperature of junction is
(Thermal conductivity of copper is 9 times that of steel) Online May 19, 2012]
(a) 90∘ C (b) 50∘ C (c) 10 C (d) 67∘ C
Sol : (a)
100∘ ∘ c
← 2− 2→
Let conductivity of steel stee1 = then from question Conductivity of copper copper =9
copper = 100∘ C
steei = 0∘ C
stee1 = copper =
2
From formula temperature ofjunction;
+
=
+
9 × 100 × + ×0×
=
9 × + ×
10
= = 90∘ C
2
42. The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K is
(0 = 5.7 × 10 Wm K ) [Online May 7, 2012]
(a) 1.7 × 10 J (b) 1.1 × 10 J (c) 2.8 × 10 J (d) 4.6 × 10 J
Sol : (a) According to Stefan’s law
=
Heat radiated per unit area in 1 hour (3600s) is
= 5.7 × 10 × (300) × 3600 = 1.7 × 10 J
43. 100g ofwater is heated from 30∘ C to 50∘ C. Ignoring the slight expansion of the water, the
change in its internal energy is (specific heat ofwater is 4184 J/kg/K): [2011]
(a) 8.4kJ (b) 84 kJ (c) 2.1kJ (d) 4.2kJ
Sol : a) = =
100
= × 4184(50 − 30) ≈ 8.4kJ
1000
44. The specific heat capacity ofa metal at low temperature (T) is given as
( kg ) = 32
400
A 100 gram vessel of this metal is to be cooled from 20 K to 4 K by a special refrigerator
operating at room temperature (27∘ C) . The amount of work required to cool the vessel is
[2011 RS]
(a) greater than 0.148kJ (b) between 0.148 kJ and 0.028 kJ
(c) less than 0.028 kJ (d) equal to 0.002 kJ
Sol : (d) Required work = energy released
Here, =∫
3.2
= 0 . 1 × 32 × ( ) ( ) =
64 × 10
= 5 × 10 = 0.002
45. A long metallic bar is carrying heat from one of its ends to the other end under steady‐state.
The variation of temperature along the length ofthe bar fi om its hot end is best
described by which ofthe following figures?[2009]
Sol: . (a) Let be the temperature at a distance from hot end of bar. Let is the temperature
ofhot end.
The heat flow rate is given by
( − )
=
⇒ − = ⇒ = −
Thus, the graph of versus is a straight line with a positive intercept and a negative slope.
The above equation can be graphically represented by option (a).
46. One end ofa thermally insulated rod is kept at atemperature and the other at . The rod is
composed oftwo sections oflength and and thermal conductivities and
respectively. The temperature at the interface of the two section is [2007]
( ) ( ) ( ) ( )
(a) ( )
(b) ( )
(c) ( )
(d) ( )
47. Assuming the Sun to be a spherical body of radius at a temperature of , evaluate the
total radiant powerd incident ofEarth at a distance fi om the Sun [2006]
(a) 4 σ (b) σ (c) σ (d)
where is the radius ofthe Earth and 0 is Stefan’s constant.

Sol : (b) From stefan’s law, totalpower radiated by Sun, = ↓
×
4
The intensity ofpower Per unit area incident on earth↑ s surface
×4
=
4
Total power received by Earth
′
= 2 × Cross— Section area of earth facing the sun = ( )
4
48. Two rigid boxes containing different ideal gases are placed on a table. Box A contains one
mole of nitrogen at temperature , while Box contains one mole ofhelium at temperature
. The boxes are then put into thermal contact with each other, and heat flows between
them until the gases reach a common fmal temperature (ignore the heat capacity of boxes).
Then, the fmal temperature of the gases, in terms of is [2006]
(a) = (b) = (c) = (d) =
Sol : (c) When two gases are mixed to gether then Heat lost by He gas = Heat gained by N2 gas
1 1 1 = 2 2 2
3 7 5
− = −
2 3 2
7 −3 =5 −5
12
⇒ 12 =8 ⇒ =
8
3
⇒ = ..
2
49. If the temperature of the sun were to increase from to 2 and its radius from to 2 , then
the ratio ofthe radiant energy received on earth to what it was previously will be [2004]
(a) 32 (b) 16 (c) 4 (d) 64
50. The temperature of the two outer surfaces ofa composite slab, consisting of two materials
having coefficients of thermal conductivity and 2 and thickness and 4 , respectively,
are and ( > ) . The rate of heat transfer through the slab, in a steady state is
( )
, with equal to [2004]
(a) (b) (c) 1 (d)
SOLUTION:’
The thermal resistance is given by
4 2 3
+ = + =
2
Amount of heat flow per second,
3 ( − )
= =
3
( )
= =
51. The earth radiates in the infia‐red region ofthe spectrum. The spectrum is correctly given by
[2003]
(a) Rayleigh Jeans law (b) Planck’s law of radiation
(c) Stefan’s law of radiation (d) Wien’s law
Sol : (d) Wein’s law correctly explains the spectrum
52. Heat given to a bodywhich raises its temperature by 1∘ C is [2002]
(a) water equivalent (b) thermal capacity
(c) specific heat (d) temperature gradient
Sol: (b) Heat required for raising the temperature of a body through 1o is called its
thermal capacity.
53. The figure shows a system of two concentric spheres of radii and are kept at
temperatures and , respectively. The radial rate offlow ofheat in a substance between
the two concentric spheres is proportional to [2005]
r  ( )
(a) l2  2  (b) ( )
(c) ( 2 − 1) (d) ( )
 r1 
Consider a thin concentric shell ofthickness ( dr) and of radius (r) and let the temperature of
inner and outer surfaces ofthis shell be and ( − ) respectively.
The radial rate of flow of heat through this elementary shell will be
[( − )− ] −
= =
( =4 2)
= −4
Since the area of the surface through which heat will flow is not constant. Integrating both sides
between the limits of radii and temperatures of the two shells, we get
2 1 2
2
= −4
1 1
2 2
−2
= −4
1 1
1 1
− = −4 [ − ]
( )
or = ( )
∝
( − )
54. Infrared radiation is detected by [2002]

(a) spectrometer (b) pyrometer
(c) nanometer (d) photometer
Sol : (b) Pyrometer is used to detect infia‐red radiation.
55. Which of the following is more close to a black body?[2002]
(a) black board paint (b) green leaves
(c) black holes (d) red roses
Sol: (a) Black body is one which absorb all incident radiation. Black board paint is quite
approximately equal to black bodies.
56. If mass‐energy equivalence is taken into account, when water is cooled to form ice, the
mass ofwater should[2002]
(a) increase (b) remain unchanged
(c) decrease (d) first increase then decrease
Sol: (c) When water is cooled at 0∘ C to form ice, energy is released from water in the form of
heat. As energy is equivalent to mass, therefore, when water is cooled to ice, its mass decreases.
57. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000
K respectively. The ratio ofthe energy radiated per second by the first sphere to that by the
second is [2002]
(a) 1: 1 (b) 16: 1 (c) 4: 1 (d) 1: 9.
Sol: (a) From stefan’s law, the energy radiated per second is given by = ↓
Here, = temperature of the body

= surface area of the body
For same material e is same. 0 is stefan’s constant
Let 1 and 2 be the temperature of two spheres. 1 and 2 be the area of two spheres.
4
= =
4
(4000) × 1 1
= =
(2000) × 4 1
58. A metallic sphere cools from 50∘ C to 40∘ C in300 s. If atmospheric temperature around is
20∘ C, then the sphere’s temperature after the next 5 minutes will be close to:
[Sep. 03, 2020 (ID]
(a) 31∘ C (b) 33∘ C (c) 28∘ C (d) 35∘ C
Sol: (b) From Newton’s Law of cooling,
− +
= −
2
Here, T1 = 50∘ C, T2 = 40∘ C
an dT = 20∘ C, t = 600S = 5 minutes
⇒ = − 20 (i)
Let be the temperature of sphere after next 5 minutes. Then
= − 20 (ii)
Dividing eqn. (ii) by (i), we get
40 − 40 + − 40
= =
10 50 + 40 − 40 50
⇒ 40 − = ⇒ 200 − 5 =
5
200
= = 33. 3∘ C
6
59. Two identical beakers A and B contain equal volumes of two different liquids at 60∘ C each
and left to cool down. Liquid in Ahas density of8 × 10 kg/m and specific heat of 2000 J
kg 1K while liquid in B has density of103 kg m and specific heat of4000 J kg 1K . Which
ofthe following best describes their temperature versus time graph schematically? (assume
the emissivity of both the beakers to be the same) [8 April 2019 I]
Sol : (b) Rate of Heat loss = =
o× × 1
− = ⇒− ∝
p× .× p
− p 10 4000
= × = ×
− p 8 × 10 2000
⇒ − > −
So, A cools down at faster rate.

60. A bodytakes 10 minutes to cool fi om60∘ C to 50∘ C. The temperature of surroundings is
constant at 25∘ C. Then, the temperature of the body after next 10 minutes will be
approximately [Online Apri115, 2018]
(a) 43∘ C (b) 47∘ C (c) 41∘ C (d) 45∘ C
Sol : . (a) According to Newton’s law of cooling,
1 − 2 1 + 2
= − 0
2
60−50 60+50
= − 25 ..... (i)
10 2
and = − 25 ..... (ii) Dividing eq. (i) by (ii),

10 60
= ⇒ = 42.85∘ C ≅ 43∘ C
(50 − )
61. Hot water cools fi om60∘ C to 50∘ C in the first 10 minutes and to 42∘ C in the next 10
minutes. The temperature of the surroundings is: [Online April 12, 2014]
(a) 25∘ C (b) 10∘ C (c) 15∘ C (d) 20∘ C
Sol : (b) By Newton’s law of cooling
− +
= −K −
t 2
where 0 is the temperature of surrounding.
Now, hot water cools from 60∘ C to 50∘ C in 10 minutes, = −K − (i)
Again, it cools fi om50∘ C to 42∘ C in next 10 minutes. = −K − (ii)
Dividing equations (i) by (ii) we get
1 55 −
=
0.8 46 −
10 55 −
=
8 46 −
460 − 10 = 440 − 8
2 = 20
0 = 10∘ C
62. A hot body, obeying Newton’s law of cooling is cooling down from its peak value 80∘ C to
an ambient temperature of30∘ C. It takes 5 minutes in cooling down from 80∘ C to 40∘ C. How
much time will it take to cool down from62∘ C to 32∘ C?
(Given In 2 = 0.693, In 5 = 1.609) [Online April 11, 2014]
(a) 3.75 minutes (b) 8.6 minutes (c) 9.6 minutes (d) 6.5 minutes
Sol : . (b) From Newton’s law of cooling,
1 −
= log
−
From question and above equation,

63. If a piece of metal is heated to temperature and then allowed to cool in a room which is at
temperature , the graph between the temperature T ofthe metal and time t will be closest
to [2013]
(a) (c)
Sol : (c) According to Newton’s law of cooling, the temperature goes on decreasing with time
non‐linearly.
64. A liquid in a beaker hastemperature (t) at time and is temperature of surroundings,
then according to Newton’s law of cooling the correct graph between
log ( − ) and is: [2012]
(a) (b) (c) (d)

Sol: . (a) According to newton’s law of cooling
=− ( − )
( )
5 = log ( )
(1)
⇒ =−
( − )
( )
And, = log ( )
(2) ⇒ ∫ ( )
=− ∫
Dividing equation (2) by (1),

⇒ log ( − )=− +
( )
Which represents an equation of straight line. = ( ) Thus the option (a) is correct.
( )
65. According to Newton’s law of cooling, the rate of cooling of a body is proportional to
( ) , where is the difference of the temperature of the body and the surroundings, and n
is equal to [2003]
(a) two (b) three (c) four (d) one
Sol : (d) From Newton’s law of cooling − dQ ( ) On solving we get, time taken to cool down
∞
from62∘ C dt to 32∘ C, = 8.6 minutes.
Thermodynamics
Thermodynamics is a branch of science which deals with exchange of heat energy between bodies and
conversion of the heat energy into mechanical energy and vice-versa.
Some Definitions
(1) Thermodynamic system
Surrounding
Gas System
Fig. 14.1
 It is a collection of an extremely large number of atoms or molecules

 It is confined with in certain boundaries.
 Anything outside the thermodynamic system to which energy or matter is exchanged is called its surroundings.
Thermodynamic system may be of three types
Open system :
It exchange both energy and matter with the surrounding.
Closed system :
It exchange only energy (not matter) with the surroundings.
Isolated system :
It exchange neither energy nor matter with the surrounding.
(2) Thermodynamic variables and equation of state :
A thermodynamic system can be described by specifying its pressure, volume, temperature, internal
energy and the number of moles. These parameters are called thermodynamic variables. The relation
between the thermodynamic variables (P, V, T) of the system is called equation of state.
For m moles of an ideal gas, equation of state is PV = mRT and for 1 mole of an it ideal gas is
PV = RT
(3) Thermodynamic equilibrium :
In steady state thermodynamic variables are independent of time and the system is said to be in the
state of thermodynamic equilibrium. For a system to be in thermodynamic equilibrium, the following conditions
must be fulfilled.
 Mechanical equilibrium : There is no unbalanced force between the system and its surroundings.
 Thermal equilibrium : There is a uniform temperature in all parts of the system and is same as that of
surrounding.
 Chemical equilibrium : There is a uniform chemical composition through out the system and the
surrounding.
(4) Thermodynamic process :
The process of change of state of a system involves change of thermodynamic variables such as
pressure P, volume V and temperature T of the system. The process is known as thermodynamic process.
Some important processes are
 Isothermal process : Temperature remain constant
 Adiabatic process : No transfer of heat
 Isobaric process : Pressure remains constant
 Isochoric (isovolumic process : Volume remains constant
 Cyclic and non-cyclic process : Incyclic process Initial and final states are same while in non-cyclic process
these states are different.
 Reversible and irreversible process :
(5) Indicator diagram :
Whenever the state of a gas (P, V, T) is changed, we say the gaseous system is undergone a thermodynamic
process. The graphical representation of the change in state of a gas by a thermodynamic process is called indicator
diagram. Indicator diagram is plotted generally in pressure and volume of gas.
V1 dV V2 V
Fig. 14.6
Zeroth law of thermodynamics:
Insulating Conducting
System System System System
System System
Conducting Insulating
(A) (B)
Fig. 14.3
If two isolated bodies A and B are in thermal equilibrium independently with a third body C, then
the bodies A and B will also be in thermal equilibrium with each other.
 The zeroth law leads to the concept of temperature. All bodies in thermal equilibrium must have a
common property which has the same value for all of them. This property is called the temperature.
 The zeroth law came to light long after the first and seconds laws of thermodynamics had been
discovered and numbered. It is so named because it logically precedes the first and second laws of
thermodynamics.
 Zeroth law of thermodynamics leads to the concept of temperature(T).

 Temperature is the measure of degree of hotness or coldness of a body.
 Temperature determines the direction of flow of heat when two bodies are placed in thermal contact.
 Heat always flows from the body at higher temperature(hot body) to the body at lower temperature(cold
body).
Relation between work and heat (Joule’s law):

The amount of heat produced is directly proportional to the amount of mechanical work
done.
HW
W=JH.
W
J= H where J= Mechanical equivalent of heat.
Mechanical equivalent of heat (J):
 It is the amount of work necessary to produce unit amount of heat energy.
 J is not a physical quantity. It is simply a conversion factor between mechanical work
and its equivalent heat energy.
Values of ‘J’:
 The value of J depends on the units of work W and heat H.
 When W is in Joules, H is in Cal, then J  4.186J / cal  4.2J / cal
 When W and H both are expressed in joules, J=1.
Applications of Joule’s law :
 A metal ball falls freely on the ground from a height ‘h1’ and bounces to height ’ h2’. If the
ball absorbs all the heat energy generated, then raise in temperature of the ball is
g (h1  h2 )
 
JS
where ‘S’ is specific heat of water
 The raise in temperature of water when it falls from a height h to the ground is,
gh
  ;
JS
where ‘S’ is specific heat of water
 The height from which ice is to be dropped to melt it completely is
JL
h= g
;
where L= Latent heat of ice.
 When a bullet of mass m moving with a velocity v is stopped abruptly by a target and all
of its heat energy liberated is retained by bullet , then the increase in temperature is.
v2
 
2 JS
If the bullet absorbs x% of heat liberated, then rise in its temperature is
x  v2 
   
100  2 JS 
 If a bullet at a temperature lesser than its melting point just melts when abruptly stopped
by an obstacle and if all the heat produced is absorbed by the bullet then
1
J  mS    mL   mv 2
2
Where L= Latent heat of fusion the material of the bullet ,
S= Specific heat
  rise in temperature before it melts.
 When a block of ice of mass M is dragged with constant velocity 'v' on a rough horizontal
surface of coefficient of friction  , through a distance d, then the mass of ice melted
is,
Mgd
m ;
JL
Where m=mass of ice melted.
In order to melt all the ice completely,
JL
the block should be dragged through a distance d 
g
d JL
Now, the time taken to melt completely is given by t  
v  gv
 When a block is dragged on a rough horizontal surface of coefficient of friction  ,

then the rise in temperature of block is,
 gd
 
JS
 When a body rotating with angular speed  is suddenly stopped, if it absorbs all the heat
generated, then rise in temperature of body is
I 2
 
2 JmS
W    t 
Where I=Moment of inertia of the given body.
 A drilling machine drills a hole to a metal plate in a time t. The machine is operated by a
torque  with constant angular speed  . If the heat generated is completely absorbed
by the plate, then the raise in temperature of the plate is
t
 
JmS
W    t 
Where, m= mass of the plate
S= specific heat of the material of the given plate
PROBLEMS
1. The height of the Niagara falls is 50m. If J  4.2 107 erg / cal . Then the difference of
temperature of water at the top and bottom of the falls is
SOLUTION:
P.E. is converted into heat.
mgh  JmS t
gh 980  5000
t   7
 0.1170 C
JS 4.2 10
2. A bullet moving with a uniform velocity v, stops suddenly after hitting the target and the
whole mass melts be m, specific heat S, initial temperature 25°C, melting point 475°C
and the latent heat L. Then v is given by
1 mv 2 mv 2
(a) mL  mS (475  25 )  
2 J
(b) mS (475  25 )  mL 
2J
mv 2 mv 2
(c) mS (475  25 )  mL 
J
(d) mS (475  25 )  mL 
2J
SOLUTION:
Firstly the temperature of bullet rises up to melting point, then it melts.
Hence according to JOULES LAW
W  JQ .
1
mv 2  J .[m .c.  mL ]  J [m S (475  25 )  mL ]
2
mv 2
mS (475  25 )  mL 
2J
3. A lead bullet of mass 21 g travelling at a speed of 100 ms 1 comes to rest in a wooden

block. If no heat is taken away by the wood, then find the raise in temperature of the
wood. (Specific heat of lead = 0.03 calorie/g °C.)
SOLUTION:
kinetic energy of the bullet = heat gained by the bullet,
1 2
mv = mSD t
2
v2 (100)2
Dt = = =
2S 2´ 0.03´ 4.2´ 1000
=39.680C
4. If two balls of same metal weighing 5 gm and 10 gm strike with a target with the same
velocity. The heat energy so developed is used for raising their temperature alone, then
the temperature will be higher
(a) For bigger ball (b) For smaller ball
(c)Equal for both the balls (d) None is correct from the above three
SOLUTION:
1
Energy = mv 2 = mc Dq;
2
  v 2
Temperature does not depend upon the mass of the balls.
5. A water fall is 84 metres high. If half of the potential energy of the falling water gets
converted to heat, the rise in temperature of water will be
(a) 0.098°C (b)0.98°C (c) 9.8°C (d) 0.0098°C
SOLUTION:
As W  JQ
1
(mgh)  J  mc 
2
gh
 
2 Jc
9 . 8  84
   0 . 098 C
2  4 . 2  1000
cal
( c water  1000 )
kg  C
6. A 10kw drilling machine is used for 5 minutes to bore a hole in an aluminium block of
mass 10  103 kg. If 40% of the work done is utilised to raise the temperature of the
block, then find the raise in temperature of the aluminium block ? (Specific heat of
Aluminium = 0.9 Jkg-1 k-1)
SOLUTION:
Work done by the drilling machine in 5 min
W= power x time
W=10  103  5  60 = 3  106J
The energy utilised to raise the temperature of the block = 40% of W
40
= 3  106 ´ = 12 ´ 10 5 J
100
Heat gained by aluminium block = mass  specific heat  increase in temperature.

12 ´ 10 5 = (10 ´ 103 )´ 0.9´ D t
12 ´ 10 5
\ Dt = = 133.3°C
0.9´ 10 4
7. A lead bullet of 10 g travelling at 300 m/s strikes against a block of wood and comes to
rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature is
(Specific heat of lead = 150J/kg, K)
(a) 100°C (b)125°C (c) 150°C (d)
200°C
SOLUTION:
Since specific heat of lead is given in Joules,
hence use W Q instead of W  JQ .
1 1 
  mv 2   m.c.
2 2 
v2 (300 )2
  
4 c 4  150
 150 C .
8. A body of mass 5 kg falls from a height of 30 metre. If its all mechanical energy is
changed into heat, then heat produced will be
(a) 350 cal (b)150 cal (c) 60 cal (d) 6 cal
SOLUTION:
W  JQ
mgh  J  Q
mgh 5  9 . 8  30
Q   350 cal
J 4 .2
9. Hailstone at 0°C falls from a height of 1 km on an insulating surface converting

whole of its kinetic energy into heat. What part of it will melt (g  10 m / s 2 )
1 1 1
(a) (b) (c)  10  4 (d) All of it will melt
33 8 33
SOLUTION:
Suppose m¢ kg ice melts out of m kg
then by using W  JQ
mgh  J (m L ) .
Hence fraction of ice melts

m  gh 9 . 8  1000 1
   
m JL 4 . 18  80 33
10. Hailstones fall from a certain height. If they melt completely on reaching the ground,
find the height from which they fall. (g=10 ms-2,L = 80 calorie/g and J = 4.2 J/calorie.)
SOLUTION:
On reaching the ground, a hailstone of mass M losses potential energy which is

converted into heat energy required to melt it.
potential energy lost = heat energy required for melting the hailstone
Mgh = ML
gh = L
L
h= ,
g
80 ´ 4.2 ´ 1000
h=
10
h = 33.6´ 1000m = 33.6 km.
11. A girl weighing 42 kg eats bananas whose energy is 980 calories. If this energy is used
to go to height h find the value of h.(J=4.2J/ calorie)
SOLUTION:
Energy gained by the girl in eating bananas = 980 calories
= 980 ´ 4.2 J.
W=H (in S.I.)
980 × 4.2 = mgh
980×4.2 = 42×9.8×h
980´ 4.2
Þ h= = 10m
42´ 9.8
12. A piece of ice at 00 C falls from rest into a lake of water which is also at 00 C and 0.5%
of ice melts. Find the minimum height from which the ice falls.
SOLUTION:
Let a mass m of ice falls from height h.
Loss in potential energy = mgh,
0.5
Heat produced = H  mLice
100
since,W = JH;
0.5
mgh  J  mLice
100
0.5 Lice
h
100  g
0.5   4.2 J / cal   80 103 cal / kg 

  171.43m
100  9.8 m / s 2
13. An ice block is projected vertically up with a velocity 20 ms-1. The amount of ice that melt when it
reaches the ground if the mass of ice block is 4.2 kg.
1) 2.5 gm 2) 2.5 kg 3) 0.25 kg 4) 0.25 gm

SOLUTION:
u2
. Maximum height attained, h 
2g
W  JH  mgh  J  xLice 
14. How much will the temperature of 100g of water be raised by doing 4200 J of work in stirring the
water?
1) 0.010c 2) 0.10c 3) 10c 4) 100c

SOLUTION:
W = JH
W = JmS 
Internal energy (U) :

Internal energy of a system is the energy possessed by the system due to molecular motion and molecular
configuration.
The energy due to molecular motion is called internal kinetic energy UK and that due to molecular
configuration is called internal potential energy UP
i.e. Total internal energy U  U K  UP
 For an ideal gas, as there is no molecular attraction U p  0
3
i.e. internal energy of an ideal gas is totally kinetic and is given by U  U K 
2
RT
3
change in internal energy U  2 R T
 In case of gases whatever be the process
f
U   R T
2
 C V T
R  R(Tf  Ti )
 T 
(  1)  1
RT f  RT i

 1
(Pf V f  Pi Vi )

 1
 Change in internal energy does not depend on the path of the process.
So it is called a point function
i.e. it depends only on the initial and final states of the system,
i.e. U  U f  U i
2
i 1 f
3
U1 U2 U3

 The internal energy of ideal gas depends only on its temperature T.
When T increases U also increases and vice versa.
 Internal energy of real gases depends upon temperature, pressure and volume.
 Real gases consists of both kinetic energy and potential energy due to intermolecular
forces.
Work (dW) :
Suppose a gas is confined in a cylinder that has a movable piston at one end.
If P be the pressure of the gas in the cylinder, then force exerted by the gas on the piston of the cylinder
F = PA
(A = Area of cross-section of piston)
Piston

dx
F Gas
When the piston is pushed outward an infinitesimal distance dx,
the work done by the gas dW  F.dx  P ( A dx )  P dV
For a finite change in volume from Vi to Vf

Vf
Total amount of work done W   Vi P dV  P(Vf  Vi )
Work from PV graph :
 If we draw indicator diagram, the area bounded by PV-graph and volume axis represents the work done

P P
P
A P
V1 V2 V V1 dV V2 V
1) Work = Area = P(V2 – V1)

V2
2) Work   V1
PdV  P(V2  V1 )
P
P2
P1
V1 V2 V
Work = Area of the shown trapezium

1
 (P1  P2 ) (V2  V1 )
2
 From W  PV  P(V f  Vi )

 If system expands against some external force then V f  Vi
dW = positive
 If system contracts because of external force then V f  Vi
dW = negative
P B Positive P B Negative
work work
A
A
V V
(A) Expansion (B) Compression
 Like heat, work done is also depends upon initial and final state of the system and path adopted for the
process
P
A
1
P
A 2 B
A1
V
(A) Less area
1 B
P
V 2
A
 A1 < A2 and W1 < W2
A2
B
V
(B) More area
 In cyclic process, work done is equal to the area of closed curve.
 It is positive if the cycle is clockwise
P
C
P2
P1 A
B
V1 V2 V
Work = Area of triangle ABC
1
  (V2  V1 )  (P2  P1 )
2
 It is negative if the cycle is anticlockwise.
P
C
P2
P1 A
B
V1 V2 V
Work = Area of rectangle ABCD

= AB ´ AD
= (V2 – V1) (P2 – P1)
 SPECIAL CASE
P
P2
P1 
Work = 4
(P2  P1 ) (V2  V1 )
V1 V2 V
First Law of Thermodynamics (FLOT)

It is a statement of conservation of energy in thermodynamical process.
 According to it heat given to a system (Q) is equal to the sum of increase in its internal energy (U) and the work done (W) by the
system against the surroundings.
The differential form of first law of thermodynamics is
dQ=dU+dW,
where dQ = heat added,
dU = Increase in internal energy
dW= work done = PdV
 dQ  dU  PdV
for bulk changes Q  U  PdV 

Sign convention :
 When heat is added (flows into) to the system
dQ is + ve(+dQ)
 When heat is taken (flows out) from the system
dQ is -ve (-dQ)
 When gas expands work is done by the gas, dw is positive
(+ dW)
 When gas compresses work is done on the gas, then work done by the gas dW is negative
(-dW)
 When internal energy of system increases
dU is +ve (+dU)
 When internal energy of system decreases
dU is -ve(-dU)
NOTE :
 Itmakesnodistinctionbetweenworkandheatasaccordingtoittheinternalenergy(andhencetemperature)
of a system may be increased either by adding heat to it or doing work on it or both.
Q and W are the path functions but U is the point function.
 In the above equation all three quantities Q, U and W must be expressed either in Joule or in calorie.
Thefirstlawintroducestheconceptofinternalenergy.
Significance and limitations of first law:
 It is a consequence of law of conservation of energy.
 This law is applicable to any process in nature.
 This law is applicable to all the three phases of matter.
 First law of thermodynamics does not indicate the direction of heat transfer. It does not tell any thing about the
conditions under which heat can be transformed into work.
PROBLEMS
1. In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the
system is W. The change in its internal energy is
Q W
(a) Q + W (b) Q – W (c) Q (d) 2
SOLUTION:
Q  U  W
U  Q  W  Q  W (using proper sign)
2. Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal
energy of the system will be
(a) – 50 J (b) 20 J (c) 30 J (d) 50 J
SOLUTION:
U  Q  W  35  15  20 J
3. When heat energy of 1500J is supplied to a gas the external workdone by the gas is 525J what is
the increase in its internal energy
SOLUTION:
Heat energy supplied Q=1500J
External workdone W=525J
By 1st law of thermodynamics
Q = U+W
U  Q  W
=1500 – 525
= 975J.
4. Consider the vaporization of 1g of water at 1000C to steam at 1000C at one atmospheric pressure.
Compute the work done by the water system in the vaporization and change in internal energy of
the system.
SOLUTION:
To change a system of mass m of liquid to vapour, heat required is Q  mL
v
The process takes place at constant pressure,
the work done by the system is the work in an isobaric process. W  PV
Where V  Vvapour  Vliquid 
From first law of thermodynamics
U  Q  W  mLv  P Vvapour  Vliquid 
Latent heat of vaporization of water
Lv  22.57  105 J / kg
Q  1.00  10 3  22.57  105   2.26  103 J
 No. of moles = weight /gram molecular weight
1
 Moles of water in 1g   0.0556 mole
18
nRT  0.0556  8.315  373
Vvapour    1.70  103 m3
P 1.013  105
The density of water is 1.00  103 kg / m3  1.00 g / cm3
Vliquid  1.00  106 m3
Thus the work done by the water system in vaporization is
W  P Vvapour  Vliquid 
 1.013  105 1.70  103  1.00  106   172 J

The work done by the system is positive since the volume of the system has increased.
From first law,
U  Q  W  U  2.26 103  172  2.09 103 J

5. A thermodynamic system goes from states (i) P1 , V to 2P1 , V (ii) P, V to P, 2V. Then work done in the
two cases is
(a) Zero, Zero (b) Zero, PV1 (c) PV1 , , Zero (d) PV1 , P1 V1
SOLUTION:
Case 1  Volume = constant

 PdV  0
Case 2  P = constant
2 V1 2 V1
  V1
PdV  P  V1
dV  PV1
6. Find the change in internal energy of the system when a system absorbs 2 kilocalorie of heat and at the same time does 500 joule of
work
(a) 7900 J (b) 8200 J (c) 5600 J (d) 6400 J
SOLUTION:
Q  2k cal  2  10 3  4.2 J
 8400 J
W  500 J .
Hence from Q  U  W ,
W  Q  U
= 8400 – 500
= 7900 J
7. When 1g of water at 100°C is converted into steam at 100°C, it occupies a volume of 1671cc at
normal atmospheric pressure. Find the increase in internal energy of the molecules of steam.
SOLUTION:
1 atmosphere = 1.013 x 105 Nm-2 ;
volume of 1gm of water, V1 = 1cc = 10-6m3;
Volume of steam = 1671 cc = 1671 x 10-6 m3
External work done dW = P(V2 - V1)
= 1.013 x 105 (1671 x 10-6 - 1 x 10-6)
= 1.013 x 105 x 1670 x 10-6
= 1.013 x 167 = 169.2 J.
Latent heat of vaporisation of steam= 540 cal/g
So, heat supplied to convert 1g of water into steam,
D Q = 540 x 4.2J = 2268J
By first law of thermodynamics
D U = D Q - D W = 2268 - 169.2 = 2098.8J
8. Calculate the external workdone by the system in KCal, when 40 KCal of heat is supplied to the
system and internal energy raises by 8400 J.
SOLUTION:
dQ=dU+dW
dU = 8400 J
8400
= KCal
4200
= 2 KCal.
 40 KCal = 2 KCal+External work done
The external work done = 40 - 2 = 38 KCal

9. An ideal monatomic gas is confined in a cylindrical by a spring loaded piston of cross section 8.0 x
10-3m2. Initially the gas is at 300K and occupies a volume of 2.4 x 10-3m3 and the springs is in its
relaxed state as shown in figure. The gas is heated by a small heater untill the piston moves out
slowly by 0.1 m. The cylinder and the piston are thermally insulated. The piston and spring are
massless and there is no friction between the position and the cylinder. The final temperature of th
gass will be
(Neglect the heat loss through the load wires of the heater. The heat capacity of the heater coil is
also negligible)
1) 300K 2) 800K 3) 500K 4) 1000K

SOLUTION:
Kx
Final pressure of the gas is Pf = P0 +
A
Tf Pf Vf

Ti pi Vi
10. In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the
gas releases 20J of heat and 8J of work is done on the gas. If initial internal energy of the gas was
30J, what will be the final internal energy?
SOLUTION:
We know that, dQ = dU+dW
Since heat is released by the system, dQ = -20J.
and work is done on the gas, dW = -8J
dU = -20 - (-8)
= -20 + 8 = -12J
 U f  U i  12 J
U f  U i  12  30  12  18 J
11. A system goes from A to B via two processes I and II as shown in figure. If U1 and U2 are the
changesininternalenergiesintheprocessesIandIIrespectively,then
P
II
A B
I
V
(a) U II  U I (b) U II  U I
(c) U I  U II (d) Relation between U I and UII can not be determined
SOLUTION:
As internal energy is a point functiontherefore change in internal energy doesnot dependsupon the path
followed
i.e.U I  U II
12. Consider the melting of 1g of ice at 00C to water at 00C atmospheric pressure. Then the change in
internal energy of the system (density of ice is 920kg/m3)
1) 334 J 2) 420 J 3) 540 J 4) 680 J
SOLUTION:
Heat required to change the phase of solid Q  mLf
Work done by system at constant pressure, W  P Vliquid  Vsolid 

From first law of thermodynamics, U QW
 mL f  P Vliquid  Vsolid 
Latent heat of fusion of water,
L f  3.335  105 J / Kg
Q  1  103  3.335  105   334 J
The density of ice is 920 kg / m3.
1  10  3
V solid   1.09  10  6 m 3
920
Vliquid  1  106 m3
W  PVliquid Vsolid 
Thus work done by the system in melting is
 1.013105 1106 1.09106   9103 J
Work done by the system is negative because the system decreased in volume.
 U   Q   W  334    9  10  3   334 J
13. The equation of state for a gas is given by PV=nRT+  V, where n is the number of moles and  is
a positive constant. The intial temperature and pressure of one mole of the gas contained in a
cylinder are T0 and P0 respectively. The work done by the gas when its temperature doubles
isobarically will be (Mains 2014)
P0 T0 R P0 T0 R
1) P   2) P   3) P0T0 RIn2 4) P0T0 R
0 0
SOLUTION:
work done W = (Vf-Vi) P
nRT0
Vi 
P0  
at constant pressure as temp doubles volume double so
2nRT0
Vf 
P0  
Specific heats of a gas:

 Gases have two types of specific heats.
Specific heat at constant volume (cv)
 Specific heat at constant pressure( cp)
 Specific heat of all substances is zero at 0K.
Specific heat of gas at constant pressure  c p 
It is the heat required to rise the temperature of 1g of a gas by 1o C at constant pressure

1  dQ 
 cp   
m  dT  p
; dQ p  mc p dT
Q p  mc p T , if c p is constant.
Q p  m  c p dT , if c p depends on temperature.
Specific heat of gas at constant volume (CV)
It is the heat required to rise the temperature of 1g of a gas by 1o C at constant volume
1  dQ 
 cv   
m  dT v
dQv  mcv dT
 Qv  mcv T , stant.
Qv  m  cv dT , emperature.
SI unit of both c p , cv is J/Kg-K

CGS unit is cal/g- o C
Molar specific heats (CP,CV) of a gas
When the above specific heats c p , cv are defined per 1mole of gas, then they are said to be molar specific
heats and represented by CP , CV .
1  dQ 
These are, C p  n  dT 
 p
1  dQ  1  dU 
Cv       ,  W  0 
n  dT v n  dT 
SI unit of both molar specific heat is J/ mol-k
NOTE :
 Cp is greater than CV
CP
   ( C ,C are molar specific heats )
CV p V
 Cp - CV= R,
where R is universal gas constant
R= 8.314 J/ mol-k  2cal / mol  K
R R
CP  and Cv 
 1  1
R
 c p  cv  r   but C  M c
M
 M  c p  cv   R
Where r is specific gas constant
c p , cv are expressed in J/Kg-K.
 For a gas having f degrees of freedom,
 f   f 
C v    R, C p  1   R
2  2
2
  1
f
CP, CV and values of different gases:

CP
S.No Atomicity of gas CP CV γ=
CV
5 3 5
1. Monoatomic R R  1.67
2 2 3
7 5 7
2. Diatomic R R  1.4
2 2 5
4
3. Tri non-linear& 4R 3R  1.33
3
and poly atomic
9 7 9
4. Tri linear R R  1.29
2 2 7
γ Of mixture of gases:
 When n1moles of a gas is mixed with n2 moles of another gas. then,
n 1C v1 +n 2 C v 2
(C v ) mixture 
n 1 +n 2
n1C p1 +n 2 C p2
(C p ) mixture  (C v ) mixture  R 
n1 +n 2
C p (mixture)
γ mixture 
C v (mixture)
n1  n2 n n
Also   1  2
mixture 1  1 1  2 1
 At constant pressure, fraction of heat absorbed that is converted into internal energy is
dU C v 1
 
dQ C p γ
 At constant pressure, fraction of heat absorbed that is converted into external workdone is
dW nRdT R 1
  1
dQ nCp dT Cp γ
PROBLEMS
1. Four moles of a perfect gas is heated to increase its temperature by 20C absorbs heat of 40 cal at
constant volume. If the same gas is heated at constant pressure find the amount of heat supplied.
SOLUTION:
At constant volume dQ  nCv dT  dU  40
At constant pressure dQ = dU + nRdT = 40 + (4´ 2´ 2)= 56 cal
2. A cylinder of fixed capacity 67.2 litres contains helium gas at S.T.P. The amount of heat required
to raise the temperature of the gas by 150C is (R =8.31 J mol-1k-1)
1) 520 J 2) 560.9 J 3) 620 J 4) 621.2 J

SOLUTION:
dQ = nCVdT
Q  U  nCV T
 R 
 CV    1 
 
3. 14 g of N2 gas is heated in a closed rigid container to increase its temperature from 230C to 430C.
The amount of heat supplied to the gas is
1) 25 cal 2) 50 cal 3) 100 cal 4) 30 cal

SOLUTION:
In a closed container, V  0
4. The specific heat capacity of a metal at low temperature(T) is given as
3
1 1  T 
C p ( kJK kg )  32   . A 100 g vessel of this metal is to be cooled from 20K to 4 K by a
 400 
0
 
special refrigerator operating at room temperature 27 C .The amount of work required to cool
the vessel is
SOLUTION:
Heat required to change the temperature of vessel by a small amount dT dQ  mC p dT
3
4  T 
Total heat required Q  m  32   dT  0.001996 kJ
20
 400 
work done required to maintain the temperature of sink to T2
 Q1  Q2 
W  Q1  Q2   Q2
 Q2 
 300  20 
For T2  20 K ; W1    0.001996  0.028 kJ
 20 
For T2  4K
 300  4 
W2    0.001996  0.148 kJ
 4 
 The work required to cool the vessel from 20K to 4K is
W2-W1=0.148-0.028=0.12KJ
As temperature is changing from 20 K to 4 K work done required will be more than W1 but less than W2 .
5. 70 cal of heat is required to raise the temperature of 2 moles of an ideal gas at constant pressure from
30ºC to 35ºC. What is the amount of heat required to rise the temperature of same gas through the same
range at constant volume?(R = 2cal mole-1K-1)
1) 28 J 2) 50 Cal 3) 75 J 4) Zero
SOLUTION:
QP  nCP T ;
CP  CV  R  CV  CP  R
U  nCV T
6. Calculate the difference between the two specific heats of nitrogen, given that the density of
nitrogen at N.T.P is 1.25 g/litre and J= 4200 J/KCal., ( in KCal/kg-K)

SOLUTION:
PV  mrT
P
The difference in specific heats, r  .
T
AT NTP P = 1.013 × 105 N/m2;
T = 273 K;
10- 3 kg 1.25´ 10- 3 kg
r = 1.25´ 3 3
= 3 -6 3
= 1.25kg / m3
10 cm 10 ´ 10 m
1.013´ 105
\ r=
1.25´ 273
= 296.8J/kg K
 The difference of specific heats = 0.0768 KCal/kg-K.
7. When an ideal diatomic gas is heated at constant pressure fraction of the heat energy
supplied which increases the internal energy of the gas is
SOLUTION:
Heat used in increasing the internal energy is Q1  Cv dT ;
Heat absorbed at constant pressure to increase the temperature by dT is Q2  C p dT
Q1 Cv 1 1
   
Q2 C p C p / Cv 
Q1 5
for diatomic gas,   7 / 5;  Q  7
2
8. The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process
is : U  a  bPV Where ‘a’ and ‘b’ are constants. What is the value of the ratio of the specific
heats?
a b 1 a 1 b
1) 2) 3) 4)
b b a a
SOLUTION:
. U  a  bPV ;
But PV  RT
dU
CV  ;
dT
CP  CV  R ;
CP

CV
9. The ratio of specific heats of a gas is  . The change in internal energy of one mole of gas when the
volume changes from V to 2V at constant pressure “P” is
PV PV
1)   1 2) PV 3)   1 4) 
SOLUTION:
U  nCV T
R
But CV 
 1
and PdV  RdT
10. A quantity of heat Q is supplied to a monoatomic ideal gas which expands at constant pressure. The
fraction of heat that goes into work done by the gas is
SOLUTION:
C p dT  Cv dT  dW ;  dW   C p  Cv  dT
Fraction of heat converted into work
dW

 C p  C v  dT  1  C v  1  1
dQ C p dT Cp 
For monoatomic gas,   5 / 3

dW 1 3 2
  1  1  .
dQ  5 5
11. The P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal
monoatomic gas. The amount of heat, extracted from the source in a single cycle is
2P0 B C
P0 D
A
V0 2V0
SOLUTION:
PV
0 0  nRTA ;
2 PV
0 0  nRTB ;
2 P0 .2V0  nRTC
Heat supplied H  nCV TAB  nCP TBC
 2P V PV   4P V 2P V 
 nCV  0 0  0 0   nC P  0 0  0 0 
 nR nR   nR nR 
 3  0 0PV  5  2 P0V 0 
= n  2 R   nR   n 2 R  nR 
    
13
 PV
0 0
2
Thermodynamic processes:
Quasi-static process: A quasi-static process can be defined as an infinitesimally slow process in which the system
remains in thermal and mechanical equilibrium with the surroundings at each and every intermediate stage.
i.e., temperature, pressure are almost constant during infinitesimal small change in the state of gas. It is an ideal
process. In practice it does not occur.
Isochoric process (or) Isometric process
 It is a process in which the volume of the system remains constant.
i.e., ΔV =0 for such process ΔW =0
 In this process, the increase in internal energy is maximum where as the work done is zero.
 In this process Q  U
P
 It obeys Gay-Lussac's law,  constant
T
P
 Indicator diagram
V
dP
 Slope of isometric curve, 
dV
f
 Specific heat is CV  R
2
 Bulk modulus of elasticity K  
Isobaric process:
 It is a process in which the pressure of the system remains constant.
i.e., P  0
 It obeys Charles law ,
V
 constant
T
P
Work
V1 V2 V
dP
 Slope of isobaric curve, 0
dV
 f 
 Specific heat is C p   2  1 R
 
 Bulk modulus of elasticity K=0.
 Work done in isobaric process is given by
W  P(V2  V1 )
 nR (T2  T1 )  nR  T
 Eg:- Boiling of water into steam.
 At constant pressure, dQ:dU:dW= nCP T : nCV T : n  CP  CV  T
R R
= CP : CV : R = : :R
 1  1
 Q : U : W   :1:    1
 2
  f  2 : f : 2    1  
 f 
 For monoatomic gas  5 : 3: 2
 For diatomic gas = 7: 5 : 2
 For non-linear poly atomic gas = 4 : 3 : 1
Isothermal process:
In this process, the pressure and volume of gas change, but temperature remains constant.
Hence internal energy is also constant.
i.e., dT  0; dU  0
 The system is in thermal equilibrium with the surroundings.
 It takes place in a thermally conducting vessel. Hence heat exchanges between system and surroundings.
 In this process dQ  dW
 It is a slow process.
 It obeys the Boyle’s law i.e. PV=Constant
 Specific heat is infinity.
work
V
dP P
 Slope of isothermal curve, tan   
dV V
dP
 Isothermal bulk modulus  P
dV V
 The workdone during the isothermal change at temperature T for n moles of gas is
V 
W  2.303nRT log10  2 
 V1 
1
P
=2.303 nRT log10  P 
 2 
 Isothermal process is ideal. In nature, no process is perfectly isothermal. But we can say melting of ice, boiling
of water are approximately isothermal. In these two processes internal energy increases even temperature is
constant.
Adiabatic process:
The pressure, volume and temperature of a gas change but total heat remains constant
i.e., dQ=0 (Q=constant).
 There should not be any exchange of heat between the system and surroundings. All the walls of the container
and the piston must be perfectly insulating.
 It is a quick process.
 The internal energy changes certainly as temperature changes.
 In the adiabatic process P, V & T are related as
(i) PV  = constant
(ii) TV  1 =constant
(iii) P1 T  =constant
 In this process specific heat is zero.
P
 Indicator diagram is
work
V
dP P
 Slope of adiabatic curve, tan    
dV V
The slope of adiabatic curve is  times to that of the isothermal curve.
 The adiabatic bulk modulus of gas is  p
i.e.  times isothermal bulk modulus.
 The workdone by the system during the adiabatic expansion is
nR
W (T -T )
 1 1 2
= nCv(T1-T2)
= nCV T
Cp
=n T1  T2 

PV
1 1  P2V2
=
 1
 It takes place in a non conducting vessel. Hence no exchange of heat takes place between system and
surroundings.
 Adiabatic expansion causes cooling and compression causes heating.
 Eg:- Sudden bursting of tube of bicycle tyre, Propagation of sound in gases
 In this process dU   dW
Comparison between isothermal and adiabatic curves :
 When expanded to the same volume from the same initial state.
P
1-isothermal
1 2-adiabatic
2
V1 V2 V
Work done : W1>W2

Final pressure: P1>P2
Final temperature: T1>T2
 When compressed to the same volume from the same initial state.
P
1-adiabatic
2
2-isothermal
1
V2 V1 V
Work done: W1<W2

Final pressure : P1>P2
Final temperature: T1>T2
 When expanded to the same pressure from the same initial state.
P 1-isothermal
P1 2-adiabatic
2 1
P2
V2 V1 V
Work done : W1>W2

Final volume : V1>V2
Final temperature : T1>T2
 When compressed to the same pressure from the same initial state.
P
1-isothermal
2-adiabatic
P1 2
P2
1
V1 V2 V
Workdone : W1  W2
Final volume : V1<V2
Final temperature : T1<T2
Cyclic process :
 If a system after undergoing through a series of changes comes back to its initial state, the process is called
cyclic.
 In a cyclic process (the system finally reaches the same initial state), workdone is equal to the area enclosed
by the cycle.
It is +ve if the cycle is clockwise. It is -ve if the cycle is anticlockwise.
P P
W=+ve W=-ve
In the cyclic process as U f  U i ,
V V
U  U f  U i  0
the first law implies Q  0  W ,
i.e Q  W ,
heat supplied is equal to the work done (area of the cycle)
Comparison of P-V curves of various processes :
P
K=0 isobaric
isochoric
K=P
isothermal
K=P
adiabatic
K 
V
Free expansion :
Stop cock
Gas Vacuum
Insulated
Figure shows an insulated cylinder divided into two parts by a thin massless fixed piston.
Volume of left compartment is filled with an ideal gas and the right compartment is vacuum.
If we release the piston, gas fills the whole space of the cylinder rapidly.
In this expansion. No heat is supplied to the gas as walls are insulated. Q  0 .
As the piston is fixed no work is done by the gas, W  0
hence internal energy remains constant.U  0 ,
T is constant
Such an expansion is called "free expansion".
Polytrophic process:
 In this process the gas obeys an additional law in the form of PV x  constant,
(where x is  ve or  ve constant) along with ideal gas equation PV  nRT
 nR  T
 In this process external Work done is W=
x 1
R R
 In this process Specific heat, C=   1  x  1
PROBLEMS
1. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process
BC is adiabatic. The temperature at A,B and C are 400 k, 800k and 600 k, respectively. Choose the
correct statement. [ MAIN 2014]
P
B
800 K
600 K
A C
400 K
V
1) The change in internal energy in whole cyclic process is 250 R

2) The change in internal energy in the process CA is 700 R
3) The change in internal energy in the process AB is -350 R
4) The change in internal energy in the process BC is -500R
SOLUTION:
5R
Process A  B U  nCV T  1 800  400   1000R
2
5R
C  A ; U  nCV T  1 2
 400  600   500R
U cycle  0 , U AB  U BC  U CA  0
1000 R  U BC  500 R  0 ;  UBC  500R
2. A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that
one part has a pressure P and volume 5V and the other part has pressure 8P and volume V, the
piston is now left free. Find the new pressure and volume for the isothermal and adiabatic
process.    1.5
SOLUTION:
P L 8P
5V V
P L P
V (6V-V )
Final pressure will be same on both sides. Let it be P  , with volume V  , in the left side and  6V  V   in the
right side
(A) if the change is isothermal.
For the gas enclosed in the left chamber,
P  5V  P V  .........( i )
While for the gas in the right chamber

8 P  V  P  6 V  V  .........( ii ) 
Solving these for V  and P  , We get
30 13 48
V  V and P  P and 6V  V   V
13 6 13
(B) If the change is adiabatic. For the gas in the left chamber,

P  5V

  P V    ............(iii)
and for the gas in the right chamber

8 P V 


 P  6V  V   ..........(iv)
dividing (iv) by (iii)
3/ 2
 6V  V   8 6V 4
   3/ 2 or  1
 V  5 V 5
 
10
i.e V   V
3
Substituting it in Equation (iii)
3/ 2
 5V  3  3 3
P  P    P  1.84 P
 10V  2 2
10 8
So P  1.84 P ; V   V and 6V  V   V
3 3
3. A gas undergoes a change of state during which 100 J of heat is supplied to it and it does 20 J of work.
The system is brought back to its original state through a process during which 20 J of heat is released
by the gas. What is the work done by the gas in the second process ?
SOLUTION:
dQ1  dU1  dW1
100  dU1  20  dU1  80 J
dQ2  dU 2  dW2  dU1  dU 2 

20  80  dW2
dW2  60 J
4. During an adiabatic process, if the pressure of an ideal gas is proportional to the cube of its
temperature, find  .
SOLUTION:
For an ideal gas of one mole PV = RT
During an adiabatic process P  T 3
P = kT3 ; where k is a constant
3
 PV   
P = k   P   k  P 3V 3
 R  3
R 
P 2V3  cons tan t ;

PV3/ 2  cons tan t
Comparing it with the equation

PV  constant for an adiabatic process,
we get   3 / 2
5. Pressure versus temperature graph of an ideal gas is as shown in figure. Density of the gas at point
A is  0 . Density at B will be
P
3P0 B
P0 A
T0 T
2T0
SOLUTION:
PM P
 or  
RT T
P P0  P   3  P0
   and     
 T  A T0  T  B  2  T0
P 3 P 3 3
       B   A  0
 T B 2  T  2 2
6. When 5 moles of an ideal gas is compressed isothermally, its volume decreases from 5 litre to 1
 1 
litre. If the gas is at 27°C, find the work done on the gas  log10  5   -0.6990  .
   
SOLUTION:
In the case of 'n' moles, work done on the gas
V 
W  nRT log e  2 
 V1 
V 
W  nRT  2.3026  log10  2 
 V1 
1
 W  5  8.314  300  2.3026  log10  
5
= 5  8.314  300  2.3026  (-0.6990)
= -2.007  104J
7. One mole of a monoatomicidealgas undergoes the process AB in the given

P - V diagram. What is the specific heat for this process ?
P
6P0 B
3P0
A
V0 5V0 V
SOLUTION:
Q 1 W
Specific heat C    U  W   Cv 
T T T
For the given process
9 P0
W  4V0  18 PV
0 0 ( W= area of P – V graph)
2
Also, T  T2  T1
(6 P0 )(5V0 ) (3P0 )V0 27 PV
0 0
= - =
R R R
3
and Cv  R
2
W 3R 18PV 0 0
\ C = Cv + = +
DT 2 æ 27PV
çç 0 0 ÷
ö
çè R ÷ ÷
ø
3R 2 R 13R
= + =
2 3 6
8. Temperature of 1 mole of an ideal gas is increased from 300 K to 310 K under isochoric process.
Heat supplied to the gas in this process is Q = 25 R, where R = universal gas constant. What
amount of work has to be done by the gas if temperature of the gas decreases from 310 K to 300 K
adiabatically?
SOLUTION:
Q  nCV T
 25R = (1) (CV) (310–300)
5
CV  R
2
As the gas is diatomic   1.4
Now work done in adiabatic process
nR (T1 - T2 ) (1)(R)(310 - 300)
W= = = 25 R
g- 1 1.4 - 1
9. Three samples of the same gas A, B and C (  = 3/2) have initially equal volume. Now the volume of each
sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final
pressures are equal for all three samples, find the ratio of their initial pressures
SOLUTION:
Let the initial pressure of the three samples be PA, PB and PC
3 3
then PA V  2   2V  2 P
PB = P ;
PC (V) = P (2V)
 PA : PB : PC = (2)3/2 : 1 : 2  2 2 : 1: 2
10. Work done to increase the temperature of one mole of an ideal gas by 300C, if it is expanding under
2
the condition V  T 3 is  R  8.314 J / mol / K  (EAM-2012)
SOLUTION:
We have, V  T 2 / 3 ;
But PV = RT
3
PV  T ; PV V 2
 P V
 P  k V  P  kV 1/ 2
2k 3/ 2
work done W   PdV  k  V 1/2 dV  V
3
11. Find the molar heat capacity in a process of a diatomic gas if it does a work of Q/4 when a
heat of Q is supplied to it
SOLUTION:
5 
dU = CvdT =  R  dT
2 
2  dU 
dT 
5R
Q 3Q
dU = dQ – dW = Q – 
4 4
Now molar heat capacity
dQ Q ´ 5R
C= =
dT 2 (dU )
5 RQ 10
 R
=  3Q 
2
 4 

3
12. P-V diagram of an ideal gas is as shown in figure. Work done by the gas in the process ABCD is:
(EAM-MED-2011)
P
2P0
C D
P0 A
B
V
SOLUTION:
WAB   PV
0 0;
WBC  0 and WCD  4 PV

0 0
WABCD   PV
0 0  0  4 PV
0 0  3PV
0 0
13. An ideal monoatomic gas is taken round the cycle ABCDA as shown in the P-V diagram. Compute
the work done in this process.
5P0 A B
Pressure (P)
3P0 C
D
2V0 3V0 5V0 6V0
Volume(V)
SOLUTION:
Total work done = Area under P-V curve (parallelogram)
= Base x Height
= (6V0 – 3V0) (5P0 – 3P0)
= (3V0)(2P0) = 6P0V0 units
14. Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of
heat absorbed and the work done by the gas in process 1-2 is :
V
2
1
T
SOLUTION:
V - T graph is a straight line passing through origin. Hence,
V T or P = constant
dQ nC p T

dW nC p T  nCV T
nC p T 5R 5
  
nRT 2R 2
15. In the P-V diagram shown in figure ABC is a semicircle. The work done in the process ABC is :
P(atm)
3 C
B
1 A
V(L)
1 2
SOLUTION:
WAB is negative (volume is decreasing)
WBC is positive (volume is increasing) and since WBC  WAB
 Net work done is positive
 Workdone=area between semicircle
   pressure radius  volume radius 
1 
  1atm   litre 
2 

 atm  litre 
2
16. The volume of one mole of the gas is changed from V to 2V at constant pressure p. If  is the ratio
of specific heats of the gas, change in internal energy of the gas is ( EAM- 2014)
SOLUTION:
dU  nCv dT
R
n dT
 1
PdV
n
 1
P  2V  V  PV
n
 1
 1
17. An ideal gas mixture filled inside a balloon expands according to the relation PV2/3=constant.
What is the temperature inside the balloon
SOLUTION:
PV2/3 = constant
 nRT  2/3
  V   constant
 V 
T .V 1/3  constant
V T3
Temperature increases with increase in volume.
18. Pressure versus temperature graph of an ideal gas is as shown in figure corresponding density   
versus volume(V) graph will be:
P C
B
D
A
T
P C P
B,C
B
1) 2)
D A,D
A
V V
P C o
B C
3) D 4)
B
A A D
V V
SOLUTION:
Along process AB, CD temperature is constant ( isothermal process)
1
i.e., P  ,
V
1
 
V
  V graph will be a rectangular hyperbola.
Along BC and DA,
V is constant
 is constant
19. A tyre pumped to a pressure of 6 atmosphere suddenly bursts. Room temperature is 25oC. Calculate
the temperature of escaping air. (  = 1.4.)
SOLUTION:
From P11 T  P21 T2
Here, P1 = 6 atm ;
P2 = 1 atm ;
T1 = 273+25 = 298K;
 = 1.4
1.4
(6)(1-1.4)(298)1.4 = (1)(1–1.4) T2
(298)1.4
T21.4  (298)1.4 6 0.4 
60.4
1 1.4
 (298)1.4  1.4 (298)1.4 298
T2   0.4   0.4
 2
 6  1.4 7
(6) (6)
2
(or) log T2 = 2.4742 - (0.7782)
7
= 2.4742 - 0.2209 = 2.2533.
Anti log of 2.2533 = 178.7
 T2 = 178.7K
t2 = 178.7 - 273 = - 94.3OC.
20. 100 g of water is heated from 300C to 500C ignoring slight expansion of the water, the change in its
internal energy is (specific heat of water is 4180J/Kg/K) (JEE MAIN-11)
SOLUTION:
dQ  dU  dW ;
dW= 0  dV is neglected 
 dQ  dU  mS 
 100  10  3   4180  20   8360 J  8.4 KJ
21. P - V diagram of a diatomic gas is a straight line passing through origin. What is the molar heat
capacity of the gas in the process
SOLUTION:
P - V diagram of the gas is a straight line passing through origin. Hence,
P V
–1
PV = constant  x  1
Molar heat capacity in the process
R R
C= +
g - 1 1- x
Here,   1.4 (for diatomic gas)
R R
 C  1.4  1  1  1 or C  3 R
22. An ideal gas is taken through the cycle A  B  C  A , as shown in the figure. If the net heat
supplied to the gas in the cycle is 5 J, what is the work done by the gas in the process C  A
2 C B
3
V(m )
1 A
2
10 P(N/m )
SOLUTION:
 WAB  PV  10  2  1  10J

 WBC  0 (as V = constant)
Q  W  U ;
U  0 (process ABCA is cyclic)
 Q  W AB  WBC  WCA
 WCA  Q  W AB  WBC  5  10  0 = –5J
23. In a cyclic process shown in the figure an ideal gas is adiabatically taken from B to A, the work
done on the gas during the process B  A is 30 J, when the gas is taken from A  B the heat
absorbed by the gas is 20J. What is the change in internal energy of the gas in the process A  B .
PA 20 J
30J B
v
SOLUTION:
WBA  30J,QBA  0
; U BA  WBA  30 J
Now, DUAB = - DUBA = - 30J
24. The relation between U, P and V for an ideal gas is U = 2 + 3PV. What is the atomicity of
gas.
SOLUTION:
For an adiabatic process dQ = 0 = dU + dW
or 0 = dU + PdV –––– (1)
From the given equation dU = 3 (PdV + VdP)
Substituting dU from (1),
-PdV = 3 (PdV + VdP)
or 4P (dV) + 3V (dP) = 0
 dV   dP 
4   3  
 V   P 
On integrating, we get
In (V4) + In (P3) = constant,
In (V4P3) = constant
V4P3= constant
PV4/3 = constant

i.e.,  

i.e., gas is polyatomic.
25. If c P and c v denote the specific heats of nitrogen per unit mass at constant pressure and constant
volume respectively, then(AIEEE-2007)
R R
1) c P  cV  2) c P  cV  3) c P  cV  R 4) c P  cV  28 R
28 14
SOLUTION:
R R
Mayer formula c P  cV  
M 28
26. Five moles of hydrogen initially at STP is compressed adiabatically so that its temperature becomes
673 K. The increase in internal energy of the gas, in kilo joule is(R = 8.3 J/mol-K;  1.4 for
diatomic gas) (EAM-2014)
SOLUTION:
Work done by an ideal gas in adiabatic expansion
R  8.3 
dU  n dT  5    300   41500 J
 1  1.4  1 
27. A gas is expanded to double its volume by two different processes. One is isobaric and the other is
isothermal. Let W1 and W2 be the respective work done, then find W1 and W2
SOLUTION:
 Vf 
 
W1  Pi V f  Vi  PV
i i  1
 Vi 
 nRT  2  1  nRT
æV f ö
÷
W2 = nRT log e ççç ÷
÷ = nRT log e (2)= W1 log e (2)
çè Vi ÷
ø
28. Three moles of an ideal monoatomic gas performs a cycle 1  2  3  4  1 as shown. The gas
temperatures in different states are, T1  400 K , T2  800 K ,
T3  2400 K and T4  1200 K . The work done by the gas during the cycle is
P
2 3
1 4
T
SOLUTION:
1  2 and 3  4 are isochoric process.
Therefore, work done is zero.
Wtotal  W23  W41  P2 V3  V2   P4 V1  V4 
 nR T3  T2   nR  T1  T4 
 nR T3  T2  T1  T4 
 800nR  2400 R
29. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should
correspond respectively to:
1
2
V
SOLUTION:
In adiabatic process Slope of P-V graph,
dP P
 
dV V
Slope   (with negative sign)
From the given graph, (slope)2 >(slope)1
 2   1
Therefore, 1 should correspond to O2    1.4  and 2 should correspond to He    1.67  .
30. When a system is taken from state i to state f along the path iaf, it is found that Q = 50 cal and
W = 20 cal. Along the path' ibf Q = 36 cal. W along the path ibf is (AIEEE-2007)
a f
i b
SOLUTION:
From first law of thermodynamics,
dQ  dU  dW
For path iaf, 50  U  20

 U  30 cal
For path ibf, dW  dQ  dU  36  30  6 cal

31. A monoatomic gas undergoes a process given by 2dU + 3dW = 0, then what is the process
SOLUTION:
dQ = dU + dW
2 dU dU 1
 dQ  dU    nCv dT
3 3 3
1 3 nRdT
 n. RdT 
3 2 2
1 dQ R
C 
n dT 2
5R
;It is not isobaric as C is not equal to
2
; It is not adiabatic as C  0
It is not isothermal as C  
so it is a polytrophic process.
32. thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat  .
It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the
surroundings. its temperature increases by (in Kelvin ) (JEE MAIN-2011)
SOLUTION:
1 2
. mv  du  nCV dT ;
2
1 m R
mv 2  T
2 M  1
33. Three moles of an ideal monoatomic gas undergoes a cyclic process as shown in the figure. The
temperature of the gas in different states marked as 1, 2, 3 and 4 are 400K, 700K, 2500K and
1100K respectively. The work done by the gas during the process 1-2-3-4-1 is (universal gas constant
is R) (EAM-2013)
2
3
P 1
4
v
SOLUTION:
Process 1  2 and 3  4 are polytrophic and process 2  3 and 4  1 are isobaric.
From the graph
P
P V   K  PV 1  K  x  1
V
Work done W  W12  W23  W34  W41
nR nR
 T1  T2   P2 V3  V2   T3  T4   P1 V1  V4 
x 1 x 1
nR nR
 T1  T2   nR T3  T2   T3  T4   nR T1  T4 
x 1 x 1
nR
 T1  T2  T3  T4   nR T3  T2  T1  T4 
x 1
3R
  400  700  2500  1100  3R  2500  700  400  1100 
1  1
3R
 1100  3R 1100 
1  1
 1
 3R 1100  1    1650 R
 2
34. moles of an ideal monoatomic gas performs ABCDA cyclic process as shown in figure below. The
gas temperatures are TA  400K , TB  800K , TC  2400K and TD  1200K. The work done by the gas is
(approximately)  R  8.314 J / mol K  (EAMCET-2010)
B C
p
D
T
SOLUTION:
Processes A to B and C to D are parts of straight line graphs of form y = mx
nR
and P  T  n  3 i.e PT
V
So, volume remains constant for the graphs AB and CD.

So, no work is done during processes for A to B and C to D.
WAB  WCD  0 and WBC  P2 VC  VB 
 nR TC  TB   3R  2400  800   4800 R
WDA  P1 VA  VD   nR TA  TD 

 3R  400  1200   2400 R
Work done in the complete cycle
W  WAB  WBC  WCD  WDA

 0  4800R  0   2400 R
 2400 R  19953.6 J  20 kJ
35. An ideal monoatomic gas undergoes a cyclic process ABCA as shown in the figure.The ratio of heat
absorbed during AB to the work done on the gas during BC is:
V
2V0 B
A C
V0
T0 2T0 T
SOLUTION:
5 5nRT0
QAB  nC P T  n R  2T0  T0  
2 2
1
QBC  WBC  nR 2T0 ln     nR 2T0 ln  2 
2
QAB 5
QBC  nR 2T0 ln  2   
WBC 4 ln  2 
36. A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by
the system is :
P
3P0 C B
2P0 O
P0 D
A
v0 2v0 V
SOLUTION:
PV
0 0
WBCOB   Area of triangle BCO  
2
PV
0 0
WAODA   Area of triangle AOD  
2
Wnet  0
Second law of thermodynamics :
Clausius statement: It is impossible for a self acting machine unaided by any external agency to transfer heat
from a cold reservoir to a hot reservoir. In other words heat can’t flow by itself from a colder to a hotter
body.
Kelvin-Planck Statement: It is impossible for any heat engine to convert all the heat absorbed from a reservoir
completely into useful work. In other words 100% conversion of heat into work is impossible.
These two statements of the second law are equivalent to each other. Because, if one is violated,
the other is also automatically.
Reversible process:
A process which can be retraced back in such a way that the system passes through the states as in direct
process and finally the system acquires the initial conditions, leaving no change anywhere else, is called
reversible process. Any quasi-static process can be reversible.
Conditions for a process to be reversible:
(a) There should be no loss of energy due to conduction, convection or dissipation of energy against any
resistance, like friction, viscosity etc.
(b) No heat should be converted into magnetic or electric energy.
(c) The system must always be in thermal, mechanical and chemical equilibrium with the surroundings.(i.e the
process must be quasi-static)
Examples : In practice, there is no
reversible process. But approximately we can give the following examples.
i) The process of change of state from ice into water is a reversible process.
ii) The process of change of state from water to steam.
iii) The gradual extension and compression of an elastic spring is approximately reversible.
iV) The electrolysis process is reversible if internal resistance is negligibly small.
v) Slow compression and expansion of an ideal gas at constant temperature.
Irreversible process:
In this process the system does not pass through the same intermediate states as in the direct process.
All the processes occurring in nature are irreversible.
Examples :1) Diffusion of gas
2) Dissolving of salt in water
3) Sudden expansion or compression of gas
Heat engine :
The device, used to convert heat energy into mechanical energy is called a heat engine.
For conversion of heat into work with the help of a heat engine the following conditions required.
i) There should be a reservoir at constant higher temperature ‘ T1 ’ from which heat is extracted. It is called the
source.
ii) Working substance which undergoes thermodynamic cyclic changes(ex: ideal gas).
iii) There should be a reservoir at constant lower temperature ‘ T2 ’ to which heat can be rejected. This is
called the sink.
 The source and sink should have very high thermal capacity.
Working of heat engine :
a) Schematic diagram of heat engine
Source Engine Sink
T1 T2
Q1 Q2
W=Q1 - Q2
b) Engine derives an amount ‘Q1’ of heat from the source.
c) A part of this heat is converted into work ‘W’.
d) Remaining heat ‘Q2’ is rejected to the sink.
Thus Q1 = W + Q2 or the work done by the engine is given by W  Q1  Q2
e)The energy Q2 is unavailable in the universe, which causes increase in entropy of universe.
Efficiency of heat engine :
 Efficiency of heat engine   is defined as the fraction of total heat, supplied to the engine which is converted
into work.
W Q1  Q2 Q2
Mathematically   Q  Q1
 1
Q1
1
 According to this , efficiency is 100% if Q2 = 0, that is no heat is rejected to the cold reservoir or sink. That
is the entire heat absorbed must be converted to mechanical work , which is impossible according to Second
law of Thermodynamics.
Carnot or Reversible or Ideal heat engine:
 When the working substance is an ideal gas and it is subjected to cyclic process consisting of isothermal
expansion, adiabatic expansion, isothermal compression and adiabatic compression, then such heat engine is
called Carnot engine. The cyclic process is called Carnot cycle.
Carnot Cycle :Carnot cycle consists of the following four stages (i) Isothermal expansion (process AB), (ii)
Adiabatic expansion (process BC), (iii) Isothermal compression (process CD), and (iv) Adiabatic compression
(process DA).
P
A T1
B
Q1
D C
T2 Q2
v
The P-V diagram of the cycle is shown in the figure.
In process AB heat Q1 is taken by the working substance at constant temperature T1 and in process CD heat
Q2 is liberated by the working substance at constant temperature T2. The net work done is area enclosed by
Q2 T2
the cycle ABCDA. After doing the calculations for different processes we can show that : Q = T
1 1
Therefore, efficiency of the Carnot engine is

Q2 T
  1  1 2
Q1 T1
As T2 is always less than T1 ,   1 . i.e., the value of  can never be equal or greater than 1. When the
temperature of sink T2 =0 K, then  can be 1 or 100% . But it is impossible.
 For Carnot engine  is independent of the nature of working substance. It depends on only the temperatures
of source and sink.
 The efficiency of an irreversible engine is always less than or equal to that of reversible engine when operated
between the same temperature limits.  always ir  r
Refrigerator:
Engine
T1 T2
Q1 Q2
W=Q1 - Q2
The refrigerator is just the reverse of heat engine In refrigerator the working substance extracts an amount of
heat Q2 from the cold reservoir (Sink)
at a lower temperature T2 . An amount of external work W is done on the working substance and finally an
amount of heat Q1 is rejected to the hot reservoir at a higher temperature T1 .
 Coefficient of performance of a refrigerator
Q2 Q2
b= = Q Q [ W = Q1 - Q2]
W 1 2
Q2 T2 T2
 For Carnot refrigerator Q = T .Thus   T  T
1 1 1 2
 The relation between efficiency of a heat engine ( ) and coefficient of performance of a refrigerator (b)
1 
working between the same temperature limits is b =

  
Let 1 and 2 are the efficiencies of heat engines working between temperature limits (T1, T2 )and (T2, T3) respectively
then the efficiency of heat engine working between temperature limits T1 and T3 is
T T
1  1  2  2  1  1
T1 T1
T3 T
2  1   3  1  2
T2 T2
T2 T3 T3
1  1 1  2  
 
T1 T2 T1
T3 T3
But   1    1  
T1 T1
  =1- 1-1 1 2 
PROBLEMS
1. Efficiency of a heat engine whose sink is at temperature of 300 K is 40%. To increase
the efficiency to 60%, keeping the sink temperature constant, the source temperature must
increased by
SOLUTION:
T2 40 3
 1   1  
T1 100 5
5 5
 T1  T2  T1   300  500 K
3 3
New efficiency    60%

T2 60 2
'
 1   1  
T1 100 5
5
T1    300  750 K ; T  750 500  250K
2
2. A refrigerator, whose coefficient of performance  is 5, extracts heat from the cooling compartment
at the rate of 250 J per cycle.
(a) How much work per cycle is required to operate the refrigerator?
(b) How much heat per cycle is discharged to the room which acts as the high temperature
reservoir?
SOLUTION:
Q L250
a) As coefficient of performance of a refrigerator is defined as   Q L / W ,So W    5  50 J
(b)As Q H  Q L  W ; so Q H  250  50  300 J

3. A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered
by 62K, its efficiency increases to 1/3. Then T1 and T2 are, respectively : (JEE MAIN-2011)
SOLUTION:
T2 1 T T2 5
  1   1 2  
T1 6 T1 T1 6 .....(1)
T2  62 1 T  62
2  1    1 2
T1 3 T1 .......(2)
On solving Equation (1) and (2)
T1  372 K and T2  310 K
4. Find the efficiency of the thermodynamic cycle shown in figure for an ideal diatomic gas.
2P0 B
Pressure (P)
P0 A C
V0 2V0
Volume(V)
SOLUTION:
Sol. Let n be the number of moles of the gas and the temperature be T0 in the state A.
Now, work done during the cycle
1 1
W    2V0  V0  2 P0  P0   PV
0 0
2 2
For the heat  Q1  given during the process A  B , we have Q1  WAB  U AB
WAB = area under the straight line AB
1 3P V

2
 P0  2P0  2V0  V0   0 0
2
Applying equation of state for the gas in the state A & B.
P0 V0  2P0  2V0 
  TB  4T0
T0 TB
 5R  15nRT0 15P0 V0
 U AB  nC V T  n    4T0  T0   
 2  2 2
3 15
Q1  P0 V0  P0 V0  9P0 V0
2 2
Obviously, the processes B  C and C  A involve the abstraction of heat from the gas.
Efficiency = Work done per cycle

Total heat supplied per cycle
1
P0 V0
1
i.e., = 2 
9P0 V0 18
5. A Carnot engine, having an efficiency of   1/10 as heat engine, is used as a refrigerator. If the
work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower
temperature is
SOLUTION: (JEE MAIN-2007)
Sol. Coefficient of performance of refrigerator
1  1  1/10 Heat extracted
  9 
 1/10 workdone
 Heat extracted  9  10  90J
6. Helium gas undergoes through a cycle ABCD (consisting of two isochoric and isobaric line) as
shown in figure. Efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)
2P0 B C
P0 D
(JEE MAIN-2012) A
V0 2V0
SOLUTION:
work done in cycle
Sol. Efficiency  heat absorbed
 100
Area under P  V diagram


QAB  QBC
PV0 0
 
nCV T1  nC p T2
PV0 0 PV
  0 0
3 5 3 5
nR TB  TA   nR TC  TB   2 PV 0 0
0 0  PV  4 PV 0 0
0 0  2 PV
2 2 2 2
PV
0 0 1
   15.4%
3 5 6.5
PV
0 0  2PV
0 0
2 2
7. A Carnot engine, whose efficiency is 40% takes heat from a source maintained at a temperature of
500 K. It is desired to have an engine of efficiency 60%. Then , the intake temperature for the
same exhaust (sink) temperature must be (JEE MAIN-12)
SOLUTION:
T2 T
Sol.   1  T ; 0.4  1  2  T2  300 K
1 500
T2 300 300
0.6  1  1
 1  1  T11   750 K
T1 T1 0.4
8. A diatomic ideal gas is used in a car engine as the working substance. Volume of the gas increases
from V to 32V during the adiabatic expansion part of the cycle. The efficiency of the engine is
(JEE MAIN-2010)
SOLUTION:
T2 1 T 1 3
Sol. TV  1
 T2V2 1  TV 1.41
 T2  32V1 
1.4 1
     1 2  1 
1 1 1 1 T1 4 T1 4 4
Entropy(s):
A  The thermodynamic coordinate or parameter that gives the measure of disorder is called entropy. We cannot
measure entropy, but we can measure change in entropy during thermodynamic change. If ‘ds’ is the small
dQ
change in entropy at temperature T, then ds =
T
Where dQ is exchange of heat between system and surroundings at temperature T..
dQ
Now the total change in entropy is s  
T
Q W V 
 Change in entropy during an isothermal change is s=   2.303nR log10  2 
T T  V1 
mL
 Change in entropy during phase change is s=
T
 Change in entropy during temperature change is
sdT
s=m  , (if s is temperature dependent)
T
 In a reversible process entropy increases if heat is absorbed and vice - versa.
 Entropy of the universe always increases if system undergoes an irreversible process.
 Entropy of universe can never be zero.
 At absolute zero temperature(0K), entropy becomes zero. But it does not occur.
THEORY BITS
1. Which of the following does not characterise the thermodynamic state of matter
1) volume 2) temperature 3) pressure 4) work
KEY : 4
2. The thermal motion means
1) motion due to heat engine 2) disorderly motion of the body as a whole
3) motion of the body that generates heat 4) random motion of molecules
KEY : 4
3. Heat required to raise the temperature of one gram of water through 10 c
1) 0.001 K cal 2) 0.01 K cal 3) 0.1 K cal 4) 1.0 K cal
KEY : 1
4. Heat capacity of a substance is infinite. It means
1) heat is given out 2) heat is taken in
3) no change in temperature whether heat is taken in (or) given out 4) all of the above
KEY : 3
5. For a certain mass of a gas Isothermal relation between ‘P’ and ‘V’ are shown by the graphs at two
P
different temperatures T1 and T2 then T2

T1
v
1)T1 = T2 2)T1>T2 3)T1 < T2 4)T1  T2
KEY : 3
6. The temperature range in the definition of standard calorie is

0
1) 14.5 0 C to 15.5 0 C 2) 15.5 0 C to 16.5 0 C 3) 1 C to 2 0 C 4) 13.5 0 C to 14.5 0 C
KEY : 1
7. The pressure p of a gas is plotted against its absolute temperature T for two different constant
volumes V1 and V2, where V1 > V2 .If p is taken on y–axis and T on x–axis.
1) The curve for V1 has greater slope than the curve for V2
2) The curve for V2 has greater slope than the curve for V1
3) The curves must intersect at some point other than T = 0
4) The curves have the same slope and do not intersect
KEY : 2
8. The internal energy of a perfect monoatomic gas is
1) Complete kinetic 2) Complete potential
3) Sum of potential and kinetic energy of the molecules
4) Difference of kinetic and potential energies of the molecules
9. dU+dW=0 is valid for
1) adiabatic process 2) isothermal process 3) isobaric process 4) isochoric process
KEY : 1
10. A piece of ice at 0 0C is dropped into

water at 00C. Then ice will
1) melt 2) be converted into water 3) not melt 4) partially melt
KEY : 3
11. The temperature determines the direction of net change of
1) gross kinetic energy 2) intermolecular kinetic energy
3) gross potential energy 4) intermolecular potential energy
KEY : 2
12. A closed vessel contains some gas at a given tmperature and pressure. If the vessel is given a very
high velocity, then the temperature of the gas
1) increases 2) decreases
2) may increase or decrease depending upon the nature of the gas 4) does not change
KEY : 4
13. Unit mass of liquid of volume V1 completely turns into a gas of volume V2 at constant atmospheric
pressure P and temperature T. The latent heat of vaporization is “L”. Then the change in internal
energy of the gas is
1) L 2) L+P(V2 - V1) 3) L - P(V2-V1) 4) Zero
KEY : 3
14. Zeroth law of thermodynamics gives the concept of
1) Pressure 2) Volume 3) Temperature 4)Heat
KEY : 3
15. We need mechanical equivalent of heat because
1) it converts work into heat
2) in C.G.S system, heat is not measured in the units of work
3) in SI system, heat is measured in the units of work
4) of some reason other than those mentioned above
KEY :2
16. When we switch on the fan in a closed room. The temperature of the air molecules
1) Increases 2) Decreases
3) Remains unchanged
4) May increase or decrease depending on the speed of rotation of the fan.
KEY : 1
17. Which type of molecular motion does contribute towards internal energy for an ideal monoatomic
gas
1) Translational 2) Rotational 3) Vibrational 4) All the above
KEY : 1
18. In which of the following processes the internal energy of the system remains constant ?
1) Adiabatic 2) Isochoric 3) Isobaric 4) Isothermal
KEY : 4
19. In a given process dW=0, dQ<0 then for a gas
1)Temperature–increases 2)Volume–decreases 3)Pressure–decreases 4)Pressure–increases
KEY : 3
20. Which of the following is constant in an isochoric process
1) Pressure 2) Volume 3) Temperature 4) Mass
KEY : 2
21. How does the internal energy change when the ice and wax melt at their normal melting points?
1) Increases for ice and decreases for wax 2) Decreases for ice and increases for wax
3) Decreases both for ice and wax 4) Increases both for ice and wax
KEY : 25)1
22. On compressing a gas suddenly, its temperature
1) increases 2) decreases 3) remains constant 4) all the above
KEY : 1
23. Certain amount of heat supplied to an ideal
gas under isothermal conditions will result in
1) raise in temperature 2) doing external work and a change in temperature
3) doing external work 4) an increase in the internal energy of the gas
KEY : 3
24. When heat is added to a system at constant temperature, which of the following is possible.
1) internal energy of system increases 2) work is done by the system
3) neither internal energy increases nor work done by the system
4) internal energy increases and work is done by the system
KEY : 2
25. The first law of thermodynamics is based on the law of conservation of
1) energy 2) mass 3) momentum 4) pressure
KEY : 1
26. A liquid in a thermos flask is vigorously shaken. Then the temperature of the liquid
1) Is not altered 2) Increases 3) Decreases 4) None
KEY :4
27. The PV diagram shows four different possible paths of a reversible processes performed on a
monoatomic ideal gas. Path A is isobaric, path B is isothermal , path C is adiabatic and path D is
isochoric . For which process does the temperature of the gas decrease?
P
A
P0
D B
1
P C
2 0
V
V0 2V0
1) Process A only 2) Process C only 3) Processes C & D 4) Processes B, C & D

KEY :3
28. A given mass of a gas expands from the state A to the state B by three paths 1,2 and 3 as shown in
the figure. If W1, W2 and W3 respectively be the work done by the gas along the three paths then
A
3
P 2
1
B
V
1) W1 > W2 > W3 2) W1 < W2 < W3 3) W1 = W2 = W3 4) W1 < W2 = W3
KEY :2
29. A given system undergoes a change in which the work done by the system equals to the decrease in
its internal energy. The system must have undergone an
1) Isothermal change 2) Adiabatic change 3) Isobaric change 4) Isochoric change
KEY :2
30. When two blocks of ice are pressed against each other then they stick together because
1) cooling is produced 2) heat is produced
3) increase in pressure will increase in melting point
4) increase in pressure will decrease in melting point
KEY :4
31. A cubical box containing a gas with internal energy U is given velocity V, then the new internal
energy of gas
1) less than U 2)more than U 3) U 4) zero
KEY :3
32. In an isobaric (constant pressure) process. the correct ratio is
1) DQ : DU = 1 : 1 2) DQ : DU = 1 : g-1 3) DQ : DU = g-1: 1 4) DQ : DU = g : 1
KEY :4
33. In an isobaric process, the correct ratio is
1) DQ : DW = 1 : 1 2) DQ : DW = g : g-1 3) DQ : DW = g-1: g 4) DQ : DW = g : 1
KEY :2
34. Air in a thermally conducting cylinder is suddenly compressed by a piston, which is then maintained
at the same position. With the passage of time :
1) the pressure decreases 2) the pressure increases 3) the pressure remains the same
4) the pressure may increase or decrease depending upon the nature of the gas
KEY :1
35. Which of the following states of matter have two specific heats ?
1) solid 2) gas 3) liquid 4) Plasma
KEY :2
36. The specific heat of a gas in an isothermal process is
1) infinity 2) zero 3) negative 4) remains constant
KEY :1
37. Why the specific heat at a constant pressure is more than that at constant volume
1) there is greater inter molecular attraction at constant pressure
2) at constant pressure molecular oscillations are more violent
3) external work need to be done for allowing expansion of gas at constant pressure
4) due to more reasons other than those mentioned in the above
KEY :3
38. Two identical samples of gases are allowed to expand to the same final volume (i) isothermally (ii)
adiabatically. Work done is
1. more in the isothermal process 2. more in the adiabatic process
3. equivalent in both processes 4. equal in all process
KEY :1
39. Which of the following is true in the case of a reversible process
1) There will be energy loss due to friction
2) System and surroundings will not be in thermo dynamic equilibrium
3) Both system and surroundings retains their initial states
4) 1 and 3
KEY :3
40. The ratio of the relative raise in pressure for adiabatic compression to that for isothermal
compression is
1 1
1)  2) 3) 1   4)
 1 
KEY :1
41. Ratio of isothermal elasticity of gas to the adiabatic elasticity is
1 1
1)  2) 3) 1   4)
 1 
KEY :2
42. The conversion of water into ice is an
1) isothermal process 2) isochoric process 3) isobaric process 4) entropy process
KEY :47
43. For the Boyle’s law to hold good, the necessary condition is
1) Isobaric 2) Isothermal 3) Isochoric 4) Adiabatic
KEY :2
44. An isothermal process is
1)slow process 2)quick process 3) very quick process 4) both 1 & 2
KEY :1
45. The temperature of the system decreases in the process of
1) free expansion 2) isothermal expansion
3) adiabatic expansion 4) isothermal compression
KEY :3
46. The pressure P and volume V of an ideal gas both increase in a process
1) It is not possible to have such a process 2) The workdone by the system is positive
3) The temperature of the system increases 4) 2 and 3
KEY :2
47. Two samples of gas A and B, initially at same temperature and pressure, are compressed to half of
their initial volume, A isothermally and B adiabatically. The final pressure in
1) A and B will be same 2) A will be more than in B
3) A will be less than in B 4)A will be double that in B
KEY :3
48. In which of the following processes all three thermodynamic variables, that is pressure volume and
temperature can change
1) Isobaric 2) Isothermal 3) Isochoric 4) Adiabatic
KEY :4
49. Two steam engines ‘A’ and ‘B’, have their sources respectively at 700 K and 650 K and their sinks
at 350 K and 300K. Then
1) ‘A’ is more efficient than ‘B’ 2) ‘B’ is more efficient than ‘A’
3) both are equally efficient 4) depends on fuels used in A and B
KEY :2
50. During adiabatic expansion the increase in volume is associated with
1) increase in pressure and temperature 2) decrease in pressure and temperature
3) increase in pressure and decrease in temperature
4)Decrese in pressure and increase in temperature
KEY :2
51. A gas is being compressed adiabatically. The specific heat of the gas during compression is
1) zero 2) infinite 3) finite but non zero 4) undefined
KEY :1
52. During adiabatic compression of a gas, its temperature
1) falls 2) raises 3) remains constant 4) becomes zero
KEY :2
53. The work done on the system in an adiabatic compression depends on
1) the increase in internal energy of the system 2) the decrease in internal energy
3) the change in volume of the system 4) all the above
KEY :1
54. The ratio of slopes of adiabatic and isothermal curves is
1)  2) 1/  3)  2 4)  3
KEY :1
55. If the temperature of the sink is decreased, then the efficiency of heat engine
1) first increases then decreases 2) increases
3) decreases 4) remains unchanged
KEY : 2
56. An ideal heat engine can be 100% efficient if its sink is at
1) 0K 2) 273K 3) 00C 4) 00F
KEY :1
57. If the temperature of a source increases, then the efficiency of a heat engine
1) increases 2) decreases 3) remains unchanged 4) none of these
KEY :1
58. By opening the door of a refrigerator inside a closed room:
1) you can cool the room to a certain degree 2) you can cool it to the temperature inside the refrigerator
3) you can ultimately warm the room slightly 4) you can neither cool nor warm the room
KEY :3
59. Which of the following will extinguish the fire quickly
1) water at 1000C 2) steam at 1000C 3) water at 00C 4) ice at 00C
KEY :71)1
60. Which of the following is true in the case of molecules, when ice melts
1) K.E is gained 2) K.E. is lost 3) P.E is gained 4) P.E. is lost
KEY :3
61. When heat is added to a system then the following is not possible?
1) Internal energy of the system increases 2) Work is done by the system
3) Neither internal energy increases nor work is done by the system
4) Internal energy increases and also work is done by the system
KEY :3
62. A sink, that is the system where heat is rejected, is essential for the conversion of heat into work.
From which law the above inference follows?
1) Zeroth 2) First 3) Second 4) Both 1 & 2
KEY :3
 PV 
63. The gas law  = constant is true for
 T 
1) isothermal change only 2) adiabatic change only
3) Both isothermal & adiabatic changes 4) neither isothermal nor adiabatic change
KEY :3
64. The efficiency of a heat engine:
1) is independent of the temperature of the source and the sink
2) is independent of the working substance
3) can be 100%
4) is not affected by the thermal capacity of the source or the sink
KEY :2
65. An ideal heat engine working between temperatures TH and TL has efficiency  . If both the
temperatures are raised by 100K each, then the new efficiency of the heat engine will be:
1) equal to  2) greater than  3) less than 
4) greater or less than  depending upon the nature of the working substance
KEY :3
66. The efficiency of the reversible heat engine is  r , and that of irreversible heat engine is I . Which
of the following relation is correct?
1) r   I 2) r   I 3)  r   I 4)r  1 and I  1
KEY :3
67. In a heat engine, the temperature of the working substance at the end of the cycle is
1) equal to that at the beginning 2) more than that at the beginning
3) less than that at the beginning 4) determined by the amount of heat rejected to the sink
KEY :1
68. The adiabatic and isothermal elasticities B and B are related as :
B B
1) B   2) B   3) B  B   4) B  B  
 
KEY :1
69. For the indicator diagram given below, select wrong statement.
P
I II
V
1) Cycle - II is heat engine cycle 2) Net work is done on the gas in cycle - I
3) Workdone is positive for cycle - I 4) Workdone is positive for cycle - II
KEY :3
70. A cubical box containing a gas is moving with some velocity. If it is suddenly stopped, then the
internal energy of gas
1) decreases 2) Increases 3) remains constant
4) may increases or decreases depending on the time interval during which box comes to rest.
KEY :2
71. The ratio [Cp / Cv] of the specific heats at a constant pressure and at a constant volume of any
perfect gas
1) can’t be greater than 5/4 2) can’t be greater than 3/2
3) can’t be greater than 5/3 4) can have any value
KEY :3
72. Which of the following formula is wrong ?
R Cp R
1) Cv =   1 2)  3) Cp =  1 4) Cp - Cv = 2R
CV
KEY :4
73. A common salt is first dissolved in water and extracted again from the water. In this process,
1) entropy decreases 2) entropy increases
3) entropy becomes zero 4) entropy remains constant.
KEY :2
74. A large block of ice is placed on a table where the surroundings are at 00C
1) ice melts at the sides 2) ice melts at the top
3) ice melts at the bottom 4) ice does not melt at all
KEY :3
75. Which of the following substance at 1000C produces most severe burns ?
1) hot air 2) water 3) steam 4) oil
KEY :3
76. What energy transformation takes place when ice is converted into water
1) heat energy to kinetic energy 2) kinetic energy to heat energy
3) heat energy to latent heat 4) heat energy to potential energy
KEY :4
77. Which of the following laws of thermodynamics leads to the interference that it is difficult to
convert whole of heat into work
1) zeroth 2)second 3) first 4)both 1 & 2
KEY :2
78. The direction of flow of heat between two gases is determined by
1) Average kinetic energy 2) total energy 3) internal energy 4) potential energy
KEY : 1
79. Heat is absorbed by a body . But its temperature does not raised. Which of the following statement
explains the phenomena ?
1) only K.E. of vibration increases 2) only P.E. of inter molecular force changes
3) no increase in internal energy takes place 4) increase in K.E. is balanced by decrease in P.E.
KEY : 2
80. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 T h e
amount of work done by the gas is greatest when the expansion is
1) isothermal 2) isobaric 3) adiabatic 4) equal in all cases
KEY :2
81. The second law of thermodynamics implies:
1) whole of heat can be converted into mechanical energy
2) no heat engine can be 100% efficient
3) every heat engine has an efficiency of 100%
4) a refrigerator can reduce the temperature to absolute zero
KEY :2
82. In the free expansion of a gas, its internal energy
1) remains constant 2) increases 3) decreases
4) sometimes increases , sometimes decreases
KEY :1
83. The internal energy of an ideal gas depends upon
1) only its pressure 2) only its volume 3) only its temperature 4) its pressure and volume
KEY :3
84. In the adiabatic compression the decrease in volume is associated with
1)increase in temperature and decrease in pressure
2) decrease in temperature and increase in pressure
3) decrease in temperature and decrease in pressure
4) increase in temperature and increase in pressure
KEY : 4
85. Which of the following is true in the case of an adiabatic process where   CP / CV ?
1) P1 T   constant 2) P T 1  constant 3) PT   constant 4) P T  constant
KEY :1
86. If an ideal gas is isothermally expanded its internal energy will
1) Increase 2) Decrease 3) Remains same
4) Decrease or increase depending on nature of the gas
KEY :3
87. Heat engine rejects some heat to the sink. This heat
1)converts into electrical energy. 2)converts into light energy.
3)converts into electromagnetic energy 4)is unavailable in the universe.
KEY :4
88. For an adiabatic change in a gas, if P, V,T denotes pressure, volume and absolute temperature of
gas at any time and  is the ratio of specific heats of a gas, then which of the following equation is
true?
1) T γ P1 γ  const. 2) T 1-γ P γ  const. 3) T γ-1 V γ  const. 4) T γ V γ  const.
KEY :1
89. PV versus T graph of equal masses of H2, He and CO2 is shown in figure. Choose the correct
alternative
3
2
PV 1
T
(1) 3 corresponds to H2, 2 to He and 1 to CO2 (2) 1 corresponds to He, 2 to H2 and 3 to CO2
(3) 1 corresponds to He, 3 to H2 and 2 to CO2 (4) 1 corresponds to CO2,2 to H2 and 3 to He
KEY :1
90. If the ratio of specific heat of a gas at constant pressure to that at constant volume is  , then the
change in internal energy of the mass of gas, when the volume changes from V to 2V at constant
pressure P, is
1) R /(  1) 2) PV 3) PV / (  1) 4)  PV/ (  1)
KEY :3
91. Heat is added to an ideal gas and the gas expands. In such a process the temperature
1) must always increase 2) will remain the same if the work done is equal to the heat added
3) must always decrease
4) will remain the same if change in internal energy is equal to the heat added
KEY :2
92. First law of thermodynamics states that
1) system can do work 2) system has temperature
3) system has pressure 4) heat is form of energy
KEY :1
93. The material that has largest specific heat is
1) mercury 2) water 3) hydrogen 4) diamond
KEY :3
94. The law obeyed by isothermal process is
1) Gay-Lussac’s law 2) Charles law 3) Boyle’s law 4) Dalton’s law
KEY :3
95. Which one of the following is wrong statement.
1) During free expansion, temperature of ideal gas does not change.
2) During free expansion, temperature of real gas decreases.
3) During free expansion of real gas temperature does not change.
4) Free expansion is conducted in adiabatic manner.
KEY :3
96. Which law defines entropy in thermodynamics
1) zeroth law 2) First law 3) second law 4) Stefan’s law
KEY :3
97. For the conversion of liquid into a solid
1) orderliness decreases and entropy decreases 2) orderliness increases and entropy increases
3) both are not related 4) orderliness increases and entropy decreases
KEY :1
98. Among the following the irreversible process is
1) free expansion of gas 2)extension or compression of spring very slowly
3)motion of an object on a perfectly frictionless surface 4) all of them
KEY :1
99. Water is used in car radiators as coolant because
1) its density is more 2) high specific heat 3) high thermal conductivity 4) free availability
KEY : 2
100. Of the following specific heat is maximum for
1) Mercury 2) Copper 3) Water 4) Silver
KEY : 3
101. Heat is
1) kinetic energy of molecules 2) potential and kinetic energy of molecules
3) energy in transits 4) work done on the system
KEY : 3
102. Gas is taken through a cyclic process completely once. Change in the internal energy of the gas is
1) infinity 2) zero 3) Small 4) Large
KEY :2
103. What will be the nature of change in internal energy in case of processes shown below ?
P P
v v
P P
v v
1) + ve in all cases 2) – ve in all cases

3) – ve in 1 and 3 and + ve in 2 and 44) zero in all cases
KEY :4
104. Which of the following is incorrect regarding the first law of thermodynamics ?
1) It introduces the concept of internal energy
2) It introduces the concept of entropy
3) It is applicable to any process
4) It is a restatement of principle of conservation of energy.
KEY :2
105. The heat capacity of material depends upon
1) the structure of a matter 2) temperature of matter
3) density of matter 4) specific heat of matter
KEY : 4
106. Heat cannot flow by itself from a body at lower temperature to a body at higher temperature is a
statement or consequence of
1) Ist law of thermodynamics 2) IInd law of thermodynamics
3) conservation of momentum 4) conservation of mass
KEY :4
107. For an isothermal process
1) dQ  dW 2) dQ  dU 3) dW  dU 4) dQ  dU  dW
KEY :1
108. When thermodynamic system returns to its original state, which of the following is NOT possible?
1) The work done is Zero 2) The work done is positive
3) The work done is negative 4) The work done is independent of the path followed
KEY :2
109. Two completely identical samples of the same ideal gas are in equal volume containers with the
same pressure and temperature in containers labeled A and B. The gas in container A performs
non-zero positive work W on the surroundings during an isobaric process before the pressure is
reduced isochorically to 1/2 its initial amount. The gas in container B has its pressure reduced
isochorically to 1/2 its initial value and then the gas performs same non-zero positive work W on
the surroundings during an isobaric process. After the processes are performed on the gases in
containers A and B, which is at the higher temperature?
1)The gas in container A 2)The gas in container B
3) The gases have equal temperature.
4) The value of the work W is necessary to answer this question.
KEY :2
110. For an adiabatic process the relation between V and T is given by
1) TV   constant 2) T  V  constant 3) TV 1  constant 4) TV  1  constant
KEY :4
88. The temperature of the system decreases in the process of
1) Free expansion 2) Adiabatic expansion
3) Isothermal expansion 4)Isothermal compression
KEY :2
111. Which of the following conditions of the Carnot ideal heat engine can be realised in practice?
1) infinite thermal capacity of the source 2) infinite thermal capacity of the sink
3) perfectly non conducting stand 4) Less than 100% efficiency
KEY :4
113. A heat engine works between a source and a sink maintained at constant temperatures T1 and T2 .
For the efficiency to be greatest
1) T1 and T2 should be high 2) T1 and T2 should be low
3) T1 should be high and T2 should be low 4) T1 should be low and T2 should be high
KEY :3
114. The heat engine would operate by taking heat at a particular temperature and
1) Converting it all into work
2) Converting some of it into work and rejecting the rest at lower temperature
3) Converting some of it into work and rejecting the rest at same temperature
4) Converting some of it into work and rejecting the rest at a higher temperature .
KEY :2
115. Which of the following processes are nearly reversible
a. Heat conduction b. Electrolysis
c. Diffusion d. Change of state
1) Only a 2)Both b and d 3) Only c 4) All of the above
KEY :2
PRACTICEBITS
1. A piece of lead falls from a height of 100m on a fixed non-conducting slab which brings it to rest. If
the specific heat of lead is 30.6 cal/kg °C, the increase in temperature of the slab immediately after
collision
1) 6.72°C 2) 7.62°C 3) 5.62°C 4) 8.72°C
KEY : 1) 2
HINT: gh
mgh  JmS    
JS
2. Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find
the height from which they fall.
(g = 10 ms -2 . L = 80 calorie/g and
J = 4.2J/calorie)
1) 336 m 2) 236 m 3) 436 m 4) 536 m
KEY :2) 1
HINT:
JmLice
mgh 
100
3. From what minimum height a block of ice has to be dropped in order that it may melt completely on
hitting the ground
mgh JL J
1) mgh 2) 3) 4)
J g Lg
KEY:3) 3
HINT:
W= JH  mgh = JmLice
4. Two spheres A and B with masses in the ratio 2 : 3 and specific heat 2 : 3 fall freely from rest. If the
rise in their temperatures on reaching the ground are in the ratio 1 : 2 the ratio of their heights of
fall is
1) 3 : 1 2) 1 : 3 3) 4 : 3 4) 3 : 4
KEY: 4) 2
HINT:
mgh  mS   h S 
h1 S1 1
  
h2 S 2  2
1
5. From what height a block of ice must fall into a well so that th of its mass may be melted (g = 10
100
m/s2)
1) 300 m 2) 336 m 3) 660 m 4) none
KEY: 5) 2
HINT:
m
mgh  Lice
100
6. Two identical balls ‘A’ and ‘B’ are moving with same velocity. If velocity of ‘A’ is reduced to half
and of ‘B’ to zero, then the raise in temperatures of ‘A’ to that of ‘B’ is
1) 3 : 4 2) 4 : 1 3) 2 : 1 4) 1 : 1
KEY:1
HINT:
1  v2  v2
mS   J m  v22  v12   1  12 11
2 2
2  2 v2  v1
7. A 50kg man is running at a speed of 18kmh 1 . If all the kinetic energy of the man can be used to
increase the temperature of water from 200 C to 300 C , how much water can be heated with this
energy?
1) 15 g 2) 20 g 3) 30 g 4) 40 g
KEY:1
HINT:
1 2
W= JH  mv  JmS 
2
8. A man of 60 kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what
height he can jump by using this energy?
1) 4m 2) 20 m 3) 28 m 4) 0.2 m
KEY: 1
HINT:
. mgh=JH
9. How much work to be done in decreasing the voume of an ideal gas by an amount of
2. x 10–4 m3 at constant normal pressure of 1 x 105 N/m2
1) 28 joule 2) 27 joule 3) 24 joule 4)25joule
KEY: 3
HINT:
dW  PdV
10. Find the external work done by the system in kcal, wen 20 kcal of heat is supplied to thesystem and
the increase in the internal energy is 8400 J (J=4200J/kcal)
1) 16 kcal 2) 18 kcal3) 20 kcal 4)19 kcal
KEY:2
HINT:
dQ  dU  dW  dW  dQ  dU
11 Heat of 30 kcal is supplied to a system and 400 J of external work is done on the system so that its
volume decreases at constant pessure. What is the change in its internal energy. (J = 4200 J/kcal)
1) 1.302 x 105 J 2) 2.302 x 105 J

3) 3.302 x 105 J 4) 4.302 x 105 J
KEY: 1
HINT:
dQ  dU  dW
12. Air expands from 5 litres to 10 litres at 2 atm pressure. External workdone is
1) 10J 2) 1000J 3) 3000 J 4) 300 J
KEY:2
HINT:
W  P V2  V1 
13. Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the
internal energy of the system will be
1) – 50 J 2) 20 J 3) 30 J 4) 50 J
KEY:2
HINT:
dU  dQ  dW
14. A gas is compressed at a constant pressure of 50 N/m2 from a volume 10m3 to a volume of 4m3.
100J of heat is added to the gas then its internal energy
1) Increases by 400J 2) Increases by 200J 3) Decreases by 400J 4) Decreases by 200J
KEY:1
HINT:
dU  dQ  P V2  V1 
15. Find the change in internal energy in joule. When 10g of air is heated from 30°C to 40°C (cv = 0.172
kcal/kg K, J = 4200 J/kcal)
1) 62.24 J 2)72.24 J 3)52.24 J 4)82.24 J
KEY:2
HINT:
dUV  mcv dT
16. The temperature of 5 moles of a gas at constant volume is changed from 1000C to 1200C. The
change in internal energy is 80J. The total heat capacity of the gas at constant volume will be in
joule/kelvin is
1) 8 2) 4 3) 0.8 4) 0.4
KEY:2
HINT:
dQ  dU  PdV  dU  P  0   dU
 dQ  dU
  
 dT V dT
17. When an ideal diatomic gas is heated at constant pressure, the fraction of heat energy supplied
which is used in doing work to maintain pressure constant is
1) 5/7 2) 7/2 3) 2/7 4) 2/5
KEY:3
HINT:
dW 1
 1
dQ 
18. When a monoatomic gas expands at constant pressure, the percentage of heat supplied that increases
temperature of the gas and in doing external work in expansion at constant pressure is
1) 100%, 0 2) 60%, 40% 3) 40%, 60% 4) 75%, 25%
KEY:2
HINT:
dU 1 dW 1
 ;  1
dQ  dQ 
19. For a gas, the difference between the two specific heats is 4150J Kg-1 K-1 and the ratio of specific
heats is 1.4. What is the specific heat of the gas at constant volume in
J Kg-1 K-1?
1)8475 2) 5186 3)1660 4) 10375
KEY:4
HINT:
R
CP  CV  R  CV 
 1
20. The specific heat of air at constant pressure is 1.005 kJ/kg K and the specific heat of air at constant
volume is 0.718 kJ/kgK. Find the specific gas constant.
1) 0.287 kJ/kg K 2) 0.21 kJ/kg K 3) 0.34 kJ/kg K 4) 0.19 kJ/kg K
KEY:1
HINT:
cP  cV  r
21. The specific heat of Argon at constant volume is 0.3122kJ/kg K. Find the specific heat of Argon at
constant pressure if R = 8.314 kJ/k mole K. (Molecular weight of argon = 39.95)
1) 5203 2) 5302 3) 2305 4) 3025
KEY: 1
HINT:
R
c P  cV 
M
22. If the ratio of the specific heats of steam is 1.33 and R = 8312J/k mole K find the molar heat
capacities of steam at constant pressure and constant volume.
1) 33.5 kJ/k mole, 25.19 kJ /kg K 2) 25.19 kJ/k mole, 33.5 kJ/kg K
3) 18.82 kJ/k mole, 10.82 kJ/k mole 4) 24.12 kJ /k mole, 16.12 kJ/k mole
KEY:1
HINT:
R R
CP  ; CV 
 1  1
23. One mole of an ideal gas undergoes an isothermal change at temperature 'T' so that its volume V
is doubled. R is the molar gas constant. Work done by the gas during this change is
(2008 M)
1) RT log4 2) RT log2 3) RT log1 4) RT log3
KEY:2
HINT:
V 
W  nRT log e  2 
 V1 
24. One mole of O2 gas having a volume equal to 22.4 litres at 00C and 1 atmospheric pressure is
compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is
1)1672.5J 2)1728J 3) –1728J 4) –1572.5J
KEY:4
HINT:
V 
W  2.303nRT log10  2 
 V1 
25. The isothermal Bulk modulus of an ideal gas at pressure P is

1) P 2) P 3) P/2 4) P / 
KEY:1
HINT:
Isothermal process PdV+VdP=0
dP dV dP dP
 ; K    P
P V  dV   dP 
   
 V   P 
26. Diatomi pressure becomes 8 times the initial pressure, then the final temperature is    3 / 2
1) 6270 C 2) 5270C 3) 4270C 4) 3270C
KEY:3
HINT:

 p V 
PV  K  1   2 
p2  V1 
1
27. The volume of a gas is reduced adiabatically to of its volume at 270C, if the value of   1.4, then
4
the new temperature will be
1) 350 x 40.4K 2) 300 x 40.4K 3) 150 x 40.4K 4) None of these
KEY:2
HINT:
 1
 V1 
T2  T1  
 V2 
28. Two moles of an ideal monoatomic gas at 270C occupies a volume of V. If the gas is expanded
adiabatically to the volume 2V, then the work done by the gas will be   5 / 3
1) –2767.23J 2) 2767.23J 3) 2500J 4) –2500J
KEY:2
HINT:
nR
W T  T 
 1 1 2
29. A container of volume 1m3 is divided into two equal compartments, one of which contains an ideal gas at
300 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings.
The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature
now would be
1) 300 K 2) 250 K 3) 200 K 4) 100 K
KEY:1
HINT:
For free expansion, dU  0  dT  0  T is constant.
30. A gas at 10oC temperature and 1.013×105 Pa pressure is compressed adiabatically to half of its
volume. If the ratio of specific heats of the gas is 1.4, what is its final temperature?
1) 103oC 2) 123oC 3) 93oC 4) 146oC
KEY:1
HINT:
 1
. TV
1 1  T2V2 1
31. Find the work done by a gas when it expands isothermally at 37oC to four times its initial volume.
1) 3753J 2) 3573J 3) 7533J 4) 5375J
KEY:2
HINT:
V 
W  2.303nRT log10  2 
 V1 
32. The efficiency of a heat engine if the temperature of source 227oC and that of sink is 27oC nearly
1) 0.4 2) 0.5 3) 0.6 4) 0.7
KEY: 1
HINT:
T2
  1
T1
33. A Carnot engine takes 3  106 cal. of heat from a reservoir at 6270C, and gives it to a sink at 270c.
The work done by the engine is
1) 4.2  106 J 2) 8.4  106 J 3) 16.8  106 J 4) zero
KEY: 2
HINT:
W T  T 
  1  2  W  1  2  Q
Q T1  T1 
34. An ideal gas undergoes four different processes from the same initial state. Four processes are adiabatic,
isothermal, isobaric and isochoric. Out of 1,2,3 and 4 which one is adiabatic.
P 4
3
2
1
V
a) 4 b) 3 c) 2 d) 1
KEY: C
HINT:
For the curve 4 pressure is constant, so this is an isobaric process.
P 4
3
2
1
V
For the curve 1, volume is constant, so it is isochoric process. Between curves 3 and 2, curve 2 is steeper, so
it is adiabatic and 3 is isothermal.
35. If an average person jogs, hse produces 14.5 x 103 cal/min. This is removed by the evaporation of
sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 x 103 cal for
evaparation) is
(a) 0.25 kg (b) 2.25 kg (c) 0.05 kg (d) 0.20 kg
KEY: A
HINT:
sweat produced / minute
Amount of sweat evaporated/minute =
Number of calories required for evaporation / kg
Amount of heat produced per minute in jogging 14.5 103 145

= =   0.25kg
Latent heat (in cal/kg) 580  103 580
36. Consider P-Vdiagramforan ideal gas shown in figure.
P 1
constant
P=
V
2
V
Out of the following diagrams, which represents the T-P diagram?
T T
2
2
i) ii) 1
1
P P
T T
2 1 1 2
iii) iv)
P P
a) (iv) b) (ii) c) (iii) d) (i)
KEY: C
HINT:
According to the question given that pV= constant Hence, we can say that the gas is going through an
isothermal process.
Clearly, from the graph that between process 1 and 2 temperature is constant and the gas expands and
pressure decreases i.e., p2 <p1 which corresponds
to diagram (iii)
38. An ideal gas undergoes cyclic process ABCDA as shown in given P-Vdiagram. The amount of work
done by the gas is
P
D C
2P0
P0
A B
V0 V
3V0
a) 6P0V0 b)-2 PoVo c) + 2 P0V0 d) + 4 P0V0
KEY: B
HINT:
Consider the p-V diagram given in the question. Work done in the process ABCD = area of rectangle
ABCDA
= (AB) x BC = (3V0-V0) x (2p0-p0)
= 2V0 x p0 = 2p0V0
As the process is going anti-clockwise, hence there is a net compression in the gas. So, work done by the gas
= -2p0V0.
39. Consider two containers A and B containing identical gases at the same pressure, volume and
temperature. The gas in container A is compressed to half of its original volume isothermally while
the gas in container B is compressed to half of its original value adiabatically. The ratio of final
pressure of gas in B to that of gas in A is
1 2 2
1  1   1 
a) 2  1 b)   c)   d)  
2 1     1 
KEY: A
HINT:
Consider the p-V diagram shown for the container A (isothermal) and for container B (adiabatic).
2 2
p p
p0 1 p0 1
V0 2V0 V0 2V0
V V
container A container B
(Isothermal) (Adiabatic)
Both the process involving compression of the gas. For isothermal process (gasA) (during 1  2)
p1V1  p 2 V2

 2p  

 p 0  2V0   p 2  V0 

 p2   0  p0   2  p0
 V0 
 p2 B  2

p0
Hence,  Rate of final pressure  21
 p2 A 2p 0
where,  is ratio of specific heat capacities for the
jas.
40. Refer to the plot of temperature versus time (figure) showing the changes in the state of ice on
heating (not to scale). Which of the following is correct ?
E
Temperature(0C)
100
C D
A B
O tm time(min)
a) The region AB represent ice and water in thermal equilibrium

b) At B water starts boiling
c) At C all the water gets converted into steam
d) C to D represents water and steam in equilibrium at boiling point
KEY: A,D
HINT:
During the processA B temperature of the system is 00C Hence, it represents phase change that is transformation
of ice into water while temperature remains 00C.
BC represents rise in temperature of water from 00C to 1000C(atC).
Now, water starts concerting into steam which is represent by CD.
41. A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following
correct ?
a) The rage of cooling is constant till milk attains the temperature of the surrounding
b) The temperature of milk falls off exponentially with time
c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk
but the net flow of heat is from milk to the surrounding and that is why it cools
d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk
to the surroundings.
KEY: B,C,D
HINT:
When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and
radiation.
According to Newton's law of cooling temperature of the milk falls off exponentially. Heat also will be transferred
from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings.
42. Which of the processes described below are irreversible?
a) The increase in temperature of an iron rod by hammering it
b) A gas in a small container at a temperatrue T1 is brought in contact with a big reservoir at a higher temperature
T2 which increases the temperature of the gas
c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston
d) An ideal gas is enclosed in a piston cylinder ar-rangement with adiabatic walls. A weight w is added to the
piston, resulting in compression of gas.
KEY: A,B,D
HINT:
a) When the rod is hammered, the external work is done on the rod which increases its temperature. The
process cannot be retraced itself
b) In this process energy in the form of heat is trans-ferred to the gas in the small container by big reservoir at
temperature T2.
d) As the widht is added to the cylinder arangemer t in the form of external pressure hence, it cannot be
reversed back itself.
43. An ideal gas undergoes isothermal process from some initial state i to final state f. Choose the
correct alternatives.
a) dU = 0 b) dQ = 0 c) dQ = dU d) dQ = dW
KEY: AD
HINT:
For an isothermal process change in temperature of the system dT = 0  T = constnat.
We know that for an ideal gas dU = change in
internal energy = nCv dT = 0
[where, n is number of moles and Cv is specific heat capacity at constant volume] From first law of
thermodynamics,
dQ = dU + dW = 0 + dW  dQ = dW
44. Figure shows theP-V diagram of an ideal gas undergoing a change of state from Ato B. Four different
parts I, II, III and IV as shown in the figure may lead to the same change of state.
P
I
IV
II
III
V
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I
(d) Work done is minimum in case II.
KEY: B,C
HINT:
Change in internal energy for the process A to B
dUA  B = nCvdT = nCv(TB-TA) which depends only on temperatures at A and B. Work done for A to B,
dWA  B = ARea under the p-V curve which is maximum for the path I.
45. Consider a cycle followed by an engine (figure.) 1 to 2 is isothermal 2 to 3 is adiabatic 3 to 1 is

adiabatic Such a process does not exist because
P
2
1
V
(a) heat is completely converted to mechanical energy in such a process, which is not possible.
(b) mechanical energy is completely converted to heat in this process,which is not possible.
(c) curves representing two adiabatic processes don't intersect.
(d) curves representing an adiabatic process and an isothermal process don't intersect.
KEY:A,C
HINT:
a) The given process is a cyclic process i.e., it returns to the original state 1.
Hence, change in internal energy dU = 0
Hence, total heat supplied is converted to work done by the gas (mechanical energy) which is not possible by
second law of thermodynamics,
c) When the gas expands adiabatically from 2 to 3. It is not possible to return to the same state without being
heat supplied, hence the process 3 to 1 cannot be adiabatic
46. Consider a heat engine as shown in figure. Q1 and Q2 are heat added bath to T1 and heat taken from T2 in one
cycle of engine. W is the mechanical work done on the engine.
T1
Q1
W
Q2
T2
If W> 0, then possibilities are:

(a) Q1 > Q2 > 0 (b) Q2 > Q1 > 0 (c) Q2 < Q1 < 0 (d) Q1 < 0, Q2 > 0
KEY: A,C
HINT:
Consider the figure we can write Q1 = W + Q2
 W = Q1 - Q2 > 0 (By question)
 Q1 > Q2 > 0 (If both Q1 and Q2 are positive) We can also, write Q2< Q1< 0 (If both Q1and Q2 are
negative).
47. 'n' moles of an ideal gas undergoes a process AB as shown in the figure. The maximum temperature
of the gas during the process will be(Mains 2016)
P
2P0 A
P0 B
V0 2V0 V
9P0 V0 3P0 V0 9P0 V0 9P0 V0

1) 2) 3) 4)
4nR 2nR 2nR nR
KEY: 1
HINT:
Line equation for given graph (y = mx +c)
P0 nRT
P V  3P0 
V0 V
1  P0 2 
T   V  3P0 V   (1)
nR  V0 
dT
To get max temp 0
dV
sub in eq(1) for Tmaxz.
48. One mole diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process
BC is adiabilitic. The temperature at A, B and C are 400 K, 800K and 600K respectively. Choose
the correct statement (Mains 2014)
B 800K
P
600K
A C
400K
V
1) The change in internal energy in whole cyclic process is 250 R
2) The change in internal energy in the process CA is 700 R
3) The change in internal energy in the process AB is -350 R
4) The change in internal energy in the process BC is -500 R
KEY: 4
HINT:
U AB  nCV  TB  TA 
U BC  nCV  TC  TB 
U total  0
U CA  nCV  TA  TC 
Previous JEE Mains Questions And Solutions
Thermodynamics
1. A gas can be taken from A to 𝐁 via two different processes ACB and ADB.
When path ACB is used 60 𝐉 ofheat flows into the system and 𝟑𝟎𝐉 of work is done by the system.
If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB
is: [9 Jan. 2019 I]
(a) 40 𝐉 (b) 80 𝐉 (c) 100 𝐉 (d) 20 𝐉
SOLUTION : ( a )
𝜟𝐔 remains same for both paths ACB and ADB

𝜟𝐐𝐀𝐂𝐁 = 𝜟𝐖𝐀𝐂𝐁 + 𝜟𝐔𝐀𝐂𝐁
⇒ 𝟔𝟎𝐉 = 𝟑𝟎𝐉 + 𝜟𝐔𝐀𝐂𝐁
⇒ 𝐔𝐀𝐂𝐁 = 𝟑𝟎𝐉
𝜟𝐔𝐀𝐃𝐁 = 𝜟𝐔𝐀𝐂𝐁 = 𝟑𝟎𝐉
𝜟𝐐𝐀𝐃𝐁 = 𝜟𝐔𝐀𝐃𝐁 + 𝜟𝐖𝐀𝐃𝐁
= 𝟏𝟎𝐉 + 𝟑𝟎𝐉 = 𝟒𝟎𝐉
2. 𝟐𝟎𝟎𝐠𝐰ater i𝐬𝐡eated f𝐢𝐢 𝐨𝐦𝟒𝟎∘ 𝐂𝐭𝐨𝟔𝟎∘ 𝐂. Ignoringtheslight expansion of water, the change in its
internal energy is close to (Given specific heat ofwater = 𝟒𝟏𝟖𝟒𝐉/𝐤𝐠𝐊): [Online April 9, 2016]
(a) 𝟏𝟔𝟕. 𝟒𝐤𝐉 𝐛 𝟖. 𝟒𝐤𝐉 (c) 𝟒. 𝟐𝐤𝐉 (d) 𝟏𝟔. 𝟕𝐤𝐉
SOLUTION : (d)
Volume ofwater does not change, no work is done on or by the system 𝐖 = 𝟎
According to first law ofthermodynamics
𝐐 =△ 𝐔 + 𝐖
For Isochoric process 𝐐 =△ 𝐔
𝜟𝐔 = 𝝁𝐜𝐝𝐓 = 𝟐 × 𝟒𝟏𝟖𝟒 × 𝟐𝟎 = 𝟏𝟔. 𝟕𝐤𝐉.
3. A gas is compressed 𝐟𝐢𝐢 𝐨𝐦 a volume 𝐨𝐟𝟐𝐦𝟑 to a volume of 𝟏𝐦𝟑 at a constant pressure of 100
𝐍/𝐦𝟐 . Then it is heated at constant volume by supplying 150 𝐉 ofenergy. As a result, the
internal energy ofthe gas: [Online Apri119, 2014]
(a) increases by 250 𝐉 (b) decreases by 250 𝐉
(c) increases by50 𝐉 (d) decreases by 50 𝐉
SOLUTION : (a)
As we know,
𝜟𝐐 = 𝜟𝐮 + 𝜟𝐰 (Ist law ofthermodynamics)
⇒ 𝜟𝐐 = 𝜟𝐮 + 𝐏𝜟𝐯
or 𝟏𝟓𝟎 = 𝜟𝐮 + 𝟏𝟎𝟎 𝟏 − 𝟐
= 𝜟𝐮 − 𝟏𝟎𝟎
𝜟𝐮 = 𝟏𝟓𝟎 + 𝟏𝟎𝟎 = 𝟐𝟓𝟎𝐉
Thus the internal energy ofthe gas increases by250 𝐉
4. An insulated container ofgas has two chambers separated by an insulating partition. One ofthe
chambers has volume 𝑽𝟏 and contains ideal gas at pressure 𝑷𝟏 and temperature 𝑻𝟏 . The other
chamber has volume 𝑽𝟐 and contains ideal gas at pressure 𝑷𝟐 and temperature 𝑻𝟐 . Ifthe
partition is removed without doing any work on the gas, the final equilibrium temperature
ofthe gas in the container willbe [2008]
𝑻𝟏 𝑻𝟐 𝑷𝟏 𝑽𝟏 +𝑷𝟐 𝑽𝟐 𝑷𝟏 𝑽𝟏 𝑻𝟏 +𝑷𝟐 𝑽𝟐 𝑻𝟐
(a) (b)
𝑷𝟏 𝑽𝟏 𝑻𝟐 +𝑷𝟐 𝑽𝟐 𝑻𝟏 𝑷𝟏 𝑽𝟏 +𝑷𝟐 𝑽𝟐
𝑷𝟏 𝑽𝟏 𝑻𝟐 +𝑷𝟐 𝑽𝟐 𝑻𝟏 𝑻𝟏 𝑻𝟐 𝑷𝟏 𝑽𝟏 +𝑷𝟐 𝑽𝟐
(c) 𝑷𝟏 𝑽𝟏 +𝑷𝟐 𝑽𝟐
(d) 𝑷𝟏 𝑽𝟏 𝑻𝟏 +𝑷𝟐 𝑽𝟐 𝑻𝟐
SOLUTION : (a)
Here 𝑸 = 𝟎 and 𝑾 = 𝟎.
Therefore from first law of thermodynamics 𝜟𝑼 = 𝑸 + 𝑾 = 𝟎
Internal energy offirst vessle + Internal energy ofsecond vessel = Internal energy of combined
vessel
𝒏𝟏 𝑪𝒗 𝑻𝟏 + 𝒏𝟐 𝑪𝒗 𝑻𝟐 = 𝒏𝟏 + 𝒏𝟐 𝑪𝒗 𝑻
𝒏𝟏 𝑻𝟏 + 𝒏𝟐 𝑻𝟐
𝑻=
𝒏 𝟏 + 𝒏𝟐
𝑷𝟏 𝑽𝟏
For first vessel 𝒏𝟏 = and for second vessle
𝑹𝑻𝟏
𝑷 𝟐 𝑽𝟐
𝒏𝟐 =
𝑹𝑻𝟐
𝑷𝟏 𝑽𝟏 𝑷𝟐 𝑽𝟐
× 𝑻𝟏 + × 𝑻𝟐
𝑹𝑻𝟏 𝑹𝑻𝟐
𝑻= 𝑷 𝟏 𝑽𝟏 𝑷𝟐 𝑽𝟐
+
𝑹𝑻𝟏 𝑹𝑻𝟐
𝑻𝟏 𝑻𝟐 𝑷𝟏 𝑽𝟏 + 𝑷𝟐 𝑽𝟐
=
𝑷𝟏 𝑽𝟏 𝑻𝟐 + 𝑷𝟐 𝑽𝟐 𝑻𝟏
5. When a system is taken from state 𝐢 to state 𝒇 along the path iaf, it is found that 𝑸 = 𝟓𝟎 cal and
𝑾 = 𝟐𝟎 cal. Along the path ibf 𝑸 = 𝟑𝟔 cal. W along the path ibf is [2007]
(a) 14 cal (b) 6 cal (c) 16 cal (d) 66 cal

SOLUTION : (b)
For path 𝒊𝒂𝒇, 𝑸𝟏 = 𝟓𝟎 cal, 𝑾𝟏 = 𝟐𝟎 cal
By first law ofthermodynamics,
𝜟𝑼 = 𝑸𝟏 − 𝑾𝟏 = 𝟓𝟎 − 𝟐𝟎 = 𝟑𝟎 cal.
For path 𝒊𝒃𝒇 𝑸𝟐 = 𝟑𝟔 cal 𝑾𝟐 =?
𝜟𝑼𝒊𝒃𝒇 = 𝑸𝟐 − 𝑾𝟐
Since, the change in intemal energy does not depend on the path, therefore 𝜟𝑼𝒊𝒂𝒇 = 𝜟𝑼𝒊𝒃𝒇
𝜟𝑼𝒊𝒂𝒇 = 𝜟𝑼𝒊𝒃𝒇
⇒ 𝟑𝟎 = 𝑸𝟐 − 𝑾𝟐
⇒ 𝑾𝟐 = 𝟑𝟔 − 𝟑𝟎 = 𝟔 cal.
6. A system goes from 𝑨 to 𝑩 via two processes I and II as shown in figure. 𝐈𝐟𝜟𝑼𝟏 and 𝜟𝑼𝟐 are
the changes in internal energiesinthe 𝐓 𝟏𝐓𝐓 tively, then [2005]
𝐯
(a) relation between 𝜟𝑼𝟏 and 𝜟𝑼𝟐 can not be determined (b) 𝜟𝑼𝟏 = 𝜟𝑼𝟐
(c) 𝜟𝑼𝟐 < 𝜟𝑼𝟏 (d) 𝜟𝑼𝟐 > 𝜟𝑼𝟏
SOLUTION : (b)
Change in intemal energy is independent ofpath taken bythe process.
It only depends on initial and final states
i. e.,
𝜟𝑼𝟏 = 𝜟𝑼𝟐
7. Which ofthe following is incorrect regarding the first law ofthermodynamics? [2005]
(a) It is a restatement of the principle of conservation of energy
(b) It is not applicable to any cyclic process
(c) It does not introduces the concept of the entropy
(d) It introduces the concept ofthe internal energy
SOLUTION : 𝐛, 𝐜
First law is applicable to a cyclic process.
Concept of entropy is introduced bythe second law ofthermodynamics.
8. Three different processes that can occur in an ideal monoatomic gas are shown in the 𝐏 vs V
diagram. The paths are lebelled as 𝐀 → 𝐁, 𝐀 → 𝐂 and 𝐀 → 𝐃. The change in internal energies
during these process are taken as 𝐄𝐀𝐁 , 𝐄𝐀𝐂 and 𝐄𝐀𝐃 and the workdone as 𝐖𝐀𝐁 , 𝐖𝐀𝐂 and 𝐖𝐀𝐃 .
The correct relation between these parameters are: [5 Sep. 2020 (I)]
(a) 𝐄𝐀𝐁 = 𝐄𝐀𝐂 < 𝐄𝐀𝐃 , 𝐖𝐀𝐁 > 𝟎, 𝐖𝐀𝐂 = 𝟎, 𝐖𝐀𝐃 < 𝟎
(b) 𝐄𝐀𝐁 = 𝐄𝐀𝐂 = 𝐄𝐀𝐃 , 𝐖𝐀𝐁 > 𝟎, 𝐖𝐀𝐂 = 𝟎, 𝐖𝐀𝐃 > 𝟎
(c) 𝐄𝐀𝐁 < 𝐄𝐀𝐂 < 𝐄𝐀𝐃 , 𝐖𝐀𝐁 > 𝟎, 𝐖𝐀𝐂 > 𝐖𝐀𝐃
(d) 𝐄𝐀𝐁 > 𝐄𝐀𝐂 > 𝐄𝐀𝐃 , 𝐖𝐀𝐁 < 𝐖𝐀𝐂 < 𝐖𝐀𝐃
SOLUTION : (b)
Temperature change 𝜟𝑻 is same for all three processes 𝑨 → 𝑩; 𝑨 → 𝑪 and 𝑨 → 𝑫
𝜟𝑼 = 𝒏𝑪𝒗 𝜟𝑻 = same
𝑬𝑨𝑩 = 𝑬𝑨𝑪 = 𝑬𝑨𝑫
Work done, 𝑾 = 𝑷 × 𝜟𝑽
𝑨𝑩 → volume is increasing ⇒ 𝑾𝑨𝑩 > 𝟎

𝑨𝑫 → volume is decreasing ⇒ 𝑾𝑨𝑫 < 𝟎
𝑨𝑪 → volume is constant ⇒ 𝑾𝑨𝑪 = 𝟎
9. In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The
final pressure ofthe gas is found to be 𝒏 times the initial pressure. The value of 𝒏 is:
[5 Sep. 2020 (II)]
𝟏
(a) 32 (b) 326 (c) 128 (d) 𝟑𝟐
SOLUTION : (c)
In adiabatic process
𝑷𝒚𝐘 = constant
𝒎 𝐘 𝒎
𝑷 = constant ⋅.⋅ 𝑽 =
𝐩 𝐩
As mass is constant 𝑷 ∝ 𝐩𝐘
𝐈𝐟𝑷𝒊 and 𝑷𝒇 𝐛𝐞 the initial and final pressure ofthe gas and 𝐩𝒊 and 𝐏𝒇 be the initial and fmal
densityofthe gas.
𝑷𝒇
Then = 𝐘
= 𝟑𝟐 𝟕/𝟓
𝑷𝒊
𝒏𝑷𝒊 𝟕/𝟓
⇒ = 𝟐𝟓 = 𝟐𝟕
𝑷𝒊
⇒ 𝒏 = 𝟐𝟕 = 𝟏𝟐𝟖.
10. Match the thermodynamic processes taking place in a system with the correct conditions. In
the table: 𝜟𝑸 is the heat supplied, 𝜟𝑾 is the work done and 𝜟𝑼 is change in internal energy
ofthe system. [4 Sep. 2020 (II)]
Process Condition
(𝑫 Adiabatic (A) 𝜟𝑾 = 𝟎
(𝜫) Isothermal (B) 𝜟𝑸 = 𝟎
(III) Isochoric (C) 𝜟𝑼 ≠ 𝟎, 𝜟𝑾 ≠ 𝟎, 𝜟𝑸 ≠ 𝟎𝝎
(IV)Isobaric (D) 𝜟𝑼 = 𝟎
(a) 𝐈 − 𝐀 , 𝐈𝐈 − 𝐁 , 𝐈𝐈𝐈 − 𝐃 , 𝐈𝐕 − 𝐃
(b) 𝐈 − 𝐁 , 𝐈𝐈 − 𝐀 , 𝐈𝐈𝐈 − 𝐃 , 𝐈𝐕 − 𝐂
(c) 𝐈 − 𝐀 , 𝐈𝐈 − 𝐀 , 𝐈𝐈𝐈 − 𝐁 , 𝐈𝐕 − 𝐂
(d) 𝐈 − 𝐁 , 𝐈𝐈 − 𝐃 , 𝐈𝐈𝐈 − 𝐀 , 𝐈𝐕 − 𝐂
SOLUTION : (d)
(𝑫 Adiabatic process :
No exchange of heat takes place with surroundings.
⇒ 𝜟𝑸 = 𝟎
(𝜫) Isothermal process:
Temperature remains constant
𝒇
𝜟𝑻 = 𝟎 ⇒ 𝜟𝑼 = 𝒏𝑹𝜟𝑻 ⇒ 𝜟𝑼 = 𝟎
𝟐
No change in internal energy 𝜟𝑼 = 𝟎 .
(𝐈𝐈𝐃 Isochoric process:
volume remains constant
𝜟𝑽 = 𝟎 ⇒ 𝑾 = 𝑷 ⋅ 𝒅𝑽 = 𝟎
Hence work done is zero.

(𝟒) In isobaric process :
pressure remains constant.
𝑾 = 𝑷 ⋅ 𝜟𝑽 ≠ 𝟎
𝒇 𝒇
𝜟𝑼 = 𝒏𝑹𝜟𝑻 = 𝑷𝜟𝑽 ≠ 𝟎
𝟐 𝟐
𝜟𝑸 = 𝒏𝑪𝒑 𝜟𝑻 ≠ 𝟎
11. A balloon filled with helium (𝟑𝟐∘ 𝐂 and 𝟏. 𝟕 atm.) bursts. Immediately afterwards the expansion
of helium can be considered as : [3 Sep. 2020 (I)]
(a) irreversible isothermal (b) irreversible adiabatic
(c) reversible adiabatic (d) reversible 𝐢𝐬𝐨𝐭𝐡𝐞𝐫𝐦𝟕𝐚𝐥
SOLUTION : (b)
Bursting ofhelium balloon is irreversible and in this process
𝜟𝑸 = 𝟎 , so adiabatic.
12. An engine takes in 5 mole of air at 𝟐𝟎∘ 𝐂 and 1 𝐚𝐭𝐦, and compresses it adiabaticaly to 𝟏/𝟏𝟎𝐭𝐡 of
the original volume. Assuming air to be a diatomic ideal gas made up ofrigid molecules, the
change in its internal energy during this process comes out to be 𝑿𝐤𝐉. The value 𝐨𝐟𝑿 to the
nearest integer is [NA 2 Sep. 2020 (I)]
SOLUTION : (46)
−𝟏
For adiabatic process, 𝑻𝑽𝐘 = constant
𝜸−𝟏 𝜸−𝟏
or, 𝑻𝟏 𝑽𝟏 = 𝑻𝟐 𝑽𝟐
𝑽 𝟕
𝑻𝟏 = 𝟐𝟎∘ 𝐂 + 𝟐𝟕𝟑 = 𝟐𝟗𝟑𝐊 , 𝑽𝟐 = 𝟏𝟎𝟏 and 𝜸 = 𝟓
𝐘 −𝟏
𝐘 −𝟏
𝑽𝟏
𝑻𝟏 𝑽𝟏 = 𝑻𝟐
𝟏𝟎
𝟏 𝟐/𝟓
⇒ 𝟐𝟗𝟑 = 𝑻𝟐 ⇒ 𝑻𝟐 = 𝟐𝟗𝟑 𝟏𝟎 𝟐/𝟓 = 𝟕𝟑𝟔𝐊
𝟏𝟎
𝜟𝑻 = 𝟕𝟑𝟔 − 𝟐𝟗𝟑 = 𝟒𝟒𝟑𝐊
During the process, change in internal energy
𝟓
𝜟𝑼 = 𝑵𝑪𝑽 𝜟𝑻 = 𝟓 × × 𝟖. 𝟑 × 𝟒𝟒𝟑 = 𝟒𝟔 × 𝟏𝟎𝟑 𝐉 = 𝑿𝐤𝐉
𝟐
𝑿 = 𝟒𝟔.
13. Which of the following is an equivalent cyclic process corresponding to the thermodynamic
cyclic given in the figure? where, 𝟏 → 𝟐 is adiabatic.
(Graphs are schematic and are not to scale) [9 Jan. 2020 I]
SOLUTION : (c)
For process 𝟑 → 𝟏 volume is constant
Graph given in option (d) is wrong.
And process 𝟏 → 𝟐 is adiabatic graph in option (1) is wrong
𝒗 = constant
𝑷 ↑, 𝑻 ↑
For Process 𝟐 → 𝟑 Pressure constant i.e., 𝑷 = constant
𝑽↓𝑻↓
Hence graph (c) is the correct 𝑽 − 𝑻 graph of given 𝑷 − 𝑽 graph
14. Starting at temperature 𝟑𝟎𝟎𝐊, one mole ofan ideal diatomic gas 𝜸 = 𝟏. 𝟒 is first compressed
𝐕𝟏
adiabatically 𝐟𝐢𝐢 𝐨𝐦 volume 𝐕𝟏 to 𝐕𝟐 = . It is then allowed to expand isobarically to volume
𝟏𝟔
𝟐𝐕𝟐 . Ifall the processes are the quasi‐static then the final temperature ofthe gas (in 𝐨𝐊) is (to
the nearest integer) . „ [9 Jan. 2020 II]
SOLUTION : . (1818)
For an adiabatic process,
𝐓𝐕 𝜸𝟏 = constant
𝜸−𝟏 𝜸−𝟏
𝑻𝟏 𝑽𝟏 = 𝑻𝟐 𝑽𝟐
𝟏.𝟒−𝟏
𝑽𝟏
⇒ 𝑻𝟐 = 𝟑𝟎𝟎 × 𝑽𝟏
𝟏𝟔
⇒ 𝑻𝟐 = 𝟑𝟎𝟎 × 𝟏𝟔 𝟎𝟒
Ideal gas equation, 𝑷𝑽 = 𝒏𝑹𝑻
𝒏𝑹𝑻
𝑽=
𝑷
⇒ 𝑽 = 𝒌𝑻 (since pressure is constant for isobaric process)
So, during isobaric process
𝑽𝟐 = 𝒌𝑻𝟐 (i)
𝟐𝐕𝟐 = 𝒌𝑻𝒇 (ii)
Dividing (i) by (ii)
𝟏 𝑻𝟐
=
𝟐 𝝉𝒇
𝟎𝟒
𝑻𝒇 = 𝟐𝑻𝟐 = 𝟑𝟎𝟎 × 𝟐 × 𝟏𝟔 = 𝟏𝟖𝟏𝟖𝑲
15. Athermodynamic cycle xyzx is shown on a V‐T diagram.
𝝉
The P‐V diagram that best describes this cycle is: (Diagrams are schematic and not to scale)
[8 Jan. 2020 I]
SOLUTION : (a)
From the corresponding V‐T graph given in question,
Process 𝒙𝒚 →Isobaric expansion,
Process 𝒚𝒛 →Isochoric (Pressure decreases)
Process 𝒛𝒙 →Isothermal compression
Therefore, correspon$ng 𝑷𝑽 graph is as shown in figure
16. A litre ofdry air at STP expands a diabatically to a volume of3 litres. 𝐈𝐟𝜸 = 𝟏. 𝟒𝟎, the work
done by air is: 𝟑𝟏𝟒 = 𝟒. 𝟔𝟓𝟓𝟓 [Take air to be an ideal gas] [7 Jan. 2020 I]
(a) 𝟔𝟎. 𝟕𝐉 (b) 𝟗𝟎. 𝟓𝐉 (c) 𝟏𝟎𝟎. 𝟖𝐉 (d) 48 𝐉
SOLUTION : (b)
Given, 𝑽𝟏 = 𝟏litre,𝑷𝟏 = 𝟏 atm
𝑽𝟐 = 𝟑 litre, 𝜸 = 𝟏. 𝟒𝟎,
Using, 𝑷𝑽𝒓 =constant ⇒ 𝑷𝟏 𝑽𝐘𝟏 = 𝑷𝟐 𝑽𝐘𝟐
𝟏.𝟒
𝟏 𝟏
⇒ 𝑷𝟐 = 𝑷𝟏 × = 𝒂𝒕𝒎
𝟑 𝟒. 𝟔𝟓𝟓𝟓
𝟏
𝑷𝟏 𝑽𝟏 −𝑷𝟐 𝑽𝟐 𝟏×𝟏− ×𝟑 𝟏.𝟎𝟏𝟑𝟐𝟓×𝟏𝟎𝟓 ×𝟏𝟎−𝟑
Work done, 𝑾 = = 𝟒.𝟔𝟓𝟓𝟓
= 𝟗𝟎. 𝟏𝑱
𝜸−𝟏 𝟎.𝟒
Closest value of 𝑾 = 𝟗𝟎. 𝟓𝑱

17. Under an adiabatic process, the volume ofan ideal gas gets doubled. Consequently the mean
𝑪𝒑
collision time between the gas molecule changes 𝐟𝐢𝐢 𝐨𝐦𝝉𝟏 to 𝝉𝟐 . If = 𝜸 for this gas then a
𝑪𝐯
good estimate for is given by: [7 Jan. 2020 I]

𝜸+𝟏
𝟏 𝟏 𝜸 𝟏 𝟐
(a) 2 (b) 𝟐 (c) (d)
𝟐 𝟐
SOLUTION : (Bonus)
𝑽
We know that Relaxation time, 𝑻 ∝ (i)
𝑻
Equation of adiabatic process is 𝑻𝑽/ 𝟏 = constant

𝟏
⇒ 𝑻 ∝ 𝑽𝐘−𝟏
𝐘−𝟏
⇒ 𝑻 ∝ 𝑽𝟏+ 𝟐 using (i)
𝟏+𝐘
⇒𝑻∝𝑽 𝟐
𝟏+𝜸
𝝉𝒇 𝟏+𝐘
𝟐𝑽 𝟐
⇒ = = 𝟐 𝟐
𝑻𝒊 𝑽
18. A sample ofan ideal gas is taken through the cyclic process abca as shown in the figure. The
change in the internal energy of the gas along the path ca is—180 𝐉, The gas absorbs 250 𝐉
ofheat along the path ab and 60 𝐉 along the path 𝐛𝐜. The work down by the gas along the path
abc is: [12 Apr. 2019 I]
𝑽→
(a) 𝟏𝟐𝟎𝐉 (b) 𝟏𝟑𝟎𝐉 (c) 𝟏𝟎𝟎𝐉 (d) 𝟏𝟒𝟎𝐉
SOLUTION : (b)
𝜟𝑼𝐚𝐜 = − 𝜟𝑼𝐜𝐚 = − −𝟏𝟖𝟎 = 𝟏𝟖𝟎𝐉
𝑸 = 𝟐𝟓𝟎 + 𝟔𝟎 = 𝟑𝟏𝟎𝐉
Now 𝑸 = 𝜟𝑼 + 𝑾 or 𝟑𝟏𝟎 = 𝟏𝟖𝟎 + 𝑾 or 𝑾 = 𝟏𝟑𝟎𝐉
19. A cylinder with fixed capacity of 𝟔𝟕. 𝟐 lit contains helium gas at STP. The amount of heat
needed to raise the temperature ofthe gas by 𝟐𝟎𝐨 𝐂 is: [Given that 𝐑 = 𝟖. 𝟑𝟏𝐉 mol 𝟏𝐊𝟏]
[𝟏𝟎𝐀𝐩𝐫. 2019 I]
(a) 350 𝐉 (b) 374 𝐉 (c) 748 𝐉 (d) 700 𝐉
SOLUTION : . (c)
As the process is isochoric so,
𝟔𝟕. 𝟐 𝟑𝐑
𝐐 = 𝐧𝐜𝐯 𝜟𝐓 = × × 𝟐𝟎 = 𝟗𝟎𝐑 = 𝟗𝟎 × 𝟖. 𝟑𝟏 = 𝟕𝟒𝟖𝐣.
𝟐𝟐. 𝟒 𝟐
20. 𝐧 moles ofan ideal gas with constant volume heat capacity 𝐂𝐯 undergo an isobaric expansion
by certain volume. The ratio ofthe work done in the process, to the heat supplied is:
[10 Apr. 2019 I]
𝐧𝐑 𝐧𝐑 𝟒𝐧𝐑 𝟒𝐧𝐑
(a) 𝐂 (b) 𝐂 (c) 𝐂 (d) 𝐂
𝐯 +𝐧𝐑 𝐯 −𝐧𝐑 𝐯 −𝐧𝐑 𝐯 +𝐧𝐑
SOLUTION : (a)
At constant volume
Work done 𝐖 = 𝐧𝐑𝜟𝐓
Heat given 𝐐 = 𝐂𝐯 𝜟𝐓 + 𝐧𝐑𝜟𝐓
𝐖 𝐧𝐑𝜟𝐓 𝐧𝐑
So, =𝐂 =𝐂
𝐐 𝐯 𝜟𝐓+𝐧𝐑𝜟𝐓 𝐕 +𝐧𝐑
21. One mole ofan ideal gas passes through a process where pressure and volume obeythe relation
𝟏 𝑽𝟎 𝟐
𝑷 = 𝑷𝟎 𝟏 − 𝟐 Here 𝐏𝐨 and 𝐕𝐨 are constants. Calculate the charge in the temperature
𝑽
ofthe gas ifits volume changes from 𝐕𝐨 to 𝟐𝐕𝐨 . [10 Apr. 2019 II]
𝟏 𝐏𝐨 𝐕𝐨 𝟓 𝐏𝐨 𝐕𝐨 𝟑 𝐏𝐨 𝐕𝐨 𝟏 𝐏𝐨 𝐕𝐨
(a) 𝟐 (b) 𝟒 (c) 𝟒 (d) 𝟒
𝐑 𝐑 𝐑 𝐑
SOLUTION : (b)
𝟏 𝑽𝟎 𝟐
We have given, 𝑷 = 𝑷𝟎 𝟏 −
𝟐 𝑽
𝟏−𝟏 𝑷
When 𝐕𝟏 = 𝐕𝟎 ⇒ 𝐏𝟏 = 𝐏𝟎 = 𝟐𝟎
𝟐
𝟏 𝟏 𝟕𝑷𝟎
When 𝐕𝟐 = 𝟐𝐕𝟎 ⇒ 𝐏𝟐 = 𝐏𝟎 𝟏 − 𝟐 =
𝟒 𝟖
𝑷 𝟏 𝑽 𝟏 𝑷 𝟐 𝑽𝟐 𝑷𝑽
𝜟𝑻 = 𝑻𝟐 − 𝑻𝟏 = | − | ⋅.⋅ 𝑻 =
𝒏𝑹 𝒏𝑹 𝒏𝑹
𝟏 𝟏 𝑷𝟎 𝑽𝟎 𝟕𝑷𝟎 𝑽𝟎
𝜟𝑻 = | 𝑷 𝟏 𝑽𝟏 − 𝑷 𝟐 𝑽𝟐 | = | − |
𝒏𝑹 𝒏𝑹 𝟐 𝟒
𝟓𝑷𝟎 𝑽𝟎 𝟓𝑷𝟎 𝑽𝟎
= = 𝒏=𝟏
𝟒𝒏𝑹 𝟒𝑹
22. Following figure shows two processes A and 𝐁 for a gas. If 𝜟𝐐𝐀 and 𝜟𝐐𝐁 are the amount of
heat absorbed by the system in two cases, and 𝜟𝐔𝐀 and 𝜟𝐔𝐁 are changes in internal energies,
respectively, then: [9 April 2019 I]
(a) 𝜟𝐐𝐀 < 𝜟𝐐𝐁 , 𝜟𝐔𝐀 < 𝜟𝐔𝐁 (b) 𝜟𝐐𝐀 > 𝜟𝐐𝐁 , 𝜟𝐔𝐀 > 𝜟𝐔𝐁
(c) 𝜟𝐐𝐀 > 𝜟𝐐𝐁 , 𝜟𝐔𝐀 = 𝜟𝐔𝐁 (d) 𝜟𝐐𝐀 = 𝜟𝐐𝐁 ; 𝜟𝐔𝐀 = 𝜟𝐔𝐁
SOLUTION : . (c)
Internal energy depends only on initial and final state So,
𝜟𝐔𝐀 = 𝜟𝐔𝐁
Also 𝜟𝐐 = 𝜟𝐔 + 𝐖
As 𝐖𝐀 > 𝐖𝐁 ⇒ 𝜟𝐐𝐀 > 𝜟𝐐𝐁
23. A thermally insulted vessel contains 150 𝐠 ofwater at 𝟎∘ 𝐂. Then the air 𝐟𝐢𝐢 𝐨𝐦 the vessel is
pumped out adiabatically. 𝐀 fiiaction ofwater turns into ice and the rest evaporates at 𝟎∘ 𝐂
itself. The mass ofevaporated water will be closed to: (Latent heat of vaporization ofwater
= 𝟐. 𝟏𝟎 × 𝟏𝟎𝟔 𝐉 kg 1 and Latent heat ofFusion ofwater = 𝟑. 𝟑𝟔 × 𝟏𝟎𝟓 𝐉 kg 1) [8 April 2019 I]
(a) 𝟏𝟓𝟎𝐠 (b) 20 𝐠 (c) 𝟏𝟑𝟎𝐠 (d) 35 𝐠
SOLUTION : . (b)
Suppose amount of water evaporated be 𝐌 gram.
Then (150‐M) gram water converted into ice.
so, heat consumed in evoporation = Heat released in fusion
𝐌 × 𝐋𝐯 = 𝟏𝟓𝟎 − 𝐌 × 𝐋
𝐌 × 𝟐. 𝟏 × 𝟏𝟎𝟔 = 𝟏𝟓𝟎 − 𝐌 × 𝟑. 𝟑𝟔 × 𝟏𝟎𝟓 𝐬
⇒ 𝐌 − 𝟐𝟎𝐠
24. The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The
correct assignment ofthe processes, in the same order is given by: [8 Apr. 2019 II]
𝐕→
(a) adbc (b) dacb (c) adcb (d) dabc
SOLUTION : . (d)
𝐚 → Isobasic,
𝐛 → Isothermal,
𝐜 →Adiabatic,
𝐝 →Isochoric
25. For the given cyclic process CAB as shown for gas, the work doneis: 2 Jan. 2019 I]
𝟏 𝟐 𝟑 𝟒 𝟓 𝐕 𝐦𝟑
(a) 30 𝐉 (b) 10 𝐉 (c) 1 𝐉 (d) 5 𝐉
SOLUTION : . (b)
Total work done by the gas during the cycle is equal to area oftriangle ABC.
𝟏
𝜟𝐖 = × 𝟒 × 𝟓 = 𝟏𝟎𝐉
𝟐
26. Arigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation
between temperature and volume for this process is 𝐓𝐕 𝐱 =constant, then 𝐱 is: [11 Jan. 2019 I]
𝟑 𝟐 𝟐 𝟓
(a) 𝟓 (b) 𝟓 (c) 𝟑 (d) 𝟑
SOLUTION : (b)
Equation ofadiabatic change is𝐓𝐕 𝜸−𝟏 = constant
𝟕 𝟕
Put = 𝟓 , we get: 𝜸 − 𝟏 = 𝟓 − 𝟏
𝟐
𝐱=
𝟓
27. Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm 𝐟𝐢𝐢 𝐨𝐦𝟐𝟎∘ 𝐂 to
𝟗𝟎∘ 𝐂. Work done by gas is close to: (Gas constant 𝐑 = 𝟖. 𝟑𝟏 J/mol‐K) [10 Jan. 2019 II]
(a) 581 𝐉 (b) 291 𝐉 (c) 146 𝐉 (d) 73 𝐉
SOLUTION : (b)
𝟏
Work done,𝐖 = 𝐏𝜟𝐕 = 𝐧𝐑𝜟𝐓 = 𝟐 × 𝟖. 𝟑𝟏 × 𝟕𝟎 = 𝟐𝟗𝟏𝐉
28. One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV
diagram. The maximum temperature attained by the gas along the path BC is given by
𝟐𝟓𝐏𝟎 𝐕𝟎 𝟐𝟓𝐏𝟎 𝐕𝟎 𝟓𝐏𝟎 𝐕𝟎

(a) 𝐛 𝟐𝟓𝐏𝟎 𝐕𝟎
𝐜
(d)
𝟖𝐑 𝟒𝐑
𝟏𝟔𝐑 𝟖𝐑
SOLUTION : (a)
𝟐𝐏𝟎
Equation ofthe BC 𝐏 = 𝐏𝟎 − 𝐕 − 𝟐𝐕𝟎
𝐕𝟎
using 𝐏𝐕 = 𝐧𝐑𝐓
𝟐𝐏 𝐕 𝟐
𝐏𝟎 𝐕− 𝟎 +𝟒𝐏𝟎 𝐕
𝐕
Temperature, 𝐓 = 𝟎
(𝐧 = 𝟏 mole given)
𝟏×𝐑
𝐏𝟎 𝟓𝐕 − 𝟐𝐕 𝟐
𝐓=
𝜞 𝐕𝟎
𝐝𝐓 𝟒𝐕 𝟓
=𝟎⇒𝟓− = 𝟎 ⇒ 𝐕 = 𝐕𝟎
𝐝𝐕 𝐕𝟎 𝟒
𝐏𝟎 𝟓𝐕𝟎 𝟐 𝟐𝟓 𝟐 𝟐𝟓𝐏𝟎 𝐕𝟎
𝐓= 𝟓× − × 𝐕𝟎 =
𝐑 𝟒 𝐕𝟎 𝟏𝟔 𝟖𝐑
29. One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its
pressure at room temperature, 𝟐𝟕∘ 𝐂. The work done on the gas will be:
(a) 𝟑𝟎𝟎𝐑 𝐥𝐧 𝟔 (b) 𝟑𝜶)𝐑 (c) 𝟑𝟎𝟎𝐑 𝐥𝐧 𝟕 (d) 𝟑𝜶)𝐑𝐡𝟐
SOLUTION : (d)
𝒑𝒇
Work done on gas = 𝐧𝐑𝐓𝓵𝒏 = 𝐑 𝟑𝟎𝟎 𝓵𝐧 𝟐
𝒑𝟏
𝑷𝒇 = 𝟐𝐠𝐢𝐯𝐞𝐧
= 𝟑𝟎𝟎 Rln2
𝒑𝒊
30. 𝐧′ moles of an ideal gas undergoes a process 𝐀 → 𝐁 as shown in the figure. The maximum
temperature ofthe gas during the process will be: [2016]
𝐏
𝟗𝐏𝟎 𝐕𝟎 𝟗𝐏𝟎 𝐕𝟎 𝟗𝐏𝟎 𝐕𝟎 𝟑𝐏𝟎 𝐕𝟎

(a) (b) 𝐜 (d)
𝟐𝐧𝐑 𝐧𝐑 𝟒𝐧𝐑 𝟐𝐧𝐑
SOLUTION : . (c)
The equation for the line is
𝐏
−𝐏𝟎 −𝐏𝟎
𝐏= 𝐕 + 𝟑𝐏 [ = ,
𝐕𝟎 𝐕𝟎
𝐏𝐕𝟎 + 𝐏𝟎 𝐕 = 𝟑𝐏𝟎 𝐕𝟎 (i)

But 𝐩𝐕 = 𝐧𝐑𝐓
𝐧𝐑𝐓 𝐧𝐑𝐓
𝐏= (ii) From(i) &(ii) 𝐕𝟎 + 𝐏𝟎 𝐕 = 𝟑𝐏𝟎 𝐕𝟎
𝐕 𝐕
𝐧𝐑𝐓
From (i) & (ii) 𝐕𝟎 + 𝐏𝟎 𝐕 = 𝟑𝐏𝟎 𝐕𝟎 𝐧𝐑𝐓𝐕𝟎 + 𝐏𝟎 𝐕 𝟐 = 𝟑𝐏𝟎 𝐕𝟎 𝐕 (iii)
𝐕
𝐝𝐓
For temperature to be maximum = 𝟎 Differentiating 𝐞. 𝐪. (iii) by 𝐕’ we get
𝐝𝐕
𝐝𝐓
𝐧𝐑𝐕𝟎 + 𝐏𝟎 𝟐𝐕 = 𝟑𝐏𝟎 𝐕𝟎
𝐝𝐕
𝐝𝐓
𝐧𝐑𝐕𝟎 = 𝟑𝐏𝟎 𝐕𝟎 − 𝟐𝐏𝟎 𝐕
𝐝𝐕
𝐝𝐓 𝟑𝐏𝟎 𝐕𝟎 − 𝟐𝐏𝟎 𝐕
= =𝟎
𝐝𝐕 𝐧𝐑𝐕𝟎
𝟑𝐕𝟎 𝟑𝑷𝟎 𝟗𝐏𝟎 𝐕𝟎
𝐕= 𝑷= [From (i)] 𝐓 𝐦𝐚𝐱 = [From (iii)]
𝟐 𝟐 𝟒𝐧𝐑
31. The pressure ofan ideal gas varies with volume as 𝑷 = 𝜶𝑽, where 𝜶 is a constant. One mole of
the gas is allowed to undergo expansion such that its volume becomes„m‟ times its initial
volume. The work done by the gas in the process is [Online May 19, 2012]
𝜶𝑽 𝜶𝟐 𝑽𝟐 𝜶 𝜶𝑽𝟐
(a) 𝒎𝟐 − 𝟏 (b) 𝒎𝟐 − 𝟏 (c) 𝟐 𝒎𝟐 − 𝟏 (d) 𝒎𝟐 − 𝟏
𝟐 𝟐 𝟐
SOLUTION : (d)
Given 𝑷 = 𝜶𝑽
𝒎𝑽
Work done, 𝒘 = 𝑽
𝑷 𝒅𝑽
𝒎𝑽 𝜶𝑽𝟐
= 𝑽
𝜶 𝑽𝒅𝑽 = 𝒎𝟐 − 𝟏 .
𝟐
32. 𝒏 moles ofan ideal gas undergo a process A→ 𝑩 as shown in the figure. Maximum temperature
ofthe gas during the process is [Online May 12, 2012]
𝑷 𝑽𝟎 𝟐𝑽𝟎
𝑽→
𝟗𝑷𝟎 𝑽𝟎 𝟑𝑷𝟎 𝑽𝟎 𝟗𝑷𝟎 𝑽𝟎 𝟗𝑷𝟎 𝑽𝟎
(a) (b) 𝐜 (d)
𝒏𝑹 𝟐𝒏𝑹 𝟐𝒏𝑹 𝟒𝒏𝑹
SOLUTION : (b)
Work done during the process 𝑨 → 𝑩
= Area oftrapezium (= area bounded by indicator diagram with 𝑽‐axis)
𝟏 𝟑
= 𝟐𝑷𝟎 + 𝑷𝟎 𝟐𝑽𝟎 − 𝑽𝟎 = 𝑷𝟎 𝑽𝟎
𝟐 𝟐
Ideal gas 𝐞𝐪𝐧: 𝑷𝑽 = 𝒏𝑹𝑻
𝑷𝑽 𝟑𝑷𝟎 𝑽𝟎
⇒ 𝑻 = 𝒏𝑹 = 𝟐𝒏𝑹
33. This question has Statement 1 and Statement2. Ofthe four choices given after the Statements,
Statement 1:
In an adiabatic process, change in internal energy ofa gas is equal to work done
𝐨𝐧/𝐛𝐲 the gas in the process.
Statement 2:
The temperature ofa gas remains constant in an adiabatic process.
(a) Statement 1 is true, Statement2 is true, Statement2 is a correct explanation of Statement 1.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is false, Statement2 is true.
(d) Statement 1 is false, Statement2 is true, Statement2 is not a correct explanation of
Statement 1.
SOLUTION : (b)
In an adiabatic process, 𝟔𝐇 = 𝟎
And according to first law ofthermodynamics
𝟔𝑯 = 𝟔𝑼 + 𝑾
𝑾 = −𝟔𝑼
34. Acontainer with insulating walls is divided into equal parts by a partition fitted with a valve.
One part is filled with an ideal gas at a pressure 𝑷 and temperature 𝑻, whereas the other part is
completly evacuated. If the valve is suddenly opened, the pressure and temperature ofthe gas
willbe: [2011 𝐑𝐒]
𝑷 𝑻 𝑻 𝑷
(a) 𝟐 , 𝟐 (b) 𝑷𝑻 (c) 𝑷, 𝟐 (d) 𝟐 , 𝑻
SOLUTION : ( d )
It is the free expansion
So, 𝐓 remains constant
⇒ 𝑷 𝟏 𝑽𝟏 = 𝑷 𝟐 𝑽𝟐
𝑽
⇒ 𝑷 𝟐 = 𝑷𝟐 𝑽
𝑷
𝑷𝟐 =
𝟐
Directions for questions 35 to 37 : Questions are based on the following paragraph.
Two moles ofhelium gas are taken over the cycle ABCDA, as shown in the P‐T diagram. [2009]
„
35. Assuming the gas to be ideal the work done on the gas in taking it 𝐟𝐢𝐢 𝐨𝐦𝑨 to 𝑩 is
(a) 300 𝐑 (b) 400 𝐑 (c) 500 𝐑 (d) 200 𝐑
SOLUTION : (b)
The process 𝑨 → 𝑩 is isobaric.
work done 𝑾𝑨𝑩 = 𝒏𝑹 𝑻𝟐 − 𝑻𝟏 = 𝟐𝑹 𝟓𝟎𝟎 − 𝟑𝟎𝟎 = 𝟒𝟎𝟎𝑹
36. The work done on the gas in taking it from 𝑫 to 𝑨 is

(a) +𝟒𝟏𝟒𝐑 (b) −𝟔𝟗𝟎𝐑 𝐜 + 𝟔𝟗𝟎𝐑 (d) −𝟒𝟏𝟒𝐑
SOLUTION : (a)
The process 𝑫 to 𝑨 is isothermal as temperature is constant.
Work done, 𝑾𝑫𝑨 = 𝟐. 𝟑𝟎𝟑𝒏𝑹𝑻𝐥𝐨𝐠 𝟏𝟎𝑷𝑫 = 𝟐. 𝟑𝟎𝟑 × 𝟐𝑹 × 𝟑𝟎𝟎
𝑷𝑨
𝟏 × 𝟏𝟎𝟓
𝐥𝐨𝐠 𝟏𝟎 − 𝟒𝟏𝟒𝐑.
𝟐 × 𝟏𝟎𝟓
Therefore, work done on the gas is +𝟒𝟏𝟒𝑹.
37. The net work done on the gas in the cycle ABCDA is
(a) 279 𝐑 (b) 1076 𝐑 𝐜 1904 𝐑 (d) zero
SOLUTION : (a)
The net work in the cycle ABCDA is
𝑾 = 𝑾𝑨𝑩 + 𝑾𝑩𝑪 + 𝑾𝑪𝑫 + 𝑾𝑫𝑨
𝑷𝑩
= 𝟒𝜶)𝑹 + 𝟐. 𝟑𝟎𝟑𝒏𝑹𝑻 𝐥𝐨𝐠 + ⊲ 𝟎𝟎𝑹 − 𝟒𝟏𝟒𝑹
𝑷𝑪
𝟐 × 𝟏𝟎𝟓
= 𝟐. 𝟑𝟎𝟑 × 𝟐𝑹 × 𝟓𝟎𝟎 𝐥𝐨𝐠 − 𝟒𝟏𝟒𝑹
𝟏 × 𝟏𝟎𝟓
= 𝟔𝟗𝟑. 𝟐𝑹 − 𝟒𝟏𝟒𝑹 = 𝟐𝟕𝟗. 𝟐𝑹
38. The work of146 𝐤𝐉 is performed in order to compress one kilo mole of gas adiabatically and in
this process the temperature ofthe gas increases by 𝟕∘ 𝐂. The gas is (𝑹 = 𝟖. 𝟑 Jmol 𝟏𝐊 𝟏 )
[2006]
(a) diatomic (b) triatomic
(c) a mixture ofmonoatomic and diatomic (d) monoatomic
SOLUTION : (a)
Work done in adiabatic compression is given by
𝒏𝑹𝜟𝑻
𝑾=
𝟏−𝜸
𝟏𝟎𝟎𝟎 × 𝟖. 𝟑 × 𝟕
⇒ −𝟏𝟒𝟔𝟎𝟎𝟎 =
𝟏−𝜸
𝟓𝟖.𝟏 𝟓𝟖.𝟏
or 𝟏 − 𝜸 = − 𝟏𝟒𝟔 ⇒ 𝜸 = 𝟏 + 𝟏𝟒𝟔 = 𝟏. 𝟒
Hence the gas is diatomic.

39. Which ofthe following parameters does not characterize the thermodynamic state ofmatter?
[2003]
(a) Temperature (b) Pressure (c) Work (d) Volume
SOLUTION : (c)
Work is not a state function.
The remaining three parameters are state function.
40. An engine operates by taking a monatomic ideal gas through the cycle shown in the figure. The
percentage efficiency of the engine is close is . [NA 6 Sep. 2020 (II)]
𝐕𝐨 𝟐𝐕𝐨
SOLUTION : (19)
From the figure, Work, 𝑾 = 𝟐𝑷𝟎 𝑽𝟎

Heat given, 𝑸𝐢𝐧 = 𝑾𝑨𝑩 + 𝑾𝑩𝑪 = 𝒏 ⋅ 𝑪𝑽 𝜟𝑻𝑨𝑩 + 𝒏𝑪𝑷 𝜟𝑻𝑩𝑪
𝟑𝑹 𝒏𝟓𝑹 𝟑𝑹 𝟓𝑹
=𝒏 𝑻𝑩 − 𝑻𝑨 + 𝑻𝑪 − 𝑻𝑩 (⋅.⋅ 𝑪𝒗 = and 𝑪𝑷 = )
𝟐 𝟐 𝟐 𝟐
𝟑 𝟓
= 𝑷 𝑩 𝑽𝑩 − 𝑷 𝑨 𝑽𝑨 + 𝑷 𝑪 𝑽𝑪 − 𝑷 𝑩 𝑽𝑩
𝟐 𝟐
𝟑 𝟓
= × 𝟑𝑷𝟎 𝑽𝟎 − 𝑷𝟎 𝑽𝟎 + 𝟔𝑷𝟎 𝑽𝟎 − 𝟑𝑷𝟎 𝑽𝟎
𝟐 𝟐
𝟏𝟓 𝟐𝟏
= 𝟑𝑷𝟎 𝑽𝟎 + 𝑷 𝟎 𝑽𝟎 = 𝑷 𝑽
𝟐 𝟐 𝟎 𝟎
𝑾 𝟐𝑷𝟎 𝑽𝟎 𝟒
EfFiciency, 𝜼 = 𝑸 = 𝟐𝟏 = 𝟐𝟏
𝐢𝐧 𝑷𝟎 𝑽𝟎
𝟐
𝟒𝟎𝟎
𝜼% =
≈ 𝟏𝟗.
𝟐𝟏
41. If minimum possible work is done by a refiigerator in converting 100 grams ofwater at 𝟎∘ 𝐂 to
ice, how much heat (in calories) is released to the surroundings at temperature 𝟐𝟕∘ 𝐂 (Latent
heat ofice = 𝟖𝟎𝐂𝐚𝐥/gram) to the nearest integer? [NA 3 Sep. 2020 (II)]
SOLUTION : (8791)
Given, Heat absorbed, 𝑸𝟐 = 𝒎𝑳 = 𝟖𝟎 × 𝟏𝟎𝟎 = 𝟖𝟎𝟎𝟎 Cal
Temperature ofice, 𝑻𝟐 = 𝟐𝟕𝟑𝐊
Temperature of surrounding,
𝑻𝟏 = 𝟐𝟕𝟑 + 𝟐𝟕 = 𝟑𝟎𝟎𝐊
𝒘 𝑸𝟏 −𝑸𝟐 𝑻𝟏 −𝑻𝟐 𝟑𝟎𝟎−𝟐𝟕𝟑
Efficiency = 𝑸 = = =
𝟐 𝑸𝟐 𝑻𝟐 𝟐𝟕𝟑
𝑸𝟏 −𝟖𝟎𝟎𝟎 𝟐𝟕
⇒ = 𝟐𝟕𝟑 ⇒ 𝑸𝟏 = 𝟖𝟕𝟗𝟏 Cal
𝟖𝟎𝟎𝟎
42. A heat engine is involved with exchange ofheat of 1915 , −𝟒𝟎𝐉, +𝟏𝟐𝟓𝐉𝐚𝐧𝐝 − 𝑸𝐉, during one
cycle achieving an efficiency of 𝟓𝟎. 𝟎%. The value of 𝑸 is: [2 Sep. 2020 (II)]
(a) 640 𝐉 (b) 40 𝐉 (c) 980 𝐉 (d) 400 𝐉 +
SOLUTION : (c)
𝐖𝐨𝐫𝐤𝐝𝐨𝐧𝐞 𝑾
EfFiciency, 𝜼 = 𝐇𝐞𝐚𝐭𝐚𝐛𝐬𝐨𝐫𝐛𝐞𝐝 = 𝜮𝑸
𝑸𝟏 + 𝑸𝟐 + 𝑸𝟑 + 𝑸𝟒
= = 𝟎. 𝟓
𝑸𝟏 + 𝑸𝟑
Here, 𝑸𝟏 = 𝟏𝟗𝟏𝟓𝐉, 𝑸𝟐 = −𝟒𝟎𝐉 and 𝑸𝟑 = 𝟏𝟐𝟓𝐉
𝟏𝟗𝟏𝟓 − 𝟒𝟎 + 𝟏𝟐𝟓 + 𝑸𝟒
= 𝟎. 𝟓
𝟏𝟗𝟏𝟓 + 𝟏𝟐𝟓
⇒ 𝟏𝟗𝟏𝟓 − 𝟒𝟎 + 𝟏𝟐𝟓 + 𝑸𝟒 = 𝟏𝟎𝟐𝟎
⇒ 𝑸𝟒 = 𝟏𝟎𝟐𝟎 − 𝟐𝟎𝟎𝟎
⇒ 𝑸𝟒 = −𝑸 = −𝟗𝟖𝟎𝐉
⇒ 𝑸 = 𝟗𝟖𝟎𝐉
𝟏
43. ACarnot engine having an efficiency of is being used as a refrigerator. Ifthe work done on
𝟏𝟎
the refrigerator is 10 𝐉, the amount ofheat absorbed from the reservoir at lower temperature is:
[8 Jan. 2020 II]
(a) 99 𝐉 (b) 100 𝐉 (c) 1 𝐉 (d) 90 𝐉
SOLUTION : . (d)
𝑸𝟏 −𝑸𝟐
For carnot refiigerator Efficiency = 𝑸𝟏
Where, 𝑸𝟏 = heat lost from sorrounding

𝑸𝟐 = heat absorbed from reservoir at low temperature.
𝑸𝟏 −𝑸𝟐 𝒘
Also, =𝑸
𝑸𝟏 𝟏
𝟏 𝐰
⇒ =
𝟏𝟎 𝐐𝟏
⇒ 𝑸𝟏 = 𝒘 × 𝟏𝟎 = 𝟏𝟎𝟎𝑱
So, 𝐐𝟏 − 𝑸𝟐 = 𝒘
⇒ 𝑸𝟐 = 𝑸𝟏 − 𝒘
⇒ 𝟏𝟎𝟎 − 𝟏𝟎 = 𝑸𝟐 = 𝟗𝟎𝑱
44. A Carnot engine operates between two reservoirs of temperatures 900 𝐊 and 300 K. The engine
performs 1200 𝐉 ofwork per cycle. The heat energy(in J) delivered by the engine to the low
temperature reservoir, in a cycle, is [NA 7 Jan. 2020 I]
SOLUTION : (𝟔𝟎𝟎. 𝟎𝟎)
Given; 𝑻𝟏 = 𝟗𝟎𝟎𝑲, 𝑻𝟐 = 𝟑𝟎𝟎𝑲, 𝑾 = 𝟏𝟐𝟎𝟎𝑱
𝑻 𝑾
Using, 𝟏 − 𝑻𝟐 = 𝑸
𝟏 𝟏
𝟑𝟎𝟎 𝟏𝟐𝟎𝟎
⇒𝟏− =
𝟗𝟎𝟎 𝑸𝟏
𝟐 𝟏𝟐𝟎𝟎
⇒ = ⇒ 𝑸𝟏 = 𝟏𝟖𝟎𝟎
𝟑 𝑸𝟏
Therefore heat energy delivered by the engine to the low temperature reservoir,
𝐐𝟐 = 𝐐𝟏 − 𝐖 = 𝟏𝟖𝟎𝟎 − 𝟏𝟐𝟎𝟎 = 𝟔𝟎𝟎. 𝟎𝟎𝑱
45. Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the
other engine to produce work) between temperatures, 𝑻𝟏 and 𝑻𝟐 . The temperature of the hot
reservoir ofthe first engine is 𝑻𝟏 and the temperature ofthe cold reservoir ofthe second engine
is 𝑻𝟐 . 𝐓 is temperature ofthe sink offirst engine which is also the source for the second engine.
How is 𝑻 related to 𝑻𝟏 and 𝑻𝟐 , ifboth the engines perform equal amount ofwork?
[7 Jan. 2020 II]
𝟐𝑻𝟏 𝑻𝟐 𝑻𝟏 +𝑻𝟐
(a) 𝑻 = 𝑻 (b) 𝑻 = (c) 𝑻 = 𝑻𝟏 𝑻𝟐 (d) 𝑻 = 𝟎
𝟏 +𝑻𝟐 𝟐
SOLUTION : . (b)
Let 𝑸𝑯 = Heat taken by first engine
𝑸𝑳 = Heat rejected by first engine
𝑸𝟐 = Heat rejected by second engine
Work done by 1st engine = work done by 𝟐𝐧𝐝 engine
𝑾 = 𝑸𝑯 − 𝑸𝑳 = 𝑸𝑳 − 𝑸𝟐 ⇒ 𝟐𝑸𝑳 = 𝑸𝑯 + 𝑸𝟐
𝜽𝑯 𝜽𝟐
𝟐= +
𝜽𝑳 𝜽𝑳
Let 𝐓 be the temperature of cold reservoir offirst engine. Then in carnot engine.
𝑸𝑯 𝑻𝟏 𝑸 𝑻
= and 𝑸𝑳 = 𝑻
𝑸𝑳 𝑻 𝟐 𝟐
𝑻𝟏 𝑻𝟐
⇒𝟐= + using (i)
𝑻 𝑻
𝑻𝟏 +𝑻𝟐
⇒ 𝟐𝑻 = 𝑻𝟏 + 𝑻𝟐 ⇒ 𝑻 = 𝟐
46. A Carnot engine has an efficiency of 1/6. When the temperature ofthe sink is reduced by 𝟔𝟐𝐨 𝐂,
its efficiency is doubled. The temperatures of the source and the sink are, respectively.
[12 Apr. 2019 II]
(a) 𝟔𝟐𝐨 𝐂, 𝟏𝟐𝟒𝐨 𝐂 (b) 𝟗𝟗𝐨 𝐂, 𝟑𝟕𝐨 (c) 𝟏𝟐𝟒𝐨 𝐂, 𝟔𝟐𝐨 𝐂 (d) 𝟑𝟕𝐨 𝐂, 𝟗𝟗𝐨 𝐂
SOLUTION : (b)
𝑻
Using, 𝒏 = 𝟏 − 𝑻𝟐
𝟏
𝟏 𝑻𝟐
𝒏= =𝟏−
𝟔 𝑻𝟏
𝑻 𝑻𝟐 −𝟔𝟐
and 𝟑 = 𝟏 − 𝑻𝟏
On solving, we get 𝑻𝟏 = 𝟗𝟗∘ 𝐂 and 𝑻𝟐 = 𝟑𝟕∘ 𝐂

47. Three Carnot engines operate in series between a heat source at a temperature 𝐓𝟏 and a heat
sink at temperature 𝐓𝟒 (see figure). There are two other reservoirs at temperature 𝐓𝟐 and 𝐓𝟑 , as
shown, with 𝐓𝟏 > 𝐓𝟐 > 𝑻 > 𝑇 𝟒 ] 𝐓𝐡𝐞𝐭𝐡𝐫𝐞𝐞 engines are equally efficient if: ( 𝟏𝟎𝐉𝐚𝐧. 𝟐𝟎𝐥𝟗𝐈
𝟏/𝟑 𝟏/𝟑 𝟏/𝟑
(a) 𝐓𝟐 = 𝐓𝟏 𝐓𝟒 𝟏/𝟐
; 𝐓𝟑 = 𝐓𝟏𝟐 𝐓𝟒 (b) 𝐓𝟐 = 𝐓𝟏𝟐 𝐓𝟒 ; 𝐓𝟑 = 𝐓𝟏 𝐓𝟒𝟐
𝟏/𝟑 𝟏/𝟑 𝟏/𝟒 𝟏/𝟒

(c) 𝐓𝟐 = 𝐓𝟏 𝐓𝟒𝟐 ; 𝐓𝟑 = 𝐓𝟏𝟐 𝐓𝟒 (d) 𝐓𝟐 = 𝐓𝟏𝟑 𝐓𝟒 ; 𝐓𝟑 = 𝐓𝟏 𝐓𝟒𝟑
SOLUTION : (b)
According to question, 𝜼𝟏 = 𝜼𝟐 = 𝜼𝟑
𝐓𝟐 𝐓𝟑 𝐓𝟒
𝟏− =𝟏− =𝟏−
𝐓𝟏 𝐓𝟐 𝐓𝟑
[Three engines are equally efficient]
𝐓𝟐 𝐓𝟑 𝐓𝟒
⇒ = =
𝐓𝟏 𝐓𝟐 𝐓𝟑
⇒ 𝐓𝟐 = 𝐓𝟏 𝐓𝟑 (i) 𝐓𝟑 = 𝐓𝟐 𝐓𝟒 (ii) From(i) and(ii)
𝟐𝐓𝟒 ) /
𝐓𝟐 = (𝐓𝟏 𝟏𝟑
𝐓𝟑 = 𝐓𝟏 𝐓𝟒𝟐 𝟏𝟑 /
48. Two Carnot engines A and 𝐁 are operated in series. The first one, A receives heat at 𝐓𝟏 =
𝟔𝟎𝟎𝐊 and rejects to a reservoir at temperature 𝐓𝟐 . The second engine 𝐁 receives heat rejected
by the first engine and in turn, rejects to a heat reservoir at 𝐓𝟑 = 𝟒𝟎𝟎𝐊 . Calculate the
temperature 𝐓𝟐 if the work outputs of the two engines are equal: [9 Jan. 2019 II]
(a) 600 𝐊 (b) 400 𝐊 (c) 300 𝐊 (d) 500 𝐊
SOLUTION : (d)
𝐓𝟏 −𝐓𝟐 𝐰𝐀 𝐓𝟐 −𝐓𝟑 𝐖𝐁
𝜼𝐀 = = and, 𝜼𝐁 = =
𝐓𝟏 𝐐𝟏 𝐓𝟐 𝐐𝟐
According to question, 𝐖𝐀 = 𝐖𝐁
𝐐𝟏 𝐓𝟏 𝐓𝟐 − 𝐓𝟑 𝐓𝟏
= × =
𝐐𝟐 𝐓𝟐 𝐓𝟏 − 𝐓𝟐 𝐓𝟐
𝐓𝟏 + 𝐓𝟑 𝟔𝟎𝟎 + 𝟒𝟎𝟎
𝐓𝟐 = = = 𝟓𝟎𝟎𝐊
𝟐 𝟐
49. A 𝐂𝐚𝐫𝐧𝐨𝐭 ↑ 𝐬 engine works as a refiigerator between 250 𝐊 and 300 K. It receives 500 cal heat
from the reservoir at the lower temperature. The amount ofwork done in each cycle to operate
the refrigerator is: [Online Apri115, 2018]
(a) 420 𝐉 (b) 2100 𝐉 (c) 772 𝐉 (d) 2520 𝐉
SOLUTION : . (a)
Given: Temperature of cold body, 𝐓𝟐 = 𝟐𝟓𝟎𝐊
temperature ofhot body; 𝐓𝟏 = 𝟑𝟎𝟎𝐊
Heat received, 𝑸𝟐 = 𝟓𝟎𝟎 cal work done, 𝐖 =?
𝑻 𝑾 𝟐𝟓𝟎 𝑾
Efficiency = 𝟏 − 𝑻𝟐 = 𝑸 ⇒ 𝟏 − 𝟑𝟎𝟎 = 𝑸
𝟏 𝟐 +𝑾 𝟐 +𝑾
𝑸𝟐 𝟓𝟎𝟎 × 𝟒. 𝟐
𝐖= = 𝑱 = 𝟒𝟐𝟎𝐉
𝟓 𝟓
50. Two Carnot engines A and 𝐁 are operated in series. Engine A receives heat from a reservoir at
𝟔𝟎𝟎𝐊 and rejects heat to a reservoir at temperature T. Engine 𝐁 receives heat rejected by
engine A and in turn rejects it to a reservoir at 𝟏𝟎𝟎𝐊. If the efficiencies ofthe two engines A and

𝐁 are represented by 𝜼𝐀 and 𝜼𝐁 respectively, then what is the value of A
B
𝟏𝟐 𝟏𝟐 𝟓 𝟕
(a) (b) (c) 𝟏𝟐 (d) 𝟏𝟐
𝟕 𝟓
SOLUTION : . (d)
𝑻𝟏 −𝑻𝟐 𝑻𝟐 −𝑻𝟑 𝑻𝟏 +𝑻𝟑
Efficiency ofengine 𝑨, 𝒏𝑨 = and 𝒏𝑩 = ; 𝑻𝟐 = = 𝟑𝟓𝟎𝑲
𝑻𝟏 𝑻𝟐 𝟐
600−350
𝒏 𝟔𝟎𝟎 𝟑𝟓𝟎−𝟏𝟎𝟎 𝟕
or 𝒏𝑨 = = 𝟏𝟐
𝑩 𝟑𝟓𝟎
51. An engine operates by taking 𝐧 moles of an ideal gas through the cycle ABCDA shown in
figure. The thermal efficiency ofthe engine is: (Take 𝐂𝐯 = 𝟏. 𝟓𝐑, where 𝐑 is gas constant)
(a) 𝟎. 𝟐𝟒 (b) 𝟎. 𝟏𝟓 (c) 𝟎. 𝟑𝟐 (d) 𝟎. 𝟎𝟖

SOLUTION : . (b)
Work‐done 𝐖 = 𝐏𝟎 𝐕𝟎
According to principle ofcalorimetry Heat given = 𝐐𝐀𝐁 = 𝐐𝐁𝐂
= 𝐧𝐂𝐕 𝐝𝐓𝐀𝐁 + 𝐧𝐂𝐏 𝐝𝐓𝐁𝐂
𝟑 𝟓
= 𝐧𝐑𝐓𝐁 − 𝐧𝐑𝐓𝐀 + 𝐧𝐑𝐓𝐂 − 𝐧𝐑𝐓𝐁
𝟐 𝟐
𝟑 𝟓 𝟏𝟑
= 𝟐𝐏𝟎 𝐕𝟎 − 𝐏𝟎 𝐕𝟎 + 𝟒𝐏𝟎 𝐕𝟎 − 𝟐𝐏𝟎 𝐕 = 𝐏𝐕
𝟐 𝟐 𝟐 𝟎 𝟎
𝐖 𝟐
Thermal efficiency of engine 𝜼 = 𝐐 = 𝟏𝟑 = 𝟎. 𝟏𝟓
𝐠𝐢𝐯𝐞𝐧
52. A Carnot fieezer takes heat from water at 𝟎∘ 𝐂 inside it and rejects it to the room at a
temperature of 𝟐𝟕∘ 𝐂. The latent heat of ice is 𝟑𝟑𝟔 × 𝟏𝟎𝟑 𝐉𝐤𝐠 −𝟏 . If 5 kg of water at 𝟎∘ 𝐂 is
converted into ice at 𝟎∘ 𝐂 by the fieezer, then the energy consumed by the fieezer is close to:
(a) 𝟏. 𝟓𝟏 × 𝟏𝟎𝟓 𝐉 (b) 𝟏. 𝟔𝟖 × 𝟏𝟎𝟔 𝐉 (c) 𝟏. 𝟕𝟏 × 𝟏𝟎𝟕 𝐉 (d) 𝟏. 𝟔𝟕 × 𝟏𝟎𝟓 𝐉
SOLUTION : (d)
𝜟𝐇 = 𝐦𝐋 = 𝟓 × 𝟑𝟑𝟔 × 𝟏𝟎𝟑 = 𝐐𝐬𝐦𝐤
𝐐𝐬𝐢𝐧𝐤 𝐓𝐬𝐢𝐧𝐤
=
𝐐𝐬𝐨𝐮𝐫𝐜𝐞 𝐓𝐬𝐨𝐮𝐫𝐜𝐞
𝐓𝐬𝐨𝐮𝐫𝐜𝐞
. . 𝐐𝐬𝐨𝐮𝐫𝐜𝐞 = × 𝐐𝐬𝐢𝐧𝐤
𝐓𝐬𝐢𝐧𝐤
Energy consumed by freezer

𝐓𝐬𝐨𝐮𝐫𝐜𝐞
⋅ woutput = 𝐐𝐬𝐨𝐮𝐫𝐜𝐞 − Qsink = Qsink −𝟏
𝐓𝐬𝐢𝐧𝐤
Given: 𝐓𝐬𝐨𝐮𝐫𝐜𝐞 = 𝟐𝟕∘ 𝐂 + 𝟐𝟕𝟑 = 𝟑𝟎𝟎𝐊,

𝐓𝐬𝐢𝐧𝐤 = 𝟎∘ 𝐂 + 𝟐𝟕𝟑 = 𝟐𝟕𝟑𝐤
𝟑𝟎𝟎
𝐖𝐨𝐮𝐭𝐩𝐮𝐭 = 𝟓 × 𝟑𝟑𝟔 × 𝟏𝟎𝟑 − 𝟏 = 𝟏. 𝟔𝟕 × 𝟏𝟎𝟓 𝐉
𝟐𝟕𝟑
53. A solid body ofconstant heat capacity 1 𝐉/𝐨 𝐂 is being heated by keeping it in contact with
reservoirs in two ways :
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount ofheat.
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same
amount ofheat.
In both the cases body is brought from initial temperature 𝟏𝟎𝟎∘ 𝐂 to fmal temperature 𝟐𝟎𝟎∘ 𝐂.
Entropy change ofthe body in the two cases respectively is : [2015]
(a) ln2, 𝟐𝒍𝒏𝟐 (b) 𝟐𝒍𝒏𝟐, 𝟖𝒍𝒏𝟐 (c) ln2, 𝟒𝒍𝒏𝟐 (d) ln2, ln2
SOLUTION : (d)
The entropy change of the body in the two cases is same as entropy is a state
function.
54. A Carnot engine absorbs 1000 𝐉 of heat energy from a reservoir at 𝟏𝟐𝟕∘ 𝐂 and rejects 600 𝐉
ofheat energy during each cycle. The efficiency of engine and temperature of sink will be:
∘ ∘
(a) 20% 𝐚𝐧𝐝 − 𝟒𝟑 𝐂 (b) 40% 𝐚𝐧𝐝 − 𝟑𝟑 𝐂 (c) 50% 𝐚𝐧𝐝 − 𝟐𝟎 𝐂 (d) 70% 𝐚𝐧𝐝 − 𝟏𝟎∘ 𝐂
∘
SOLUTION : (b)
Given: 𝐐𝟏 = 𝟏𝟎𝟎𝟎𝐉 , 𝐐𝟐 = 𝟔𝟎𝟎𝐉
𝐓𝟏 = 𝟏𝟐𝟕∘ 𝐂 = 𝟒𝟎𝟎𝐊 , 𝐓𝟐 =? 𝜼 =?
𝐖
Efficiency ofcarnot engine, 𝜼 = 𝐐 × 𝟏𝟎𝟎%
𝟏
𝐐𝟐 −𝐐𝟏 𝟏𝟎𝟎𝟎−𝟔𝟎𝟎
or, 𝜼 = × 𝟏𝟎𝟎% or, 𝜼 = × 𝟏𝟎𝟎%
𝐐𝟏 𝟏𝟎𝟎𝟎
𝜼 = 𝟒𝟎%
𝐐 𝐓
Now, for carnot cycle 𝐐𝟐 = 𝐓𝟐
𝟏 𝟏
𝟔𝟎𝟎 𝐓𝟐
=
𝟏𝟎𝟎𝟎 𝟒𝟎𝟎
𝟔𝟎𝟎×𝟒𝟎𝟎
𝐓𝟐 = = 𝟐𝟒𝟎𝐊 = 𝟐𝟒𝟎 − 𝟐𝟕𝟑
𝟏𝟎𝟎𝟎
𝐓𝟐 = −𝟑𝟑∘ 𝐂
55. The above p‐v diagramrepresents the thermodynamic cycle ofan engine, operating with an
ideal monatomic gas. The amount ofheat, extracted from the source in a single cycle is [2013]
𝟏𝟑 𝟏𝟏
(a) 𝐩𝟎 𝐯𝟎 (b) 𝐩𝟎 𝐯𝟎 (c) 𝐩𝟎 𝐯𝟎 (d) 𝟒𝐩𝟎 𝐯𝟎
𝟐 𝟐
SOLUTION : (b)
Heat is extracted from the source in path DA andAB is
𝟑 𝑷 𝟎 𝑽𝟎 𝟓 𝟐𝑷𝟎 𝑽𝟎
𝜟𝑸 = 𝑹 + 𝑹
𝟐 𝑹 𝟐 𝑹
𝟑 𝟓
⇒ 𝑷𝟎 𝑽𝟎 + 𝟐𝑷𝟎 𝑽𝟎
𝟐 𝟐
𝟏𝟑
= 𝑷 𝟎 𝑽𝟎
𝟐
56. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a
temperature of 𝟓𝟎𝟎𝐊. It is desired to have an engine of efficiency 60%. Then, the intake
temperature for the same exhaust (sink) temperature must be: [2012]
(a) efficiency of Camot engin𝐞 caimot 𝒌 made larger than 50%
(b) 𝟏𝟐𝟎𝟎𝐊 (c) 𝟕𝟓𝟎𝐊 (d) 𝟔𝟎𝟎𝐊
SOLUTION : (c)
𝑻
The efficiency ofthe carnot’s heat engine is given as 𝜼 = 𝟏 − 𝑻𝟐 × 𝟏𝟎𝟎
𝟏
When efficiency is 40%,

𝐓𝟏 = 𝟓𝟎𝟎𝐊; 𝜼 = 𝟒𝟎
𝑻𝟐
𝟒𝟎 = 𝟏 − × 𝟏𝟎𝟎
𝟓𝟎𝟎
𝟒𝟎 𝟐 𝑻
⇒ 𝟏𝟎𝟎 = 𝟏 − 𝟓𝟎𝟎
𝟐𝑻 𝟔𝟎
⇒ 𝟓𝟎𝟎 = 𝟏𝟎𝟎 ⇒ 𝑻𝟐 = 𝟑𝟎𝟎𝐊
When efficiency is 60%, then

𝟔𝟎 𝟑𝟎𝟎 𝟑𝟎𝟎 𝟒𝟎
= 𝟏− ⇒ =
𝟏𝟎𝟎 𝑻𝟐 𝑻𝟐 𝟏𝟎𝟎
𝟏𝟎𝟎×𝟑𝟎𝟎
⇒ 𝑻𝟐 = ⇒ 𝑻𝟐 = 𝟕𝟓𝟎𝐊
𝟒𝟎
57. The door of a working refrigerator is left open in a well insulated room. The temperature of air
in the room will [Online May 26, 2012]
(a) decrease (b) increase in winters and decrease in summers
(c) remain the same (d) increase
SOLUTION : (d)
In a refrigerator,
the heat dissipated in the atmosphere is more than that taken from the cooling chamber,
therefore the room is heated. If the door of a refrigerator is kept open.
58. This question has Statement 1 and Statement2. Ofthe four choices given after the Statements,
Statement 1:
An inventor claims to have constructed an engine that has an efficiency of 30% when
operated between the boiling and fieezing points of water. This is not possible.
Statement 2:
The efficiency of a real engine is always less than the efficiency of a Carnot engine
operating between the same two temperatures. [Online May 19, 2012]
(a) Statement 1 is true, Statement2 is true, Statement2 is not the correct explanation of
Statement 1.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is false, Statement2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of
Statement 1.
SOLUTION : (d)
According to Camot’s theorem
‐ no heat engine working between two given temperatures of source and sink can be more
efficient than a perfectly reversible engine
i.e. Carnot engine working between the same two temperatures.
𝑻𝟐
Efficiency of Carnot’s engine, 𝒏 = 𝟏 −
𝑻𝟏
where, 𝑻𝟏 = temperature of source

𝑻𝟐 = temperature of sink
𝟏
59. A Carnot engine operating between temperatures 𝑻𝟏 and 𝑻𝟐 has efficiency 𝟔. When 𝑻𝟐 is
𝟏
lowered by62 𝐊 its efficiency increases to 𝟑. Then 𝑻𝟏 and 𝑻𝟐 are, respectively: [2011]
(a) 𝟑𝟕𝟐𝐊𝐚𝐧𝐝𝟑𝟏𝟎𝐊 (b) 𝟑𝟑𝟎𝐊𝐚𝐧𝐝𝟐𝟔𝟖𝐊 (c) 𝟑𝟏𝟎𝐊𝐚𝐧𝐝𝟐𝟒𝟖𝐊 (d) 𝟑𝟕𝟐𝐊𝐚𝐧𝐝𝟑𝟏𝟎𝐊

SOLUTION : (d)
𝑻 𝟏
Efficiency ofengine 𝜼𝟏 = 𝟏 − 𝑻𝟐 = 𝟔
𝟏
𝑻 𝟓
⇒ 𝑻𝟐 = 𝟔 (i)
𝟏
When 𝑻𝟐 is lowered by 𝟔𝟐𝐊, then

𝑻𝟐 −𝟔𝟐 𝑻 𝟔𝟐 𝟏
Again, 𝜼𝟐 = 𝟏 − = 𝟏 − 𝑻𝟐 + 𝑻 = 𝟑 (ii)
𝑻𝟏 𝟏 𝟏
Solving (i) and (ii), we get,

𝟓
𝑻𝟏 = 𝟑𝟕𝟐𝐊 and 𝑻𝟐 = 𝟔 × 𝟑𝟕𝟐 = 𝟑𝟏𝟎𝐊
60. A diatomic ideal gas is used in a Carnot engine as the working substance. Ifduring the
adiabatic expansion part ofthe cycle the volume ofthe gas increases from 𝑽 to 32 𝑽, the
efficiency ofthe engine is [2010]
(a) 𝟎. 𝟓 (b) 𝟎. 𝟕𝟓 (c) 𝟎. 𝟗𝟗 (d) 𝟎. 𝟐𝟓
SOLUTION : ( b )
𝜸−𝟏
For adiabatic expansion 𝑻𝟏 𝑽𝐘−𝟏
𝟏 = 𝑻𝟐 𝑽𝟐
𝒈−𝟏 𝒈−𝟏
⇒ 𝑻𝟏 𝑽 = 𝑻𝟐 𝟑𝟐𝑽
𝑻𝟏
⇒ = 𝟑𝟐 𝜸−𝟏
𝑻𝟐
𝟕
For diatomic gas, 𝜸 = 𝟓
𝟐
𝜸−𝟏 =
𝟓
𝑻𝟏 𝟐
= 𝟑𝟐 𝟓 ⇒ 𝑻𝟏 = 𝟒𝑻𝟐
𝑻𝟐
𝑻 𝑻 𝟏 𝟑
Now, efficiency = 𝟏 − 𝑻𝟐 = 𝟏 − 𝟒𝑻𝟐 = 𝟏 − 𝟒 = 𝟒 = 𝟎. 𝟕𝟓.
𝟏 𝟐
61. ACarnot engine, having an efficiency of 𝜼 = 𝟏/𝟏𝟎𝐚𝐬𝐡𝐞𝐚𝐭 engine, is used as a refrigerator. If

the work done on the system is 10 𝐉, the amount of energy absorbed from the reservoir at lower
temperature is [2007]
(a) 𝟏𝟎𝟎𝐉 (b) 99 𝐉 (c) 90 𝐉 (d) 1 𝐉
SOLUTION : (c)
The efficiency (𝜼) of a Carnot engine and the coefficient ofperformance (𝜷) of a refrigerator are
𝟏−𝜼
related as 𝜷 = 𝜼
𝑸𝟐
Also, 𝜷 = 𝑾
𝟏 − 𝒏 𝑸𝟐
𝜷= =
𝒏 𝑾
𝟏
𝟏 − 𝟏𝟎 𝑸𝟐
𝜷= 𝟏
= .
𝑾
𝟏𝟎
is independent of path taken by the process.
𝑸𝟐
⇒𝟗=
𝟏𝟎
⇒ 𝑸𝟐 = 𝟗𝟎 J.
62. The temperature‐entropy diagram of a reversible engine cycle is given in the figure. Its
efficiency is [2005]
𝟏 𝟏 𝟐 𝟏
(a) 𝟒 (b) 𝟐 (c) 𝟑 (d) 𝟑
SOLUTION : (d)
𝟏
𝑸𝟏 = area under 𝐁𝐂 = 𝑻𝟎 𝑺𝟎 + 𝑻𝟎 𝑺𝟎
𝟐
𝑸𝟐 = area under AC = 𝑻𝟎 𝟐𝑺𝟎 − 𝑺𝟎 = 𝑻𝟎 𝑺𝟎

and 𝑸𝟑 = 𝟎
𝑸𝟐 𝑻𝟎 𝑺𝟎 𝟏
=𝟏− =𝟏−𝟑 =
𝑸𝟏 𝑻𝟎 𝑺𝟎 𝟑
𝟐
63. Which of the following statements is correct for any thermodynamic system 𝑹𝒆𝒋𝒆𝒄𝒕 [2004]
(a) The change in entropy can never be zero
(b) Internal energy and entropy are state functions
(c) The internal energy changes in all processes
(d) The work done in an adiabatic process is always zero.
SOLUTION : (b)
Intemal energy and entropy are state ffinction,
they are independent ofpath taken.
64. “Heat cannot by itselfflow 𝐟𝐢𝐢𝐎 𝐦 a body at lower temperature to a body at higher temperature”
is a statement or consequence of [2003]
(a) second law ofthermodynamics (b) conservation of momentum
(c) conservation of mass (d) first law ofthermodynamics
SOLUTION : (a)
This is a consequence of second law of thermodynamics
65. A Carnot engine takes 𝟑 × 𝟏𝟎𝟔 𝐜𝐚𝐥 ofheat 𝐟𝐢𝐢𝐨𝐦 areservoir at 𝟔𝟐𝟕∘ 𝐂, and gives it to a sink at
𝟐𝟕∘ 𝐂. The work done by the engine is [2003]
(a) 𝟒. 𝟐 × 𝟏𝟎𝟔 𝐉 (b) 𝟖. 𝟒 × 𝟏𝟎𝟔 𝐉 (c) 𝟏𝟔. 𝟖 × 𝟏𝟎𝟔 𝐉 (d) zero
SOLUTION : (b)
Here, 𝑻𝟏 = 𝟔𝟐𝟕 + 𝟐𝟕𝟑 = 𝟗𝟎𝟎𝐊
𝑻𝟐 = 𝟐𝟕 + 𝟐𝟕𝟑 = 𝟑𝟎𝟎𝐊
𝑻 𝟑𝟎𝟎 𝟏 𝟐
EfFiciency 𝜼 = 𝟏 − 𝑻𝟐 = 𝟏 − 𝟗𝟎𝟎 = 𝟏 − 𝟑 = 𝟑
𝟏
𝑾
But 𝜼 = 𝑸
𝑾 𝟐 𝟐 𝟐
= 𝟑 ⇒ 𝑾 = 𝟑 × 𝑸 = 𝟑 × 𝟑 × 𝟏𝟎𝟔 = 𝟐 × 𝟏𝟎𝟔 cal
𝑸
= 𝟐 × 𝟏𝟎𝟔 × 𝟒. 𝟐𝐉 = 𝟖. 𝟒 × 𝟏𝟎𝟔 𝐉
66. Which statement is incorrectQ [2002]
(a) All reversible cycles have same efficiency
(b) Reversible cycle has more efficiency than an irreversible one
(c) Carnot cycle is a reversible one
(d) Carnot cycle has the maximum efficiency in all cycles
SOLUTION : (a)
All reversible engines have same efficiencies if they are working for the same temperature
ofsource and sink.
If the temperatures are different, the efficiency is different.
67. Even Carnot engine cannot give 100% efficiency because we cannot [2002]
(a) prevent radiation (b) fmd ideal sources
(c) reach absolute zero temperature (d) eliminate fiction
SOLUTION : (c)
In Carnot’s cycle we assume frictionless piston, absolute insulation and ideal source
and sink(reservoirs).
𝑻
The efficiency of carnot’s cycle 𝜼 = 𝟏 − 𝑻𝟐
𝟏
The efficiency of carnot engine will be 100% when its sink 𝑻𝟐 is at 𝟎 K.

The temperature 𝐨𝐟𝟎𝐊 (absolute zero) cannot be realised in practice
so, efficiency is never 100%.
Kinetic Theory of Gases
Gas
In gases the intermolecular forces are very weak and its molecule may fly apart in all
directions. So the gas is characterized by the following properties.
 In gases, molecules are far apart from each other and mutual attractions between them are
negligible.
 They can easily compressed and expand.
 It has no shape and size and can be obtained in a vessel of any shape or size.
 It expands indefinitely and uniformly to fill the available space.
 It exerts pressure on its surroundings.
GAS LAWS :
BOYLE'S LAW:
At constant temperature, the volume of a given mass of gas is inversely proportional to its
pressure.
P1 P2
Increase
pressure
Decrease
V1 pressure V2
(A) (B)
Let us consider an ideal gas in a container with piston , it has initial volume V 1 at pressure P 1. . if
piston pushed inword at constant temperature , pressure increased to P 2 then its volume decreases
to V 2. therefore volume is inversely proportional to pressure at constant temperature
1
i.e. V
P
const an t
P
V
PV = constant
PV
1 1  PV
2 2
NOTE :
mass m
From Density   
volume V
m
volume V   (Density of the gas)
Then PV = constant
m 
PV = P    constant

Here mass m = constant
P
   constant
P1 P2

1 2
NOTE :
In PV = K, the value of the constant 'K' depends on
1. Mass of gas 2. Temperature of gas 3. System of units

Boyle's law generally holds good only at low pressure and high temperatures.A gas which obey
Boyle's law under all conditions of temperature and pressure is called ideal gas.
Boyle's law can be experimentally verified by Quill's tube (or) Boyle's law apparatus.
The graphs drawn between P & V at constant temperture of a gas are called isotherms
Graphical representation :
If m and T are constant
P - V Graph :
V
(A)
PV - P and PV - V Graph :
PV PV
P V
(B) (C)
Charle’s law :
If the pressure remaining constant, the volume of the given mass of a gas is directly proportional to
its absolute temperature.
P
P
Increase
volume
T2 > T1
Decrease
V2
T1 V1 volume
(A) (B)
Let us consider an ideal gas in a container with piston , it has initial volume V 1 at absolute temperature T1.
. At constant pressure , temerature of gas increased to T 2 then its volume increases to V 2. therefore volume
of gas is directly proportional to its absolute temperature at constant pressure.
i.e., V  T
V  cons tan t (T )
V V1 V
 constant  2
T T1 T2
NOTE :
mass m
From Density   
volume V
m
volume V   (Density of the gas)
m
V

V
Then T
 constant
V m
  constant
T T
T  constant
1T1  2T2
Graphical representation: If m and P are constant
V 1/V V
T T 1/T
(A) (B) (C)
1/P P
T 1/T
(D) (E)
NOTE :
1
If the pressure remains constant, the volume of the given mass of a gas increases or decreases by 273 . 15
of
its volume at 0°C for each 1°C rise or fall in temperature.
 1 
Vt  V0  1  t .
 273 . 15 
Vt
V0
This is Charle’s law for centigrade scale.

t(°C)
– 273.15 O
Gay-Lussac’s law or pressure law :
The volume remaining constant, the pressure of a given mass of a gas is directly proportional to its
absolute temperature.
Let us consider an ideal gas in a container with piston , it has initial pressure P 1 at absolute
temperature T1. . At constant volume , temerature of gas increased to T 2 then its pressure
increases to P2. therefore PRESSURE of gas is directly proportional to its absolute temperature at
constant volume .
PT
P
 constant
T
P1 P
 2
T1 T2
Graphical representation :
If m and V are constants
P P/T P/T
T T or 1/T P or 1/P
(A) (B) (C)
1/P P
T 1/T
(D) (E)
Grahm’s law of diffusion :

When two gases at the same pressure and temperature are allowed to diffuse into each other, the
rate of diffusion of each gas is inversely proportional to the square root of the density of the gas
1 1
i.e. rate of diffusion r 
 M
  is the density of the gas
M  is the molecular weight of the gas
r1 2 M2
 
r2 1 M1
If V is the volume of gas diffused in t sec then

V
rate of diffusion r  t
r1 V t
 1 2
r2 V2 t1
Dalton’s law of partial pressure :

The total pressure exerted by a mixture of nonreacting gases occupying a vessel is equal to the
sum of the individual pressures which each gases exert if it alone occupied the same volume at a given
temperature.
For n gases
P  P1  P2  P3  ..... Pn
where
P = Pressure exerted by mixture
P1 , P2 , P3 , ...... Pn  Partial pressure of component gases.
Ideal Gas :
A gas which obey Boyle's law under all conditions of temperature and pressure is called ideal gas.
 Real gases obey gas laws only at low pressure and high temperatures. All Gases are real gases only.
 Attraction between the molecules of perfect gas is zero.
 Ideal or perfect gas obey gas laws at all temperatures and pressures without any limitations.
 Hydrogen or Helium behaves closely as perfect gas. Hence it is preferred in constant volume gas
thermometers.
Assumption of Ideal Gases (or Kinetic Theory of Gases):
Kinetic theory of gases relates the macroscopic properties of gases (such as pressure, temperature etc.) to
the microscopic properties of the gas molecules (such as speed, momentum, kinetic energy of molecule etc.)
Actually it attempts to develop a model of the molecular behaviour which should result in the observed
behaviour of an ideal gas. It is based on following assumptions :
 Every gas consists of extremely small particles known as molecules. The molecules of a given gas are
all identical but are different than those of another gas.
 The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.
 Their size is negligible in comparison to intermolecular distance (10–9 m)
 The volume of molecules is negligible in comparison to the volume of gas. (The volume of molecules is
only 0.014% of the volume of the gas).
 Molecules of a gas keep on moving randomly in all possible direction with all possible velocities.
 The speed of gas molecules lie between zero and infinity
 The gas molecules keep on colliding among themselves as well as with the walls of containing vessel.
These collisions are perfectly elastic.
 The time spent in a collision between two molecules is negligible in comparison to time between two
successive collisions.
 The number of collisions per unit volume in a gas remains constant.
 No attractive or repulsive force acts between gas molecules.
 Gravitational attraction among the molecules is ineffective due to extremely small masses and very
high speed of molecules.
 Molecules constantly collide with the walls of container due to which their momentum changes. The
change in momentum is transferred to the walls of the container. Consequently pressure is exerted by
gas molecules on the walls of container.
 The density of gas is constant at all points of the container.
Equation of State or Ideal Gas Equation :
The equation which relates the pressure (P) volume (V) and temperature (T) of the given state of an
ideal gas is known as ideal gas equation or equation of state.
From Boyle’s Law
1
V
P
.........(1)
From charles Law
VT ............(2)
From equation (1) and (2)
T
V
P
T
V K
P
PV
 K (cons tan t )
T
PV PV
1 1
 2 2
T1 T2
For 1 mole of gas K  R (universal gas constant. )
PV
T
R (constant)
PV = RT
where R = universal gas constant.

For n moles of gas PV  nRT
Universal gas constant (R) :

A gas constant per one mole of gas is called as Universal gas constant
 it is same for all gases

 The value of "R" does not depend on the mass of gas or its chemical formula.
 The fact that R is a constant for all gases is constant with Avagadro's hypothesis that "equal volumes of
all gases under same conditions of temperature and pressure contains equal number of molecules".
PV Pressure  Volume Work done

R  
nT n  Temperature n  Temperature
At S.T.P. the value of universal gas constant is same for all gases
J cal ~ 2 cal litre  atm
R= 8.31
mole  kelvin
 1.98
mole  kelvin mol  kelvin
 0 . 8221
mole  kelvin
.
Joule
It’s unit is
mole  kelvin
Dimension : [ML2 T 2 1 ]
Specific gas constant (r) :

A gas constant per unit mass of gas is called as specific gas constant
 it is different for different gases
It is represented by per gram gas constant
R
i.e., r
M
.
Since the value of M is different for different gases. Hence the value of r is different for different gases.
R
e.g. It is maximum for hydrogen rH 2 
2
Joule
It’s unit is
kg  kelvin
Dimension : [L2 T 2 1 ]
ideal gas equation PV  nRT
m
But No of moles n 
M
m
PV  RT
M
PV  mrT
Boltzman’s constant (k) :
A gas constant per one molecule of gas is called as Boltzman’s constant .
 it is same for all gases
R 8 . 31
i.e., k  
N 6 . 023  10 23
 1 . 38  10 23 J / K
J
unit is
kelvin
dimension : [ML2 T 2 1 ]
ideal gas equation PV  nRT
no.of molecules N
But No of moles   avagadro's number = N
A
N
n
NA
N
PV  RT
NA
PV  Nk BT
Different forms of gas equation
Quantity of gas Equation Constant
1 mole gas PV = RT R = universal gas constant
n mole gas PV = n RT R = universal gas constant
1 molecule of gas PV  Nk BT k B = Boltzmann’s constant
1 gm gas PV  mrT r = specific gas constant
Real Gases :
The gases actually found in nature are called real gases. They do not obeys gas Laws.
For exactly one mole of an ideal gas

PV T
1 V K
RT P
PV
 K (cons tan t )
T
KR
PV  nRT
PV
Plotting the experimentally determined value of RT
for exactly one mole of various real
gases as a function of pressure P, shows a deviation from identity.
PV
 The quantity RT
is called the compressibility factor and should be unit for an ideal gas.
2 CH4
N2 2 200 K 500 K
H2
1.5
1.5
1000 K
PV CO2
1 PV
Ideal gas 1
RT Ideal gas
RT
0.5
0.5
0 200 400 600 800 1000

0 300 600 900 1200 P(atm)
P (atm)
Fig. 13.8
Fig. 13.7
 A real gas behaves as ideal gas most closely at low pressure and high temperature. Also can actual gas
can be liquefied most easily which deviates most from ideal gas behaviour at low temperature and
high pressure.
Equation of state for real gases :
It is given by Vander Waal’s with two correction in ideal gas equation. The it know as Vander
Waal’s gas equation.
(i) Volume correction :
Due to finite size of molecule, a certain portion of volume of a gas is covered by the molecules
themselves. Therefore the space available for the free motion of molecules of gas will be slightly less than the
volume V of a gas.
Hence the effective volume becomes (V – b).
(ii) Pressure correction :
Due to intermolecular force in real gases, molecule do not exert that force on the wall which they
would have exerted in the absence of intermolecular force. Therefore the observed pressure P of the gas
will be less than that present in the absence of intermolecular force.
 a 
Hence the effective pressure becomes  P  2  .
 V 
(iii) Vander Waal’s gas equations
 a 
For 1 mole of gas  P  2  (V  b)  RT
 V 
 an 2 
For n moles of gas  P  2  (V  nb)  nRT
 V 
Here a and b are constant called Vander Waal’s constant.
Dimension : [a] = [ ML5T 2 ] and [b] = [L3]
Units : a = N  m4 and b = m3.

Pressure of an Ideal Gas :
Y
Y

vx vy v

vx X
X vx
vz
L Z
Z
Consider an ideal gas (consisting of N molecules each of mass m) enclosed in a cubical box of side L.
Let Any molecule of gas moves with velocity v in any direction and collide with wall A1 and
rebounds

where v  v xî  v y ˆj  v z kˆ
v  v x2  v y2  v z2 .
Due to random motion of molecule v x  vy  v z
v 2  3v x2  3v y2  3v z2
Time between two successive collision with the wall A1

Distance travelled by molecule between two successive collision
t 
Velocity of molecule
2L
t 
vx
1 v
Then the number of collision per second. n   x
t 2 L
This molecule collides with the shaded wall (A1) with velocity vx and rebounds with velocity v x .
The change in momentum of the molecule p  ( mvx )  (mvx )  2mvx
As the momentum remains conserved in a collision,
the change in momentum of the wall A1 is p  2mv x p  2mvx
After rebound this molecule travel toward opposite wall A2 with velocity vx , collide to it and again
rebound with velocity vx towards wall A1.
Force exerted by a single molecule on shaded wall is equal to rate at which the momentum is transferred
to the wall by this molecule.
p 2mvx mv 2
i.e. FSingle molecule    x
t (2 L / vx ) L
m
The total force on the wall A1 due to all the molecules Fx   vx2
L
m 2 mN 2
F X= (v x  v x2 2  v x2 3  ...)  vx
M 1 L
v x2  mean square of x component of the velocity..

Now pressure is defined as force per unit area, hence pressure on shaded wall
Fx mN 2 mN 2
Px   vx  vx
A AL V
For any molecule, the mean square velocity v 2  v x2  v y2  v z2
by symmetry v x2  vy2  vz2
v2
v x2  v y2  v z2 
3
1 mN 2 1 m N 2
Total pressure inside the container P v  vrms (where vrms  v 2 )
3 V 3 V
Relation between pressure and kinetic energy :
1 mN 2
As we know P
3 V
v rms
but M = mN = Total mass of the gas

1 M 2
P 3 V
v rms
M
Density of gas 
V
1 2
Pressure P  3
 v rms ... (i)
1 2
Kinetic Energy K.E = Mvrms
2
K .E 1M 2 1 2
 K.E. per unit volume = E  2  V  vrms   vrms ...(ii)
V 2
2
From (i) and (ii), we get P  E
3
i.e. the pressure exerted by an ideal gas is numerically equal to the two third of the mean kinetic energy of
translation per unit volume of the gas.
Effect of mass on pressure :

1 mN 2
Pressure P vrms
3 V
(m N )T
P [As v rms
2
T ]
V
If volume and temperature of a gas are constant
P  mN
Pressure  (Mass of gas).
i.e. if mass of gas is increased, number of molecules and hence number of collision per second
increases . so pressure will increase.

Effect of on pressure :
(m N )T
P
V
If mass and temperature of a gas are constant.
P  (1/V),
i.e., if volume decreases, number of collisions per second will increase due to lesser effective
distance between the walls resulting in greater pressure.

Effect of on pressure :
(m N )T
P
V
If mass and volume of gas are constant,
P  (v rms ) 2  T
i.e., if temperature increases, the mean square speed of gas molecules will increase and as gas
molecules are moving faster, they will collide with the walls more often with greater momentum resulting in
greater pressure.
Average translational kinetic energy of a gas:
Let M be the molecular mass and V be the molar volume of a gas. Let m be the mass of each molecule.
Then
3 3 3
 Mean K.E. per mole of a gas, E  PV  RT  k B N T
2 2 2
3
 Mean K.E. per molecule of a gas E  k BT
2
3 RT
 K.E of 1 gram of gas 
2 M
Note :
 Kinetic energy per molecule of gas does not depends upon the mass of the molecule but only depends
upon the temperature of the gas.
3
It is given as E  k BT or E  T
2
 Molecules of different gases say He, H 2 and O2 etc. at same temperature will have same translational
kinetic energy though their rms speeds are different
 Kinetic energy per mole of gas depends only upon the temperature of gas.
 Kinetic energy per gram of gas depend upon the temperature as well as molecular weight (or masso f
one molecule ) of the gas
3 kT T
KEgram   KEgram 
2m m
 From the above expression it is clear that higher the temperature of the gas, more will be the gas, more
will be the average kinetic energy possessed by the gas molecules. At T=0, E=0, i.e. at absolute
zero the molecular motion ceases.
 The kinetic interpretation of temperature gives the relation between the average kinetic energy of a gas
molecule and absolute temperature.
i.e E   3 / 2  NK BT  E  T
 Average kinetic energy is independent of pressure, volume and the nature of the ideal gas
PROBLEMS
1. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During
an observation time of 1 second, an atom travelling with the root–mean–square speed parallel to one of
the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other
25
atoms. Take R = J/mol–K and k = 1.38 × 10–23 J/K.
3
(a) Evaluate the temperature of the gas.
(b) Evaluate the average kinetic energy per atom.
(c) Evaluate the total mass of helium gas in the box.
SOLUTION :
Volume of the box = 1m3,
Pressure of the gas = 100 N/m2.
Let T be the temperature of the gas
1
(a) Time between two consecutive collisions with one wall = sec
500
2I
This time should be equal to v , where  is the side of the cube.
rms

1
2 vrms 
50 0
vrms = 1000 m/s
3 RT
 = 1000
M
1 0 00 2 M 10 6  3  1 0 3 
T   1 6 0K
3R  25 
3 
 3 
3 3
(b) Average kinetic energy per atom = kT = [(1.38 × 10–23) 160J = 3.312 ×–21 J
2 2
m
(c) From PV = nRT = RT,
M
PVM
Mass of helium gas in the box m=
RT
100  1   4  10 3 
Substituting the values, m = = 3.0 × 10–4 kg
 25  
  160 
3
2. Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the masses of
molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively. Find
the temperature of the mixture.
SOLUTION :
3
Total energy of molecules of first gas = n 1 kT1 ,
2
3
Total energy of molecules of second gas = n 2 kT2
2
Let temperature of mixture be T
3
then total energy of molecules of mixture = k (n 1  n 2 )T
2
3 3
(n 1  n 2 )kT  k(n 1 T1  n 2 T2 )
2 2
T
 n1T1  n2T2 
 n1  n2 
3. 1 kg of diatomic gas is at a pressure of 8  104 N / m 2 . The density of the gas is 4kg / m3 .

The energy of the gas due to its thermal motion is
SOLUTION :
5
Energy of diatomic gas due to its thermal motion is  PV
2
5 m
 P
2   
4. Gas at a pressure P0 is contained in a vessel. If the masses of all the molecules are halved and
their speeds are doubled, the resulting pressure P will be equal to
p0
1) 4P0 2) 2P0 3) p0 4)
2
SOLUTION :
1 2
Pressure P    v rms 
3
2
P  m  v rms 
2
P2 m2  v2 
  
P1 m1  v1 
m 2
P2 2  2v 
  
P1 m v 
P2  2 P0
Various Speeds of Gas Molecules:

The motion of molecules in a gas is characterised by any of the following three speeds.
Root mean square speed :
It is defined as the square root of mean of squares of the speed of different molecules
v12  v 22  v 32  v 42  ....
i.e. vrms   v2
N
From the expression of pressure
1 2
P  v rms
3
3P 3 PV 3 RT 3 kT
vrms    
 Mass of gas M m
Mass of gas
where  
V
 Density of the gas ,
M=   (mass of gas),
pV  RT ,
R= kN A ,
k Boltzmann’s constant,
M
m= NA = mass of each molecule.
With rise in temperature rms speed of gas molecules increases as
v rms  T .
With increase in molecular weight rms speed of gas molecule decreases as

1
v rms  .
M
e.g., rms speed of hydrogen molecules is four times that of oxygen molecules at the same temperature.
rms speed of gas molecules is of the order of km/s
e.g., at NTP for hydrogen gas

3 RT 3  8 . 31  273
(vrms )    1840 m / s .
M 2  10 3
3
 rms speed of gas molecules is  times that of speed of sound in gas, as
3 RT
rms speed of gas v rms 
M
RT
speed of sound in gas v s 
M
3
v rms  vs

 rms speed of gas molecules does not depends on the pressure of gas (if temperature remains constant)
because P   (Boyle’s law)
if pressure is increased n times then density will also increases by n times but vrms remains constant.
 Moon has no atmosphere because vrms of gas molecules is more than escape velocity (ve).
A planet or satellite will have atmosphere only if vrms  ve
 The molecules of gases will escape out from a planet. If the temperature of planet
M v e2
T
3R
vrms  ve  ; ve = escape velocity of planet
M = molecular mass of gas

 At T = 0; vrms = 0
i.e. the rms speed of molecules of a gas is zero at 0 K. This temperature is called absolute zero.
(2) Most probable speed :
The particles of a gas have a range of speeds. This is defined as the speed which is possessed by
maximum fraction of total number of molecules of the gas. e.g., if speeds of 10 molecules of a gas are 1, 2,
2, 3, 3, 3, 4, 5, 6, 6 km/s, then the most probable speed is 3 km/s, as maximum fraction of total molecules
possess this speed.
2kBT 2RT 2PV 2P

Most probable speed vmp  m

M

massof gas


(3) Average speed : It is the arithmetic mean of the speeds of molecules in a gas at given temperature.
v 1  v 2  v 3  v 4  .....
v av 
N
and according to kinetic theory of gases
8kBT 8RT 8PV 8P

Average speed vavg  m

M

  massof gas 

 .
Relation between av,rms and mp :
Average speed vavg  0.92vrms
Most probable speed v mp  0.816v rms
v rms : v avg : v mp  1.73 :1.60 :1.41
Clearly, vrms  vavg  vmp

PROBLEMS
1. The root-mean-square (rms) speed of oxygen molecules (O2) at a certain absolute
temperature is  . If the temperature is doubled and the oxygen gas dissociates into
atomic oxygen, the rms speed would be
1)  2) 2 3) 2 4) 2 2
SOLUTION :
3 RT
rms speed of gas v rms 
M
V T
M
V2 T M
 2 1
V1 T1 M 2
V2 2T 2M

V1 T M
V2
2
V1
V2  2V
2. Calculate the temperature at which the oxygen moleules will have the same rm,s velocity as the hydrogen
molecules at 1500 C . Molecular weight of oxygen is 32 and that of hydrogen is 2
SOLUTION :
3RT
Vrms 
M
AsV1rms  V2 rms , therefore T M
T1 M 1 M 
 ; T2  T1  2 
T2 M 2  M1 
 32 
T2  423   , T2  6768 K
2
t 2  6768  273  64950 C
3.Two moles of an ideal gas X occupying a volume V exerts a pressure P. The same pressure is exerted by one
of another gas Y occupying a volume 2V. if the molecular weight of Y is 16 times the molecular weight of
X, find the ratio of the rms speeds of the molecules of X and Y.
SOLUTION :
3PV  m 
Vrms    n
nM  M 
V1rms V1 n2 M 2 V1 rms V  1  16
 ;   
V2 rms V2 n1 M 1 V2 rms 2V  2  1
V1 rms
2
V2 rms
4. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to
the r.m.s. speed of a helium gas atom at 200 C ? Atomic mass of argon  39.9u and that of
helium =4.0u
SOLUTION :
Here, atomic mass of argon, M1  39.9u
atomic mass of helium, M2  4.0u
Suppose that the r .m.s speed of argon gas atoms at temperature T is equal to the r .m.s . speed of
Helium gas atoms at 200 C i.e., at T  20  273  253K .
Suppose that u r .m.s and ur.m.s are ur.m.s speeds of argon and helium gas atoms at temperatures T
and T’ respectively
3RT
Now. ur.m.s =
M
3RT 3R  253
 ur.m.s  and ur.m.s 
39.9 4.0
Since ur.m.s = ur.m.s , we have
3RT 3R  253

39.9 4.0
253  39.9
(or) T  2,523.7 K
4.0
5. Your are given the following group of particles, ni represents the number of molecules with
speed vi
ni 2 4 8 6 3

vi ms 1  1.0 2.0 3.0 4.0 5.0
calculate (i) average speed (ii) rms speed
SOLUTION : (i) average speed

n1v1 +n 2 v2 +n 3 v3 +n 4 v 4 +n 5 v5
v ang 
 n1 +n 2 +n 3 +n 4 +n 5 
 2 1   4  2   8  3   6  4    3  5  73
   3.17 ms 1
 2  4  8  6  3  23
(ii) root mean square speed is
 n1v12 +n 2 v 22 +n 3 v 32 +n 4 v 24 +n 5 v 52 
v rms =  
 n1 +n 2 +n 3 +n 4 +n 5 
1
2
 2 1  4  4  8  9  6 16  3  25 
   3.36ms 1
 2 4863 
6.. If the molecular weight of two gases are M1 and M2 then at a temperature the ratio of root mean square
velocity v1 and v2 will be
M1 M2 M1  M 2 M1  M 2
(a) M2 (b) M1 (c) M1  M 2 (d) M1  M 2
SOLUTION :
3RT
Vrms 
M
1
Vrms 
M
V2 M2

V1 M1
7. The r.m.s. velocity of a gas at a certain temperature is 2 times than that of the oxygen molecules at
that temperature. The gas can be
(a) H2 (b) He (c) CH 4 (d) SO 2
SOLUTION :
1
v rms 
M
v1 M2

v2 M1
1 M2
   M 2  16 .
2 32
Hence the gas is CH 4 .
8. At what temperature will the rms speed of oxygen molecule will be sufficient for escaping from
earth? (Ve= 11.2km/s,m=2.76x10-26kg and k=1.38x10-23J/K)
1) T  91.2 2) 9.36  104 K
3) 0.36  104 K 4) 5.36  104 K
SOLUTION :
If the temperature is T,
according to kinetic theory of gases translational KE  3 kBT

2
The oxygen molecule will escape from earth
3 1
if kT  mve2
2 2
mve2
i.e. T 
3k
2
2.76  1026  11.2  103 
T
3  1.38 1023
T  8.36  104 K
9. If the density of hydrogen at STP is 0.09kgm 3 , calculate the rms velocity of its molecules
at 00 C .
1) 2.84  103 ms 1 2) 1.84  103 ms 1
3) 0.84  103 ms 1 4) 2  103 ms 1
SOLUTION :
1 2
P    vrms 
3
3P
 vrms 

Here P=76cm of Hg = 1.013 105 Nm 2
  0.09kg m 3
 3  1.013  10 5 
; v rms  = 1.84 103 ms 1
0.09
10. At what temperature is the root mean square speed of oxygen molecules equal to the r.m.s
speed of carbon dioxide molecules at -230C, molecular weight of oxygen =32 and that of carbon
dioxide = 44.
1) +91.20C 2) 91.20C 3) +112.20C 4) 112.20C
SOLUTION :
3RT
Vrms 
M
3RT ' 3R  250

V 'rms  
M 44
Vrms=V’rms
3RT 3R  250

32 44
0
T  91.2 C
11. Your are given the following group of particles, ni represents the number of molecules with
speed vi
ni 2 4 8 6 3

vi ms 1  1.0 2.0 3.0 4.0 5.0
most probable speed is .....

SOLUTION :
By definition,velocity belongs to more molecules is most probable speed=3.0 ms1.
12. If the density of hydrogen at STP is 0.09kgm 3 , calculate the rms velocity of its molecules
at 00 C .
SOLUTION :
1 2
P    vrms 
3
3P
 vrms 

Here P=76cm of Hg = 1.013 105 Nm 2
  0.09kg m 3 ;
 3  1.013  10 5 
v rms 
0.09
v rms  1.84 103 ms 1

13. When the temperature of a gas is raised from 27°C to 90°C, the percentage increase in the r.m.s.
velocity of the molecules will be
(a) 10% (b) 15% (c) 20% (d) 17.5%
SOLUTION :
3 RT v2 T2 (273  90 )
vrms  v    1 .1
M 1 T1 (273  30 )
v 
% increase   2  1   100  0 . 1  100  10 %
v
 1 
14. The r.m.s speed of oxygen molecule (O2) at a certain temperature T is V. If on increasing the
temperature of the oxygen gas to 2T, the oxygen molecules dissociate into atomic oxygen, find
the speed of the oxygen atom
1) 2V 2) V 3) V/2 4) 3V
SOLUTION :
3KT
Vrms 
m
3K  2T 
V 'rms 
m/2
3K4T
V 'rms 
m
V’rms = 2Vrms
15. Calculate the temperature at which root mean square velocity of SO2 gas molecules is same as
that of O 2 molecules at 1270C. Molecular weights of O2 and SO2 are 32 and 64 respectively
1) 5270C 2) 8000C 3) 5000c 4) 6270c
SOLUTION :
3RT
root mean square velocity of SO2 V rms  M
3RT '
root mean square velocity of O2 V rm s =
M'
Vrms  V 'rms
3RT 3RT '

=
M M'
T '  527 0 C
16. If three molecules have velocities 0.5, 1 and 2 kms-1 respectively, calculate the relation between
the root mean square speed and average speed
1) Vrms = Vavg/2 2) Vrms = Vavg 3) Vrms = 1.134 Vavg 4) Vrms = 2Vavg
SOLUTION :
V12  V22  V32

Vrms  = 1.323
3
V1  V2  V3
Vavg   1.67km / s
3
Vrms = 1.134 Vavg
17. If the r.m.s. velocity of a gas at a given temperature (Kelvin scale) is 300 m/sec. What will be the r.m.s.
velocity of a gas having twice the molecular weight and half the temperature on Kelvin scale =
(a) 300 m/sec (b) 600 m/sec (c) 75 m/sec (d) 150 m/sec
SOLUTION :
3 RT T
vrms   vrms 
M M
v2 M 1 T2
 
v1 M 2 T1
1 1
 
2 2
v1 300
v2 
2

2
=150m/sec
18. At what temperature is the rms speed of an atom in an argon gas cylinder equal to the rms
speed of a helium gas atom at 200 C ? Atomic mass of argon  39.9u and that of helium =4.0u
SOLUTION :
3RTar 3RTHe
Given that  v rm s  A r =  v rm s H e 
M ar M He
3RTar 3R(253)
; 
40 4
T 253

40 4
T  2530 K
T  2530  273
T  2257 0C
19. Two vessels having equal volume contain molecular hydrogen at one atmosphere, and helium
at two atmospheres respectively. What is the ratio of rms speeds of hydrogen molecule to that
of helium molecule, if both the samples are at the same temperature
SOLUTION :
According to kinetic theory of gases,
3RT
v rms =
M
 vrms H M He
=
 v rms He MH
vH   4 / 2 v He
v H  2  v He 
20. Three vessels of equal capacity have gases at the same temperature and pressure. The first
vessel contains neon (monoatomic), the second contains chlorine (diatomic) and the
third contains uranium hexafluoride (polyatomic)
(a) Do the vessels contain equal number of respective molecules?
(b) Is the root mean square speed of molecules the same in the three cases ? If not, in which
case is v rm s largest?
SOLUTION :
(a) Yes, the vessels contain equal number of respective molecules. It is because, equal volume of all the
gases under same temperature and pressure contain equal number of molecules (Avogadro’s hypothesis)
3 kT  v 1
(b) The root mean square speed is given by v rms = rm s 
m m
Since vrms .depends upon mass of the molecule of the gas, it will not be same in the three cases.
Since v rm s is inversely proportional to the square root of the mass of the molecules of the gas, it will be
largest for the gas, whose molecules are lightest. Therefore v rm s .will be largest for neon gas.
DEGREE OF FREEDOM (f)
• The number of possible ways in which a system can possess internal energy is called degrees of freedom.
OR
• The number of independent ways in which a molecule or an atom can exhibit motion or have energy is
called it's degrees of freedom.
OR
• The number of independent coordinates required to specify the dynamical state of a system is called it's
degrees of freedom.
For example
(a) Block has one degree of freedom, because it is confined to move in a straight line
and has only one transistional degree of freedom.
f=1
f=1
(b) The projectile has two degrees of freedom becomes it is confined to move in a plane and so it has
two translational degrees of freedom.
f= 2
(c) The sphere has two degrees of freedom one rotational and another translational.
Similarly a particle free to move in space will have three translational degrees of
freedom.
f=2
1 K2 7
Note : In pure rolling sphere has one degree of freedom as KE = mv2 (1+ )= mv2
2 R2 10
• The degrees of freedom are of three types :

(i) Translational degrees of freedom :
The maximum number of translational degrees of freedom can be three.
These are
1 2 1 2 1 2
mvx , mv y , mvz .
2 2 2
(ii) Rotational degrees of freedom :
The maximum number of rotational degrees of freedom can be three. The number of degrees of freedom
in this case depends on the structure of the molecule.
1 1 1
These are I x x2 , I y y2 , I z z2
2 2 2
(iii) Vibrational degrees of freedom:
Their numbers depend on atoms in the molecule and their arrangement. These degrees of freedom are
considered at a very high temperature.
Note : At room temperature only translational and rotational degrees of freedom are taken into account.
1. Monoatomic gas: The degrees of freedom of monoatomic gas molecules are due to three independent
translational motions along x, y and z axis. The degrees of freedom are
1 2 1 2 1 2
mvx , mv y , mvz .
2 2 2
2. Diatomic (or) Linear polyatomic gas:
Ix, Iy  0
Iz  0
The molecule has three degrees of freedom of translational and two degrees of freedom of rotational
these are
1 2 1 2 1 2 1 1
mvx , mv y , mvz , I xx2 , I y y2 .
2 2 2 2 2
Note : If vibrational degrees of freedom are taken into account, then total number of degrees of freedom of
diatomic molecule becomes 7 at high temperature. These are
1 2 1 2 1 2 1 1 1 1
mvx , mv y , mvz , I xx2 , I y y2 ,  v2 , kr 2 .
2 2 2 2 2 2 2
1 2 1
Here  v corresponds to kinetic energy of vibration (  is the reduced mass) and kr 2 corresponds
2 2
to potential energy of vibration (k is the force constant, r is the separation between the atoms)
3. Triatomic or non linear polyatomic gas:
The molecule of polyatomic gases like CO2 , H 2O, NH 3 , CH 4 , etc. has more than two atoms.
It has 3 translational and 3 rotational degrees of freedom as shown in figure Apart from this, if such a
molecule has v vibrational modes, it will have additional 2v vibrational degrees of freedom, each vibrational
mode contributing 2 vibrational degrees of freedom.
Thus, total number of degrees of freedom of a polyatomic gas molecule,
i.e., f  3  3  2v   6  2v 
Ato micity o f gas Trans latio nal Rotational To t al

y
Monoatomic
3 0 3
Ex. Ar, Ne, Idea l gas etc x
z
Diatomic
3 2 5
Ex. O 2 , Cl2 , N2 etc.
Triatomic (line ar)

Ex. CO 2, C 2 H 2 3 2 5 O=C=O
Tr iatomic (Non–linear)
or Polyatomic 3 3 6
Ex. H 2 O, NH 3 , C H 4
At high temperatures a diatomic molecule has 7 degrees of freedom. (3 translational, 2 rotational and 2
vibrational)
Law of equipartition of energy :
According to this law, for a system in thermal equilibrium, the total energy of a dynamic system is equally
distributed among its various degrees of freedom.
1
The energy associated with each degree of freedom is K BT per molecule
2
1
or RT per mole.
2
For a molecule with f degrees of freedom
1 f
Energy related with each degree of freedom = kT Energy related with all degree of freedom = kT
2 2
RT fR T
Energy per mole U  f  
2 2
U fR
Molar specific heat at constant volume Cv  
T 2
fR f 
Molar specific heat at constant pressure Cp =Cv  R   R  R   1
2 2 
Cp  2
Ratio of specific heat    1  
Cv  f 
4  f vib
Note:  poly  3  f ,
vib
f vib  number of modes of vibrations (at high temperatures)
Cv ,Cp and γ for different gases:

Monoatomic gas: Degrees of freedom f = 3
RT 3RT
Kinetic energy per mole E  U  3  
2 2
U 3R
T 2
5R
Molar specific heat at constant pressure C p =C v  R 
2
Cp 5R/2 5
Ratio of specific heat   =  1.66
Cv 3R/2 3
Diatomic Gas : Degrees of freedom f = 5
RT 5RT
Kinetic energy per mole U  5  
2 2
U 5R
Molar specific heat at constant volume C v  
T 2
7R
Molar specific heat at constant pressure C p =C v  R 
2
pC 7R / 2 7
Ratio of specific heat   C  5R / 2  5  1.4
v
Triatomic (or) polyatomic Gas:
Degrees of freedom f = 6
RT
Kinetic energy per mole U  6   3RT
2
U
Molar specific heat at constant volume C v   3R
T
Molar specific heat at constant pressure C p = C v  R  4 R
Cp 4R 4
Ratio of specific heat      1.33
Cv 3R 3
Specific heat capacity of solids
A solid consists of a regular array of atoms in which each of the atoms has a fixed equilibrium position.
As such in a solid, an atom has no translational or rotational degrees of freedom. It has only three
vibrational modes along three mutually perpendicular directions. Since each vibrational mode contributes
two vibrational degrees of freedom, an atom in a solid has six vibrational degrees of freedom. Further, as
1
energy associated per degree of freedom per atom is k BT ,
2
total inernal energy per mole ( N A atoms) of solid, i.e.,
1   R 
U  6  k BT  N A  3RT  as k B  
2   NA 
dU d
Thus, C    3RT  or
dT dT
C  3R  6cal / mol C º  25 J / mol K
The specific heat capacity at high temperatures (usually above 300K) is the same for all solids and is
approximately equal to 3R or 6 cal/mol Cº or 25 J/mol K.
Specific heat capacity of Water

A water molecule  H 2O  consists of 2 hydrogen atoms and 1 oxygen atom. Treating water like a solid,
each atom in its molecule has 6 degrees of freedom and as such its each molecule has 3  6  18
degrees of freedom. According to the law of equipartition of energy,
1
energy associated per molecule per degree of freedom  k BT
2
1 
internal energy associated per mole of water, U  18  k BT  N A  9 RT
2 
dU d
Tus, C    9 RT   9 R
dT dT
PROBLEMS
Cp
01. Each molecule of a gas has f degrees of freedom. The ratio C   for the gas is
V
f 1 2 ( f 1)
1) 1  2) 1  3) 1  4) 1 
2 f f 3
SOLUTION :
U fR
T 2
fR f 
Molar specific heat at constant pressure Cp =Cv  R   R  R   1
2 2 
Cp  2
Ratio of specific heat    1  
Cv  f 
2 A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is
the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 °C ? (R = 8.31 J
mo1–1 K–1).
SOLUTION :
1 mole of any gas occupies 22.4 litres
so 44.8 litres of helium contains 2 moles.

Since the volume of cylinder is fixed , the heat required is CV
Here Helium is monoatomic gas n  3/ 2

3
 CV  R
2
so heat required = nCV dT  2 1.5R 15  45R  3745
CP
3.. The ratio of two specific heats CV of CO is
(a) 1.33 (b) 1.40 (c) 1.29 (d) 1.66
SOLUTION :
Co is diatomic gas, for diatomic gas
7 5
CP  R and C V  R
2 2
CP 7 R / 2
 di    1 .4
CV 5 R / 2
CP
4. Considering the gases to be ideal, the value of  
CV for a gaseous mixture consisting of = 3 moles
of carbon dioxide and 2 moles of oxygen will be ( O2  1 .4 ,  CO 2  1 . 3)
(a) 1.37 (b) 1.34 (c) 1.55 (d) 1.63
SOLUTION :
 1 1   3  1 .3 2  1 .4
 2 2 
 1  1  2  1 (1 .3  1) (1 .4  1)
 mix    1 .33
1 2 3 2
 
 1  1  2  1 (1 .3  1) (1 .4  1)
5. One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific
heat of the mixture at constant volume is
3
(a) 8 (b) 2
R (c) 2R (d) 2.5 R
SOLUTION :
1C V1   2 CV2
(C V )mix 
1   2
3 5
1 R 1 R
 2 2  2R
1 1
 3 5 
 (CV )mono  R, (CV )di  R 
 2 2 
6. A gas has molar heat capacity C = 37.55J m ole -1 K -1 , in the process PT=constant. Find the
number of degrees of freedom of the molecules of the gas.
SOLUTION :
PT = K(constant) ...... (i)
RT RT
But PV  RT  P   T  K
V V
dV 2 R
  PT  K  ...... (ii)
dT P
dQ  dU  dW
; C dT  C vdT  PdV
 dV  dV
C  Cv  P   as, C  Cv  P
 dT  dT
 2R 
C  Cv  P    C  Cv  2 R
 P 
 Cv  C  2 R ........ (iii)
fR
But Cv  ....... (iv) from (iii) and (iv)
2
fR 2  C  2R 
 C  2R  f 
2 R
2  37.55  2  8.3
 5
8.3
7. What is the total kinetic energy of 2g of Nitrogen gas at temperature 300 K.
SOLUTION :
Given, mass of Nitrogen gas= 2 g
molecular waight M W  28, gm
Temperature T  300 K ,
R  8.31 107 erg mol 1 K 1
3 RT 3RT
Kinetic energy of 2g of N 2  2  2 M  M
W W
3  8.31 107  300

 erg  267 107 erg
28
8. Calculate the total kinetic energy of one kilo mole of Oxygen gas at 27 0 C
SOLUTION :
oxygen is a diatomic gas, its molecules have 5 degrees of freedom.
5
Therefore, the total kinetic energy of a molecule of Oxygen is kBT .
2
As, 1kgmole of Oxygen has N molecules,
the total kinetic energy of one kgmole of Oxygen at temperature T is
5 5
RT   8.31 103  300
2 2
= 6.23  106 Joule / Kg  mole
9. Estimate the average thermal energy of a helium at (i) room temperature (27 0 C ) (ii) the
temperature on the surface of the sun (6,000 K) and (iii) the temperature of 107 K . Given,
k  1.38  1023 J K 1
SOLUTION :
Here, k  1.38  1023 JK 1
(i) T  27  273  300 K
3 3
 K .E  k T  1.38  10 23  300 ,
2 2
= 6.21 1021 J
(ii) T=6,000 K,
3 3
 K .E .  k T   1.38  10 23  6, 000 ,  1.242 1019 J
2 2
(iii) T  107 K
3 3
 K .E .  k T   1.38  1023  6, 000 ,
2 2
 2.07  10 16 J
Mean free path    :
 The mean free path of a gas molecule may be defined as the average distance travelled by the molecule
between two successive collisions.
2
1 5
4
3
Let 1 , 2 ,...........n be the distance travelled by a gas molecule during n collisions, then the mean free
path of gas molecule is given by
1  2  ...........  n

n
During the collision, a molecule of a gas moves in a straight line with constant velocity. The statistical study
of heat gives the mean free path as following
1

2 nd 2
Where d is the diameter of molecule, n is the number of molecules per unit volume.
v rms
Collision frequency f 

 Mean free path depends on nature of molecule and with increase in n (number density) it decreases.
N P 1 K BT
 Here n  V 
K BT Hence,   2
2 d P
 
P T
1 m m
  2
 2

2 nd 2  m.n  d 2 d 2 
1
 As    and  m ,
 
 m
Brownian motion :
It provides a direct evidence for the existence of molecules and their motion. The zigzag motion of gas
molecules is Brownian motion because it occurs due to random collision of molecules. But this motion
cannot be seen. However, the zigzag motion of pollen grains   10 m  can be seen under a microscope
15
PROBLEMS
1. Estimate the mean free path for a water molecule in water vapour at 373K. The diameter of the molecule
is 2  10 10 m , and at STP number of molecules per unit volume is 2.7  1025 m 3
SOLUTION :
The number density(n) is inversely proportional to absolute temperature
1
 n
T
n373 273
 
n273 373
273
n373  n273 
373
273
n373  2.7 10 25   2  10 25 m 3
373
Given d= 2  10 10 m
1
Hence , mean free path  
2 d 2 n
1
  4  107 m
= 10 2
2  3.14   2  10   2  10 25
2. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing
nitrogen at 2 atm and temperature 170 C . Take the radius of nitrogen molecule to be 1A0 . ( Molecular
mass of nitrogen = 28u)
SOLUTION :
Here, P  2  1.013  105  2.026  105 Nm 2 ;
T  17  273  290 K ; M  28 g  28  103 kg

molecular diameter
. d  1 2  2 A0  2 1010 m;
Now , mass of a nitrogen molecule,
M 28  103
m   4.65  1026 kg
N 6.02  1023
Also, volume occupuied by the nitrogen gas
RT 8.31 290
V   1.19  102 m3
P 2.026  105
Therefore, density of the nitrogen gas,
M 28  101
  2
 2.353kgm 1
V 1.19  10
 m
Now,  
2 d 2 
4.65  1026
 2

2  2  1010   2.353
 1.11  107 m
3P 3  2.026  105
Now, vrms  
 2.353
 508.24ms 1
Therefore, collision frequency,
vrms 508.24
f   7
 4.58 109 S 1
 1.11 10
3. The collision frequency of nitrogen molecule in a cylinder containing nitrogen molecule in a cylinder
containing at 2.0 atm pressure and temperature 170 C . (Take radius of a nitrogen molecule is 1.0 A0 )
SOLUTION :
1
mean free path  
2 d 2 
K BT
  p  nK BT 
2 d 2 p

1.38 10   290
23
 1.110  7
1.414 3.14  2 10  2.026 10 
10 5
3 K BT 3  1.3823  290
Vrms    5.1102 m / s
m 28  1.66  1027
 collision frequency
Vrms5.1 102
f   7
 4.6  109 s 1
 1.110
4. Estimate the mean free path for a water molecule in water vapour at 373K. The diameter of the
molecule is 2 10 10 m , and at STP number of molecules per unit volume is 2.7  1025 m 3
SOLUTION :
The number density(n) is inversely proportional to absolute temperature
1
 n
T
n373 273
 
n273 373
273
 n373  n273 
373
273
n373  2.7 10 25   2  10 25 m 3
373
Given d= 2  10 10 m ;
1
Hence, mean free path   7
2 d 2 n  2.81 10 m
5. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting
all vibrational modes, the total internal energy of system is
1) 5RT 2) 11RT 3) 3RT 4) 7RT
SOLUTION :
Total internal energy of system
f1 f
 U oxygen  U arg on  1 RT   2 2 RT
2 2
5 3
 2 RT  4 RT  5 RT  6 RT  11RT
2 2
[As f1  5 (for oxygen) and f 2  3 (for argon)]
6. A container has 1 mole of nitrogen at 270C. The pressure in side the container is 2 atmosphere.
Assuming the molecules move with root mean square speed. Then find The number of collisions
per second which the molecules make with unit area of the container wall.
SOLUTION :
a) The number of molecules n present per unit volume at pressure P and temperature T.
P 2  1.05 105
n   5.07  1025
kT 1.38 1023  300
3RT 3N A kT
rms  
M M
3  6.02  1023  1.38 1023  300

 3
 516.75ms 1
28  10
1
Number of collisions/m2 =  n   rms
6
1
  5.07  1025  516.75  4.37  1027
6
7. Each molecule of nitrogen gas heated in a vessel to a temperature of 5000 K has an aver age
energy E1. Some molecules of the gas escape into atmosphere at 300 K. Due to collision with
air molecules, average kinetic energy of the nitrogen molecule changes to E2.
Find the ratio E1/E2.
1) 7/6 2) 3/7 3) 6/7 4) 7/3
SOLUTION :
At T = 5000 K ( HIGH TEMPERATURE )
f = 7
1 1
E1  f  KT  7  K  5000
2 2
= 17,500 K
At T = 300 K ( LOW TEMPERATURE )
f = 5
1 1
E 2  f  KT  5  K  300
2 2
= 750 K
E1 / E2 = 7/3
8. A vessel contains two non-reactive gases neon and oxygen . The ratio of their partial pressures
is 3:2. Estimate the ratio of number of molecules Atomic mass of Ne=20.2 u, molecular mass
of O2  32.0 u
3 2 3 5
1) 2) 3) 4)
2 3 5 2
SOLUTION :
Since V and T are CONSTANTS
we have PV
1  1 RT and P2V   2 RT
P1 1
 
P2 2 where 1 and 2 refer to neon and oxygen respectively
P1 3
Given P  2
2
1
  3
2 2
N1 N2
By definition 1  N and  2  N
A A
where N1 and N 2 are the number of molecules of 1 and 2

 The ratio of number of molecules
N1 1 3
 
N 2 2 2
9. A vessel contains two non-reactive gases neon and oxygen . The ratio of their partial pressures
is 3:2. Find mass density of neon and oxygen in the vessel. Atomic mass of Ne=20.2 u, molecular
mass of O2  32.0 u
SOLUTION :
Since V and T are CONSTANTS
we have PV
1  1 RT and P2V   2 RT
P1 1
 
P2 2
where 1 and 2 refer to neon and oxygen respectively
P1 3
Given P  2
2
1 3
 
2 2
If 1 and  2 are mass densities of 1 and 2 respectively,
1 m1 / V m1
we have   m / V  m
2 2 2
m1 m2
But we can also write 1  M and 2  M
1 2
m1 1 M 1 3 20.2
      0.947
m2 2 M 2 2 32.0
10: A vessel is filled with a gas at a pressure of 76cm of mercury at a certain temperature. The
mass of the gas is increased by 50% by introducing more gas in the vessel at same temperature.
Find out the resultant pressure of the gas.
SOLUTION :
According kinetic theory of gases,
1m 2
pressure P   vrms  .
3V
Given, T= constant
so vrms = constant.
also V=onstant
P m
P2 m 2

P1 m1
P2  m1  50 m1  3
;  
76 m1 2
3
P2   76  114cm of Hg
2
11. The total number of air molecules in a room of capacity 20 m3 at a temperature of 27 0 C and
1 atm pressure is.
SOLUTION :
Given ;
Volume (V) = 20 m3
Temperature (T) = 27+273=300K
Pressure (P) = 1 atm = 1.01 105 Pa
PV
Total number of air molecules N  K T
B
1.01 105  20
N
1.38  1023  300
N  4.87  1026
12. At what temperature, the mean kinetic energy of O2 will be the same for H 2 molecules at – 73°C
(a) 127°C (b) 527°C (c) – 73°C (d) – 173°C
SOLUTION :
Mean kinetic energy of molecule depends upon temperature only. For O2 it is same as that of H 2 at
the same temperature of 73 C .
13. A cylinder of fixed capacity 22.4 litres contains helium gas at standard temperature and pressure.
What is the amount of heat needed to raise the temperature of the gas in the cylinder by 30 °C
? (R = 8.31 J mo1–1 K–1).
SOLUTION :
1 mole of any gas occupies 22.4 litres.
Since the volume of cylinder is fixed , the heat required = nCV dT
3
Here Helium is monoatomic gas so CV  R
2
Heat required  1  1.5 R  30  45 R  374.5 J
14. A pressure cooker contains air at 1 atm and 30°C. If the safety value of the cooler blows when the
inside pressure  3 atm, then the maximum temperature of the air, inside the cooker can be
(a) 90°C (b) 636°C (c) 909°C (d) 363°C
SOLUTION :
P1 T
Since volume is constant,Hence  1
P2 T2
1 (273  30 )

3 T2
T2  909 K  636 C
15. A container has 1 mole of nitr ogen at 270C. The pressure in side the container is 2 atmosphere.
Assuming the molecules move with root mean square speed. If the container is made thermally
insulated and moves with constant speed 0 . If it stops suddenly, the process results in the rise
of the temperature of the gas by 10C. Calculate the speed 0 .
SOLUTION :
1
Kinetic energy  m02
2
Heat energy gained  Cv T  Cv 1  Cv and CP  Cv  R
CP R
1 
Vv Cv
R R
 1  i.e. Cv 
Cv  1
1 2 R 1
Cv  mv0  mv02
2  1 2
2R
which gives V0  m   1
 
2  8.31
 3
 38.5ms 1
28 10  1.4  1
16. Figure shows a cylindrical tube of radius r and length l, fitted with a cork. The friction coefficient
between the cork and the tube is  . The tube contains an ideal gas at temperature T, and
atmosphereic pressure P0. The tube is slowly heated; the cork pipe out when temperature is
doubled. What is normal force per unit length exerted by the cork on the periphery of tube ?
Assume uniform temperature throughout gas at any instant.
d I
d N
P 0
 N
P g a s
SOLUTION :
Since volume of the gas is constant
Pi Pf

Ti T f
 Tf 
Pf  Pi    2 Pi  2 P0
 Ti 
The forces acting on the cork are shown in the figure, in equilibrium.
P0  A   N  2 P0 A
P0 A
N

N is the total normal force exerted by the tube on the cork; hence contact force per unit length is
dN N PA
  0
dl 2 r 2 r
17. One mole of an ideal monatomic gas requires 210 J heat to raise the temperature by 10 K, when heated
at constant temperature. If the same gas is heated at constant volume to raise the temperature by 10 K
then heat required is
(a) 238 J (b) 126 J (c) 210 J (d) 350 J
SOLUTION :
(Q )P  C P T and (Q )V  C V T
3
R
(Q)V CV 3
  2 
(Q)P C P 5
R 5
2
 3 5 
(CV )mono  2 R, (C P )mono  2 R 
 
3 3
(Q)V   (Q )P   210  126 J
5 5
18. A nitrogen molecule at the surface of earth happens to have rms speed for that gas at 0ºC.
If it were to go straight up without colliding with other molecules, how high would it rise? (Mass
of nitr ogen molecule ,m = 4.65  10-26 kg,R = 8.3 J/mol/K)
SOLUTION :
The molecule goes to a height h till its entire K.E is converted into P.E
1 2
mgh  m  v rms  ;
2
2
 vrms  3RT
h 
2g M .2g
3  8.31 273
h
28  10 3  2  9.8
19. Certain amount of an ideal gas are contained in a closed vessel. The vessel is moving with a constant
velocity v. The molecular mass of gas is M. The rise in temperature of the gas when the vessel is
suddenly stopped is (  C P / C V )
Mv 2 Mv 2 (  1) Mv 2 Mv 2
(a) 2 R(  1)
(b) (c) 2 R (  1)
(d) 2 R(  1)
2R
SOLUTION :
1
If m is the total mass of the gas then its kinetic energy  mv 2
2
When the vessel is suddenly stopped then total kinetic energy will increase the temperature
of the gas.
1 m R
Hence mv 2   Cv T  Cv  T [As Cv  ]
2 M  1
m R 1
T  mv 2
M  1 2
Mv 2 (  1)
T  .
2R
20. The energy of a gas/litre is 300 joules, then its pressure will be
(a) 3  10 5 N / m 2 (b) 6  10 5 N / m 2
(c) 10 5 N / m 2 (d) 2  10 5 N /m2

SOLUTION :
Energy  300 J / litre  300  10 3 J / m 3
2 2  300  10 3
P E  2  10 5 N / m 2
3 3
21. Two thermally insulated bulbs, filled with air and connected by a short tube containing a valve,
initially closed. The pressures, volumes and temperatures in the two vessels are P1,V1,T1 and
P2,V2,T2 respectively. Find the P, T values after opening the valve.
SOLUTION :
When the valve is opened, the air flows from the bulb at higher pressure to the bulb at lower pressure.
In equilibrium both the vessels have the same pressure.=P
Total volume of the air, V = V1 + V2
After mixing of air, total number of moles, n = n1+n2
After mixing of air, total number of moles, n = n1+n2 Let the common temperature attained be T.
Hence P(V1+V2)=(n1+n2)RT
P
 n1  n2  RT
..........(1)
V1  V2 
The combined system is thermally insulated; hence Q = 0;
system does no mechanical work, since dV = 0, W = pdV = 0.
From first law of thermodynamics, Q = dU + W
Hence dU = 0.
There is no change in internal energy.
The internal energy U of an ideal gas is given by
nRT PV
U  nCT    1    1
   
n1RT1 n2 RT2
U initial  
  1    1
 R 
   n1T1  n2T2  .........(2)
  1 
U final 
 n1  n2  RT
   1 ..........(3)
U initial  U final
n1T1  n2T2   n1  n2  T ]
 nT n T 
T   1 1 2 2  .......(4)
 n1  n2 
PV
1 1  PV
2 2

 PV
1 1 PV
2 2
   T1T2  PV 2 2
1 1  PV

T2    PV 2 2T1 
  1 1T2  PV
 T1
From eqn. (1)
 n1  n2  RT R
P   n1  n2  T
V1  V2  V1  V2
From eqn. (4)
 n1  n2  T  n1T1  n2T2
 n1T1  n2T2    1 1 2 2 
R PV  PV
P
V1  V2  V1  V2 
as PV
1 1  n1 RT1
and PV
2 2  n2 RT2
Maxwell’s Law (The distribution of molecular speeds).
a. The vrms gives us a general idea of molecular speeds in a gas at a given temperature. This doesn’t mean
that the speed of each molecule is v rms .
b. Maxwell derived an equation giving the distribution of molecules in different speeds as follows:
3/2 mv 2
 m  2

2 k BT
dN  4 N   ve dv
 2 k BT 
Where dN = Number of molecules with speeds between v and v+dv ,
m = mass of the gas molecule
kB  Boltzmann’s constant N = total number of molecules
dN
c. Graph between ( number of molecules at particular speed) and v ( speed of these molecules).
dv
dN
From the graph it is seen that is maximum at most probable speed.
dv
3RT 8RT 2 RT
  
M M M
dN
dv 12
3
1vmp
2vavg
3vrms
v
v (v+dv)
From the graph

1) The area under the graph represents total number of molecules.
2) The shape of the curve is such that area (shown shaded enclosed by its portion on right side of vmp is
more than the area on the left side of vmp . Thus, the number of molecules having speeds less than vmp is
less than the number of molecules having speeds more than vmp .
Note: 1.If 1 moles of CV1 , 2 moles of CV2 ..... are mixed, then by conservation of energy
U 1  U 2  ......  U
1CV T  2CV T  ......  (1  2  ...)CV T
1 2 mix
n1CV1  n2CV2  ......

 CVmix  and CPmix  CVmix  R
n1  n 2  ....
2. If 1 moles of  1 and 2 moles of  2 .... are mixed then by conservation of energy we have
U 1  U 2  .....  U
1CV T  2CV T  ......  (1  2  ...)CV T

1 2 mix
 R   R  R
1    2    ..  ( 1  2  ..)
 1 1    2 1   mix  1
1 2 ( 1  2  ...)
  .... 
1 1  2 1  mix  1
1 2 1   2
For two gases   1    1  
1 2 m ix  1
PRACTICE BITS
1. Three closed vessels A, B and C are at the same temperature T and contain gases which obey the
Maxwellian distribution of velocities. Vessel A contains only O 2 , B only N 2 and C a mixture of equal
quantities of O 2 and . If the average speed of the molecules in vessel A is , that of the molecules in vessel
B is , the average speed of the molecules in vessel C is
(a) (V1  V2 ) / 2 (b) V1 (c) (V1 V 2 )1 / 2 (d) 3 kT / M
KEY:b
2. Given graph gives variation of PV/T with P for 1gm of oxygen at two different
temperatures T1 and T2 . If density of oxygen is 1.427 kg m3 . The value of PV/T at point A
and relation b w T1 & T2 is
PV
T
Ideal
A
gas
T1
T2
P
1) 0.256J/K and T1<T2 2) 8.314 J/K and T1>T2
3) 8.314 J/K and T1<T2 4) 0.256J/K and T1>T2
KEY:4
m PV m
HINT : PV   RT  RT   R
M T M
Real gas behave as ideal gases at low pressure and high temperature, so, T1  T2
03. A gas has volume V and pressure p. The total translational kinetic energy of all the
molecules of the gas is
3 3
1) pV only if the gas is monoatomic 2) pV only if the gas is diatomic
2 2
3 3
3) > pV if the gas is diatomic4) pV in all cases
2 2
KEY:4
04. Three closed vessels A, B are C are at the same temperature. Vessel A contains only O2,
B only N2 and C a mixture of equal quantities of O2 and N2. If the average speed of O2
molecules in vessel A is 1 , that of N2 molecules in vessel B is 2 , the average speed of O2
molecules is vessel C is
1)
1

 
2 1 2  2) 1 3) 
1 2 4)
3kT
M
KEY:2
05. If the pressure in a closed vessel is reduced by drawing out some of the gas, the mean free
path of the two molecules
1) Increases 2) decreases 3) Remains unchanged
4) Increases of decrease according to the nature of the gas
KEY:2
6. The temperature of the gas consisting of rigid diatomic molecules is T = 300K. Calculate the
angular root mean square velocity of a rotating molecule if its moment of inertia is equal
to I  2.1 1039 g cm2 .
1) 6.3  1012 rad / sec 2) 6.8  1012 rad / sec
3) 3.6  1012 rad / sec 4) 3.2  1012 rad / sec

KEY:1
1 1
HINT : I  2  2. kT
2 2
KEY:1
07. The temperature of a gas is -68°c. To what temp should it be heated so that the root mean
square velocity of the molecules is doubled.
1) 547°c 2) 820°c 3) 1092 K 4) 547 K

KEY:1
8. N Molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the
same vessel, which are maintained at a temperature T. The mean square velocity of molecules of B
type is denoted by  2 and the mean square of the ‘X’ component of the velocity of A type is denoted by
2
 , then  2 is
2
1) 2 2) 1 3) 1 3 4) 2 3
KEY:4
HINT :
Mean kinetic energy of two types of molecules should be equal.
1 1 2 2
so,
2
 
m 3 2   2m  v 2  2 
2 v 3
09. The pressure exerted on the walls of the container by a gas is due to the fact that gas molecules
are
1) Losing their kinetic energy at walls 2) Stricking to the energy
3) Changing their moment due to collision with the walls
4) Getting accelerated towards the wall
KEY:3
10. A gas mixture consists of 2 moles of oxygen and 4 moles of argon

at temperature T. Neglecting all vibrational modes, the total
internal energy of the system is :
1) 4 RT 2) 15 RT 3) 9 RT 4) 11 RT
KEY:4
11. The figure given below shows the plot of versus P for oxygen gas at two different temperatures.
Read the following statements concerning the curves given below
i) The dotted line corresponds to the ‘ideal’ gas behaviour

ii) T1 > T2
iii) The value of at the point, where the curves meet on the yaxis is the same for all gases.
Which of the above statement is true ?
1) i only 2) i and ii 3) all of these 4) none of these
KEY:3
12. Three closed vessels A, B and C at the same temperature T and contain gases which obey
the Maxwellian distribution of velocities. Vessel A contains only contains only contains a
mixture of equal quantities of. If the average speed of the molecules in vessel A is and
that of the molecules in vessel B is, the average speed of the molecules in vessel C is
1) 2) 3) 4)
KEY:2
13.1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K in
figure. One face of the cube (EFGH) is made up of a material which totally absorbs
any gas molecule incident on it. At any given time
1) The pressure on EFGH would be zero

2) The pressure on all the faces will be equal
3) The pressure of EFGH would be double the pressure on ABCD
4) The pressure on EFGH would be half that on ABCD.
KEY:4
14. A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP.
The vessel is being carried by a rocket which is moving at a speed of 500 in
vertical direction. The pressure of the gas inside the vessel as observed by us on
the ground
1) remains the same because is very much smaller than of the gas
2) remains the same because motion of the vessel as a whole does not affect the
relative motion of the gas molecules and the walls
3) will increase by a factor equal to where was the original mean square velocity
of the gas
4) will be different on the top wall and bottom wall of the vessel
15. A vessel of volume V = 5.0 litre contains 1.4 gm of nitrogen at temperature T=1800 K. Find the
pressure of the gas if 30% of its molecule are dissociated into atom at this temperature.
1) 0.54 x 105 N/m2 2) 1.94 x 105 N/m2
3) 2.62 x 105 N/m2 4) 3.75 x 105 N/m2
KEY:2
 RT P1 1
HINT : P  , 
M V P2 1  2
16. The pressure of an ideal gas varies according to the law P = P0 – AV2 where P0 and A are positive
constants. What is the highest temperature that can be attained by the gas.
P0 P0 P0 P0
1) 2)
nR A nR 2A
2P0 P0 2P0 P0
3) 4)
nR 2A 3nR 3A
KEY:4
HINT : P  P0  AV 2
nRT
  P0  AV 2 at max
V
dT
temperature  0  P0  3 AV 2
dV
17. The mass 15 gram of Nitrogen is enclosed in a vessel at 300K. What heat must be supplied to
it to double the rms velocity of its molecules
1) 10J 2) 10KJ 3) 103J 4) 102J
KEY:2
HINT :since v2  T  T2  1200K
m R
Heat  H   ncvdT  T2  T1 
M   1
18 At what absolute temperature T, is rms speed of a hydrogen molecule equal to its escape velocity
from the surface of the moon? The radius of moon is R, g is the acceleration due to gravity on
moon’s surface, m is the mass of a hydrogen molecule and k is the Boltzmann constant.
mgR 2mgR 3mgR 2mgR
1) 2) 3) 4)
2k k 2k 3k
KEY:4
3KT
HINT : vrms  v  2 gR
m e
19 Find the number of degrees of freedom of molecules in a gas whose molar heat capacity at
constant pressure is equal to Cp=29 J/(mol.K)
1) 3 2) 4 3) 5 4) 6
KEY:3
R  2
HINT :C p    1 ;    1  f 
 
20 Determine the gas temperature at which the root mean square velocity of hydrogen molecules
exceeds their most probable velocity by v  400 m / s
1) 384K 2) 342K 3) 300K 4) 280K
KEY:1
RT
HINT : vrms  v p   3  2   v
M
2
M  v 
T    384k
R  3  2 
21 Two chambers one containing m1 gram of a gas at pressure P1 and other containing m2 gram of
same gas at pressure P2 are put in communication with each other. If temperature remains
constant, the common pressure reached will be,
P1 P2 (m1  m2 ) m1m2 ( P1  P2 )
1) Pm  P m 2) Pm  P m
1 2 2 1 1 2 2 1
P1 P2 m1 m1m2 P2
3) Pm  P m 4) Pm  P m
1 2 2 1 1 2 2 1
KEY:1
P
HINT : From Boyle’s law,  Const  K

P1 m1 m1 K
1  , (or) V1  
K 1 P1
m2 K
llly, V2 
P2
m m  m  m2
V  V1  V2  K  1  2  ;   1
 P1 P2  V1  V2
P1 P2 (m1  m2 )
P  K 
Pm1 2  P2 m1
22. How many degrees of freedom does a gas molecule have under standard conditions if gas has
density 1.3 kg/m3 and the velocity of sound propagation in the gas is v=330m/s?
1) 2 2) 3 3) 4 4) 5
KEY:4
p
HINT : v

2
  1 where f is degrees of freedom
f
23 In crude model of a rotating diatomic molecule of chlorine (Cl2), the two Cl atoms are 2.0  10 10 m
apart and rotate about their centre of mass with angular speed   2.0  1012 rad / s . What is
the rotational kinetic energy of one molecule of Cl2, which has a molar mass of 70.0g/mol?
1) 2.32 1021 J 2) 2.32  1021 J
3) 2.32 1021 erg 4) 2.32  1021 erg

KEY:1
HINT :

m m
cl cl
70  103
m  5.81 1026 kg and
2  6.02  1023
2.0 1010
r  1010 m ;
2
M.I =2( mr2 ) = 2mr2
2
I  2  5.811026 1010 
1 2
 KR  I   2.32  10 21 J
2
24. An ideal gas undergoes a process in which PV–a = constant, where V is the volume occupied by the
gas initially at pressure P. At the end of the process, rms speed of gas molecules has become a1/2
times of its initial value. What will be the value of CV so that energy transferred in the form of heat
to the gas is 'a' times of the initial energy.
a 2
 1 R  a  1 R
2
 a  1 R  a  1 R
1) 2) 3) 4)
a2 1  a  1
2
 a  1  a  1
KEY:4
HINT : Q  aU1
f
nC T2  T1   a nRT1 (1)
2
but v rms T  T2  aT1
(as rms speed became a times).
af
From (1)  C  a  1 T1  RT1
2
afR  2C  R
C  here, f  v  C  Cv  (for polytrophic process)
2  a  1  R  1 n
where n = a
THEORY BITS
1. Which of the following statements are true regarding the kinetic theory of gases ?
1) The pressure of the gas is directly proportional to the average speed of the molecules
2) The root mean square speed of the molecules is directly proportional to the pressure
3) The rate of diffusion is directly proportional to average speed of the molecules
4) The average kinetic energy per molecule is inversely proportional to the absolute temperature
KEY:3
2. At a given temperature if Vrms is the root mean square velocity of the molecules of a gas and Vs
 Cp 
be the velocity of sound in it, then these are related as    
Cv 

3  3
1) vrms  vs 2) vrms   vs 3) vrms   vs 4) vrms     vs
 3  
KEY:2
3.The temperature of an ideal gas is increased from 120K to 480K. If at 120K, the root
mean square velocity of the gas molecules is v then at 480K, it becomes
1) 4v 2) 2v 3) v/2 4) v/4
KEY:2
4. At a given volume and temperature the pressure of a gas:
1) varies inversely as its mass 2) varies inversely as the square of its mass
3) varies linearly as its mass 4) is independent of its mass
KEY:3
5. The root mean square speed of the molecules of a gas at absolute temperature T is
proportional to
1) 1/ T (2) T (3) T (4) T 2
KEY:2
6. If gas molecules undergo inelastic collision with the wall of the container:
1) the temperature of the gas will decrease 2) the pressure of the gas will increase
3) neither the temperature nor the pressure will change 4) the temperature of the gas will increase
KEY:3
7. Which of the following methods will enable the volume of an ideal gas to be made four times
greater? Consider absolute temperature
1) Quarter the pressure at constant temperature
2) Quarter the temperature at constant pressure
3) Half the temperature, double the pressure
4) Double the temperature, double the pressure
KEY:1
8. If k is the Boltzmann constant, the average kinetic energy of a gas molecule at absolute
temperature T is
1) k T / 2 (2) 3 k T / 4 (3) k T 4) 3 k T / 2
KEY: 4
9. A gas has volume V and pressure P. The total translational kinetic energy of all the molecules
of the gas is
1) 3/2 PV only if the gas is monoatomic 2) 3/2 PV only if the gas is diatomic
3) > 3/2 PV if the gas is diatomic 4) 3/2 PV in all cases
KEY:4
10. The root mean square speed of the molecules of an enclosed gas is v. what will be the root mean
square speed if the pressure is doubled, the temperature remaining the same?
1) v/2 (2) v (3) 2v (4) 4v
KEY:2
11. A vessel contains a mixture of 1 mole of oxygen and two moles of nitrogen at 300K. The
r atio of the r otational k inetic ener gy per O 2 molecule to that of per N2 molecule is
1) 1:1 2) 1:2 3) 2:1
4) depends on the moment of inertia of the two molecules
KEY:1
12. On any planet, the presence of atmosphere implies (Crms = root mean square velocity of
molecules and ve = escape velocity)
1) Crms  ve 2) Crms  ve 3) Crms  ve 4) Crms  0
KEY:1
13. Choose the correct statement. When the temperature of a gas is increased
1) the kinetic energy of its molecules increases
2) the potential energy of its molecules increases
3) the potential energy decreases and the kinetic energy increases; the total energy remaining unchanged
4) the potential energy increases and the kinetic energy decreases; the total energy remaining unchanged
KEY:1
14. The number of molecules per unit volume (n) of a gas is given by
P kT P RT
1) 2) 3) 4)
kT P RT P
KEY:1
15. The number of molecules of N2 and O2 in a vessel are same. If a fine hole is made in the vessel then
which gas escapes out more rapidly?
1) N2 2) O2 3) both 4) sometimes N2 and sometimes O2
KEY:1
16. The following four gases are at the same temperature . In which gas do the molecules have the
maximum root mean square speed?
1) Hydrogen 2) Oxygen 3) Nitrogen 4) Carbon dioxide
KEY:1
17. Two vessels having equal volume contain molecular hydrogen at one atmosphere and helium at
two atmospheres respectively. If both samples are at the same temperature, the rms velocity of
hydrogen molecules is:
1)equal to that of helium 2) twice that of helium 3) half that of helium 4) 2 times that of helium
KEY:4
18. E0 and Eh respectively represent the average kinetic energy of a molecule of oxygen
and hydrogen. If the two gases are at the same temperature, which of the following statements
is true?
1) E0  Eh (2) E0  Eh (3) E0  Eh
4) Nothing can be said about the magnitude of E0 and Eh as the information given is insufficient.
KEY:2
19. If the Avogadro’s number was to tend to infinity; the phenomenon of Brownian motion
would:
1) remain completely unaffected
2) become more vigorous than that observed with present finite values of Avogadro’s number, for all sizes
of the Brownian particles
3) become more vigorous than that observed with the present finite value of Avogadro’s number, only for
relatively large Brownian particles
4) become practically unobservable as the molecular impact would tend to balance one another, for
practically all sizes of Brownian particles
KEY:4
20. The root mean square speed of a group of N gas molecules, having speeds v1 , v2 ,......., vN is:
1 2 1
1)  v1 v2 .......vN  b) v1
2
 v22 .......  vN 2 
N N
1 2 2
3)
N
 v1  v2 .......  vN2  4)  v1  v2  .......  v N 
2
KEY:3
21. The average kinetic energy of a molecule of a gas at absolute temperature T is
proportional to
1) 1/ T (2) T (3) T (4) T 2
KEY:3
22. The relation between rms velocity, vrms and the most probable velocity, vmp , of a gas is:
3 2 2
1) vrms  vmp 2) vrms  vmp 3) vrms  vmp 4) vrms  vmp
2 3 3
KEY:2
23. Some gas at 300K is enclosed in a container. Now the container is placed on a fast moving train.
While the train is in motion, the temperature of the gas:
1) rises above 300K 2) falls below 300K
3) remains unchanged 4) becomes unsteady
KEY:3
24. The average energy for molecules in one degree of freedom is :

3 kT 3
1) kT (2) (3) kT (4) kT
2 2 4
KEY:2
25. Two balloons are filled, one with pure He gas and other by air, respectively. If the pressure and
temperature of these balloons are same, then the number of molecules per unit volume is:
1) more in the He filled balloon 2) same in both balloons
3) more in air filled balloon (4) in the ratio of 1:4
KEY:2
26. Which of the following phenomena gives evidence of the molecular motion?
1) Brownian movement 2) Diffusion 3) Evaporation 4) All the above
KEY:4
27. On colliding in a closed container the gas molecules:
1) transfer momentum to the walls 2) momentum becomes zero
3) move in opposite directions 4) perform Brownian motion
KEY:4
28. The root mean square velocity, vrms , the average velocity vav and the most probable velocity,,
vmp of the molecules of the gas are in the order:
1) vmp  vavg  vrms 2) vrms  vavg  vmp
3) vavg  vmp  vrms 4) vmp  vrms  vavg

KEY:2
29. The temperature of a gas is raised while its volume remains constant, the pressure exerted by
the gas on the walls of the container increases because its molecules.
(1) Lose more kinetic energy to the wall
(2) Are in contact with the wall for a shorter time
(3)Strike the wall more often with higher velocities
(4)Collide with each other with less frequency.
KEY:3
30. At upper atmosphere, an astronaut feels:
1) extremely hot 2) slightly hotter
3) extremely cool 4) slightly cooler
KEY:4
31. The average distance travelled by a molecule of gas at temperature T between two successive
collisions is called its mean free path which can be expressed by (P is pressure of gas, K is
Boltzmann constant, d diameter of molecule)
1 P KT KP
(1) (2) 2 d 2T (3) (4) 2 d 2T
2 d 2 P 2 d 2 P
KEY:3
32. Which of the following statements about kinetic theory of gases is wrong
1)The molecules of a gas are in continuous random motion
2) The molecule continuously undergo inelastic collisions
3) The molecules do not interact with each other except during collisions
4) The collisions amongst the molecules are of
short duration
KEY:2
33. The root mean square speed of gas molecules
1) is same for all gases at the same temperature
2) depends on the mass of the gas molecule and its temperature
3) is independent of the density and pressure of the gas
4) depends only on the temperature and volume of the gas
KEY:2
34. Consider a gas with density  and c as the root mean square velocity of its molecules contained
in a volume. If the system moves as whole with velocity v, then the pressure exerted by the gas
is
1 2 1 2 1 2 1
1)  c
3
2)   c  v 
3
3)
3

 cv  4)
3
 2
 c v 
KEY:1
35. Choose the only correct statement from the
following:
1) The pressure of a gas is equal to the total kinetic energy of the molecules in a unit volume of the gas.
2) The product of pressure and volume of a gas is always constant.
3) The average kinetic energy of molecules of a gas is proportional to its absolute temperature.
4) The average kinetic energy of molecules of a gas is proportional to the square root of its absolute
temperature.
KEY:3
36. If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the
molecules
(1) Decreases (2) increases (3) Remains unchanged
(4) Increases or decreases according to the nature of the gas
KEY:2
37. If P is the pressure of the gas then the KE per unit volume of the gas is
P 3P
1) 2) P 3) 4) 2P
2 2
KEY:3
38. At absolute zero temperature, the kinetic energy of the molecules:
1) becomes zero (2) becomes maximum 3) becomes minimum (4) remains constant
KEY:1
39. Choose the correct statement from the following:
1) The average kinetic energy of a molecule of any gas is the same at the same temperature.
2) The average kinetic energy of a molecule of a gas is independent of its temperature.
3) The average kinetic energy of 1g of any gas is the same at the same temperature.
4) The average kinetic energy of 1g of a gas is independent of its temperature.
KEY:1
40. Two gases of equal mass are in thermal equilibrium. If Pa , Pb and Va and Vb are their respective
pressures and volumes, then which relation is true
(a) Pa  Pb ; Va  Vb (b) Pa  Pb ; Va  Vb
Pa P
(c)  b (d) Pa Va  Pb Vb KEY:d
Va Vb
PREVIOUS JEE MAINS QUESTIONS AND SOLUTIONS
1. Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and
temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in
number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a
volume 2V is given by P2 . The ratio P2 /P1 is —. [NA Sep. 06, 2020 (I)]
Sol : (5) Using ideal gas equation, =
⇒ = × 250 [.⋅ 1 = 250K] (i)
5
2 (2 1 = × 2000 [.⋅ 2 = 2000K] (ii) Dividing eq. (i) by(ii),
4
4 × 250 1
= ⇒ =
2 5 × 2000 5
= 5.
2. The change in the magnitude ofthe volume ofan ideal gas when a small additional pressure
is applied at a constant temperature, is the same as the change when the temperature is
reduced by a small quantity at constant pressure. The initial temperature and pressure ofthe
gas were 300 K and 2 atm. respectively. If | |= | |, then value of in (K/atm. ) is[NA Sep.
04, 2020(II)]
Sol : (150) In first case,
From ideal gas equation
=
+ = 0 (As temperature is constant) =− (i)
In second case, using ideal gas equation again
=−
=− (ii)
Equating (i) and (ii), we get
=− ⇒ =
Comparing the above equation with | |= | | , we have

300K
= = = = 150K/atm
2atm
3. The number density of molecules of a gas depends on their distance from the origin as ,
( )= . Then the total number of molecules is proportional to:[12 April 2019 II]
(a) 0
−3/4
(b) 0
1/2
(c) 0
1/4
(d) 0
−3
Sol :. (a) = ∫ p( )
= ×4
/
∝ a
4. A vertical closed cylinder is separated into two parts by a frictionless piston ofmass m and
ofnegligible thickness. The piston is free to move along the length ofthe cylinder. The length
ofthe cylinder above the piston is 1 , and that below the piston is 2, such that 1 > 2. Each part
of the cylinder contains n moles ofan ideal gas at equal temperature T. Ifthe piston is stationary,
its mass, m, will be given by: (R is universal gas constant and g is the acceleration due to
gravity) [12 Jan. 2019 II]
(a) (b) (c) + (d)
Sol : (d) Clearly from figure,

P2 A = P1 A + mg
⋅ ⋅
or, = + mg
ℓ ℓ
P2 A ↑ Img
⇒ nRT − = mg
ℓ
nRT −
m=
g ℓ ⋅ℓ
5. Thetemperamre ofan open room ofvolume 30m increases from 17∘ C to 27∘ C due to sunshine.
The ahnospheric pressure in the room remains 1 × 10 Pa. Ifn and nf are the number of
molecules in the room before and after heating, then nf − ni will be : [2017]
(a) 2.5 × 10 (b) −2.5 × 10 (c) −1.61 × 10 (d) 1.38 × 10
Sol : (b) Given: Temperature = 17 + 273 = 290
Temperature = 27 + 273 = 300
Atmospheric pressure, 0 = 1 × 105 Pa
Volume of room, 0 = 30 3
Difference in number of molecules, − =?

Using ideal gas equation, = ( ),
N0 = Avogadro′ s number
⇒ = (N )
− = 0 0
( )
( )N0
1 × 10 × 30 1 1
= × 6.023 × 10 −
8.314 300 290
= −2.5 × 10
6. For the P‐V diagram given for an ideal gas,
v
out of the following which one correctly represents the T‐P diagram? [Online April 9, 2017]
(b) (c)
Sol : (c) From P‐V graph,

P∝ , T = constant and Pressure is increasing from 2 to 1 so option (3) represents
correct T‐P graph.
7. There are two identical chambers, completely thermally insulated from surroundings. Both
chambers have a partition wall dividing the chambers in two compartments. Compartment 1 is
filled with an ideal gas and Compartment 3 is filled with a real gas. Compartments 2 and 4 are
vacuum. A small hole (orifice) is made in the partition walls and the gases are allowed to expand
in vacuum.
Statement‐l: No change in the temperature of the gas takes place when ideal gas expands in
vacuum. However, the temperature ofreal gas goes down (cooling) when it expands in vacuum.
Statement‐2: The internal energy of an ideal gas is only kinetic. The internal energy of a real
gas is kinetic as well as potential. [Online April 9, 2013]
(a) Statement‐l is false and Statement‐2 is true.
(b) Statement‐l and Statement‐2 both are true. Statement‐2 is the correct explanation of
Statement‐l.
(c) Statement‐l is true and Statement‐2 is false.
(d) Statement‐l and Statement‐2 both are true. Statement‐2 is not correct explanation of
Statement‐l.
Sol : a) In ideal gases the molecules are considered as point particles and for point particles,
there is no internal excitation, no vibration and no rotation. For an ideal gas the internal energy
can only be translational kinetic energy and for real gas both kinetic as well as potential energy
8. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the
gas molecules inside will [2002]
(a) increase (b) decrease (c) remain same
(d) decrease for some, while increase for others
Sol : :. 8. (c) The centre ofmass ofgas molecules also moves with lorry with uniform speed. As
there is no relative motion of gas molecule. So, kinetic energy and hence temperature remain
same
9. Number of molecules in a volume of 4 cm of a perfect monoatomic gas at some temperature
and at a pressure of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at )
is 4 × 10 erg, = 980cm/s , density of mercury = 13.6g/cm ) [Sep. 05, 2020 (I)]
(a) 4.0 × 10 (b) 4.0 × 10 (c) 5.8 × 10 (d) 5.8 × 10
Sol : (c) Given: K.E. mea = = 4 × 10
= 2 cm ofHg, = 4cm
p 2 × 13.6 × 980 × 4
= = = 4 × 10
× 10
10. Nitrogen gas is at 300∘ C temperature. The temperature (in K) at which the rms speed ofa H2
molecule would be equal to the rms speed of a nitrogen molecule, is . (Molar mass ofN gas 28 g);
[NA Sep. 05, 2020 (II)]
Sol : Room mean square speed is given by
3
=
Here, = Molar mass ofgas molecule = temperature ofthe gas molecule We have given
N2 = H2
3 N2 3 H2
=
MN2 H2
573
⇒ = ⇒ = 41K
2 28
11. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127∘ C. At 2 atm
pressure and at 227∘ C, the rms speed ofthe molecules will be: [9 April 2019 I]
(a) 100m/s (b) 80 √5m/s (c) 100 √5m/s (d) 80 m/s
3
Sol : (c) =
(273 + 127) 400 4 2
= = = = =
(273 + 237) 500 5 √5
√5 √5
2 = 1 = × 200 = 100√5m/s.
2 2
12. If10 gas molecules each ofmass 10 kg collide with a surface (perpendicular to it) elastically
per second over an area 1 m with a speed 10 m/s, the pressure exerted by the gas molecules
2
will be ofthe order of: [8 April 2019 I]

(a) 10 N/m (b) 10 N/m (c) 10 N/m (d) 10 N/m
Sol : (Bouns) Rate ofchange ofmomentum during collision
− (− ) 2
= =
×( )
so pressure =
×
10 × 2 × 10 × 10
= =2
1×1
13. The temperature, at which the root mean square velocity of hydrogen molecules equals their
escape velocity from the earth, is closest to: [8April 2019II] [Boltzmann Constant B = 1.38 ×
10 −23
J/K Avogadro Number NA = 6.02 × 10 /kg Radius ofEarth: 6.4 × 10 m Gravitational
26
acceleration on Earth = 10ms ]

(a) 800K (b) 3 × 10 K (c) 10 K (d) 650K
Sol : (c) rs =
3
= 11.2 × 103
3
or = 11.2 × 103
3×138×10−23
or = 11.2 × 103 = 10 K
2×10−3
14. Amixmre of2 moles ofhelium gas (atomic mass = 4u), and 1 mole ofargon gas (atomic mass
= 40u) is kept at 300 K in a container. The ratio oftheir rms speeds
Vrms (he1ium)
is close to: [9 Jan. 2019 I]
Vrms (argon)
(a) 3.16 (b) 0.32 (c) 0.45 (d) 2.24
Sol : (a) Using =

V (He) M 40
= = = 3.16
V (Ar) M 4
15. Nmoles ofa diatomic gas in a cylinder are at atemperature T. Heat is supplied to the cylinder
such that the temperature remains constant but n moles ofthe diatomic gas get converted into
monoatomic gas. What is the change in the total kinetic energy ofthe gas? [Online April 9, 2017]
(a) nRT (b) 0 (c) nRT (d) nRT
Sol : (a) Energy associated with N moles ofdiatomic gas,
5
Ui = N RT
2
Energy associated with n moles ofmonoatomic gas
= n RT
3
Total energywhen n moles ofdiatomic gas converted into monoatomic (Uf ) = 2n RT+(N‐n)
2
RT
= nRT + NRT
Now, change in total kinetic energy ofthe gas
1
U=Q= nRT
2
16. In an ideal gas at temperature T, the average force that a molecule applies on the walls of a
closed container depends on T as Tq . A good estimate for q is: [Online April 10, 2015]
(a) (b) 2 (c) 1 (d)
Sol : (c) Pressure, =
( )
or, =
If the gas mass and temperature are constant then
∝( ) ∝
So, force ∝ ( ) ∝
i. e., Value of =1
17. A gas molecule of mass M at the surface ofthe Earth has kinetic energy equivalent to 0∘ C. If it
were to go up straight without colliding with any other molecules, how high it would rise?
Assume that the height attained is much less than radius ofthe earth. (kB is Boltzmann
constant). [Online Apri119, 2014]
(a) 0 (b) (c) (d)
Sol : d) Kinetic energy ofeach molecule,

K.E. = K T
In the given problem,
Temperature, T = 0∘ C = 273K
Height attained by the gas molecule, h =?
K.E. = K (273) =
K.E. = P. E. ⇒ = Mgh
or h =
18. At room temperature a diatomic gas is found to have an r.m.s. speed of1930 ms . The gas is:
(a) H2 (b) C1 (c) O2 (d) 2
Sol : (a) C =
× . ×
(1930)2 =
3 × 8.314 × 300
M= ≈ 2 × 10 kg
1930 × 1930
The gas is H2 .
19. In the isothermal expansion of 10g of gas fi om volume V to 2V the work done by the gas is 575J.
Mat is the root mean square speed of the molecules of the gas at that temperature? [Online April
25, 2013]
(a) 398m/s (b) 520m/s (c) 499m/s (d) 532m/s
3pv
Sol : (c) vrms = massofthegas
20. A perfect gas at 27∘ C is heated at constant pressure so as to double its volume. The final
temperature of the gas will be, close to [Online May 7, 2012]
(a) 327∘ C (b) 2 )∘ C (c) 54∘ C (d) 3 )∘ C
Sol : (a) Given, 1 =
2 =2
∘
1 = 27 + 273 = 300K
2 =?
From charle’s law
= (Pressure is constant) or, =
2 = 600K = 600 − 273 = 327∘ C
21. A thermally insulated vessel contains an ideal gas of molecular mass and ratio of specific
heats . It is moving with speed and it’s suddenly brought to rest. Assuming no heat is lost
to the surroundings, its temperature increases by: [2011]
( ) ( )
(a) (b) (c) (d) ( )
Sol : (c) As, work done is zero.

So, loss in kinetic energy = heat gain by the gas
1
= =
2 1 −1
1
=
2 Y−1
22. Three perfect gases at absolute temperatures 1 , 2 and 3 are mixed. The masses ofmolecules
are 1 , 2 and 3 and the number of molecules are 1, 2 and 3 respectively. Assuming no loss
of energy, the final temperature of the mixture is : [2011]
( )
(a) (b) (c) (d)
Sol :P1 V1 + P2 V2 + P3 V3 = PV
+ +
+ +
=
1 1 + 2 2 + 3 3
mlx =
1 + 2 + 3
23. One kg of a diatomic gas is at a pressure of 8 × 10 N/m . The density of the gas is 4kg/m . What
is the energy of the gas due to its thermal motion?[2009]
(a) 5 × 10 J (b) 6 × 10 J (c) 7 × 10 J (d) 3 × 10 J
Sol :. (a) Given, mass = 1 kg
Density = 4 kg m−3
Volume = = m
Internal energy ofthe diatomic gas
5 5 1
= = × 8 × 10 × = 5 × 10
2 2 4
Alternatively:
K.E = = = × [⋅.⋅ = ]
5 5 1 × 8 × 10
= = × = 5 × 10 J
2 2 4
24. The speed ofsound in oxygen ( 2) at a certain temperature is 460 ms . The speed ofsound in
helium (He) at the same temperature will be (assume both gases to be ideal) [2008]
(a) 1421ms (b) 500 ms (c) 650 ms (d) 330 ms
Sol : (a) The speed of sound in a gas is given by = ∝ [As and is constant]
= ×
14 4
= × = 0.3237
32 167
O2 460
He = = = 1421m/s
0.3237 0.3237
25. At what temperature is the r. m. s velocity of a hydrogen molecule equal to that of an oxygen
molecule at 47∘ C? [2002]
(a) 80 K (b) −73K (c) 3 K (d) 20 K
Sol : (d) RMS velocity ofa gas molecule is given by
3
rms =
Let be the temperature at which the velocity ofhydrogen molecule is equal to the velocity
ofoxygen molecule.
3 3 × (273 + 47)
=
2 32
⇒ = 20K
26. Molecules of an ideal gas are known to have three translational degrees of freedom and two
rotational degrees of freedom. The gas is maintained at a temperature of T.
The total internal energy, U ofa mole ofthis gas, and the value of ( )( ) are given,
respectively, by: [Sep. 06, 2020 (D]

(a) U = RT and = (b) U = 5RT and =
(c) U = RT and = (d) U = 5RT and =
Sol :. (c) Total degree offreedom = 3+2= 5
Total energy, = =
And = =1+ =1+ =
27. In a dilute gas at pressure and temperature , the mean time between successive collisions of
a molecule varies with is: [Sep. 06, 2020 (II)]
(a) (b) (c) (d) √
√
Sol :. (b) Mean free path, =
√
where, = diameter ofthe molecule
= number ofmolecules per unit volume But, mean time ofcollision, =
rms
3
But rms =
1
= ⇒ ∝
√
28. Match the Cp /Cv ratio for ideal gases with different type of molecules : [Sep. 04, 2020 (I)]
Column‐I C olumn‐II
Molecule Type Cp /Cv
(A) Monatomic (I) 7/5
(B) Diatomic rigid molecules (II) 9/7
(C) Diatomic non‐rigid molecules (III) 4/3
(D) Triatomic rigid molecules (IV) 5/3
(a) (A) − (IV), (B) − (II), (C) − (I), (D) − (III)
(b) (A) − (III), (B) − (IV), (C) − (II), (D) − (I)
(c) (A) − (IV), (B) − (I), (C) − (II), (D) − (III)
Sol :. (c) As we know,
= = 1 + , where = degree offreedom
(A) Monatomic, =3
2 5
=1+ =
3 3
(B) Diatomic rigid molecules, =5
2 7
=1+ =
5 5
(C) Diatomic non‐rigid molecules, =7
2 9
=1+ =
7 7
(D) Triatomic rigid molecules, =6
2 4
=1+ =
6 3
29. A closed vessel contains 0.1 mole of a monatomic ideal gas at 200 K. If 0.05 mole of the same gas
at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be
close to [NA Sep. 04, 2020 (I)]
Sol : (266.67) Here work done on gas and heat supplied to the gas are zero.
Let Tbe the fmal equilibrium temperature ofthe gas in the vessel.
Total internal energyofgases remain same.
i.e., 1 + 2 = 1 + 2
or, 1 1 + 2 2 =( 1 + 2)
⇒ (0.1) (200) + (0.05) (400) = (0.15)
800
= = 266.67K
3
30.
Consider a gas oftriatomic molecules. The molecules are assumed to be triangular and made
Of mass less rigid rods whose vertices are occupied by atoms. The internal energy ofa mole of
the gas at temperature is: [Sep. 03, 2020 (I)]
(a) (b) (c) (d) 3
Sol : (d) Here degree of fiieedom, = 3 + 3 = 6 for triatomic nonlinear molecule.
Internal energy ofa mole ofthe gas at temperature ,
6
= = =3
2 2
31. To raise the temperature ofa certain mass ofgas by 50∘ C at a constant pressure, 160 calories of
heat is required. When the same mass ofgas is cooled by 100∘ C at constant volume, 240 calories
of heat is released. How many degrees of freedom does each molecule of this gas have
(assume gas to be ideal)? [Sep. 03, 2020 (II)]
(a) 5 (b) 6 (c) 3 (d) 7
Sol :b) Let and be the specific heat capacity ofthe gas at constant pressure and volume.
At constant pressure, heat required
=
⇒ 160 = ⋅ 50 (i)
At constant volume, heat required
=
⇒ 240 = ⋅ 100 (ii)
Dividing (i) by(ii), we get
160 50 4
= ⋅ ⇒ =
240 100 3
= = = 1 + (Here, = degree of fiieedom)
⇒ = 6.
32. A gas mixture consists of3 moles ofoxygen and 5 moles of argon at temperature . Assuming the
gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) ofthe
mixture is: [Sep. 02, 2020]
(a) 15 (b) 13 (c) 20 (d) 11
Sol : (a) Total energy ofthe gas mixture,
1 1 1 2 2 2
mlx = +
2 2
5 5
=3× + ×3 = 15
2 2
33. An ideal gas in a closed container is slowly heated. As its temperature increases, which ofthe
following statements are true? [Sep. 02, 2020 (II)]
(1) The mean free path of the molecules decreases
(2) The mean collision time between the molecules decreases
(3) The mean free path remains unchanged
(4) The mean collision time remains unchanged
(a) (2) and (3) (b) (1) and (2) (c) (3) and (4) (d) (1) and(4)
Sol : (a) As we know mean : ee path
1
=
√2
Here, = no. ofmolecule

= volume ofcontainer
= diameter ofmolecule
But = =
⇒ = =
1
=
√2
For constant volume and hence constant number density of gas molecules is constant.
So mean free path remains same.
As temperature increases no. of collision increases so relaxation time decreases
34. Consider two ideal diatomic gases A and B at some temperature T. Molecules ofthe gas A are
rigid, and have a mass m. Molecules of the gas B have an additional vibrational mode, and have
a mass . The ratio of the specific heats (CAv and CBv ) ofgas A and B, respectively is: [9 Jan 2020 I]
(a) 7: 9 (b) 5: 9 (c) 3: 5 (d) 5: 7
Sol : (d) Specific heat of gas at constant volume
1
= ; = degree of freedom
2
For gas A (diatomic)
f = 5(3translational +2 rotational)
5
=
2
For gas B (diatomic) in addition to (3translational +2 rotational) 2 vibrational degree of
freedom.
7
= Hence = =
2
35. Two gases‐argon (atomic radius 0.07nm, atomic weight 40) and xenon (atomic radius 0.1nm,
atomic weight 140) have the same number density and are at the same temperature. The ratio
of their respective mean free times is closest to: [9 Jan 2020 II]
(a) 3.67 (b) 1.83 (c) 2.3 (d) 4.67
Sol : (Bonus) Mean free path of a gas molecule is given by
1
=
√2
Here, = number of collisions per unit volume
= diameter of the molecule
If average speed of molecule is v then
Mean free time, =
1 1
⇒ = =
√2 √2 3
3
=
√
∝ = ×
40 0.1
= × = 1.09
140 0.07
36. The plot that depicts the behavior of the mean free time (time between two successive
collisions) for the molecules of an ideal gas, as a function of temperature ( ) , qualitatively, is:
(Graphs are schematic and not drawn to scale) [8 Jan. 2020 I]
meanfreepath 1
Sol :. (c) Relaxation time ( ) ∝ ⇒ ∝
speed v
and, ∝√
1
∝
√
Hence graph between v/s is a straight line which is correctly depicted by graph shown in
√
option (c).
37. Consider a mixture of moles of helium gas and 2 moles of oxygen gas (molecules taken to be
rigid) as an ideal gas. Its value will be: [8 Jan. 2020 II]
(a) 19/13 (b) 67/45 (c) 40/27 (d) 23/15
Sol : (a) Helium is a monoatomic gas and Oxygen is a diatomic gas.
3 5
For helium, 1
= and 1
=
2 2
5 7
For oxygen, 2
= and 2
=
2 2
+
=
+
. +2 . 19 × 2
⇒ = =
. +2 . 2(13 )
⋅. ( )( ) mixture =
38. Two moles ofan ideal gas with = are mixed with 3
moles of another ideal gas with = . The value of for the mixture is: [7 Jan. 2020 I]
(a) 1.45 (b) 1.50 (c) 1.47 (d) 1.42
Sol :d) Using, mixture

=
+
⇒ + =
−1 −1 −1
3 2 5
⇒ + =
−1 −1 −1
9 2×3 5 5
⇒ + = ⇒ −1=
1 2 −1 12
17
⇒ = = 1.42
12
39. Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid).
What is the molar specific heat ofmixture at constant volume? (R = 8.3J/mo1K) [12 April 2019 I]
(a) 19.7 J/mo1L (b) 15.7 J/mo1K (c) 17.4J/mo1K (d) 21.6J/mo1K
+ 2
Sol : (c) [ ] min =
1 1 2
1+ 2
2× +3×
=
2+3
= 2.1R = 2.1 × 8.3 = 17.4J/mol − k
40. A diatomic gas with rigid molecules does 10 J ofwork when expanded at constant pressure. What
would be the heat energy absorbed by the gas, in this process? [12 April 2019 II]
(a) 25 J (b) 35 J (c) 30 J (d) 40 J
Sol : (b) = = =(
/ )
=
or = 1− =
×
or = = = 35J
41. A25 × 10 m volume cylinder is filled with 1 molofO gas at room temperature (300K) . The
molecular diameter of O2 , and its root mean square speed, are found to be 0.3 nm and 200 m/s,
respectively. What is the average collision rate (per second) for an O2 molecule? [10 April 2019 I]
(a) −10 (b) ∼ 10 (c) ∼ 10 (d) ∼ 10
Sol : (c) V = 25 × 10 m , N = 1mole ofO
T = 300K
Vr = 200m/s
1
=
√2N r
〈 〉
Average time 1 = = 200. N r . √2
√2 × 200 × 6.023 × 10
= . × 10 × 0.09
25 × 10
The closest value in the given option is = 10
42. When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its
temperature increases by T. The heat required to produce the same change in temperature, at
a constant pressure is: [10 April 2019 II]
(a) (b) (c) (d)
Sol :. (c) Amount ofheat required (Q) to raise the temperature at constant volume
Q = nC T (i)
Amount ofheat required Q1 at constant pressure
Q1 = nCP T (ii)
Dividing equation (ii) by(i), we get
=
7 7
⇒ =( ) . .⋅ = =
5 5
43. An HC1 molecule has rotational, translational and vibrational motions. Ifthe rms velocity ofHCl
molecules in its gaseous phase is , m is its mass and kB is Boltzmann constant, then its
temperature will be:[9 April 2019 I]
(a) 6 (b) 3 (c) (d)
Sol : (a) =3
or =6
44. The specific heats, C and Cv of a gas of diatomic molecules, A, are given (in units ofJmo1 k )
by29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30
and 21. Ifthey are treated as ideal gases, then: [9April 2019II]
(a) A is rigid but B has a vibrational mode.
(b) A has a vibrational mode but B has none.
(c) A has one vibrational mode and B has two.
(d) Both A and B have a vibrational mode each.
29 30 10
Sol : (b) = =
22
= 1.32 < 1.4 (diatomic) and =
21
=
7
= 1.43 > 1.4
Gas A has more than 5‐degrees of fiieedom.
45. An ideal gas occupies a volume of 2 m3 at a pressure of3 × 10 Pa. The energy ofthe gas:
[12 Jan. 2019 I]
(a) 9 × 10 J (b) 6 × 10 J (c) 10 J (d) 3 × 10 J
Sol : (a) Energy ofthe gas, E
= nRT = PV
f
= (3 × 10 )(2) = f × 3 × 10
2
Considering gas is monoatomic i. e., f = 3
Energy, E = 9 × 10 J
46. An ideal gas is enclosed in a cylinder at pressure of2 atm and temperature, 300 K. The mean
time between two successive collisions is 6 × 10 s. Ifthe pressure is doubled and temperature is
increased to 500 K, the mean time between two successive collisions wiil be close to: [12 Jan.
2019 II]
(a) 2 × 10 s (b) 4 × 10 s (c) 0.5 × 10 s (d) 3 × 10 s
Sol : (b) Using, =
√T no. ofmo1ecules
t∝ .⋅. n =
P Vo1ume
√
or, = × ≈ 4 × 10
× √
47. A gas mixture consists of3 moles ofoxygen and 5 moles ofargon at temperature T. Considering
only translational and rotational modes, the total internal energy of the system is : [11 Jan.
2019 I]
(a) 15 RT (b) 12 RT (c) 4 RT (d) 20 RT
Sol : (a) U = n RT + n RT
Considering translational and rotational modes, degrees offreedom f1 = 5 and f2 = 3
5 3
u = (3RT) + × 5RT
2 2
U = 15RT
48. In a process, temperature and volume of one mole of an ideal monoatomic gas are varied
according to the relation VT = K, where K is a constant. In this process the temperature ofthe
gas is increased by T. The amount of heat absorbed by gas is (R is gas constant): [11 Jan. 2019
II]
(a) R T (b) KR T (c) R T (d) T
Sol : (a) According to question VT = K
we also know that PV = nRT
PV
⇒T=
nR
PV
⇒V = k ⇒ PV = K
nR
C= + C (For polytropic process)
R 3R R
C= + =
1−2 2 2
Q = nC T
= × T [here, n = 1 mole]
49. Two kg ofa monoatomic gas is at a pressure of 4× 10 N/m The density of the gas is 8 kg/m
What is the order of energy of the gas due to its thermal motion? [10 Jan 2019 II]
(a) 10 J (b) 10 J (c) 10 J (d) 10 J
Sol : (c) Thermal energy ofN molecule
3
=N kT
2
= RT = (nRT) = PV
3 m 3 2
= P = P
2 p 2 8
3 2
= × 4 × 10 × = 1.5 × 10 J
2 8
therefore, order = 10 J
50. A 15g mass of nitrogen gas is enclosed in a vessel at a temperature 27∘ C. Amount of heat
transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3J/K
mole] [9 Jan. 2019 II]
(a) 0.9kJ (b) 6 kJ\ (c) 10 kJ (d) 14 kJ
Sol :c) Heat transferred,
Q = nC T as gas in closed vessel
To double the rms speed, temperature should be 4 times i. e., T′ = 4T as vrms = 3 /
15 5 ×
Q= × × (4 − )
28 2
7
.⋅. = = & − =
5
or, Q = 10000J = 10kJ
51. Two moles ofan ideal monoatomic gas occupies a volume V at 27∘ C. The gas expands
adiabaticallyto a volume 2 V. Calculate(1) the final temperature ofthe gas and(2) change in its
internal energy. [2018]
(a) (1) 189 K (2) 2.7kJ
(b) (1) 195 K (2) −2.7kJ
(c) (1) 189 K (2) −2.7kJ
(d) (1) 195 K (2) 2.7kJ
Sol : (c) In an adiabatic process
T = Constant or, T1 V1 −1 = T2 V2 −1
For monoatomic gas =
300
(300) V2/3 = T2 (2V)2/3 ⇒ T2 = (2)2/3
T2 = 189K(final temperature)
Change in internal energy U = n R T
3 25
=2 (−111) = −2.7kJ
2 3
52. Two moles of helium are mixed with n with moles of hydrogen. If = for the mixture, then the
value ofn is [Online Apri116, 2018]

(a) 3/2 (b) 2 (c) 1 (d) 3
Sol : (b) Using formula,
n1 1 n2 2
+
1 −1 2 −1
mixture
= ( )mix ( ) = n1 n2
+
l −1 2 −1
3
Putting the value ofn = 2, n2 = n, ( )mix ( ) =
2
5 7
1
= , 2
= and solving we get, n = 2
3 5
53. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is
observed that Cp − Cv = a for hydrogen gas C − C = b for nitrogen gas T1i correct relation
between a and b is: [2017]
(a) a = 14b (b) a = 28b (c) a = (d) a = b
Sol : (a) As we know, − = where and are molar specific heat capacities
or, − =
For hydrogen ( = 2) − = =
2
For nitrogen ( = 28) − = =
28
= 14 or, = 14
54. An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant
pressure Cp and at constant volume (Cv ) is: [Online April 8, 2017]
(a) 6 (b) (c) (d)
Sol : (d) The ratio of specific heats at constant pressure Cp and constant volume (Cv )
C 2
= = 1+
C f
where fis degree offreedom

C 2 7
= 1+ =
C 5 5
55. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C
remains constant. If during this process the relation ofpressure P and volume V is given by
PV =constant, then n is given by(Here CP and CV are molar specific heat at constant pressure
and constant volume, respectively) : [2016]
(a) n = (b) n = (c) n = (d) n =
Sol : . (d) For a polytropic process

C=C + C−C =
R R
1−n= 1− =n
C−C C−C
C−C −R C−C −C +C
n= =
C−C C−C
C−C
= C −C
C−C
56. Using equipartition of energy, the specific heat (in Jkg K ) of aluminium at room temperature
can be estimated to be (atomic weight of aluminium = 27) [Online April 11, 2015]
(a) 410 (b) 25 (c) 1850 (d) 925
Sol : (d) Using equipartition of energy, we have
6
=
2
3 × 1.38 × 10 × 6.02 × 10
=
27 × 10
C = 925J/kgK
57. Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 10 atm. at room
temperature (300 K). Taking R = 8.0JK mole , 1 atm = 10 Pa and NAvogadro = 6 × 1023 mole−1 ,
the mean distance between molecules of gas in an evacuated vessel will be of the order of.
(a) 0.2 n (b) 02mm (c) 0.2 cm (d) 0.2nm
Sol :. (b)
58. Figure shows the variation in temperature ( T) with the amount of heat supplied (Q) in an
isobaric process corresponding to a monoatomic (M), diatomic (D) and a polyatomic(P) gas. The
initial state ofall the gases are the same and the scales for the two axes coincide. Ignoring
vibrational degrees of freedom, the lines , and respectively correspond to : [Online April 9,
2013]
(a) P, M and D (b) M, D and P (c) P, D and M (d) D, M and P

Sol : (b) On giving same amount of heat at constant pressure, there is no change in temperature
for mono, dia and polyatomic.
No. ofmo1ecu1e. s
( Q)P = Cp T =
Avogedro’sno
or T ∝
.
59. A given ideal gas with y = = 1.5 at a temperature . If the gas is compressed adiabatically to
one‐fourth of its initial volume, the final temperature will be [Online May 12, 2012]
(a) 2√2 (b) 4 (c) 2 (d) 8
Sol : (c) = constant
1′ −1 Y−1
1 1 = T2 2
/
⇒ ( ) /= T 1
4
[⋅.⋅ = 1.5, = , = and 2 = ]
4
12
4
2 = / =2
60. If and denote the specific heats ofnitrogen per unit mass at constant pressure and
constant volume respectively, then [2007]
(a) − = 28 (b) − = 28 (c) − = 14 (d) − =
Sol : (b) According to Mayer’s relationship
− = , as per the question ( − ) =
⇒ − = 28
Here M = 28 = mass ofl unit ofN
61. A gaseous mixmre consists of16 g ofhelium and 16 g of oxygen. The ratio ofthe mixture is
[2005]
(a) 1.62 (b) 1.59 (c) 1.54 (d) 1.4
Sol : (a) For mixture ofgas specific heat at constant volume
1 1
+ 2 2
=
1 + 2
62. One mole ofideal monatomic gas ( = 5/3) is mixed with one mole of diatomic gas ( = 7/5) .
What is for the mixture? Y Denotes the ratio of specific heat at constant pressure, to that at
constant volume [2004]
(a) 35/23 (b) 23/15 (c) 3/2 (d) 4/3
Sol :No. ofmoles ofhelium,
16
1 = = =4
4
Number ofmoles ofoxygen,

16 1
2 = =
32 2
5
3 1 1 5 6 + 9
=
4 × 4+ + × = 4
2 2 2 2 2
×
= = and
×
Specific heat at constant pressure
+ 1
= −4 × 5 4+ 2
1 1 2 2
= 1 7
( 1 + 2) 2 + ×
2 2
10 + 9 47
= =
2 18
47 18
= × = 1.62
18 29
63. During an adiabatic process, the pressure ofa gas is found to be proportional to the cube ofits
absolute temperature. The ratio for the gas is [2003]
(a) (b) 2 (c) (d)
Sol : (d) ∝ ⇒ = constant ....(i)
But for an adiabatic process, the pressure temperature relationship is given by
1−Y Y
= constant
⇒ =constt. (ii)
From (i) and(ii) = −3 ⇒ = −3 + 3 ⇒ =

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PHYSICS ( 1 ST YEAR ) IIT MATERIAL

2. UNITS AND MEASUREMENTS

3. MOTION IN A STRAIGHT LINE

6. WORK , ENERGY AND POWER

7. SYSTEM OF PARTICLES AND ROTATIONAL MOTION

10. MECHANICAL PROPERTIES OF SOLIDS

11. MECHANICAL PROPERTIES OF FLUIDS

12. THERMAL PROPERTIES OF MATTER

14. KINETIC THEORY

Some important conversions:

Accuracy and precision of instruments :

Measurement- Measurement- Measurement- Average

A 2.25cm 2.27cm 2.26cm 2.26cm

B 2.252cm 2.250cm 2.251cm 2.251cm

C 2.250cm 2.250cm 2.251cm 2.250cm

Imperfection in Experimental technique or Procedure:

It is a pure number having no units.

x3  1.33  1.34  0.01; x4  1.33  1.35  0.02;

x5  1.33  1.32  0.01; x6  1.33  1.36  0.03;

x7  1.33  1.30  0.03; x8  1.33  1.33  0.00;

Diameter of wire = M.S.R +( C.S.R x L.C)

1 1 1 R ' R1 R2

R ' R1 R2

'2 R1 '2 R2

12. Force F = Mass  acceleration  M 1 L1T 2  N

28. Moment of inertia I=Mass  (radius of gyration)2  M 1 L2T 0  kgm2

68. Specific resistance (or

70. Current density

 Specific heat, Specific gas constant L2 T 2  1 

 Thermal capacity, Entropy, Boltzmann constant, Molar thermal capacity, M L2 T 2  1 

 Wave number, Power of a lens, Rydberg’s constant  L1 

 Electric field and potential gradient  MLT 3 A1 

 Rydberg’s constant and propagation constant  M 0 L1T 0 

 Solar constant and poynting vector  ML0T 3 

 Dimensions of M,L,T in  are 0,2,0

  ab    M 1 L5T 2   M 0 L3T 0    M 1 L8T 2 

  ab    M 1 L5T 2   M 0 L3T 0    M 1 L8T 2 

The side of cube = 10 mm + 1 0.1 mm  1.01 cm

Accuracy, precision, types of errors and combination of errors

A  bl  l b  10cm 2 , key-1

Significant figures & Rounding off

Sol : T  F xW yV z ; M 0 L0T 1 [ MLT 2 ]x [ ML2T 2 ] y [ LT 1 ]z

TOPIC-1…..Unit of Physical Quantities

Density of material in new system

2. A metal sample carrying a currentalong X-axis with density Jx is subjected to a

sol. (b) According to question Ey ∝ Jx BZ

TOPIC-2…..Dimensions of Physical Quantities

Resistance, 𝑙‐length, 𝐸‐Electric field, 𝐵‐magnetic field and 𝜀0 , 𝜇0 , ‐ free space

Thus, 𝑥, 𝑦, 𝑧 will have the same dimension of speed.

4. Dimensional formula for thermal conductivity is (here 𝐾 denotes the temperature):

5. A quantity x is given by (𝐼𝐹𝑣 2 /𝑊𝐿4 ) in terms of moment of inertia 𝐼, force 𝐹, velocity

Dimension of Energy, 𝐸 = ML2 T −2

𝜇0 𝑐 → M𝐿T −2 A−2 × LT1 =ML2 T 3 A2

(a) A−1 T1 M1 L3 (b) AT 2 M −1 L−1 (c) AT −3 ML3/2 (d) A2 T 3 M −1 L−2

(a) [𝐿A−2 ] (b) [A−1 ] (c) [LTA] (d) [LT 2 ]

sol. (c) Let t ∝ Gx hy Cz

17. The dimensions of stopping potential 𝑉0 in photoelectric effect in units of Planck’s

vacuum, is: [8 Jan. 2019 II]

sol. (d) Dimension ofA ≠ dimension of(C)

= [M −1 L−3 T 3 ][I2 ] = [M −1 L−3 T 3 I2 ]