Ct22esphysolv (1&2)

Units & Dimensions 11
1
11. (3) Speed of light c 
0  0
Dimensions  L T 1 
1. (3)
12. (1) Number of significant figures 4 are 3,
2. (4)
because 103 is decimal multiplier..
3. (2)
13. (4)  3.20  4.80 105  8.00 105
LdI
4. (3) From the formula,   Number of significant is 3.
dt
14. (4) 3.8 106  4.2 105
henry- ampere
 
second   3.8 101  4.2  105
5. (3)
  0.38  4.2 105
6. (4)
v v.e v.e (ve)t   4.58105

7. (3) R    2
I I .e e .e e Rounding off to one place of decimal.
t
The sum  4.6 105
[ve][t ] Joule-sec  h 
[ R ]    .
[e 2 ] (charge) 2  e 2  15. (2) Subtraction is correct upto one place of
decimal, corresponding to the least number of
8. (1)
decimal places.7.26 - 0.2 = 7.06 = 7.1 J.
9. (2) E  hv   ML2T 2    h  T 1  16. (4)
  h    ML2T 1  . Total mass = (2.3 + 0.02015 + 0.02017) kg

= 2.34032, upto one decimal place
10. (4) Magnetic moment of a current carrying
coil is defined as the product of current in the = 2.3 kg.
coil with the area of coil
17. (1)
 
M  IA 18. (4)
2
Thus, dimensions of [ M ]  [ A][ L ] 19. (1) Mean time period T = 2.00 sec
 [ L2 A] Mean absolute error T  0.05sec.
2 Units & Dimensions
To express maximum estimate of error, the time constant G is 6.6  1011 Nm2 kg 2 . Dimensional
period should be written as  2.00  0.05 sec.
formula of G is  M 1 L3T 2  .
20. (2) When two quantities are divided, their
maximum fractional or relative errors are added
[ E]  ML2T 2 
up 10. (2) [ x]    LT 1
[ P]  MLT 1 
X A B
hence:  
X A B 11. (4)
4 12. (4) Angular momentum = mvr

21. (1) V   r 3
3  {MLT 1 ][ L]  [ ML2T 1 ].
 % error in volume
L L 1 T
 3  % error in radius. 13. (3)     A1 
RCV  R  CV Q  
3  0.1
 100 2 2
 dp  [ MLT .L ]
5.3 14. (2)  dx    [ML2T 2 ].
  [ L]

22. (3) g  4 2 15. (4) Electromotive force = Potential difference
T2
2 2
 W  ML T 
 2%  2  102 V   [ ML2T 3 A1 ]
 q  AT 
T
 2%  3  102  M   ML1T 2 
T 16. (3) Given, mass × pressure =
density  ML3 
g  2T
    2  102  2  3  102
g  T
  ML2T 2 
 8  102  8% Which represent work.

17. (4) If the number is less than 1, the zero (s)
on the right of decimal point and before the first
1. (4) non - zero digits are not significant.
2. (2) In 0.06900, the underlined zeros are not
significant. Hence, number of significant figures
3. (1)
is four.
4. (3)
18. (4) Options (2) and (3) have four significant
dv figures; (1) has two and (4) has three significant
5. (2)   /
dx figures.
6. (3) 19. (1)
7. (2) 20. (2)
8. (4) 21. (3) a  b  1.00, a  b  0.84. In case of
9. (4) Gravitational constant G has a constant multiplication and division result will contain least
value and dimensions. In SI system value of number of significant figure. In 0.08 there is only
gravitational one significant figure.
22. (1) Given, L= 2.331 cm  m 

 3
= 2.33 (correct up to two decimal places)and  m 
B=2.1 cm =2.10 cm
 2 102  3 102
 L+B=2.33 + 2.10 = 4.43 cm = 4.4 cm
Since minimum significant figure is 2.  5 102  5%
23. (4) Radius of the sphere, r = 1.41 cm (3 4 2 l
significant figures) Volume of the sphere, 33. (2) T  2 l / g  T 2  4 2l / g  g 
T2
4 4 3
Here % error in
V   r 3   3.14  1.41
3 3 1mm 0.1
l  100   100  0.1%
100 cm 100
 11.736cm3
Round off up to 3 significant figure = 11.7cm3 0.1

and % error in T  100  0.05%
2 100
24. (2)
25. (2)  % error in g = % error in l + 2 (% error in T)
26. (1)  0.2 %

27. (2)
28. (4) All given measurements are correct upto
two decimal places. As here 5.00 mm has the 1. (4)
smallest unit and the error in 5.00 mm is least 2. (3) The units for these quantities are given by
(commonly taken as 0.01 mm if not specified),
hence, 5.00 mm is most precise. Nm2
29. (2) kg 2 c2 c
 2
  .
30. (2) When two quantities are multiplied, their Nm kg 2 kg
maximum relative errors are added up c2
X A B In other words, this has units of charges divide

hence:   by mass.
X A B
3. (1)
1
31. (4)  E  mv 2 4. (2)
2
 % Error in KE L
5. (3) [t ]    and [t ]  [ RC ]
R
= % error in mass + 2  % error in velocity
6. (3) Time period, T  p a  b E c
 2   2  3  8%
m m
or T  kp a  b E c
32. (3)   3
V  k, is a dimensionless constant.
m According to homogeneity of dimensions,
Given:  2%  2  102
m LHS = RHS
 a b
T    ML1T 2   ML3   ML2T 2 
c
 1%  1 102

T    M a b c   L a 3b  2c  T 2a 2c   ML2T 2 

   ML2T 3 I 2 
Comparing the powers we obtain   
IT I
a bc 0 11. (2) According to the method of dimensional

analysis the dimension of each term on both sides
a  3b  2c  0 of an equation must be same.
2a  2c  1
 T  kc x G y h z ( K  dimensionless constant )
On solving, we get
 [ M 0 L0T 1 ]  [ LT 1 ]x [ M 1 L3T 2 ] y [ ML2T 1 ]z
5 1 1
a   ,b  , c 
6 2 3  [ M 0 L0T 1 ]  [ M  y  z Lx  3 y  2 z T  x  2 y  z ]
7. (3) y  z  0 (i)
 M 0 L0T 1    M x   M yT 2 y    M x  yT 2 y 
x  3y  2z  0 (ii)
Equating powers of M and T. x  2 y  z  1 (iii)
x  y  0,  2 y  1 On solving Eqs. ( i ), ( ii ) and ( iii )
1 1 5 1
or y  ,x  0 x ,y z
2 2 2 2
1 Hence, dimension of time are [G1 2 h1 2 c 5 2 ]

or x
2
12. (1) This problem can be solved with
8. (1) Note carefully that every option has Gh dimensional analysis. All of the answers have
and c5 , the form
Gh   M 1 L3T 2   ML2T 1    M 0 L5T 3  t  1.23G 1/ 2  1/ 2 r n
The dimensions of t are [T], the dimensions of

c    LT 1  G are [ L]3 [ M ]1[T ]2 , the dimensions of  are
1/ 2 [ L]3 [ M ], and the dimensions of r are [ L] . So
 Gh 
 5   T 
c  [T ]  ([ L]3 [ M ]1[T ]2 )1/ 2 ([ L ]3 [ M ])1/ 2 [ L ]n
9. (2) By principle of dimensional homogeneity
[T ]  [T ][ L]n
 a 
 V 2   [ P] n0
 
The time to collapse does not depend on the
2 1 2
[ a]  [ P ][V ]  [ ML T ]  [ L ]  [ ML T ] 6 5 2
size of the cloud at all!
10. (3) Resistance 13. (3) The final result should be 3 significant
figures.
potential difference V W
R   14. (1) In multiplication or division, the final result
current i qi should retain as many significant figures as are
there in the original number with the least
[Dimensions of work] significant figures. In this question, density should

[Dimensions of charge] [Dimensions of current] be reported to two significant figures.
4.237 g T r h
Density   1.6498   0
2.5cm3 T r h
As rounding off the number, we get density g , 2 and 10 3

are constant 
Percentage error
 1.7gcm 3
15. (4) T  102  0.01 102  0.01 

100    100
T  1.25 102 1.5 102 
V0 4
16. (1) R0    4 = (0.8 + 0.689)
I0 1
 1.489  1.489%  1.5%
R V I
  1
R V I 20. (4) g  4 2 .
T2
0.25 0.1 g l T
   0.1625   100   100  2  100
4 1.0 g l T
R  4  16.25%
l T
 100  2. 100
17. (2) Given, A  1.0m  0.2m, B  2.0  0.2m l T
Let, Y  AB  1.0 2.0  1.414m 21. (1) The current voltage relation of diode is
I   e1000V /T  1 mA, (given)
Rounding off to two significant digit Y = 1.4 m
When, I  5mA, e1000V /T  6mA
Y 1  A B 
   1000
Y 2 A B  Also, dI   e
1000V / T
 dv (By exponential
T
1  0.2 0.2  0.6 function)
   
2  1.0 2.0  2  2.0
1000
  6mA    0.01  0.2mA
0.6Y 0.6 1.4 300
 Y    0.212
2  2.0 2  20
Rounding off to one significant digit
1. (4)
Y  0.2m
m
Thus, correct value for 2. (3) bt 3  m  b   ms 3
t3
AB  Y  Y  1.4  0.2m 3. (3)
uv f u v   u  v  0 idl sin 
18. (4) f  ,    4. (3) From Biot-Savart’s law, B 
uv f u v uv 4 r2
v u u v 4 Br 2
    0 
v u uv uv idl sin 
 Unit of magnetic permeability
rhg
19. (2) Surface tension, T   103
2 teslam2
  WbA1m1
Relative error in surface tension, Am
2
 J   Ft   MLT .T  1  Fr 2  MLT 2 .L2
5. (1)  
 A   A   L2  Also,      Q 2    ML3T 4 A2
       0   A2T 2
  ML1T 1      
  0   M 2 L4T 6 A4 .
3
 0 
6. (4) A  A e t
10. (1) [ ]  ML1T 1
Power is dimensionless
[ E ]  ML2T 2
 t 3  (T  )
[ve ]  M 0 LT 1
   T 3 
[  ]  M 1 L3T 0
7. (4) As P is a number then a and b have same
units 2 2
 Energy   ML T  2 2
aE 11. (1)  Mass   M   [ L T ]


 M  L T     
t
 Pressure   ML1T 2  2 2
a  ML T 2 2
M  Density    ML3   [ L T ]

L T     
T
 Force   MLT 2  2 2
1 2
a  (M L T )  b 3  Linear mass density    ML1   [ L T ]
   
1 e2 12. (3) Given, y  r sin t  hx 
8. (3) F
4 0 r 2
where, t  angle
2
e 
    [ F ][r 2 ]  [ MLT 2 .L2 ]  [ ML3T 2 ] 1
0     T 1 
T  
1  1   1  ( angle is dimensionless)
 ch    LT 1  ML2T 2  T    ML3T 2 
     
Similarly kx = angle
2
 e 
 0 0 0
 M LT 1
 ch k   T 1 
 0  x  
1  T 1 
and  c.    1    LT 1 
 0 0 k  L 
 
9. (3) From  B.dl  0inet 1 q1 q2
13. (2) From Coulomb ’s law , F  4 R 2
0
 Weber   Volt  Sec   ( Joul / coul )  Sec 

[ 0 ]     
 Am   Am   Am  q1q2
0 
 N  m  Sec   N 
4 FR2
2  2
    2   MLT A
 ( A  Sec)( A  m)   A  On substituting the units, we get
2 Breadth b  (10.1  0.1)cm

0 
C2  AT 
N  m  M L T  2   L 2
2
Area A  l  b  16.2  10.1  163.62cm2
Rounding off to three significant digits, area
as 4 is dimensionless
1
A  164 cm2
  M L  3 T 4 A 2 
 A l b 0.1 0.1
   
1 A l b 16.2 10.1
14. (2)  0.05 =0.0500
20
1.01  1.62 2.63
 
15. (4) Here 16.2 10.1 163.62
Length of the cube, L  1.2 102 m 2.63
 A  A
163.62
Volume of the cube, V  (1.2  102 m)3
2.63
6  163.62   2.63cm2
 1.728 10 m 163.62
as the result can have only two significant  A  3cm 2 (By rounding off to one significant
figures therefore, on rounding off, we get figure)
V  1.7 106 m3
Area, A  A   A  (164  3)cm2
16. (1) Volume V  l  b  t
20. (2) Percentage error in A
3
 12  6  2.45  176.4cm 1 

  2  1  3  3  1 2   2  %  14%
V  1.764 102 cm3  2 
Since, the minimum number of significant figure 21. (3) A  2.5ms 1  0.5ms 1 , B  1.10s  0.01s
is one in breadth, hence volume will also contain
only one significant figure. Hence, x  AB   2.5 0.10  0.25m
V  2  102 cm3 .
x A B
 
17. (1) D   4.23  0.01 cm x A B
0.5 0.01 0.05  0.025 0.075
d   3.89  0.01 cm    
2.5 0.10 0.25 0.25
t   D  d  / 2   4.23  0.01   3.89  0.01 x  0.075  0.08m
Rounding off to two significant figures.
  4.23  3.89   0.01  0.01
AB   0.25  0.08 m
  0.34  0.02 / 2cm   0.17  0.01 cm
22. (1) Let x  M a LbT c
18. (4) R1  24  0.5 
X  aM bL cL 
 100       100
R2  8  0.3  X  M L T 
RS  R1  R2   32  0.8    a   b  c  %
19. (2) Given that 23. (3) Measured time period of 100 oscillations
are 90 sec, 91 sec, 95 sec and 92 sec.
l  (16.2  0.1)cm
Mean value of time, Impulse is equal to change in momentum.

90  91  95  92 2. (1) Sum of 436.32, 227.2 and 0.301 is
tm   92sec
4 663.821. Because figure. 8 is more than 5, so 1
add in 663.
Absolute error in measurement
 663 + 1 = 664.
t1  tm  t1  2sec
3. (1)
t2  tm  t2  1sec Mass 4.237 g
 Density    1.694 g / cm3
Volume 2.5 cm3
t3  tm  t3  3sec
 1.7 g / cm3
t4  tm  t4  0sec
4. (2) Given, molar of one mole of hydrogen
Mean absolute error
 22.4 L  22.4 103 m3
2 1 3  0
tmean   1.5sec d 1010
4 Diameter of hydrogen molecule (r)  
2 2
But the least count of the measuring clock is 1
sec.  0.5  1010 m
So it cannot measure up to 0.5 second so we 4
have to round it off. So mean error will be 2 Volume of one molecule of hydrogen   r 3
3
second.
Hence mean time 92  2sec 4 3
 3.14   0.5 1010 
3
V V  V
24. (3) R  R  R 
I I  I  5.234 1031 m3
Number of molecules in one mole of hydrogen
 
 R  V  1  V / V  = Avogadro number (N)
R 1    
 R  I  1  I   6.023  1023
 I 
 Atomic volume of one mole of hydrogen
 R   V   I 
        3  3 %  6% = Number of molecules in one mole of
 R   V   I  hydrogen  Volume of one molecule of hydrogen
 6.023 1023  5.234 1031
 3.152 107 m3
1. (1) Because, dimension formula of tension
same as force Molar volume 22.4 103
Now,   7.1104
 [ MLT 2 ]
Atomic volume 3.154 107
5. (2) Area of object  1.7cm2  1.75  104 m2

force
and surface tension   [ ML0T 2 ]
length Area of image  1.55m2
Work and torque, both are product of force and Area of image
length.  Areal magnification  Area of object
1.55  4.234 1.005  0.0201

  8857
1.75  104  0.0855289
Linear magnification  Area magnification Rounding-off up to three significant figures, we

get
 8857  94.1 Volume of the sheet  0.0855m3
4. (1) The speed of light in vacuum (3) = 1(new
unit of length s 1 )
1. (3) Here, x = 5 cm
Time taken by light to reach the earth
x1  5  4.9  0.1cm
t  8min 20s
x2  5  4.805  0.195 cm
  8  60  20 s  500s
x3  5  5.25  0.25 cm
 Distance between the sun and the earth
x4  5  5.4  0.4 cm
 Speed of light × Time
Hence, option (3) is most accurate. x  ct
2. (1) Young modulus Y  1.9 1011 N / m2  1 new unit of lengths-1   500s
1N  105 dyne,1m2  104 cm 2 = 500 new unit of length

5. (3) Magnification of microscope = 100
0.9 1011 105
So in CGS Y  dyne / cm2 Observed width of the hair = 3.5 mm
104
Y  1.9  1012 dyne / cm2 Observed width

Magnification =
Real width
3. (4) Given, length(l) = 4.234 m
Observed width 3.5
Breadth (2) = 1.005 m Real width = 
magnification 100
Thickness (t) = 2.01 cm = 0.0201 m
= 0.035 mm
Area of sheet (1)
 2l  b  b  t  t  l 
1. (1) The relative error in surface area is
  4.234 1.005  1.005  0.0201
s r
 2
  0.0201 4.234   s r
The relative error in volume is
 2  4.3604739
V r
2  3
 8.7209478m V r
As thickness has least number of significant V 3
  
figure 3, therefore rounding -off area up to three V 2
significant figure, we get 2. (1)
2
Area of sheet (1)  8.72m
A  3P 2Q 1 R S 
Volume of sheet (V)  l  b  t  100        100
A  P Q 2 R S 
1
 3  0.5  2  1   3  1.5
Dimensions of R   ML2T 3 A2 
2
2 2
A Dimensions of   MLT A 
 100  6.5%  
A Dimensions of   M 1 L3T 4 A2 
3. (1) Let mass, related as M  T x C y h z   M 2 L4T 6 A4    R 2 

x y z
M 1 L0T 0  T '  LT
1 1
  M 1L2T 1  7. (2) t  rb S c/2 d a/4
M 1 L0T 0  M z Ly  2 z  T x  y  z and t  d 1/ 2 . S 1/2

z=1
a 1 
  a2 
y + 2z = 0 x-y-z =0 4 2 

1 c
y = -2 x+2-1=0 and    c  1
2 2 
x = -1
Now [d ]  ML3
1 2 1
M  T C h 
and [ S ]  MT 2
4. (1) Dimension of A  dimension of (C) 1 1

Hence A - C is not possible. [M 0 L0T 1 ]  Lb .[MT 2 ] 2 [ML3 ]2
5. (4) Let 0 related with e, m, c and h as 3 3

b  0b  .
follows. 2 2
0  ke a m b h d
a b c d
 MLT 2 A2    AT   M   LT 1   ML2T 1  V
1. (5) R
On comparing both sides we get I
a  2 (i) log e  R   log e  V   log e  I 
b  d 1 (ii) R V I
  
c  2d  1 (iii) R max V I
a  c  d  2 (iv)
 2 0.2 
    100%
By equation (i), (ii), (iii) & (iv) we get,  50 20 
a  2, b  0, c  1, d  1.   .04  .01  100%
 h   .05  100%  5%
 0    2 
 ce  
2. (3) If C  aiˆ  bjˆ then
6. (1) Dimensions    MLT 2 A2     
AC  A B
Dimensions of    M 1 L3T 4 A2  a  b 1 (1)
11
   
B C  A B I
 V  RI 
2a  b  1 (2) VA
Solving equations (1) and (2) we get


 L  I  V 
W W

 ML2T 2  q it
 2
1
a  ,b 
2   I T     L 
 
3 3
 1 4 5   M 1 L3T 3   I 2    M 1 L3T 3 I 2 
C   
9 9 9
6. (3) Let unit ‘u’ related with e, a0 h and c as
3. (1) A  4 r 2
follows.
a b c d

dA
2
dr  u    e   a0   h   c 
A r
Using dimensional method
4
V   r3 a b c
 M 1 L2T 4 A2    A1T 1   L  ML2T 1   LT 1 
d
3
a = 2, b = 1, c = -1, d = -1
dV dr dV  1 dA  3  dA 
3   3    7. (3) According to question
V r V 2 A  2 A 
E y  J x Bz
dV 3

V 2  Constant of proportionality
Ey C m2
4. (2) Given P  a1/ 2b2 c3 d 4 K  
Bz J x J x As
Maximum relative
 E I 
P 1 a b c d  As B  C  speed of light  and J  Area 
 2 3 4  
P 2 a b c d
8. (3) Angular momentum  m  v  r
1
  2  2 1  3  3  4  5  32%  ML2T 1
2
Q ML2T 2
5. (3) We know that resistivity Latent heat L    L2T 2
m M
RA
 Charge
 Capacitance C   M 1 L2T 4 A2
P.d
1 
Conductivity  resistivity  RA
12 Vectors & Scalars
Now the difference vector is nˆd  nˆ1  nˆ2 or

nd2  n12  n22  2n1n2 cos  1  1  2cos(1200 )
 nd2  2  2(1/ 2)  2  1  3  nd  3
1. (1)      
   9. (4) A  B  A  B  B  B
A  2B  3C  (2iˆ  ˆj )  2(3 ˆj  kˆ)  3(6iˆ  2kˆ) 
It is only possible when B is a null vector..
 20iˆ  5 ˆj  4kˆ   
10. (2) Let C  A  B  3iˆ  ˆj  7kˆ  5iˆ  ˆj  9kˆ
2. (3)   
C  A  B  8iˆ  2 ˆj  16kˆ
3. (3) As the multiple of ĵ in the given vector is
zero therefore this vector lies in XZ plane and 8
The direction cosine is
projection of this vector on y-axis is zero. 324
4. (3) The component of force in vertical 11. (2) As two vectors are same so the angle
direction between them is zero.
 
1 12. (3) A  B  0    900

 Fcos  Fcos 60  5   2.5N
2 13. (2) As, R2  P2  P2  2P2 cos
5. (3) F  F12  F22  2F1F2 cos900  F12  F22  R 2  2 P 2 (1  cos )
6. (4) F1  F2  F3  4iˆ  6 ˆj  F3  0  
R 2  4 P 2 cos 2  R  2 P cos
 2 2
 F3  4iˆ  6 ˆj  
14. (2) A  B  0  sin   0    0
7. (3) The resultant makes less angle with the
vector of greater magnitude. Two vectors will be parallel to each other.
8. (2) Let n̂1 and n̂2 are the two unit vectors, 15. (4)
 
then the A B 15 1
 16. (1) sin        300
sum is ns  nˆ1  nˆ2 or ns2  n12  n22  2n1n2 cos AB 5 6 2
 
 1  1  2cos  17. (4) Given, A  B  AB cos   6
 
Since it is given that ns is also a unit vector,, and A  B  AB sin   6 3
therefore
1
AB sin  6 3
1  1  1  2cos   cos      1200    3
2 AB cos  6
Vectors & Scalars 13
 
or tan   3 6. (2) A  B  4iˆ  3 ˆj  6iˆ  8 ˆj  10iˆ  5 ˆj
 
and   60 A  B  (10) 2  (5) 2  5 5
   
18. (2) Given, A  B  3 A  B 5 1 1
tan       tan 1  
10 2 2
 AB sin   3 AB cos 7. (1)
   
or 0 r  a  b  c  4iˆ  ˆj  3iˆ  2 ˆj  kˆ  iˆ  ˆj  kˆ
tan   3    60

r iˆ  ˆj  kˆ iˆ  ˆj  kˆ
rˆ   
r 12  12  (1)2 3
1. (4)  
8. (1) If A and B are parallel
 
2. (1) Resultant of vectors A and B  
   A B  0
R  A  B  4iˆ  3 ˆj  6kˆ  iˆ  3 ˆj  8kˆ
 kˆ(2 p  20)  0

R  3iˆ  6 ˆj  2kˆ
p = -10

R 3iˆ  6 ˆj  2kˆ 3iˆ  6 ˆj  2kˆ 
Rˆ      B  5iˆ  10 ˆj
R 32  62  (2)2 7
 B  25  100  5 5
3. (1) A  iˆ  ˆj  A  12  12  2  
9. (1) 2 A  3B  2(iˆ  ˆj  2kˆ)  3(2iˆ  ˆj  kˆ)
Ax 1
cos     cos 45   45.  4iˆ  5 ˆj  7kˆ
A 2
 
 Magnitude of 2 A  3B  (4)2  (5)2  (7)2
 16  25  49  90
4. (4) 10. (2) Given, P  iˆ  ˆj  kˆ , then we have
P  Q  iˆ
The X component of force F is
Q  iˆ  iˆ  ˆj  kˆ  ˆj  kˆ
Fx  Fcos30  F 
3

3
F  
2 2 11. (2) A  B  A2  B2  2 AB cos 
The Y component of force F is Given that A2  B2  2 AB cos   A  B

1 1 Squaring on both sides
Fy  Fsin 30  F   F
2 2
A2  B2  2 AB cos   A2  B 2  2 AB
5. (2) Magnitude of unit vector  1
cos   1    0
 (0.5)2  (0.8)2  c2  1 12. (3) Let the given vectors be
 
By solving we get c  0.11 A  2iˆ  3 ˆj  8kˆ and B  4iˆ  4 ˆj   kˆ
Dot product of these vectors should be equal to 

B  (1)2  32  42  1  9  16  26
zero because they are perpendicular.
   
 A  B  8  12  8  0  8  4    1/ 2 A  B  2(1)  3  3  (1)(4)  3
13. (1) Let the given vectors be  

A.B 3
   
A  iˆ  ˆj  kˆ ; B  iˆ  ˆj B 26
 A  12  12  12  3, B  12  12  2 17. (4) For the resultant,

 
A B  2 R 2  R 2  R 2  2 R 2 cos 
   R 2  2R 2  2 R 2 cos 
A B 2  2
cos      
AB 3 2  3  1
 1  cos 
2
2 1 1
sin   1  cos 2   1    1
3 3 3 or cos    or   120
2
 1  18. (4) Let, A and B be the two forces.
  sin 1  
 3
    A2  B2  5
14. (3) A B  A B
or A2  B2  25 (i)
 2  2
Squaring on both sides, A  B  A  B and A2  B2  2 AB cos120  13
        or 25  2 AB  (1/ 2)  13
 ( A  B)  ( A  B)  ( A  B)  ( A  B)
or 2 AB  24 (ii)
           
 A A  2A B  B  B  A A  B  B  2A B Solving eq’s (i) and (ii), we get
  A=3N
 4 A  B  0  4 AB cos  0
and B =4N
 cos  0    90  
19. (4) Using, A  B  AB cos
   
15. (1) ( A  B) is perpendicular to ( A  B) . Thus  
A B
     cos  
( A  B)  ( A  B)  0 AB
   
or A2  B  A  A  B  B 2  0. (iˆ  2 ˆj  2kˆ)  iˆ 1 1
or cos   
(12  22  22 )1/ 2 9 3
Because of commutative of dot product
   
A B  B  A 20. (1) As, AB cos   AB
 A2  B 2  0 or A  B cos   1    0
Thus the ratio of magnitudes A / B  1 B sin 

21. (1) tan  
A  B cos 
 
  A B
16. (2) The projection of A on B     Angle made by the resultant vector with
B 
A
 
2 2 1 28. (1) Given, A  B  0
  60, A  B  1  tan   
1 3  
1    A B (Since, A  B  AB cos )
2
 
 
Now, A B  1
 1 
  tan 1  
 3 or AB sin   1
 AB sin90  1 or AB  1
22. (4) From the property of vector product, C
must be perpendicular to the plane formed by  A  1 and B  1
 
vector A and B. So, A and B are perpendicular unit vectors.
     
23. (2) A  B and B  A are anti parallel to each 29. (4) a  b  ab sin  sin  can’t be greater
other. So the angle will be  .
than 1
   
24. (3) ( A  B)  ( B  A)      
30. (4) Since, ( A  B)  ( B  A), so C   D
 AB sin  (nˆ )  AB sin  (nˆ ) ,  
i.e., C and D are antiparallel to each other..
 sin   sin  2sin  0
 sin   0,   0 or 
   
25. (3) If A  B  0, A is perpendicular to B.
   
If A  C  0, A is perpendicular to C.
   1. (4)
So A is perpendicular to both B and C.
   
Also B  C is perpendicular to both B and C.
  
Hence A is parallel to B  C. From  ACD
  
26. (4) AD  AC  CD (i)
           
( A  B)  ( A  B)  A  A  A  B  B  A  B  B From ADE
      
 0  A B  B  A  0 AD  AE  ED (ii)
      Adding (i) & (ii)
 B  A  B  A  2( B  A)     
2AD  AC  CD  AE  ED
 
27. (3) Vector perpendicular to A and B,    
 AC  AF  AE  AB
   
A  B  AB sin  nˆ  CD  AF 
  Adding AD on both sides    
 ED  AB 
 unit vector perpendicular to A and B
       
A B 3AD  AB  AC  AD  AE  AF
nˆ   
A  B sin   
( AD  2 AO)
 (Triangle) since the two vectors are having same

2. (2) Let the component of A makes angles
 ,  and  with x,y and z axis respectively then magnitude and the third vector is 2 times that
of either of two having equal magnitude, i.e.,
   
the triangle should be right angled triangle
cos 2   cos 2   cos 2   1 Angle between A and B,   90
1 Angle between B and C,   135

 3cos2   1  cos  
3
Angle between A and C,   135
A 2
 A x =A y =Az =Acos   7. (1) Aˆ  Bˆ  ( Aˆ  Bˆ )  ( Aˆ  Bˆ )
3
3. (3)  Aˆ  Aˆ  Aˆ  Bˆ  Bˆ  Aˆ  Bˆ  Bˆ
  
A  3iˆ  2 ˆj  kˆ, B  iˆ  3 ˆj  5kˆ, C  2iˆ  ˆj  4kˆ
 1  Aˆ  Bˆ  Aˆ  Bˆ  1

A  32  (2) 2  12  9  4  1  14  2  2cos  2(1  cos )
   
B  12  (3)2  52  1  9  25  35  2  2sin 2   4sin 2
 2 2
 8. (1) According to Lami’s theorem
C  22  12  (4)2  4  1  16  21
P Q R
 
2 2
As B  A  C therefore ABC will be right sin  sin  sin 
angled triangle.      
9. (3) Here, A  B  C and B  C  A

4. (2) A  2iˆ  4 ˆj  5kˆ  
Since C  A therefore

 A  (2)2  (4)2  (5)2  45 B 2  C 2  A2  2CA cos(90)
 
2 4 5 Also, C  A
 cos   ,cos   ,cos   .
45 45 45
5. (4) If n identical forces each with angular B 2  2 A2  B  2 A
2 Now, A2  B2  2 AB cos  C 2  A2
separation between any adjacent vectors is
n
then the resultant force is always equal to zero. 1
Therefore we find cos  
2
3
This gives  
4
6. (4) 10. (4) A  3N , B  2 N
then R  A2  B 2  2 AB cos
From triangle law, three vectors having R  9  4  12 cos  (i)

summation zero should form a closed triangle. Now A  6 N , B  2 N then
2 R  36  4  24 cos  iˆ ˆj kˆ
1  2 2 1  iˆ  10 ˆj  18kˆ
From (i) and (ii) we get cos      1200 6 3 2
2
11. (1) R 2  F12  F22  2 F1 F2 cos   

Unit vector perpendicular to both A and B
Substituting, F1  ( x  y), F2  ( x  y) and
iˆ  10 ˆj  18kˆ iˆ  10 ˆj  18kˆ
 
R  ( x 2  y 2 ) , we get 12  102  182 5 17

14. (1) Given OA  a  3iˆ  6 ˆj  2kˆ and
 ( x2  y2 ) 
1
  cos   2 2  
 2( x  y )  OB  b  2iˆ  ˆj  2kˆ
12. (2) The vector diagram is
iˆ ˆj kˆ
 
 (a  b )  3 6 2
2 1 2
 (12  2)iˆ  (4  6) ˆj  (3  12)kˆ
 10iˆ  10 ˆj  15kˆ
 B sin  
 
tan     a  b  425  5 17
 (i)
 A  B cos  
Where  is the angle made by the vector 1   5 17

  Area of  OAB  a b  sq.unit.
 2 2
( A  B) with A .
B sin 
Similarly, tan   (ii)
A  B cos  
1. (4) A  Ax iˆ  Ay ˆj  Az kˆ, B  Bx iˆ  B y ˆj  Bz kˆ
Where  is the angle made by the vector
   
  A  B   Ax  Bx  iˆ   Ay  By  ˆj   Az  Bz  kˆ
( A  B) with A .
  Ax   B x , Ay   B y , Az   B z
Note that the angle between A and ( B) is
(180   ). Adding (i) and (ii), we get 2. (3) Diagonal of the hall  l 2  b2  h2
B sin B sin   102  122  142

tan   tan   
A  B cos A  B cos
 440  20.9  21m.
2 AB sin 
 2 3. (1)
( A  B 2 cos 2  )

  4. (2) Unit vector parallel to A
13. (3) A  2iˆ  2 ˆj  kˆ and B  6iˆ  3 ˆj  2kˆ 
A i  2 j  2k
    
C  A  B  (2iˆ  2 ˆj  kˆ)  (6iˆ  3 ˆj  2kˆ) A 1 4  4

i  2 j  2k 9. (4) Here, a  b  c and c  a  b

3 
Let  be angle between a and b .
B  4  36  4  7
 c  a2  b2  2ab cos
(i  2 j  2k )
 Required vector  7 a  b  a 2  b2  2ab cos
3
5. (4) Here all the three forces will not keep Squaring both sides, we get
the particle in equilibrium so the net force will
(a  b)2  a 2  b2  2ab cos 
not be zero and the particle will move with an
acceleration. a 2  b2  2ab  a 2  b2  2ab cos
6. (4) B  72  (24)2  625  25 2ab  2ab cos  0 or 1  cos   0
1 0
Unit vector in the direction of A will be cos   1 or   cos (1)  0 .
3iˆ  4 ˆj
Aˆ  10. (3)
5
11. (3) Force F is in the x-y plane so a vector
 3iˆ  4 ˆj  along z-axis will be perpendicular to F.
So required vector  25  5   15iˆ  20 ˆj
   
A  B (2iˆ  3 ˆj )  (iˆ  ˆj ) 2  3 5
12. (1)   
7. (4) According to definition of cross-product B 2 2 2
   
C  D is perpendicular to both A and B. 13. (4) The two vectors which are perpendicular
   to each other will have their dot product equal
i.e., A  (C  D)  0
to zero.
    14. (1) The vector diagram is
or AC  A D  0
 
or A(component of C along A ) + A (component
 
of D along A) = 0
2sin 60 3
tan 45  
8. (3) A  2cos60 A  1
A 1  3  A  3 1
As the metal sphere is in equilibrium under the 15. (1)

   
   A  B  3( A  B )  AB sin   3 AB cos 
effect of three forces is T  P  W  0
From the figure T cos   W (i)  tan   3    60

T sin  P (ii)    
16. (4) A  B  3 ( A  B)
From equation (i) and (ii), we get P  W tan 
and T 2  P2  W 2 AB sin   3 AB cos  tan   3

     
   60 3. (2,4) Given, A  B  C  0 , then A, B and C are
   in one plane and are represented by the three
Now R  A  B  A2  B 2  2 AB cos  sides of a triangle.
    
1 (1)  B  ( A  B  C )  B  0  0
 A2  B 2  2 AB    ( A2  B 2  AB )1/ 2
2      
or B A B B  BC  0
     
17. (2) Area of parallelogram  A  B  A B  B  C (i)
       
Let A  ˆj  3kˆ, B  iˆ  2 ˆj  kˆ  ( A  B)  C  ( B  C )  C ; It cannot be zero.
      
iˆ ˆj kˆ
If B || C , then B  C  0 , then ( B  C )  C  0
  
C  A  B  0 1 3  7iˆ  3 ˆj  kˆ Thus, option (1) is correct
1 2 1      
(2) ( A  B)  C  ( B  C )  C  0

Hence area  C  49  9  1  59 sq.unit       
If B || C , then B  C  0 , then ( B  C )  C  0
Thus, option (2) is false.

   
1. (1) (3) ( A  B)  D  AB sin  . The direction of D is
A B (iˆ  ˆj )  (iˆ  ˆj ) 11 perpendicular to the plane containing A and B.
cos   0
AB 1  1  1  (1)
2 2 2 2 2     
( A  B)  C  D  C . Its direction is in the plane
   90 of A, B and C. Thus, option (3) is correct.
2. (2) (4) If C 2  A2  B2 . then angle between A and B

  
3. (2) If r makes an angle  with x-axis, then is 90.  ( A  B)  C  0
component of r along x-axis  r cos  .   
Since ( A  B) is perpendicular to C
It will be maximum if cos   max  1    0 , 0
i.e., r is along positive x-axis. Thus, option (4) is false.

 
    4. (2,4) When A is perpendicular to B , then
4. (1,2) If A  B  A , then either B  0 or A
 
  A  B  A2  B2  2 AB cos900  A2  B2
and B will be antiparallel, where, B  2 A .
 
A  B  A2  B2  2 AB cos900  A2  B2
     
1. (4) A scalar quantity has the same value for When A  0 or B  0 then A  B  A  B
observers with different orientations of the axes.
2. (2)
20 Motion in One Dimension
s
 s1
s/2 s1 2
t1  , t2  , t3 
v0 v1 v2
1 vs
1. (3) Total distance covered = Final reading - Given t 2  t 3  s1  2  v  v  ,
1 2
Initial reading  6285  6225  60 km.
s
2. (2) Displacement of a particle may be zero, t 2  t3 
2  v1  v 2 
because final position of the particle may
coincide with its initial position. s
3. (2) The body completes for and half rotations The average speed  v  t  t  t
1 2 3
and hence the displacement is equal 2R (shortest
distance between the initial and final position) 2v0  v1  v2 
 v 
4. (4) Let S be the total distance travelled by  v1  v2  2v0 
the particle. 8. (2)
s 9. (2) vA  tan 30  and vB  tan 45 

Average speed  t  t  ....
1 2
vA tan30 1/ 3 1
   
s/n s/n vB tan 45 1 3
t1  , t2  ......
v1 v2
10. (4) Let l be the total length of the train and
s v ' be the velocity of the midpoint of train
v

1 s s 
   .... l
n  v1 v2  v '2  u 2  2a (i)
2
Distance 1 km (ii)
v avg    2 km/hr v ' u '2  2a.l
5. (2) Time 1
hr
2 v2  u 2
From (1) and (2) we get v ' 
2
6. (3) The average or Mean speed is
11. (1)
R
 v  = 0.5 m/s 12. (4) s  1.2m, v  640ms 1
T
7. (2) Let S be the total distance travelled by a = ?; u = 0; t = ?
the particle. Let t1, t2 and t3 be the times taken
2as  v 2  u 2
by the particle in the three parts.
8  64 103
 2a 1.2  640  640  a 
3
Motion in One Dimension 21
v 15 17. (4) According to given relation acceleration

v  u  at  t    103
a 4 a  t  
 3.75 102 s  4ms. dv dv
As a  t   
13. (4) Let initial velocity of the bullet be u dt dt
Since particle starts from rest, its initial velocity
u is zero i.e., At time t = 0, velocity = 0
After penetrating 1 cm its velocity becomes
2
v t t 2
  dv    t   dt  v   t.
From v  u  2as
2 2 0 0 2
2 18. (1) Distance covered by first body in
u
   u  2a 1
2
2
  5th second after start,
3u 2 3u 2 a1 9a
 2a  a S5  0   2  5  1  1
4 8 2 2
Further it will penetrate through distance x and 5th second of first body will be 3rd second of
stops. second body. Distance covered by the second
For distance x, v  0, u  u / 2, s  x, a  3u 2 / 8 body in the 3rd second after its start.
u
2
 3u 2  a2 5a
2 2
From v  u  2as  0     2   x
S3  0   2  3  1  2
2 2 2
 8 
1 Since S5  S3  a1 : a2  5 : 9
 x  cm
3 19. (3) Given that
dv
14. (4) Acceleration a   0.1 2t  0.2t. v 2  a  bx
dt
Which is time dependent, i.e., non-uniform dv dx
acceleration 2v b
dt xt
15. (3) In the given question, acceleration at
dv b
3    Constant
t = 0 is 3   4  0 and velocity is also zero at dt 2
4
20. (1)
t = 0,
21. (3) The ball has zero initial speed and smaller
So, particle will remain at rest. Hence, v = 0
average speed during the time of flight to the
4 passing point. So the ball must travel a smaller
and x m
3 distance to the passing point than the ball your
friend throws.
16. (2) From equation of motion
22. (2) During first half, velocities are lesser as
v2  u 2  2as compared to velocities during second half. So
more time is taken in first half.
Given, v  100m/s, u  200m/s
23. (4)
s  10cm  10  102 m
24. (3) Suppose v be the velocity of the body
2 2 after falling through half the distance. Then
 100    200   2  a  10  102
40
4 2
s  20m, u  0 and g  10m / s 2
a  15  10 m/s 2
v 2  u 2  2 gh  02  2  10  20
29. (2) For particle 1 moving with constant
velocity, s1  50t
 v  20m / s
For particle 2 moving with constant
25. (3) Let u be the initial speed
1 2 1 2 2
Given that acceleration, s2  at   10t  5t

2 2
1  1  When s2  s1  125
u  10(5  )  2 u  10(6  ) 
2  2 
The above equations
u  45  2u  110
5t 2  50t  125
u  65ms 1
On solving, we get
1 1
26. (4) h  ut  gt 2  81  12t   10  t 2
2 2

t  (5  5 2)s  t = 5 1  2 s 
  
 t  5.4sec 30. (1) vb  vb , p  v p
27. (1) Let h is the initial height of the balloons
  u   v  v  u
and u be the initial velocity of the bomb with
respect to ground. 31. (2) Let after a time t, the cyclist overtake
1 2
1 1
h  ut  gt 2  h  gt 2  ut
the bus. Then 96   2  t  20  t
2
2 2
 t2  20t  96  0
1
 gt 2  ut  h  0
2 20  400  4  96
t 
2 1
u  u 2  2 gh
t . 20  4
g   8 sec and 12 sec.
2
So as u increases, t increases
28. (2) Let t is the total time taken by the ball.
a 1. (3) When the drunkard walks 8steps forward

From, D n = u +  2n  1 we have, total and 6 steps backward, the displacement in the
2
distance travelled in last 2 seconds of fall is first 14 steps  8m - 6m  2 m
D  D t  D  t 1 Time taken for first 14 steps = 14 s
Remaining distance he has to cover is 8 m. So if
 g   g  he makes 8 steps forward then he falls in
 0   2t  1   0  2  t  1  1
 2   2  the pit. Total time he takes is 14 s+8 s = 22 s
g 10 2. (1) When location of a particle has changed,
  4t  4    4  t  1 it must have covered some distance and
2 2
undergone some displacement.
or, 40  20  t  1 or t  2  1  3s
3. (1) Total time of motion is 2 min 20 sec =
Distance travelled in t second is 140 sec.
1 1 As time period of circular motion is 40 sec, in
s = ut + at  0   10  32  45m
2
2 2 140 sec, athlete will complete 3.5 revolution, i.e.,

he will be at diametrically opposite point, i.e., Avg vel  Avg speed

Displacement = 2R.
4. (3) The distance travelled is  R Avg. vel
 1
Avg. speed
The displacement is 2R
Displacement 2 Displacement
 9. (1) Velocity of particle  Total time
Distance 
5. (4) Let s be the total distance. t1 and t2 are
Diameter of circle 2 10
the times in which the body covers distances   4m / s.
5 5
2s 3s
and
5 5 10. (1) If t1 and 2 t 2 are the time taken by particle
respectively. to cover first and second half distance
respectively.
2s / 5 3s / 5
v1  v2 
t1 t2 x/2 x
t1   (1)
s 5v1v2 3 6
Average speed  
t1  t2 3v1  2v2
x1  4.5t 2 and x 2  7.5t 2
6. (4) Let 2x is the total distance traveled by
the particle x x
so, x1  x2   4.5t 2  7.5t 2 
Average Speed 2 2
Total distance 2x 2 v1 v2 x
   t2  (2)
Total time x x v1  v2 24

v1 v2
x x x
Total time t = t1  2t2   
6 12 4
Total displacement
Average velocity 
Total time Total distance x
vavg    4m/sec
It comes back to its initial position Total time x
4
Total displacement is zero. Hence, average
velocity is zero. So, average speed = 4 m/sec.
Displacement 11. (2) 12. (3)
7. (4) Velocity =
Time 13. (2) Distance travelled by the particle is
Displacement = 25 m x  40  12t  t 3
Distance 75m We know that, speed is rate of change of
Time =   5s
speed 15m / s
dx
distance i.e., v 
25m dt
Velocity   5m / s
5s
d
v0 g
v 
dt
 40  12t  t 3   0  12  3t 2
8. (4)  v g
2 2
But final velocity v = 0
As the distance is always either greater or equal
to displacement. 12  3t 2  0
12 3u 2 u2
 t2   4  t  2s  6a  a
3 4 8
Hence, distance travelled by the particle before Further it will penetrate through distance x
coming to rest is given by
3
For distance x, v  0, u  u / 2, s  x, a  u 2 / 8
x  40  12  2    2 
2 2
2 2 u u
 40  24  8  64  8  56m From v  u  2as  0     2    x
2 8
   
At t  0, x1  40m
 x  1 cm .
t  2 s, x2  56m
20. (1) Given : Initial velocity of a body u = 0
 S  x2  x1  16m Let s be the distance covered by a body in time
14. (1) Total distance to be covered for crossing t.
the bridge = length of train + length of bridge 1 1
 s  ut  at 2 or s  at 2
= 150 m +850 m = 1000 m 2 2
Distance 1000  s  t2
Time    80sec
Velocity 45  5
18 21. (2) u  0, a  2ms 2 ; v1  0  2  4
15. (1) s  5t 2  6t v1  8ms 1

ds Now,
  10t  6  0
dt a  2ms 1 ; v  u  at  0  8  2t  t  4s
3 1
t  s S1   2  42  16m
5 2
dy 4
16. (3) Velocity, v    t  16 8
dt 3 To come to rest t   4s
2
For body to be at rest, v  0
1
S2  8  t   2  t 2  8  4  42  16m
4 2
 t  16  0  t  12 sec. Total distance travelled in forward direction is
3
32 m
17. (2) Conceptual
Return path
1
18. (4) s  at 2  s  t
2 S  32m, u  0, a  2ms 2
19. (1) Let initial velocity of the bullet  u
1 1
S  at 2  32   2  t 2
u 2 2
After penetrating 3 cm its velocity becomes
2
t  32  4 2 s
From v 2  u 2  2as
2
u 2  After a time (8  4 2)s the particle is again
   u  2a (3)
2 at the origin.
22. (2) 4 -2
v1  10 m/s, t = 5/sec, v 2 = 20 m/s, a = ? Put u = 0, a = ms ,n = 3
3
v2  v1 20-10 4 4 10
a = = = 2 ms -2 d  0   2  3  1   5  m
t 5 3 2 6 3
From the formula v1 = u1  a t, we have 27. (3) Before collision the initial velocity is

vi   2 gh1 ˆj , after collision the velocity is
10  u  2  3  u  4 m/sec 
v f  2 gh2 ˆj
23. (3) Distance covered by the body 5th second.
 
a a 9a  v  vi
a f
 2 gh2 + 2 gh1 ˆj
S5th  ut   2n  1  0   2  5  1  t

t
2 2 2
and distance covered in 5 second,  h1  h2 

a  2g  
1 1 25a t
S5  ut  at 2  0  a  25 
2 2 2
 1500 2 m / s 2 (upward)
9 S5th
  .
S5 25 28. (2)
29. (4) Once the ball has left the thrower’s hand,
0 dv t
it is a freely falling body with a constant, non-
24. (3)   2.5 dt
6.25 0
v zero, acceleration of a   g .
0 30. (2) Value of g is small on the moon
2 v  2.5t
6.25
31. (2)
2 6.25  2.5t 32. (3) f  mk
t = 2s
mg  mk  ma
dv
25. (3)  bt  dv  bt  a  g  k (where k is proportionality
dt constant).
bt 2 Acceleration is same for both bodies
v  K1
2
1
33. (2) h  ut  at 2  10  u  1  5
At t  0, v  v0  K1  v0 2
dx 1 2 1 bt 3 (1) 2  u  15ms 1 ; V  u  at
Again  bt  v0
 x   v0t  K2
dt 2 2 3  15  10 1  5ms 1
At t  0, x  0  K 2  0 34. (2)
1 2 2 4u 2
v 2  u 2  2 gh   3u    u   2 gh  h 
 x  bt 3  v0 t g
6
35. (1) Consider the downward direction as
26. (3) Distance travelled in the nth second is
positive
a
given by d = u +  2n-1 Given u = 0, a = g
2
1 In T/3 second, vertical distance moved is given

Dn  0  a  2n  1)  1 T 
2
1 gT 2 h
2
by h '  g    h '   
2 3 2 9 9
1
D1 : D2 : D3  g  2 1  1 : h 8h
2  Position of ball from ground  h  
9 9
1 1
g  2  2   1 : g  2  3  1 41. (3) For man: S  (Vmin t ) For bus:
2 2
=1 : 3 : 5 1
S  45   2.5  t 2
2
36. (3) The net displacement of the body is equal
to zero. Vmin t  45  1.25t 2 ;
37. (1) The distance travelled in last second. Here t  6 s
g 1 90
S Last  u   2t  1   9.8  2t  1  4.9  2t  1 Vmin  6  45  1.25  36  Vmin   15ms 1
2 2 6
and distance travelled in first three seconds, 42. (4) Let speed of the third car be v km h -1
1 Then, relative speed of third car with respect to
SThree  0   9.8  9  44.1m
2 two cars   v  30  km h -1
According to problem S Last  SThree Two cars are separated by 5 km and third car
crosses them in 4 min.
 4.9  2t  1  44.1  2t  1  9
5  60
 t  5sec.
 v  30   75  v  45 km h -1
4
38. (1) VA  VB 43. (1)
1 2 Total length 50  50 100

39. (4) h gt1 Time     4sec
2 Relative velocity 10  15 25
1 2 44. (2)
2h  gt2
2 45. (2) Let the velocity of the scooter be v ms 1 .
1 2 Then (v  10)100  1000 or v  20ms 1
3h  gt3
2
46. (4) Let speed of elevator be ve
t1 : (t2  t1 ) : (t3  t2 )
L L
t1   1min 
2h 4h 2h 6h 4h ve ve
:  : 
g g g g g Let speed of person on stationary elevator be
1: ( 2  1) : ( 3  2) L L
vP then t2  v  3min  v
P P
2 1
40. (1) We have s  ut  gt when the escalator is moving:
2
L L 3
t3   min  45s
1 ve  vP L L 4
 h  gT 2 u  0 
2 1min 3min
v 3
 vdv    2 x  3x  dx
2
1 2.7  2
47. (3) S r  ar t 2  t   0.7 s 2 0
2 11
v2  4 3
S r  relative displacement of the bolt w.r.to the   x 2  x3  0
2
elevator
ar  relative acceleration of bolt w.r.to the v 2  4  36
elevator. v 2  32
Velocity cannot be imaginary that means particles
cannot move from 0 to 3m. Particle will stop
dx dx before reaching 3m.
1. (4) v x   x    dt
dt x 5. (1) If a stone is dropped from height h then
1 2
By integrating both sides  x 1/ 2 dx    dt h gt (i)
2
If a stone is thrown upward with velocity u then
 x  t2
1 2
2. (4) Let the initial velocity of ball be u. h  ut1  gt1 (ii)
2
u If a stone is thrown downward with velocity u
Time of rise t1  g  a and height reached then
u2 1 2
 h  ut2  gt2 (iii)
2( g  a) 2
From (i) (ii) and (iii) we get
Time of fall is given by
1 2 1 2
1 u2 ut1  gt1  gt (iv)
( g  a)t22  2 2
2 2( g  a)
1 1
ut2  gt22  gt 2 (v)
u 2 2
 t2 
( g  a)( g  a)
Dividing (iv) and (v) we get
1 1
 t2  t1 because g  a  g  a 1
ut1 2 
g t 2  t12 
 
ut2 1
1 g  t 2  t22 
3. (2) v  180  16 x  2 2
t1 t 2  t12
dv dv dx or  
a  . t2 t 2  t22
dt dx dt
1 1
dv 1  By solving t  t1t2 .
v  (180  16x) 2  (180  16x) 2 (16)  8ms 2
dx 2
4. (4) Given that a  2 x  3x 2 6. (2) v  x   bx2n
dv dv dx dv dv
  v  2 x  3x 2 av  bx 2 n {b  2n  x 2 n 1}  2b 2 nx 4 n 1.
dt dx dt dx dx
7. (4) Let the car accelerate at rate  for time

t1 then maximum velocity attained,
v  0  t1  t1
Now, the car decelerates at a rate  for time

1
 t  t1  and finally comes to rest. Then, vAt  at 2  d  vB t
2
0  v  t  t1   0  t1  t  t1 a 2
t   vB  vA  t  d  0
2
 1
 t1  t ; S1   t12 The cars meet when
 2
1 2 2 a
S2   t2 v B
 vA   4 d  0
2 2
2
1 v  v A   2ad
 S 1  S2   t12   t22  B
2
2
d
v A
 vB 
1  2
S t 2a
2 
2
8. (3) If the body starts from rest and moves v A
 vB 
The cars will not collide if d 
with constant acceleration then the ratio of 2a
distance in consecutive equal time intervals is
11. (2)
1
S1  a(42 )
2 Where u, u1 , u2 and u3 are the speeds at the
positions shown
1
S2  a (82  4 2 )
2 u  u1 u u u  u3
v1  , v2  1 2 , v3  2
2 2 2
1
S3  a (12 2  82 )
2 u  u2 u u
 v1  v2  , v2  v3  1 3
2 2
S1 : S 2 : S3  1: 3 : 5.
u2  u  a  t1  t2  & u3  u1  a  t2  t3 
1
9. (2) Sn  g sin   2n  1 ,
2 v1  v2 t1  t2
 
v2  v3 t2  t3
1
Sn 1  g sin {2  n  1  1}
2 12. (1) Acceleration of a body along AB
is g cos 
Sn 2n  1

Sn 1 2n  1 Distance travelled in time
10. (3) Let d is the initial separation between the 1
t sec  AB  ( g cos  )t 2
cars . 2
When the two cars meet each other. From

1
From (i) & (ii)
ABC , AB  2 R cos  2 R cos  g cos t 2
2 1 2 1
ut1  gt1  ut2  gt22
2 2
4R R
 t2  or t  2
g g 1
 u  t1  t2   g  t22  t12 
2
Here ‘t’ is independent of ‘  ’
g g gt t
13. (4) After time t, if y is the height of the body,  h  t1  t1  t1  gt12  2 1
2 2 2
1 2 2h
Then, y  ut  gt  Required time t   t1t2
2 g
u 2 gy   8  16  8 2
t  1  1  2 
g u 
15. (3) Let time taken by ball to reach maximum
height be T.
2u 2 gy
t  1 2
g u Also v  u  gT
At maximum height, v  0
2 gy g 2 (t )2
 2 1
u 4u 2 So, u  gT
2g g2 u  2  9.8  19.6 m s-1

 2 ( yB  yA )  2 (t A2  tB2 )
u 4u hence, the speed should be more than
19.6 m s-1 .
Where yB  y A  h
16. (1) The average speed =
8( yB  y A ) 8h
g  2H Total distance
(t A  t B ) (t A  t B2 )
2 2 2
=
T Time of flight
14. (3) For downward motion,
u2
1 2 H= 2g (maximum height reached by the body)
h  ut1  gt1 (i)
2
2u
For upward motion, T= g (Time of flight)
u2
1 2.
h  ut2  gt22 (ii) 2g u
2  
Average speed  2u  2
 )
 g 
17. (3) v  4t  t 2
Average velocity is
5 5
2
 vdt  (4t  t ) dt
v  0
 0
 5 m/s
5 5 3
18. (2) Given t  x  3

2
x  t  3  x   t  3
v  2  t  3 D istance 8cm
Avg speed    4cm / s
tim e 2s
at t = 0s x = 9m
D isplacement
t = 6s x = 9m Avg velocity  0
tim e
For t < 3s v is (-) ve 21. (1) As the trains are moving in the same
For t >3s v is (+) ve direction, so the initial relative speed  v1  v2  and
The motion of the body from t = 0s to t = 6s is by applying retardation final relative speed
shown in the figure below. becomes is zero.
v1  v2
From v  u  at  0   v1  v2   at  t  .
a
22. (2) Time taken by man to go from M to P
and then P to N = time taken by float to go from
M to N.
Let u is the velocity of river and v is the velocity
99 of the man w.r.to the still water.
The average speed in 6s is =  3m / s
6
19. (2) For first part
u = 0, t = T and acceleration = a
1 2 1 2
 v  0  aT  aT and S1  0  aT  aT D D D
2 2  
v  u 2(v  u ) 2u
For second part,
v
u = aT, retardation  a1 , v =0 and time taken Simplify to get 3
u
 T1 (let) 23. (2) At time t
 0  u  a1T1  aT  a1T1 Velocity of A, vA  u  gt upward
2 2 2
u 1aT Velocity of B, vB  gt downward
and from v 2  u 2  2aS2  S2  
2a1 2 a1
If we assume that height h is smaller than or
1  aT  equal to the maximum height reached by A, then
S2  aT  T1  As a1  
2  T1  at every instant vA and vB are in opposite
1 2 1 directions.
aT  aT  T1
S1  S2 2 2
 vav  
T  T1 T  T1  VAB  vA  vB
1
aT T  T1 
1
2  aT .
T  T1 2
20. (3)
 u  gt  gt [Speeds in opposite directions get The relative motion is shown in the below
added] diagram
u .
24. (3) We use the relative motion equation for
displacement
1
S r  Sr0  ur t  ar t 2
2
Sr -displacement of ball w.r.to lift 1
sr  ur t  ar t 2
2
Take the origin at the base of the lift
100  5(t )  t  20s
 S r0  0
27. (2) This problem can be solved very easily
(Since the body is projected from the floor itself) with the help of relative velocity. As the two
    trains approach each other the time taken to
Sr  Sbl , ur  vbl  vjˆ meet each other is
   
ar  abl  ab  al s r  vr t
  km 
ar   giˆ  ajˆ  (a  g ) ˆj 100km   20 t
 h   t  5h
When the ball strikes the floor again
So the fly travels for 5h with a constant speed

S r  0 (relative displacement) of 40km/h
1 km
0  vt ˆj  (a  g )t 2 ˆj s fly  40  5h  200km
2 h
2v
t
ag
25. (1) We can solve the problem with the help 1. (3) x  t 1
of relative velocity. Let are  vr  x &  vr  y the x
Squaring both sides, we get,
and y components of relative velocity of the 2
x   t  1  t 2  2t  1
bullet.
dx
l Differentiating it w.r.t time t, we get,  2t  2
(vr ) X  v2 cos  v1 ; ( Sr ) X  l ; t  dt
v2 cos  v1 (i)
dx
b Velocity, v   2t  2.
(vr ) y  v2 sin  ; ( S r ) y  b; t  (ii) dt
v2 sin 
vA tan  A tan 30 1/ 3 1
v2 l sin  2. (1)     .
From (i) & (ii) b  v cos  v vB tan B tan 60 3 3
2 1
26. (3) u AB  10iˆ  5 ˆj  5iˆ 2h

3. (2) Time of fall 
g
aAB  5iˆ  5iˆ  0
h
u BB  0, aBB  0 Time taken by the sound to come out 
c
8. (2) Let the car accelerate at rate  for time

2h h  2 1
Total time  g
  h
c
  t1 then maximum velocity attained,
 gh c 
v  0   t1   t1
1 dx 1
4. (1) x v  2 Now, the card decelerates at a rate  for
t 5 dt  t  5
time (t  t1 ) and finally comes to rest. Then ,
Acceleration,
0  v   (t  t1 )  0   t1   t   t1
dv 2 3/2
a  3
 a   velocity  .
dt  t  5 
 t1  t
 
5. (2) Given that t  x2  1 
 v t
 
x2  1  t 2
9. (2) Let car start from point A from rest and
dx moves up to point B with acceleration f
2x  2t  xv  t
dt
Velocity of car at point B,   2 f S
dv dx
x  v 1 [As  2  u 2  2as ]
dt dt
Car moves distance BC with this constant
ax  1  v 2 velocity in time t
1  v2 x2  t 2 1 x  2 fS .t [As s = ut] (i)

a   3
x x3 x
So the velocity of car at point C also will be
6. (4) According to problem 2 fS and finally car stops after covering
Distance travelled by body A in 5th sec and distance y.
distance travelled by body B in 3rd sec of its 2
motion are equal.

Distance CD  y 
 2 fS  
2 fS
 2S (ii)
2  f / 2 f
a1 2 a
0
2
 5  42   0  2 [32  22 ]
2 So, the total distance AD = AB + BC + CD =
45S [Given]
a1 5
9a1  5a2    S  x  2S  15S  x  12S
a2 9
Substituting the values of x in equation (i) we
dt 1 get
7. (2)  2x    v 
dx 2x   x  2 fS .t  12 S  2 fS .t  144 S 2  2 fS .t 2
dv dv dx 1 2
a   . S  ft .
dt dx dt 72
dv v.2 v0 v
av   2.v.v2  2v3 10. (1)   t1  1
dx  2x   2 t1 
 Retardation  2v3 . 0v v v

  t2  
t2  
t1  As area’s of two triangles are same.

  ,
t2  x1  x2
1 2 13. (4) Maximum height attained  u 2

x1
 t1
 2 ;
x2 ( t )t  1  t 2 14. (3) Given v  6t 2  6t 3
1 2 2
2
dv
 12t  18t 2
t1   dt
As  , t2  t1
t2  
d 2v
  12  36t
dt 2
1 2
x
 t1 
 1  2  When v is maxima or minima
x2  2 1 2 2  dv 2
  t1   2 t1
 2   12t  18t 2  0  t  s, t  0
dt 3
x1  t1 d 2v
    12 (minima)
x2  t2 dt 2 t  0
11. (3) Assume that the passenger will catch the
bus after time t. d 2v
 12 (maxima)
dt 2 t  2 s
When the passenger catches the bus d  S1  S2 3
1 vmin  6(0)2  6(0)3  0

Displacement of bus  S1   at
2
2
15. (4)
Displacement of passenger  S2   ut
1 2
s1  gt
2
1 u  u 2  2ad
d  at 2  ut  t  1
2 a s2  g (2t )2  4 s1
2
Passenger will catch when t is real
1
i.e., u 2  2ad s3  g (3t ) 2  9 s1
2
umin  2ad s1 : s2 : s3  s1 : 4 s1 : 9 s1  1: 4 : 9
12. (1) The possible graph between v and t is 16. (1) Consider the origin at the top of the tower
shown. and downward is positive and upward is negative
S0  0
v0 v0
 ;     
t t
u =20 m/s (Going towards    ve direction) nu 2 1 n 2u 2

 H   g 2
g 2 g
a = g (Accelerated motion)
1  2 gH  nu 2 (n  2)
s  20t  gt 2
2 18. (3) Time taken for the first drop to reach
When it hits the ground 1
2
ground is s  gT
2
s  50 m
29
T 
1 10
50  20t  10t 2
2
T 2
Time gap for each drop  
t 2  4t  10  0 3 10
4  56 1 2 1 2
t  14  2 Distances of 2nd ,3rd drops are gt2 , gt3
2 2 2
Method-2 2T T
Where t2  , t3 
If we do the problem graphically we can avoid 3 3
quadratic equation. When big numbers are The distances from the root are 4m and 1m.
involved solving quadratic equation is difficult.
19. (2) Time of ascent
The graph between v and t is a straight line
u
 1s   1  u  10 ms 1
dv g
since a   g  constant
dt u2 102
height (h)    5m
The velocity when the body touches the ground 2 g 2 10
is v2max  202  2 10  50  vmax  10 14 1 2

gt
Total distance 2 gt
Total distance travelled = Area of the graph 20. (1) Avg.velocity=  
total time t 2
1

50  20  t   t  10 14  20
2
 Given that
gt g
  t  1s
2 2
10 Velocity after reaching ground =gt = g
t 
2  14
on rationalising we get t  14  2 21. (3) The motion of the body is shown below
17. (1) Time to reach the maximum height
u
t1 
g
If t 2 be the time taken to hit the ground

 
1
 H  ut2  gt22
 v  v 40 ˆj  30î
a 2 1   4 ˆj  3î
2 t 10s

But t 2 = nt1 (given) a  5m / s 2
22. (2) Given that v=10-5t 1 0

t
k 2t 2  
v     v0 + - v0 kt  dt 
2 t0  0  4  
The average velocity is v 
 (10  5t)dt
0
20 1 k 2t03 t 2
v  v0t0   v0 k 0 
t0  12 2
1 2 2
v    10 dt  5 t dt )   5m / s

2 0 0   k 2t 2 v kt 
v  v0  0  0 0 
23. (1) Average acceleration  12 2 
 t1+t2 
v0
  Function of a dt  v 
3
= 0
( t1+t2 )  0 26. (3) Given that the ratio of the times is

t1 t1 +t2 t1 : t 2 : t 3  1: 8 :1
 a1dt+
0

t1
a2 dt
a1t1+a2 t2 let t1=k, t2=8k, t3=k, a is the acceleration and
 a  =
t1+t2 t1+t2 retardation in first and third parts
1 ak 2
24. (4) The velocity of the body is v= u+at s1= at12 = ,v1=at1=ak
2 2
As the velocity function is given
s2  v1t2  8ak 2
The average velocity is
1
t
s3  v2 t3  at3 2 (v2  v1  ak )
 vdt 1t
2
v  0
  u+at  dt
t 0 t 0 s3 
ak 2
2
1 t t
 at s1+s2 +s3 9ak 2 9ak
   udt+a  tdt  =u+ v  = = =54 km/h
t 0 0  2 10k 10k 10
(Given that v1= ak= 60 km/h)
25. (1) Given that a = -k v
27. (3) Effective speed of the bullet
v t
dv dv
=-k v   = -k  dt = speed of bullet + speed of police jeep
dt v0 v 0
 180m / s  45km / h  180  12.5 m / s

k
2  v0 - v  =+kt  v= v0 - t
2  192.5m / s
The time at which the velocity becomes zero is Speed of thief’s jeep = 153km/h = 42.5m/s.
Velocity of bullet w.r.to thief’s car
2 v0
t0 =
k = 192.5-42.5 = 150m/s.
t0 28. (2) Distance b/w the cars A and B remains
 vdt constant. Let the distance ‘x’
The average speed v  0
Velocity of C w.r.t. A and B V = 45 + 36 =
t0
81km/h
k 2t 2 5
v = v0 +  v0 kt Distance  81  6.75Km.
4 60
1 1 Sr  H
29. (4) d rel  arel t 2  1.5  (10  2)t 2
2 2
1
Sr  ur t  ar t 2
1 1 2
1.5   12  t 2  t  sec
2 2 H  (u2  u1 )t  0
30. (3) In the time 10s the displacement of the B H
is 100m and its velocity is 20m/s. By the time A t 
u2  u1
starts its motion B is ahead of A by 100 m. the
time after which A and B meet is 33. (4)
1 Method – 1
vB t  vA t  2t 2  100m
2 When the two cars meet each other
vB  20m / s, vA  40m / s
S A  SB  75
2
t  20t  100  0
10t + 5t = 75
t  10( 2  1) t = 5 sec.
31. (3) We use the concept of relative motion Method-2 (Relative frame method)
Let v1 & v2 are velocities of the two trains We can solve with respect to any car, we do it
with respect to car B
We know that the displacement equation in
relative frame is  
v AB  v A  v B  10iˆ  5iˆ  15iˆ
1 Here car B is at rest and car A is moving with
Sr  ur t  ar t 2  ar  0 
2 15 m/s w.r.to B
S r  relative displacement, We can transfer the picture into relative frame
as shown in the figure
When moving in opposite direction
100   v1  v2  20 (i)
When moving in same direction
100   v1  v2  40 (ii) 1
sr  ur t  ar t 2 ( ar  0)
u r  relative velocity, ar  relative acceleration 2
By solving (i) & (ii) we get 75  15t  t  5sec.
v1  15 / 4 m / s v2  5 / 4 m / s d
34. (3) Time to meet the cars : t  v  v
1 2
Distance travelled by bird in this time

v3 d 10  2000
s  v3t    400 m
32. (1) v1  v2  20  30 
ur  u2  u1 1. (4) a 2 t 2
ar  ( g )  ( g )  0 For t  2
t 2  2t 10  t 
v  1  cos  m / s
a  2t 2  a  4t  2
dv = (4 - t)dt t
Velocity is maximum when cos  1
v  4t  t / 22
2
at t = 2, v = 6 m/s. i.e., when t = 2sec.
For t > 2
 Maximum speed  vmax at t  2sec
t 2 t 2
Method 2
a 2t 2t
v t
 dv   tdv
6 2
t t2
v  6  t 2 / 2 2  v  4
2
At t = 4, v = 12 m/s.
2. (293) Parachute bails out at height H from
ground. Graph of a - t is shown below
Velocity at point A t
Velocity   a dt = Area under a - t graph.
v  2 gh  980 m / s
0
The velocity at ground v1  3m / s Whose magnitude is maximum at t = 2s.

Acceleration  2m / s 2
 At t = 2 s, speed is maximum.
2. (52) Displacement of the body is same when
it is at that height in two cases
1
s  ut  at 2 (taking origin at the base)
2
1
u 2  v2 10.2  ut  (10)t 2
H  h  2
2 2
980  9 u  u 2  204
 t 
 242.75 10
4
 H  242.75  h
u  u 2  204 u  u 2  204
t1  , t2 
 242.75  50  293m 10 10
t  10  t2  t1  u 2  204  50
1. (2) Method1 u  52 m / s
t t
  
v   adt    5sin t  dt
0 0 2 
10  t  1. (2) In the graph 2, for one value of

  cos  1 m / s displacement, there are two timings. As a result
 2 
of it, for one time, the average velocity is positive station and the hotel = 10 km
and for other time is equivalent negative. Due
 Displacement of the taxi = 10 km
to it, the average velocity can vanish for two
time intervals. Distance travelled by the taxi = 23 km
2. (3) Speed of police van 7
Time taken by the taxi = 28 min  h
15
 5 
 v  = 30 km/h  253 m / s
p 1 km / s  m / s 
 18 
Total distance travelled
Avg speed 
Total time taken
160 23
Speed of thief’s car  v  = 192 km/h  m / s.  = 49.3 km/h
3  7 / 15
Muzzle speed of bullet  vB  =150 m/s Displacement
Average velocity 
Total time
The effective speed of the bullet
10
vB'  vB  v p  = 21.43 km/h
 7 / 15
25 475
 150   m/ s
3 3
Speed of the bullet with which it hits the thief’s 1. (3) Initial velocity of the car (u) = 126 km/h
car
5 5
 126  m / s = 35 m/s 1km / h  m / s 
 475 160  18  18 
   m / s  105m / s
 3 3 
Final velocity of the car (v) = 0
3. (2) Length of each train,
Distance travelled (s) = 200 m
A  B  400m
From equation of motion, v2  u 2  2as
Initial velocities of both trains,
2
v A  vB  42km / h = 20 m/s v 2  u 2 0   35  3.06m / s 2
or a  
2s 2  200
Distance travelled by train A in 50 s
 Retardation of motion,
S A  vA  t v  u  at
S A  20  50  1000 m v u  0  35
t  = 11.4 s
Distance travelled by train B in 50 s, a  49 / 16
 Car will stop after 11.4 s
1
S B  u B t  ag t 2
2 2. (1) Let jet airplane be moving upwards right
1 2
(+ve direction) with velocity v J and ejected gases
S B  20  50   1  50  be moving downwards (-ve direction) with
2
= 1000 + 1250 = 2250 m velocity v g while observer be at rest on the
Original distance between the two trains ground i.e., v0  0
 SB  S A  vJ  500km / h
= 2250 - 1000 = 1250 m v gJ  1500 km / h  v g  v J  1500 kmph
4. (4) Given, shortest distance between the  vg  1500  500  1000 kmh 1
Therefore, relative velocity of the ejected gases  6  3v 

with respect to the observer is 1000 km/h, in ln    3t
 6 
downward direction.
6  3v 3t
3. (3) Speed of car A  vA   36 km / h = 10 m/s e
6
v
1  e 3t  v  2 1  e3t 
2
1. (1) Displacement = distance (as the body is

Speed of car B and car C travelling in one direction though out time)
u A  uC  54km / h  15m / s
 r2  rv rt
Relative velocity of car B w.r.t., car A = Area  
2 2
uBA  uB  u A  15  10  5m / s   10  2
  10  m
Relative velocity of car C w.r.t., car A 2
uCA  uC  u A  15   10   25m / s Here rv  velocity radius, rt  times radius
Distance between car A and car B = 1km = 2. (1) From t = 0s to 6 s the body is travelling in
1000m one direction
Time taken by car C to travel distance AC = From t = 0s to 12s the body changes its direction
1000m at t = 6s
1000 1 1

25
 40s Distance    2  4  2  4   2  4 
2 2 
Let car B starts to accelerate with an
acceleration a, Using equation of motion. 1 1 
   2  4  2  4   2  4   32m
1 2 2 
s  ut  at 2
2
1 1 
1 Displacement    2  4  2  4   2  4 
s  uBAt  at 2 2 2 
2
1 1 1 
2    2  4  2  4   2  4  0
1000  5  40  a   40   2 2 
2
 a  1m / s 2 Displacement 0
Average velocity =  0
time 12s
Therefore, car B should accelerate with an
acceleration 1m / s 2 Distance 32m 8
Average speed=   m/ s
time 12s 3
dv
4. (1) Given ,  6  3v 3. (3) Here the velocity is (+)ve throughout the
dt
motion. So the lift will be going in upward
dv direction only. The area of the graph is
  dt
6  3v
1 1
Integrating it, we have hmax   2  3  8  3   2  3 =30 m
2 2
1 v
The area of the graph is equal to the maximum
ln[6  3v]0  t
3 height reached by the lift.
3
x2
16 t  x22  x12  16
4. (1) Change in velocity v   a dt  x dx  3 0 dt  2  3 t
x1
0
3
v  Area of the graph t  122  42   12sec
32
1 11. (1) After each bounce the velocity decreases
v   1 4  2  4
2 and hence displacement decreases gradually in
each step. Here the displacement is measured
v  0  10m / s (initial velocity is zero) from the point where it is released.
v  10m / s 12. (2) When the bat hits the ball first the ball
5. (1) The correct graph is (1). Other graph decelerates and then it accelerates. In both
shows more than one velocity of the particle at cases the bat exerts force in backward direction
single instant of time which is not pratically only so a=(+) ve throughout the motion. It
possible. increases from 0 and reaches maximum value
and then decreases to zero.
6. (1) The slope of the tangent drawn to the
curve is zero at five instants. 13. (3)
Method 1
7. (4) A lives closer to school than B since
xQ  x P 1
x  at 2 (0  t  t0 ) (i)
2
A starts from the school earlier than B
Equation (i) represents a parabola (concave
Slope of (x-t) graph for B is more than for A upward) with vertex at (0,0) .
so B walks faster than A For the deceleration motion,
A and B both reach home at same time dv
 a
B overtakes A on the road once dt
v t
x
8. (1) Instantaneous velocity v    dv  a  dt
t at0 t0
xA 4m  v  at0  a(t  t0 )

From graph, vA  t  8s  0.5 m/s
A
dx
Again,  v  at0  a(t  t0 )
xB 4m dt
and vB  t  8s  0.5 m/s
B x t t
  dx  at0  dt  a  (t  t0 )dt
i.e., v A  vB  0.5 m/s 1 2
2
at0 t0 t0
9. (3) Conceptual 1 1
 x  at02  at0 (t  t0 )  a(t  t0 )2 (ii)
10. (4) The equation of the straight line is y = 2 2
mx Equation (ii) represents a parabola (convex
 1 upward) with vertex at  2t0 , at02 
2 
1  2 x
   option (3) is correct
v  12  4 
  Method 2
16 The particle accelerates and decelerates at the
 vx 
3 same rate for the same interval of times starting
from rest and hence it’s velocity should be zero 17. (1) From t = 0s to 2s the velocity increases
at t  2t0 . That is the slope of the tangent is equal and from t = 2s velocity decreases and finally
to zero. becomes zero. Throughout the motion v is (+)ve
This is true for option (3) 18. (2) The relation between v and s is
14. (4) The equation of the straight line is v = ms
y  y1  m( x  x1 ) 10  0
m 1
10  0
4600  900
m v =s
0.6
on differentiating the equation w.r.t. time t
x1  0.6, x  s
dv ds
y1  900, y  v 2 
dt dt
3700 a=v
v2  900  ( s  0.6)
0.6
19. (2) The positions of the particles as a function
on differentiating the equation w.r.t. t we get of time t are
dv 3700 ds 1 2
2v  x1  u1t  gt
dt 0.6 dt 2
dv 3700
a   3084 1 2
dt 1.2 x2  u2 t  gt
2
1 1
15. (2) v  Area   4  4   2  4  4m / s x1  x2  (u2  u1 )t (i)
2 2
16. (1) We know that change in velocity of a As the first particle strikes the ground at 6s so
body v   adt = Area of the graph the eq-(i) is valid until 6s. After 6s the first
particle comes to rest and second particle will
If vi  v f  v  0  Area be in motion until 10s. So the graph is a straight
1 1 line from 0s to 6s and from 6s to 10s it is a
3  2+ 2  2   b  h  0 parabola.
2 2
From the triangle we can write 20. (2) Consider that two particles are moving
as shown in the diagram
2 h v1  v cos iˆ  v sin  ˆj
tan    1
2 b
v2  atiˆ
1
 8   b2  0
2 v12  v1  v2  (v cos   at )iˆ  v sin  ˆj
b=4&h=4
v12  (v cos  at )2  v2 sin2 
So the time t = 5 + 4 = 9 sec
7. (1) In graphs 2, 3 & 4 the time is going in

v12  v2  a2t 2  2(v cos )at backward direction which is not possible so the
possible graph is 1.
d v12 2a 2t  2v(cos  )a
 8. (4) Slope of both lines are positive and equal,
dt 2 v 2  at 2  2at (v cos  )
So v A  v B
2a 2 t  2v(cos  )a  0
Displacement
9. (1) Velocity =
v cos Time
t
a 40  60
  2m / s
10
10. (3) The slope of the graph is negative at this
1. (2) As velocity is positive the body is moving point.
in only one direction throughout the time. In this
11. (2) The velocity of the body as a function of
case displacement and distance are same.
time is v = u-gt. The graph is a straight line
Distance = Displacement = Area of the graph having (-)ve slope. The correct option is (2).
1  1 
12. (2) Throughout the motion the body is under
   2  4   (4  4)    2  4  acceleration but with decreasing a value. The
2  2 
velocity reaches maximum at the end
= 24 m
1
vmax  v  Area   10  11  55m / s
2. (3) The given graph is between s and t 2
Total distance =20+20+20+20 = 80m 13. (1) When the body is at a height equal to d
the velocity is zero and the magnitude of velocity
Total displacement = 0 (Finally the body position
increases non linearly and reaches maximum
is zero at t = 16 s)
when it is about to hit the ground and after the
80m collision it re-bounces with velocity in opposite
Average speed   5ms 1
16s direction and decreases slowly. The sign of
Average velocity = 0 velocity changes after collisions and its
magnitude decreases.
3. (1)
14. (1) Velocity at any time t is
Displacement  (4  2  2  2  2  2)SI units
given by
Distance  (4  2  2  2  2  2)SI units
v = u +at
4. (2) Height = Area under velocity time graph
v  v0  gt
1 1
=   30  0    3  0   10  4  3  straight line with negative slope. During its
2 2
return path the velocity should have opposite
1 1 sign. The correct graph is (1).
=  30  3   10 1  45  5  40 m .
2 2
15. (2) Downward motion
5. (3) Speed can never be negative, so graph v = -gt
(3) is not possible
The velocity of the rubber ball increases in
6. (1) The slope of the graph changes, downward direction and we get a straight line
 dx  between v and t with a negative slope.
decreases continuously   v  and finally it
 dt 
becomes zero. 21
We get y  h  gt
2
The graph between y and t is a parabola with y 10s to 20s since the area of triangles are same.
= h at t= 0. As time increases y decreases. From 0s to 10s the particle is in acceleration
and 10s and 20s the particle is in deceleration
Upward motion
but it is moving in same direction so
The ball suffers elastic collision with the
displacement is non zero.
horizontal elastic plate therefore the direction
of velocity is reversed and the magnitude 21. (2) Displacement during acceleration and
remains the same. Here retardation is from t=20s to t=40s
v = u - gt, where u is the velocity just after 1
collision. Displacement = Area   20  3  20 1  50m
2
As t increases, v decreases. We get a straight
22. (2) The slope of the tangent (speed) to the
line between v and t with negative slope.
curve is constant from A to B and from C to D
1 2
23. (3) Acceleration is constant for sometime so
Also y  ut  gt
2 the velocity increases and then velocity remains
16. (3) For the graphs (i) and (iv), slope is constant as a = 0. Again a is (+)ve and
constant, hence the velocity is constant and
constant so velocity increases further.
hence acceleration is zero.
24. (2) Equation of line is
For the graph (iii), the particle’s velocity first
decreases and then increases. It means negative y  y1  m( x  x1 )
acceleration is involved in this motion. For graph
(ii) velocity increases and hence it is an v0
x1  x0 , y1  0, m 
accelerated motion. x0
17. (4) As the car starts from rest the velocity is x  x, y  v
given by
v0
v ( x  x0 )
1 x0
v  at  x  at 2
2
dv v v  v 
The graph between x & t is parabolic and its  a  0 (v)  0  0 ( x  x0 ) 
dt x0 x0  x0 
slope (v) should increase. When v =constant
(a=0) then x & t graph is a straight line with v02
constant slope. So the correct graph is (4). a  x  x0 
x02
dv dv The graph between a and x is a straight line
18. (2) Acceleration a  v
dt dx having + ve slope and negative Y-intercept.
From the graph, when 25. (2)
dv  
v  4m / s,  tan 60  3 v A  10iˆ, v B  20cos60iˆ  20sin 60 ˆj  10iˆ  10 3 ˆj
dx   
v AB  v A  v B  10iˆ  (10iˆ  10 3 ˆj)  10 3 ˆj
 a  4 3m / s 2  
 v BA  v AB  10 3 ˆj
19. (3) v 2  02  2as
1
26. (2) x1  at 2 and x2  ut
v 2  2as 2
1
20. (3) As the graph is symmetric about t-axis  x1  x2  at 2  ut
average value of a from 0s to 20s is zero. The 2
average speed is same from 0s to 10s and from 1
y  at 2  ut. ( y  x1  x2 )
2
This equation is a parabola.

dy d2 y
 at  u and 2  a 1. (3)
dt dt
d2y  50  10  x  10  x ms 1
As  0 i.e., graph should have a minima at Velocity v  10 
dt 2 200 5
u for 0  x  200
t .
a
So the correct option is (2) and v  50 ms 1 for 200  x  400.
1 dv dv
27. (2) y1  10t  gt 2 As acceleration a  v ,
2 dt dx
1
y2  40  gt 2
2 1 x  x 
for 0  x  200, a  10     2   ms 2
5 5  25 
y1  y2  30t (straight line)
but stone with 10 m/s speed will fall first and and for 200  x  400, a  0 ms 2 .
the other stone is still in air. Therefore path will
be straight line first and it becomes parabolic till 2. (3) a=-C
other stone reaches the ground. vdv
 C
dx
vdv   Cdx
1. (1) Distance travelled by car in
1 675 v2
15sec  (45)(15)  m , Distance travelled  Cx  K
2 2 2
by scooter in 15seconds  30 15  450 v2 K
x 
Let car catches scooter in time t; 2C C
3. (2)
675
 45(t  15)  30t
2
337.5  45t  675  30t  15t  337.5
 t  22.5sec
2. (4) 2
 1 2 2
 1   u  u  2 as
 n
  n  1 2   2n  1 
2as  u 2 1    u2  2 
  n    n 
 
Let n be the number of planks required to stop

the bullet.
aCB  2m / sec 2 0  u 2  2ans

u2 u2 n2
1 n  
200   2t 2 2as  2n  1  2n  1
2 u2  2 
 n 
t  10 2 s
Motion in Two Dimensions & Circular Motion 45
vb  velocity of bird
vt  velocity of train
vbt  velocity of bird w.r.to train

1. (3) 
vt  40iˆ
2. (4) The velocity of the particle, V  kyiˆ  kxjˆ   
vbt  vb  vt  40iˆ  40 ˆj
dx dy
  ky.  kx 
dt dt vbt  40 2 NE direction
dy dy dt kx
    5. (3) When a force of constant magnitude is
dx dt dx ky
perpendicular to the velocity of particle acts on,
ydy  xdx  y 2  x 2  c work done is zero and hence change in kinetic
energy is zero.
This is the required general equation.
6. (3) When the two particles collide then their
3. (1) xy position vectors are equal.
ax  2m / s 2 , ay  0    
r1  v1t  r 2  v 2t
u x  8m / s 2 , u y  15m / s
 3iˆ  5 ˆj    4iˆ  3 ˆj  2
vx  8  2t , v y   15
   
 5iˆ  3 ˆj   iˆ  7 ˆj 2 
v  vx iˆ  v y ˆj   8  2t  iˆ  15 ˆj
4. (2) Consider the traditional coordinate system  8

as shown
7. (2) Let  is the angle made by the velocity
of man w.r.to y - axis.

Given vb  40 ˆj The time for crossing is
46 Motion in Two Dimensions & Circular Motion
d 11. (2) Let v be the speed of boatman in still

t water.
v2  u 2
60
 5  v  13 m/s
v  52
2
8. (1) From the standard equation


vb   vr  vbr sin   iˆ  vbr cos ˆj

 vb   4  5sin   iˆ  5cos ˆj
Resultant of v and u should be along AB.
For shortest distance 4  5sin  0 Components of vb (absolute velocity of boatman)
4 along x and y directions are,
 sin  
5 vx  u  v sin 
 is the angle with respect to y – axis
and v y  v cos 
the angle with respect to river flow (positive x –
 vy
4 Further, tan 450 
axis) is  sin 1  
2 5 vx
9. (1) Let vr is the velocity of river flow.. v cos

or 1 
u  v sin 
u u
v 
sin   cos  2 sin (  45)
v is minimum at,   45  90 or   45
u
2
and vmin 
3   5  vr2  vr  2m / s 2
10. (1) Assume that river flow is along +x-axis 12. (2) Horizontal component of rain’s velocity
and the swimmer swims along +y-axis w.r.to will be equal to velocity of wind which is 2 m/s
water in north direction. If cyclist goes towards north
with velocity 2 m/s, then w.r.t ‘him’ rain’s
Given that horizontal component of velocity will be zero,
 and he will see only vertical component.
v r  6iˆ
 
 13. (2) Given vrG  v r  30 ˆj
v mr  9 ˆj ( perpendicular to river flow)

   v w  10iˆ
v mr  v m  v r
  

v m  v r  v mr 

v m  6iˆ  9 ˆj
vm  117km / h.
  
v rw  v r  v w  30 ˆj  10iˆ h g h2  d 2
u  
h2  d 2 2 u2
 v
vrw  10iˆ  30 ˆj , tan   b
vr g  h2  d 2 
 ys  h 
2u 2
10 1
tan   
30 3 g  h2  d 2 
1 2
and y f  h  gt  h 
10 1 2 2u 2
tan      tan 1 w.r.to. vertical direction
30 3
and ys  y f the stone always hits the fruit
  
14. (1) vr  vb  vrb
17. (3)

vb v 18. (2) The rate of change of velocity is equal to
From fig tan   
vr u acceleration due to gravity (g).
2v0
v 19. (2) T
 tan   g
u
v  2v0u0
1
  tan  
R  u0T 
g
u 
  20. (1) Time to reach max. height  tm
15. (3) vw  20iˆ, vc  20 ˆj
Time to reach back to ground  tm
Here we have to look for velocity of wind w.r.t
car. Total time of flight T f  tm  tm
  
So vw / C  vw  vc  20 iˆ  20 ˆj T f  2t m
16. (4) Let after time t, ys and yf be respective 21. (3) The initial velocity vector is
heights of stone and fruit. 
u  u cos iˆ  u sin  ˆj
u cos  t  d
The velocity vector after time t is

d h d2 2
v  u cos iˆ   u sin   gt  ˆj
t  
d u  
u
h  d2 When v is perpendicular to u

v.u  0
u 2 cos 2  u 2 sin 2   u sin  gt  0
u
t
g sin 
2u2
22. (4) Given H  2 g  u  40  2  10
1
 ys  u sin   t  gt 2
2 u  800  20 2
u 2 sin 2 u2 100 10 1

R R sin 2  sin 2  2

g g 10  1000
3
u 2 800
Rmax    80m 1
g 10    is very small 
2000
23. (3) For angle h 100
   h  m  5cm.
R 2000
u 2 sin (90  2 ) u 2 cos 2
(45   ), R   29. (2) Let the speed of the projectile
g g
be 125ms 1 at time t. Then
For angle
u 2 sin (90  2 ) u 2 cos 2 v 2  vx2  vy2
(45   ), R  
g g
Here, vx  u x  u cos 45
24. (4)
1
25. (2) As the two velocity vectors become  10 2   10ms 1 (1)
2
perpendicular the horizontal components are
equal at both positions. and vy  u y  gt
v sin   u cos vy  u sin 45  10t
1
 10 2   10t  10  10t (2)
2
Squaring and adding (1) and (2), we get
2 2
vx2  v y2  10   10  10t 
2
 v  u cot   125  2
 10   10  10t 
2
26. (2) At the maximum height potential energy

is maximum and hence its kinetic energy is or 10  10t 
2
 125  100  25
minimum at which the body covers half of the
range along x-axis  10  10t    5
27. (2) Let the angle be  , then at any time ‘t’
10  5 10  5
in x - direction, vx  u cos , in y - direction, Hence, t1  and t2 
10 10
v y  u sin   gt
 10  5   10  5 
 t2  t1      1s
v x  v y ,  u sin   gt  u cos   10   10 
30. (4)
u
 u sin   cos   gt  t  sin   cos 
g 2h 2  2h 
31. (4) xv , 2x  v '
28. (2) Let h be the height of the target and the g g
projectile is aimed at the top of the the height h From the above two eq’s
that makes an angle  with the horizontal
direction. v '  2v
32. (1) The situation is shown in the adjoining  90 

 2(r )sin  
figure.  2 
2 1
 2  1 
T 2
4  2 cm
  [As T  60 s ]
60 2 30 s
u sin   9.8sin 30o  4.9 m s 39. (4) Electrostatic force provides necessary
centripetal force for circular motion of electron.
From the relation, h  u sin  t  1 2  gt 2 d 2
40. (4) Angular acceleration   22
dt 2
We get, 29.4  4.9t  1 2   9.8t 2  t  2sec
4 v2
33. (4) Horizontal velocity of ball and person are 41. (1) Given that a  
r2 r
same so both will cover equal horizontal distance
in a given interval of time and after following 2
v
the parabolic path the ball falls exactly in the r
hand which threw it up. 2m
34. (4) Vertical component of velocity is Momentum of the particle is P  mv 
r
unchanged relative to trolley and ground.
42. (4) As time periods are equal therefore ratio
2v sin   2 
Now time of flight  of angular speeds will be 1 : 1.    .
g  T 
Time of flight remains unchanged relative to 43. (3) The earth rotates west to east. So the
trolley & ground. Hence right choice is (4). velocity of train 2 increases. ac  v 2 /R . Hence
cenpripetal acceleration of train 2 is more.
35. (3) Displacement, velocity and acceleration
change continuously with respect to time
because of change in direction.
1. (2)
36. (2)
2. (4) From figure,
37. (1) As body covers equal angle in equal time  
intervals its angular velocity and hence OA  0iˆ  30 ˆj, AB  20iˆ  0 ˆj
magnitude of linear velocity is constant. 
BC  30 2 cos15 iˆ  30 2 sin 45 J
38. (4) In 15 seconds hand rotate through 90
 30iˆ  30 ˆj
 Net diaplacement
 Change in velocity,,
 
v  2v sin       
2 OC  OA  AB  BC  10iˆ  0 ˆj OC  10m.
3. (3) For particle A

vAB  10 2
y=x
The angle made by velocity of car A with respect
dy dx to positive x-axis as seen by car B is
  vAy  vAx
dt dt
10
tan    1    315
& vB  3iˆ 10
6. (1) Consider the traditional coordinate system
& vxA  vB
as shown
 vxA  3iˆ  v y A  3 ˆj
 
 v A  3iˆ  3 ˆj  v A  3 2m / s.
4. (4) The displacement of the person is shown

in the figure

Given v b  20 ˆj
 vb  velocity of bird
S1  5î
 vt  velocity of train
S2  10iˆ  10jˆ
   vbt  velocity of bird w.r.to train
S  S1  S2  S  325
 
5. (2) In this problem we v AB write and are v t  20iˆ
represented them on a coordinate system   
v bt  v b  v t  20iˆ  20 ˆj

v bt  20 2 NE direction
  
7. (3) v B  v BA  v A
  
 5iˆ  12 ˆj  3iˆ  2 ˆj 

v B  8iˆ  10 ˆj
Total displacement
  8. (1) Average velocity =
v A  10iˆ , v B  10 ˆj time
   2m
v AB  v A  v B  10iˆ  10 ˆj   2ms 1
1s
9. (2) Given, x  ct 2 14. (3) Given AB = velocity of boat = 8 km/h
dx AC = Resultant velocity of boat = 10 km/h

 vx   2ct
dt
Also, y  bt 2
dy
 vy   2bt
dt
 v  vx2  vy2  2t c 2  b 2 2 2
 BC = Velocity of river  AC  AB
10. (4) Position vector of the body, r  3iˆ  7 ˆj
 (10)2  (8)2  6km/h
 Acceleration of the body
15. (1)
a  4iˆ and t  3s Here,
1 2 1 x x 2x x x 2 xv
Using x  u0t  at  0  3   4  3
2
t1    , t2    2
2 2 v v v v   v   v  2
2 xv 2x
 x  18 t2  2

or     2 
New Coordinates of the body are (21, 7) v 2 1  2  v 1  2 
  v   v 
d t1
11. (2) t t2 
2
v u 2
or 2 or t2  t1
1
v2
16. (1) Horizontal component of rain’s velocity
will be equal to velocity of wind which is 4 m/s
in north direction. If cyclist goes towards north
with velocity 4 m/s, then w.r.t ‘him’ rain’s
horizontal component of velocity will be zero,
15 1
   u  3 km/h and he will see only vertical component.
60 5  u2
2
17. (2) Here, vR  25ms 1 , vW  10 ms 1

12. (4) Here v  0.5 m sec.u  ?
 Velocity of rain w.r.t
vr v 1 woman: vRW  vR  vW
So sin    r  or u  0.25ms 1
v 0.5 2
Let vRW makes an angle  with vertical then
vW 10
tan     0.4
vR 25
13. (2) Let vr is the velocity of river flow..
2
5  4  vr2  vr  3m / s
She should hold her umbrella at an angle of 26. (3)

  tan 1 (0.4) with vertical towards south. 
27. (3) ui  u cos iˆ  u sin  ˆj

u f  u cos iˆ
 
u  u f  u sin  ˆj
18. (3)

u  u sin 
The velocity of the rain and the wind are
  28. (3) For   45
represented by the vectors Vr and Vw as shown
in the figure. To protect himself from the rain the u 2 sin 2 45 u 2  1 
boy should hold his umbrella in the direction of H max 
2g

4g  sin 45  
 2

resultant velocity vR makes with the vertical, then
V 12 u 2 sin 90 u 2 R u2 4g
tan   w   12  R  ;    2  4  R  4H
or   tan 1   g g H g u
Vr 35  35 
19. (3) The horizontal component of rain should 29. (2) As

have same direction and magnitude as the
u 2 sin 2  H sin 2 1 sin 2 30 1 / 4 1
velocity of man. H  1    
2g H 2 sin  2 sin 2 60 3 / 4 3
20. (1)
30. (2) Horizontal velocity remains same
30 cos 60  v cos 45
 v  15 2m / s
31. (4) Given that R  4 3H
u 2 sin 2 u 2 sin 2 
4 3
Flag will flutter in the direction of wind w.r.t g 2g
steamer:
   cot   3

vWS  vW  v S  18iˆ  18 ˆj  8iˆ    32. (1) v '  v0 cos

vWS  18 ˆj
v0 1
21. (3) 22. (3) 23. (3) 24. (3)  v0 cos   cos      60
2 2
25. (3) Let the intial and final velocities of the
projectile are 33. (4) The velocities are v1  vcos î  vsin ˆj
V i  u cos i  u sin  j and v2  vcos î  vsin ˆj
V f  u cos i  u sin  j
  v  v2  v1  2vsin ˆj
u - intial speed
u 2 sin 2  2u sin 
Given that V i  V f  0    45 0 34. (2) H and T
2g g
4u 2 sin 2  1 2
or T2  38. (1) y1  u1 sin 1t  gt and x1  u1 cos 1 .t
g2 2
T 2  4u 2 sin 2   2g 8 1
  2 2  y2  u2 sin  2 t  gt 2 and x2  u2 cos  2 .t
H  g  u sin  g 2
8H 2H y2  y1 u2 sin  2  u1 sin 1
or T2  T  2    constant
g g x2  x1 u2 cos  2  u1 cos 1

35. (3) u i  u cos θiˆ  u sin θjˆ 39. (3) Since, vertical displacement is same, as
 well as initial velocity in vertically downward
u f  u cosθiˆ direction is zero for both the bodies. Both the
   bodies reach the ground simultaneously.
u  u f  u i  u sin θjˆ
vx 2 gH
 40. (4) tan     1    45
u  u sin θ vy 2 gH
36. (4) For projectile A 41. (2)

42. (4) The horizontal velocity of the stone will
u A2 sin 2 45
Maxiumum height, H A  2g
be the same as that of the train. Vertical motion
is accelerated motion. Thus, the resultant motion
For Projectile B will be along a parabolic path.
uB2 sin 2  43. (4) 44. (2)
Maxiumum height, H B  2g 45. (1) Because velocity is always tangential and
centripetal acceleration is radial.
Given that H A  H B
46. (1) When speed is constant in circular motion,
2 2 
u A sin 45 uB sin  2 2 it means work done by centripetal force is zero.

2g 2g mv 2
47. (2) F . For same mass and same speed
r
sin 2  u A2
 if radius is doubled then force should be halved.
sin 2 45 uB2
48. (4)
2
u  49. (3)  W  FS cos    90
sin 2    A  sin 2 45
 uB  50. (2) Due to centrifugal force.
2 2
 1   1  1 v2
sin 2        51. (4) Centripetal acceleration, ac 
 2  2 4 R
Where v is the speed of an object and R is the
1 1 radius of the circle.
sin      sin 1    30o
2 2
It is always directed towards the centre of the
37. (1) circle. Since v and R are constants for a given
2u sin  2u sin  uniform circular motion, therefore the magnitude
T 2  u sin   g
g g of centripetal acceleration is also constant.
However, the direction of centripetal
u 2 sin 2  g 2 acceleration changes continuously. Therefore,
H   5m
2g 2g a centripetal acceleration is not a constant vector.
52. (4) For a particle performing uniform circular 58. (2) In time t, particle has rotated an
motion, magnitude of the acceleration remains angle    t
constant.
The net displacement is
2
d
53. (3) Angular acceleration   21
dt 2
54. (3) Since, n  2,   2  2  4 rad/s 2
2 2 25
So acceleration   r  (4 )  m/s 2  4 2
100
2
s  y2   a  x 
mv 2
55. (1) F . If m and n are constants then
r y  a sin  a sin  t
1 x  a cos   a cos  t
F
r 2 2
s  a sin t    a  a cos t 
F1  r2 
   t
F2  r1   s  2a sin
2
56. (4)
1. (2) y  ax2
dy dx
  (2ax)  2acx
dt dt
As momentum is vector quantity
 Change in momentum  dx 
  c 
 dt 
P  2mv sin( / 2)
 2mv sin(90)  2mv d2y
Further a   2acx  2ac2
dt 2
But kinetic energy remains always constant so
 
change in kinetic energy is zero. 2. (2) v B  2v A (i)
57. (4) In a circular motion   
vCA  xiˆ  vC  v A  xiˆ (ii)
2   
Centripetal acceleration
v
A vCB  xjˆ  vC  v B  xjˆ (iii)
r
From Eq’s (i), (ii) & (iii)
Velocity is doubled acceleration becomes
we get vc  2 x iˆ  xjˆ
4.v 2
 4A
r x 1
tan   
Ratio of after and before change 2x 2
4A 4 1
    tan 1   south of East
A 1 2
 
 4 :1
3. (2) x  kt  
 4
v y  5 
y  kt  k t 2  4 2  42  (2) 2 
 2 

By eliminating t vz  5 
 42  42  (2)2 
x x2
yk  k 2 10 10 5
k k vx  vy  vz 
3 3 3

y x x2 dT 10 10   5 
k  2(1)   2(1)    2(2)  
dt 3  3   3 
4. (1) As the motion is in plane we first find the
speed of the body and then we find its distance. dT
 20C / sec
x  a sin(ωt ) dt
y  a(1 cosωt ) 6. (1) For particle P, motion between A and C

be an accelerated one while between C and B
v x  aωcosωt a retarded one. But in any case horizontal
component of it’s velocity will be greater than
v y  aωsin ωt
or equal to v on the other hand in case of particle
Q, it is always equal to v. horizontal displacement
v 2  vx2  v y 2  a 2ω2
of both particles are equal,so tP  tQ
v  a ω (The speed is constant) 
7. (4) VMW  Velocity of swimmer w.r.t. water
ds
v   aω 
dt VW  Velocity of flow
t
 ds   aωdt
0
s  aωt
5. (3) Given that T  x 2  y 2  z 2
dT dx dy dz
 2x  2 y  2z
dt dt dx dt 10
sin    30
(x, y, z)=(1, 1, 2) 20
dx dy dz  Angle with flow  90  30  120

, , are the component of velocity of the
dt dx dt 8. (1) Let v be the river velocity and u the
bird along x, y and z axis velocity of swimmer in still water. Then
Given that the bird is flying with speed 5m/s
 W 
along the vector 4iˆ  4 ˆj  2kˆ t1  2  
2 2
 u v 
v x  5 cos α; v y  5 cos β; v z  5 cos γ W W 2uW 2W
t2    2 2 and t3 
 u v u v u v u
 4 
vx  5 
 42  42  (2) 2  Now we can see that t12  t2 t3
9. (4) The representation of the motion of the 

vrM   3iˆ  vry ˆj
person is shown in the figure. The person wants
to reach point B from point A in minimum By representing case (ii) geometrically
amount of time.
3
 tan 45 
vry
 vry  3

 vr  3iˆ  3 ˆj
vr  3 2 Km / h
11. (1) Suppose velocity of rain

The person runs from A  C and then he swims v R  vx iˆ  v y ˆj
from C  B . The person takes a total time of and the velocity of the man
2
d  ( w  x) 2 
x v m  uiˆ
t 
v1 v2
 Velocity of rain relative to man
When ‘t’ becomes maximum or minimum.   
v Rm  v R  v m   vx  u  iˆ  v y ˆj
dt 1 2(w  x)(1)
  0 According to given condition the rain appears
dx v1 22 d 2  (w  x)2
to fall vertically, So  vx  u  must be zero.
v2 ( w  x)
  vx  u  0 or vx  u
v1 d  (w  x)2
2
v2
 cos
v1
10. (1) By taking the traditional coordinate system

let the velocity of rain w.r.t ground is

vr  vrx iˆ  vry ˆj
When he doubles his speed,
vry   vry 
v 'm  2ujˆ
(Rain has vertically downward component)   
Now v Rm  v R  v 'm
Case – (i)
    
 vx iˆ  v y ˆj  2ujˆ
vM  3iˆ
  vx  2u  iˆ  v y ˆj  uiˆ  v y ˆj
 vrx  3km / h

 The v Rm makes an angle  with the vertical
v r  3iˆ  vry ˆj
Case – (ii)

vM  6iˆ
u u 2 2
tan    vy  u cos   vy  u 2 cos2  (i)
vy tan  5
Thus the velocity of rain
H 
 v y2  u 2 sin 2   2 g   (ii)
u ˆ  2 
v R  v x iˆ  v y ˆj  ui  j.
tan 
H 
12. (4) Given that v is the velocity of plane with 02  v y2  2 g   (iii)
respect to wind.  2 
Consider the motion geometrically. The triangle From eq (i) and (ii) and (iii) we get   60
rule is given in the figure.
 14. (3)
v PW - Velocity of plane with respect to wind
u 2 sin 2 45
H 2g tan 45 1
tan    2  
R 2u sin 45 cos 45 2 2
 2
vW - Velocity of wind w.r.t ground 2g

v P - Velocity of plane w.r.t ground 1
   tan 1  
v 2
pw
2
v v
p
2
w
2
 15. (4)
v P  v2  u 2
Let u and  be the velocity of projection and,

Time for going from A to B is t1  angle of projection respectively
v2  u 2
The returning motion from B to A 1
Then, h  (u sin )t  gt
2
 2
v P  v2  u 2 (similarly)
 2u sin   2h
 t 2   t  0
  g  g
Time for going from B to A is t2 
v  u2
2
  2 gh  
2
2 u sin    
 t1  1  1   
Total time  g   u sin   
v2  u 2  
13. (2) Given that
  2 gh  
2
u cos  
2 2
v y  vx2 u sin    
5 And  t2  g 1 1   u sin   
   

vx  u cos 
Now v  eˆ  0
2u sin  2 gh
 t '  t1  t2  1 2 2
g u sin   {u cosα  sinβ  (u sinα - gt )cosβ}kˆ =0
( t ' - is the time elapsed between A and B) u sin(α  β)
t=
g cosβ
u 2 sin 2 
But, H  2g
 u sin   2 gH
Method-2
(From the frame of inclined plane)
8H h 8( H  h)
t  1  Component of velocity normal to the inclined
g H g
plane is zero and acceleration perpendicular to
Method-2 inclined plane is g cosβ
Let T be the time of flight
 u sin(α -β)  ( g cosβ )t  0
2u sin 
T 
g u sin(α  β)
t=
g cosβ
4u 2 sin 2  8  u 2 sin 2   8 H
T2    
g  2 g 
17. (3) Time of flight  3  7  10sec
g2 g
90  u cos  4 (i)
Hence we can write in similar way. For the time
elapsed between A and B to travel a height (H- R  u cos  10 (ii)
h)
8( H  h) 90 4
2
t1  
g R 10
 R  225m
8( H  h)
Thus, t  18. (1) From the diagram y=x tan α
g
16. (3)
Method-1
(From ground)
y=(R - x)tan β
y y yR
tan α  tanβ   
x R  x x( R  x )
Unit vector along the inclined plane
eˆ  (cosβ iˆ  sinβ ˆj )  x
 y  x tan  1  
 R 
Velocity vector after time t
 we know the equation of projectile
v  u cosα iˆ  (u sinα - gt ) ˆj
yR dh d  gR 2 
tan    R tan   2 sec2   0
x ( x  R) d 
d  2vo 

 tan   tanα  tanβ gR tan 
0  1  R  Vo2 / g tan 
19. (1) Method - I vo2
Take the origin at the point of projection and By putting R values in equation –(2) we get
downward direction as negative.
vo
sin  
When the body reaches the ground 2v  gh
2
o
R  vo cos t (i)
20. (1) Let u be the velocity when projected with
1 angle 2θ , then equating the horizontal velocities
h  vo sin t  gt 2 (ii)
2 in both the cases, we get
By eliminating t we get
u cos 2θ
 gR 2  v cosθ  u cos 2θ  v 
h  R tan   2 sec2  cosθ
 2v  o
1
Now we should solve this for R we get a very Where, secθ 
cosθ
complicated expression
 v  u cos2θsecθ
R tan  2vo2 2vo2 h
R 
2
  0 Using cos 2θ  2cos 2 θ  1 ,
g sec  g sec  g sec2 
2 2
and, u  4ms 1
Rv 2 sin 2 2vo2 h cos 2 
R2  o  0
g g v  4  2cos2 θ  1 secθ  v  4  2cosθ  secθ 
vo2 sin 2 sin 2 (2)vo4 8vo2 h cos 2  21. (4) Given that y  cx  ax 2
 
g g2 g
R The angle of projection is
2
dy
tan    x  0
dR dx
By doing  0 (From maxima and minima)
d dy
  c  2ax  c
dx
vo
sin  
We get 2v  gh
2
o
   tan 1  c 
Method - II 22. (1) Area in which bullet will spread   r 2

In this method we can avoid solving quadratic
v2
dR dR dh For maximum area, r  Rmax 
equation  this expression is zero if g
d  dh d 
[When θ = 45ο ]
dh
0
d 2
2  v2   v4
so from equation – (3) Maximum area  R max     2
g g
23. (3) If the ball hits the nth step, then horizontal Horizontal component of acceleration of the
distance traversed  nh . projectile w.r.to platform = -a
Here, velocity along horizontal direction  u . 1
 (u cos )T  aT 2  S
Velocity along vertical direction  0 2
(S –horizontal displacement of the particle w.r.to
 nb  ut
platform)
1
nh  0  gt 2 When the particle reaches the same point the
2
2
relative displacement is S=0
nb 1  nb 
From (i), t  ,  nh  g    g
v 2  u     tan 1  
 a 
2hu 2
n 26. (2) Ball will strike the point D if velocity of
gb 2
particle with respect to platform is along AD or
25 component of its relative velocity along AB is
24. (1) vc  45 km/h  m/s
2 zero.
For the resultant motion to be upwards. 
v b  u cosî  u sinˆj
v cos  vc  0 
v Plate  viˆ
vc 25 / 2 1 
cos         120 v Plate   u cos  v  î+usin ˆj
v 25 2
25. (1) Method-1 So, u cos   v
2u sin  v
T=time of flight  g (w.r.to ground or   cos 1  
u
frame) vy gt 10  10 1
27. (3) tan     
Distance transferred by the platform in time T vx vx 500 5
= Horizontal distance covered by the particle in
1 1
time T i.e.,   tan with horizontal
5
If u be the velocity of projection with respect to
28. (2) Horizontal component of velocity,
the platform and v0 be the velocity of the
platform at the instant of projection, u
uH  u cos 60o 
2
1
v0T  aT 2  (v0  u cos  ) T
2 ut
AC  uH  t 
1 2u cos  2
 a  u cos 
2 g
and AB  AC sec30
g
   tan 1  
 a   ut   2  ut
   
Method-2  2  3  3
(From the frame of platform) 29. (3)
2u sin  Method-1
T
g (From ground)
Body – 1
x y
x1o  0 y1o  0
u1x  40 u1y  40
a1x  0 a1y   g
1
r1x  40t r1y  40t  gt 2
2
 1 
Unit vector along the inclined plane r 1  r i  r j
1 1 r 1  40ti   40t  gt 2  j
x y  2 
eˆ  (cosβ iˆ  sinβ ˆj ) Body – 2
x y
Velocity vector after time t
 x2o  100 y2o  0
v  u cosα iˆ  (u sinα - gt ) ˆj
u2x  0 u2 y  30
Now v  eˆ  0
a2x  0 a2 y   g
 {u cosα  sinβ  (u sinα - gt )cosβ}kˆ =0
1
r2 x  100 r2 y  30t  gt 2
u sin(α  β) 2
t=
g cosβ r 2  r2x i  r2 y j
Method-2
 1 
(From the frame of inclined plane) r 2  100i   30t  gt 2  j
 2 
Component of velocity normal to the inclined
r12  r1  r 2
plane is zero and acceleration perpendicular to
inclined plane is g cosβ r12   40t  100  i  10t j
2 2
 u sin(α -β)  ( g cosβ )t  0 r12  distance   40t  100  10t 
u sin(α  β) (Don’t expand the square terms)
t= d 2 d 2
g cosβ d r12  40t 100  10t 
 dt dt
30. (2) Method-I dt 2
2  40t 100  10t 
2
(Position vector method) 2 40t 100 40  210t 10

 0
2 2
Take the common origin at the first body and 2  40t 100  10t 
consider the traditional coordinate system
  40t  100 40  10 10t   0
Let r 1 and r 2 are the position vectors of the
40
two bodies. 160t  400  10t  0  t  sec
17
T  2.4sec
 Minimum distance is possible between them
at
t = 2.4 sec
The minimum distance is given by
2 2
By observing the above three diagrams we can
r12 min
  40  2.4   100   10  2.4   24.3m represent the resultant motion in the figure shown
below.
If the bodies really meet then we get r12 min
0
Method-II:
Relative Frame Method
From the given information we can write the
velocities of the two bodies
v1  v1 x i  v1 y j v 2  v2 x i  v2 y j
All the three particles meet at the centroid
Let us draw the motion of bodies w.r.t body- 2 From the symmetry of the diagram at any given
instant the particles lie on an equilatral triangle
To get the shortest distance we should draw a
whose side length is decreasing from L.
perpendicular line from the particle-2 to the
relative path followed by particle-1. The To find the time of meeting we can consider
perpendicular distance is the shortest distance any two particles let A and B be the particles.
possible between the two particles. The relative velocity between the two particles
along the line that joins the two particles is given
by
Vr  VAr  VBr  u  (u cos60 )
3u
Vr 
2
VAr & VBr are the components of velocities of A
v12  v1  v 2  40i  10 j
and B along the line that joins A & B.
v 22  0
r12 min
sin    r12 min
 100sin 
100
10m / s 1 1
tan    tan    sin  
40m / s 4 17
100
r12 min
  24.3m
17 Sr S L 2L
Vr  t r  
31. (4) The following successive diagrams t Vr 3u 3u
represent the motions of the three particles after 2
small intervals of times 32. (4) The magnitude of acceleration is contant
in (A) and decreasing in (B).
In (A)  r constant, at  0;
v2
v constant, at  constant
R
In (B)  r is increasing, v constant
v2
at  0; ar  decreasing
R
33. (2)
Let  is the radius of curvature when V makes
v v2
 an angle  with horizontal  
R a cos 
v cos 30

v sin  v cos   v cos30  cos  
v
3 / sin 
v3
v sin 2  
 av cos30
3
then v  vmin  v0 cos 30 (At the highest point 
v  0.6m / s
is minimum)
dx 3v2
34. (1) x  2t  vx  2  min 
dt 4a
36. (2) The radius of curvature is
3/2
  dy 2 
1    
   dx  
dy d2 y
2
y  2t  vy   4t
dt dx2
vy 4t
 tan     2t y  3x2
vx 2
Differentiating with respect to time we get, dy d2 y
 6x, 2  6
dx dx
 sec   ddt  2
2
dy d2 y
x  y 0  0& 2  6
dx dx
d
 1  tan 2   2   1/ 6
dt
37. (3) The path followed by the particle is a
d 2 d parabola, as shown in figure. The normal
 2
; at t  2s is
dt 1  4t dt acceleration at the highest point is g.
d 2 2
 2
 rad / s.
dt 1  4  2  17
35. (1) For convenience rotate the diagram so

that the acceleration comes down. Now it is a
normal projectile problem where a =g
V2 cos 2  1
 x  u cos t , y  u sin  t  gt 2
g 2
38. (2) Let  be the angle made by the velocity
x 1 1 
of the particle w.r.to horizontal direction at half  cos   ,sin    y  gt 2 
of the maximum height. ut ut  2 
The radius of curvature is By eliminating  we get
g 2t 4
x 2  y 2  gt 2 y   u 2t 2
4
By comparing the above equation with the
equation of circle having radius r and centre
having coordinates (O,  )
x 2  ( y   )2  r 2
V2  gt 2
 (1) where  , r  ut
g cos  2
2 2 H
 v sin     u sin    2 g   (2)
 2
1. (4) The displacements of the body are
u 2 sin 2 
H (3)
2g  6 ˆ 6 ˆ  4 ˆ 4 ˆ
S1  i j , S2  i j
V cos   u cos  (4) 2 2 2 2
From Eq’s 1, 2, 3 & 4 we get

3/ 2
 sin 2   3/ 2
u 2 1  u 2 1  cos2  
2 
  
2 2 g cos
g cos 

39. (3) Given V  iˆ  ˆj and a  iˆ  ˆj  kˆ
V2

ar
V  Vx2  Vy2  2
ar is the component of acceleration    10 ˆ

S  S 1  S 2  2iˆ  j
perpendicular to V. 2
ar  a z  1 Angle at which resultant displacement makes
with the east direction
 2
y  component 5 2
40. (3) Take the origin at the point of projection. tan   
x  component 2
The x and y coordinates of a particle that make
an angle  with horizontal directions are   tan 1 (5)
The net distance S  ( 2) 2  (5 2) 2 dy C t 2

 C2 t  y  2
dt 2
S  52
x C
   1
2. (2) Angle between V & a is obtuse and y C2
acceleration is directed towards left.   
7. (3) v BW  v BG  v RG  6iˆ  8 ˆj

3. (4) Given: v  3tiˆ  4tjˆ ms 1
L
Acceleration, 8. (1) 10  (i)
 v
 dv d
a  (3tiˆ  4t ˆj )  3iˆ  4 ˆj
dt dt L L
12.5   (ii)
 v u2 2
v 1  u 2 / v2
a  32  4 2  5.0 ms 2
4. (1) Given y = 3x + 4 10 L v 1  u 2 / v2
From (i) and (ii)  
12.5 v L
 v y  3v x
4 122
vx  1, ax  1 a y  3a x  1 2
5 v
v y  3, a y  3 v  10, 16 122 122 16 9
 1 2  2  1 
25 v v 25 25
a  10
12 3 12  5
y-x graph gives the shape of path of  v   20 m/s
5. (1) v 5 3
particle 9. (2) Horizontal component of velocity of rain
dx drop = velocity of man = 10 km/h.
 8 sin 2 t
dt
 v sin 30  10
x t
 v  20km / h
`  dx   8 sin 2 t dt
8 0
10. (2)
8 This problem is complex if we try to do with
x 8 [cos2 t ]t0
2 normal method we take
12  x 
x  8  4 1  cos 2 t   cos 2 t  vr  vrx iˆ  vry ˆj (as discussed in previous
4
problem)
dy We need two conditions to find two unknown
 5 cos 2 t
dt quantities
y t 5 Case – (i) Man is moving with 8 km/h
 dy  5  cos 2 t dt  y  sin 2 t
0 0 2  
vM  8iˆ, vr  vrx iˆ  vry ˆj
2 2
  x  12  y

   2
1
vrM  (vrx  8)iˆ  vry ˆj
 4  5
 
2 Given that with respect to him the rain drops
are falling vertically down.
dx C t2
6. (1)  C1t  x  1
dt 2  vrx  8  0 vrx  8
Case – (ii)
 v y  20 m / s
Man is moving with 12 km/h in same direction.
 v y - is the velocity two seconds before reaching
vM  12iˆ vr  8iˆ  vry ˆj vry   vry
the maximum height
v 20 4
At that point tan 53  v  v  3
y
x x
 vx  15m / s
12. (2)
u 2 sin 2  2u sin   3
 R  u cos   (1)
vrM  4iˆ  vry ˆj g  g  g
4 2u sin  3
tan 30   vry  4 3 T 
vry (2)
g g

vrg  vr  8iˆ  4 3 ˆj
1
 3
From (1) & (2) we get   tan  2 
vr  4 7km / h  
11. (2) Method-1 13. (2) The horizontal range is the same for the
angles of projection  and (90   )
Let u be the velocity of projection and α be the
angle of projection
2u sin 
t1 
Now, time taken to reach the highest point g
u sinα
  Ta ( T time of assent) 2u sin(90   ) 2u cos
g a t2  
g g
v sin 53  u sin α - g (Ta  2) 2u sin  2u cos 2  u 2 sin   2
t1t2      R
 u sin(α  g g g  g  g
 v sin 53  u sin α - g   2
 g  u 2 sin 2
where R  g
 v sin 53  2 g (i)
Also, Hence t1t2 R (as R is constant)
v sin 53  u cos α (ii) 14. (1) We known the equation of projectile as
 x
u cos α 3  y  x tan  1  
  cot 53   R 
2g 4
Where x and y can be any point on the projectile
3 we can take (x,y) = (4m, 4m)
 u cos α  10m / s  15m / s
2
 4 9
Method-2 4  4 tan  1    tan  
 18  7
Two seconds before maximum height
sin   9 / 130
0  v y  g (2)
18. (1) Assume that the particles are projected

cos   7 / 130
from the origin as shown.
u 2 sin 2
R For particle one we can write its x and y-
g
coordinates x  u cos 1t , y  u sin 1t
gR
u2   u  13.6m / s
2sin  cos  y2 x2
sin 2 1  cos2 1  
15. (1) u 2t 2 u 2t 2
1  x 2  y 2  u 2t 2 (i)
y  4 x  x2
2 For second particle x  u cos 2t , y  u sin 2 t
dy dx dx
4 x y2 x2
dt dt dt sin 2 2  cos2 2  
u 2t 2 u 2t 2
dy dx  x 2  y 2  u 2t 2 (ii)
At x0  4  4(4)  16m / s
dt dt
From the above two equation we can conclude
Initial angle of projection is that all the particles will have same equation
dy / dt 16 x 2  y 2  u 2 t 2 by having ut as the radius. All
tan    4 the particles lie on a circle of radius ut.
dx / dt 4
If u be the initial velocity then 19. (1)
u cos  4
 u sin   tan  u cos   16
u 2 sin 2 2( u cos  )(u sin  )

R 
g g
 x
From equation of trajectory y  x tan  1  
2(4)(16)  R
R  12.8 m
10
 6 2
16. (4) y  ax  bx 2 , for height to be maximum:  3  6 tan  1    tan  
 24  3
dy a 20. (1)
 0 or a  2bx  0 or x 
dx 2b 21. (3) The horizontal component of velocity
2 2
 a   a  a v x  u  10 m s
ymax  a    b   
 2b   2b  4b

17. (1) Given that v  iˆ  2 ˆj
u x  1& u y  2
 xt (1)
1 vy vy
y  2t  10t 2  (2)

Now, tan   tan 45 
vx

10
 v y  10 m s
2
From (1) and (2), y  2 x  5 x 2 22. (2) v0 sin   v (i)
v0 sin   u (ii)
Time of flight depends on vertical component of

velocity.
2v0 sin  2v
T 
g g
23. (3) Time of flight remains unchanged relative

to trolley & ground. Hence right choice is (3).
2
Vertical component of projectile relative to trolley, u2 10sin 37 
H   3m
v yT  v sin  2 a 2 g cos53
27. (3) Method-I
And vertical component is unchanged relative
to trolley and ground. We use the position vector method to find the
shortest distance possible between the bodies.
2u sin  We write the position vectors of the two bodies
Now time of flight  g w.r.to the common origin. The magnitude of
Time of flight remains unchanged relative to relative position between two bodies always
represents the separation or distance between
trolley & ground. Hence right choice is (3).
the bodies. When the separation becomes
24. (1) The motion of the train will affect only minimum then
the horizontal component of the velocity of the
d
ball. Since, vertical component is same for both r AB  0 (from maxima and minima)
dt
observes, the hm will be same, but R will be
different. By solving the above equation we get the time
25. (1) The ball strikes D if the velocity of the (t) value and by keeping it in r AB we get least
ball w.r.to the plank is along AD. separation between the bodies.

i.e v b  u cos iˆ  u sin  ˆj 2 1
We know that, s  so  ut  at
2

v b  viˆ Here s can be treated as position also

v bP   u cos  v  iˆ  u sin  ˆj r A  rAx i  rAy j
 u cos  v  0 1
rAx  0  10t  0t 2  10t
2
v 1
cos  Similarly rA  50  0  t    0 t 2  50
u y
2
So time taken to reach D is rBy  15t j rB yo  0, aBy  0
 
t
L

L r B  100i  15t j
v u cos
r AB  r A  r B  10t  100  i   50  15t  j
26. (2) Maximum height from inclined plane is 2 2
r AB  10t  100   50  15t 
d From the above two equations we get

When separation is minimum r AB  0 200
dt d m
13
d 2 10t  100 10   2  50  15t  15 Here we can do the problem with respect to
r AB  0
dt 2
2 10t  100    50  15t 
2 particle A also.
28. (4) Let t is the time at which these two
2 10t  10010  2  50  15t  15  0 particles meet each other.
70
t 
13
2 2
 70   15  70 
r AB   10   100    50  
min
 13   13 
200 d
 v22 t 2  v12 t 2  d 2 t 
13 v  v12
2
2
Here the time t corresponds to minimum separation 29. (4) Radial acceleration
only since the maximum separation will be infinite
v2 (30)2
as they are moving continuously. ar    1.8 ms2
Method-2 r 500
Relative velocity method: Tangential acceleration at  2 m/sec 2
Resultant acceleration
a 2  ar2  at2  2at at cos 
Here   90
a 2  (1.8)2  (2) 2  0  3.24  4  7.24
a  7.24  2.7 ms 2
30. (1) x  a 1  cos t   a cos t    x  a 
Here we consider the motion of body A w.r.to B, y  a sin t
that is the body B will be at rest and body A will
be moving with relative velocity in the plane. 2
 x  a   y 2  a2
v A  10i , v B  15 j 31. (4) The particle is moving in circular path
 OQ  100m  From the figure mg  R sin
 
 OP  100m  PQ 
mv 2
v AB  10i  15 j , v BB  0  R cos
r
The perpendicular drawn to the relative path from
the body (2) assumed to be at rest gives the shortest
distance of approach
15 3 2 3
tan     cos    sin  
10 2 13 13
From the triangles
50 3 d
tan    & sin   From Eqs. (i) and (ii), we get
100  PQ 2 PQ
rg r Tangential acceleration at  a
tan   and tan  
v2 h  Resultant acceleration
2 2
v (0.5)
h   0.25 m  2.5 cm a '  ar2  at2  2ar at cos
g 10
32. (2) The acceleration vector shall change the But here   90  cos  cos900  0
component of velocity u along the acceleration 2

2 2  v2  2
and a '  ar  at    a
vector. r
 
v2

r 35. (3) V  k1iˆ  k2 xjˆ
an
dx dy
Radius of curvature rmin means v is minimum and  Vx  k1 ,  Vy  k 2 x
dt dt
an is maximum.
x t
The body follows the projectile path and at the  dx   k dt  x  k t  V

 
1
1 y  k1 k 2 t
height point v0 sin 60 becomes Zero.
y t
k1k2t 2
 dy   k1k2tdt  y 
  2
k2 x 2
From the above two equations we get y  2k1
The radius of curvature

Since ‘a’ remains constant   dy 3/2 
1    
  dx  
 2 cos 60  m = 0.5m.


rmin  d2 y
2
dx2
33. (4) The path followed by the particle is a
parabola, as shown in figure. The normal dy k2 x dy d2y k
 , at x  0  0, 2  2
acceleration at the highest point is g. dx k1 dx dx k1
   k1 / k2 (At the origin )

Method-II
t V2
The radius of curvature is   a
r
2
 1 
5
v cos   2 
2 2
    1.25m
g 10
v2
34. (3) Radial acceleration ar   
r V and a are shown in figure where

V  k1î  k 2ˆj

 dV dx ˆ
a  k2 j  k1k2 ˆj
dt dt
From the above figure ar  a sin 
Vx
V  Vt  sin  
V
V2 V3
 
a sin  aVx
(10 2)2  g 
at x = 0, y= 0 Vx  V  k1 , a  k1k2   20 2  Since ar  
ar  2
   k1 / k2 (At the origin) With respect to ground the velocity and
acceleration are
36. (1) Consider the point of projection as the
origin and the xy plane is taken on horizontal
surface
V2 V2
 
ar g cos 
Vx
cos 
V
Let d is the distance traveled by the particles V3
  39m
after time t gVx
For the first particle
x  d cos  ,y  d sin
u 2 sin 2  u 2 sin 2 
x2  y2  d 2 1. (20) R ,H 
g 2g
For the second particle
x  d cos  ,y  d sin  H max at 2  90
x2  y2  d 2 u2
H max 
We can prove that for any particle x 2  y 2  d 2 2g
So all the particle lie on a circle of radius (d = u2
ut)  10  u 2  10 g  2
2g
37. (4), 38. (4) With respect to trolley the velocity
u 2 sin 2 u2
and acceleration are R  Rmax 
g g
10  g  2 u2
Rmax   20 metre  Rmax 
g g
Given, R  16km  16 103 m


2. (5) Given: v  4.0 iˆ  5.0tjˆ ms 1
g  100 m/s 2
Acceleration,
  u 2  16  103  10
 dv d
a  (4.0iˆ  5.0t ˆj )  5.0 ˆj ms 2
dt dt  u  400m/s

a  (5.0)2  5.0 ms2
 ˆ
3. (5) v  l  2 Jˆ u 2 sin 2  15
1. (3) Here, 50  g
 x t ...(i)
u2 50 50
1 or    100
y  2t  (10t 2 ) ...(ii) g sin 30 1 / 2
2
From (i) and (ii) u 2 sin 2  45 u 2
R   100 m
g g
y  2 x  5 x2
2. (2) Change in momentum is the product of
force and time.
2sin   p 
1. (7 2 ) p  mg 
g  F  
 t 
 
Given u  3iˆ  4 ˆj, a  0.4iˆ  0.3 ˆj, t  10s  2mv sin   2mv sin 45
  
v  u  at  3iˆ  4 ˆj  (0.4iˆ  0.3 ˆj) 10  7iˆ  7 ˆj 2mv
  2mv
 2
 v  72  72  7 2 units
 
2. (3)  and  
t t 1. (4) Given, initial velocity (u)  40 m/s
 Height of the hall (H )  25 m
 
t2
Let the angle of projection of the ball be  , when
but   constant maximum height attained by it be 25m.
1 (2)2 Maximum height attained by the ball
So, 
1   2 (2  2)2
u 2 sin 2 
1 1  H
or   2 3 2g
1   2 4 1
3. (400) For a body moving in projectile motion, (40)2 sin 2 
25 
the horizontal range covered is 2  9.8
25  2  9.8
u 2 sin 2 or sin 2  
R 1600
g
 0.3063

For maximum range   45 ,sin 2  1 or sin  0.5534
 sin33.6  1 
 KE  p 2   constant 
or   33.6  2m
 
u 2 sin 2 And Kinetic energy
 Horizontal range ( R)  g
2
(40) sin 2  33.6

9.8
1600  sin 67.2

9.8
1600  0.9219
Y  150.5 m
9.8
1
v02 sin 2   m u 2  g 2t 2  2ugt sin 
 
2
2. (1,2) Height, h  2 g i.e., h  sin 2 
 v  v ˆj  v iˆ 
h sin 2 1  y x 
 1  1
h2 sin 2  2
v 2  u 2  2 gy
 
So, sin 2 1  sin 2  2
or 1   2
2v0 sin 
Time of flight, T  g
or T  sin  1
2
Kinetic energy  m u  2 gy
 
2
T sin 1
 1  1
T2 sin  2 2 1
Intercept on y-axis  mu
or T1  T2 2
2
u 2 sin 2 1 x
Horizontal range, R  KE  m
 
g 2 t 
or R  sin 2 3. (1) If air resistance is taken into consideration

then range and maximum height, both will
R1 sin 21
  1 decrease.
R2 sin 2 2
1
or R1  R2 4. (3) y   gt 2 , x  v0 t
2
Total energy of each particle will be equal to
KE of each particle at the time of its projection.  g 
 y    2  x2
 2v0 
1. (3) dy  g 
m    2  x
1 dx  2v0 
2
2. (3) Kinetic energy  mv
2
2 g2 2
1  p2  1 2 K .E   gt   v02  x  v02
 m  p v02
2  2 
 m  2m
5. (3) s  t3  t
ds 
 
^ ^
Speed, v   3t 2
dt 1. (1) Here, v  K y i x j
dv dx ^ dy ^
 
^ ^
and rate of change of speed, at   6t i j  K yi  x j
dt
dt dt
Tangential acceleration at t = 2s,
dx dy
at  6  2  12ms 2  Ky and  Kx
dt dt
2
and at t  2 s, v  32  12ms 1
dy dy / dt Kx
 
v2 144 2 dx dx / dt Ky
Centripetal acceleration, ac   ms
R 20
ydy  xdx
Net acceleration  a12  ai2  14ms2
 ydy   xdx
 y 2  x 2  constant
1. (2) 2. (3)

2. (3) The angle transveresed in time is
3. (1) v  u  gt 2
 g  2  
4. (1) y  x tan    2 2 x   t  
 2u cos   2 2
dy  g  
m   tan    2 2 x i.e., at t  , the position of two particles is
dx  u cos   2
 gv  shown in the figure.
m  tan    2 0 2  t
 u cos  
1
5. (1) h  H  gt 2
2
6. (2)
 
 The relative velocity v A -v B is
So, vr  2ωR sin(ωt)
 
^ ^
^
 R 1 i  R 2  j    R  R  i
2 1
t  0, vr  0
3. (3) Speed, V  constant (from question)
At t = T/2, vr  0
Centripetal acceleration,
So two half cycle will take place.
V2  Angle with downstream  90  30  120.

a
r 3. (1) Arc length  radius  angle
ra  constant   
So, B  A  A 
Hence graph (3) correctly describes relation
between acceleration and radius.
4. (4) 2u x  40  u x  20
1
50  2u y   10  22  u y  35
2
4. (3) Let ‘t’ be the time taken by the bullet to
uy 35 7
tan     hit the target.
u x 20 4
 700 m  630 ms 1t
7
  tan 1   700m 10
4  t 1
 sec
630 ms 9
For vertical motion, u  0
 1 2 1  10  500
2
1. (2) If x  position vector, then velocity  h gt   10     m  6.1m
2 2  9 81

 dx
v  0.5t 2 i  3t j  9k (given) Therefore, the rifle must be aimed 6.1 m above
dt the centre of the target to hit the target.

2 2 2
u2 u2
 x   0.5t 2 idt   3t.jdt
  9kdt
  5. (2)  Rmax   40  u  20m / s
0 0 0
g 10
 4 2  20sin
 x  i  6j  18k T  4sin 
3 10
 2 R  ux  T   20  20cos   4sin 
 x   4 / 3  62  182  19
  80  sin   sin  .cos 
 If the angle x makes with y axis is β,
dR
then cos β  6/19. For maximum R, 0
d
2 1
 tan β  19/6   1  18/6  3  2cos2   cos  1  0  cos   2
 β  tan 1  3 .    60
2. (4)
6. (2)
Given, u s  12 m /s and ur  6 m /s  x  4
y  x tan  1    h  4  1 1  
 R  10 
6 1
sin       30
12 2 h  2.4m
76 Newton’s Laws of Motion
 tan   tan 45
   45
13. (2) 50 g sin 37  mg

1. (4) Inertia means resistance to change. It is
the property of the body by virtue of which it 3
m  50sin 37  50  5  30kg
cannot change by itself its state of rest or of
uniform motion. 14. (1) For equilibrium of forces, the resultant of
2. (3) 3. (2) 4. (4) 5. (3) two (smaller) forces should be equal and opposite
to third one.
6. (2) When a sudden jerk is given to C, an
impulsive tension exceeding the breaking tension 15. (4) By displacing T1 & T2 we can represent
develops in C first, which breaks before this the vectors as shown in the figure.
impulse can reach A as a wave through block.
7. (2) When the elevator moves downward with
an acceleration then pseudo force acts upward.
Which will reduce the weight of the person.
8. (1)
dp d 16. (3) F   m1  m2  m3  g sin 
9. (4) F  (a  bt 2 )  2bt
dt dt
The force on m3 is
 F t
10. (3) If 1 N and 2 N act in the same direction m3 F
N  m3 g sin  
and 3 N acts in opposite direction, then  m1  m2  m3 
equilibrium is possible.
17. (1) The force diagram is
11. (2) In this case, one 2 kg wt on the left will
act as the support for the spring balance. Hence
its reading will be 2 kg.
12. (4) The free body diagram of the rope is
shown in the figure.
50
tan   1
50
From Lami’s theorem
F3 3N 3N
 
sin 90 sin(90   ) cos
Newton’s Laws of Motion 77
3 24. (1)
F3  (i)
cos
F3 4N 4
 
sin 90 sin(180   ) sin 
4
F3  (ii)
sin 
From Eqs (i) & (ii)
4
  tan 1   that makes with (-) ve y-axis.
 3 Let T be the tension in the string
Let a be acceleration of the system.
F3  32  42  5
The equations of motion are
18. (2)
Ma  Mg sin53  T (i)
Resultant of forces be zero.
19. (3) The surfaces of rod and wedge that are and Ma  T  Mg sin37 (ii)
in contact are as shown below.
Adding (i) and (ii), we get
Mg (sin 53  sin 37)

a
2M
 0.98 ms 2  1ms 2
25. (1) T = tension in the string
 Applied force F = 2T
v2  v1 cos (90-θ)  v2  v1 sin 
T = F/2 (1)
20. (4) The two surfaces of the blocks are shown
below.
Tf
Acceleration of block  .
m
F
f
The velocity components must be same along  a 2
m
the normal.
26. (2) The tension in the string is
v1 sin   v2 cos
2 
21. (3) 22. (3) T   3  y  3 g  a 
3 
23. (1) The acceleration with which the system
2 
should move so that the block does not move T    3  y   3  15 
3 
w.r. to the wedge is g tan  . The required force
is F  (M  m) g tan  T  75  10y
27. (3) u y  40 m / s , Fy  5 N , m  5 kg 1
S  ut  at 2
2
Fy
So ay   1m/s 2 ,( Asv  u  at ) 1 1
m 2
4  a  4   a =  0.5
2 2
 v y  40  1  t  0  t  40 sec
 s  4m and t = 4s given 
28. (2) Spring balance reading  2T
a = gsinθ   k  g  cos θ
2m1m2 g
Where T  m  m , m1  1Kg and m2  5Kg
1 2 0.9
 k   0.5
29. (3) Let T be the tension in the string 3
39. (4) As the block is held by the string,
1
T  Mg sin 300  15  10   75 N .
2
40. (1) Here net driving force
The newton’s 2nd law for the bead is

  
Fext  Fs  mar
mg  ma  T  ma
 T  mg
30. (1) N  m ( g  a)  4N mg mg
 mg   downward
31. (1) 32. (1) 33. (2) 34. (1) 2 2
35. (3) Hence friction will act upward and its magnitude
mg
36. (4) In the given condition the required should be f  . If the block ‘m’ is stationary
centripetal force is provided by frictional force 2
between the road and the tyre. the friction between m and the wall should be
static.
mv 2
  mg  v   Rg f  f lim
R
37. (3) Here,   0.8. Let F be horizontal force mg mg 1

  .N    (mg )    .
that the boy is applying on the pole. 2 2 2
Frictional force, 41. (3)

f   N  mg 42. (4) Let l be the length of the incline
mg 40 10 1 2
N   500 N  F  N  500 N For smooth plane l  g sin(45)t1
 0.8 2
38. (2) Coefficient of static friction, For rough plnae
1 1
 s  tan 30ο   0.577  0.6 l  ( g sin 45   g cos 45)t22
3 2
Given t2  3t1 4. (4)

From the above equations we get 5. (4) As the rain drop is falling down with a
constant speed, its acceleration, a  0. Hence,
t12  1   2  1   2
   t2     9t1 net force on the drop is equal to zero. As the
2  2 2  2 2 cork is floating on the surface of water, its weight
8
is balanced by the upthrust. Hence, net force
  on the cork is zero. As the car is moving with
9
constant velocity, its acceleration a  0. Hence,
43. (3) Let T be the tension in the branch of net force on the car, F  ma  0. Whenever a
the tree when monkey is descending with body is thrown vertically upwards gravitational
acceleration a then mg  T  ma; and Given pull of earth gives it a uniform acceleration,
a  g in downward direction. Hence, net force
that T  0.75mg
on the pebble  mg .
mg
  ma 6. (2) 7. (1) 8. (2)
4
9. (3) Initially due to upward acceleration
g apparent weight of the body increases but then
a it decreases due to decrease in gravity.
4
10. (2)
44. (1) For mass M1 (4 kg)
dP
T  M 1 g sin 37  1M 1 g cos37  M 1a (i) 11. (3) F
dt
For mass M 2 (2 kg) If F  0 then P will change
M 2 g sin 37  T  2 M 2 g cos37  M 2 a (ii) 12. (2)
Adding and the substituting the given values, 13. (3) Let F is the force
we get
F  m1a1 & F  m2 a2
2 g  0.6  4 g  0.6  0.25  2 g
F
0.8  0.75  4 g  0.8 F  (m1  m2 )a  a 
m1  m2
 (4  2)a
F aa
a  1 2
g (1.2  2.4  0.4  2.4)  6a F F a1  a2

a1 a2
or a  1.3 ms 2
14. (1) For jumping he presses the spring
using (ii) platform, so the reading of spring balance
increases first and finally it becomes zero.
2  9.8  0.6  T  0.25  2  9.8  0.8  2  1.3
15. (4)
or T  11.76  3.98  2.6  5.29 N 16. (3) Swimming is a result of pushing water in
the opposite direction of the motion and water
exerts force inforward direction.
1. (2) 17. (4) According to Newton’s Second Law:
2. (3) Particle will move with uniform velocity Force = rate of change of linear momentum.
due to inertia.
dP
3. (2) 18. (4) F
dt
If F  0 then P will change As 4kg is in equilibrium

19. (1) A cricket player lower his hands while
catching a ball to increase the time so as to T  4 g  T1 ,  T1  2 g  19.6N
decrease the force exerted by the ball on 29. (2) The two surfaces which are in contact
cricketer’s hands. This is not an example of are shown below. The velocity components must
Newton’s third law of motion. be same along the normal.
20. (4) P  5t 2  t  5
dp
F  10t  1  5(a)
dt
a t
21. (2)
22. (1) As the spring balances are light, so v1  v2 cos 
tension in both the springs will be, same and
equal to weight of block suspended. 30. (2) The two surfaces are as shown below.
23. (3) Let T1 be the tension in the spring and T2
be the tension in the right string
v1 cos  v2 cos  90   
v1 cos  v2 sin 
in equilibrium. Acc to Lami’s theorem v1  v2 tan 
200 T1 31. (2) 32. (2) 33. (1)

sin  90  sin  90  53 34. (2) The force acting on the block are shown
in the figure.
200 T
 1
1 cos53
T1  120 N  Kx  K  3000 N / m
24. (2) The string has to balance both the blocks
5 kg and 4 kg. The tension is T   5  4  g  90N
25. (3) Net force on the pulley is 2T  2mg w.r to the wedge the block is in dynamic
equilibrium.
26. (3) 27. (2)
28. (4) From the given figure N sin   ma
N cos   mg
T  6g
From the above equations 37. (1) T3  (m1  m2  m3 )a

a  36  (1  8  27)a  a  1m / s 2
tan  
g
Now T2  ( m1  m2 ) a  (1  8)  1  9 N
a  g tan
38. (4)
35. (2) Let T is the tension is the string and a is
the acceleration of both the blocks. By applying
Newton’s 2nd law on both the blocks.
P
Acceleration of the system 
mM
MP
The force exerted by rope on the mass  .
m M

F 6iˆ  8 ˆj
39. (2) Acceleration  in the
T - 20 = 2a (i) m 10
direction of the force and displacement
  1 2 1  6iˆ  8 ˆj 
s  ut  at  0   100  30iˆ  40 ˆj
2 2  10 
4
So the displace ment is 50 m along tan-1 3 with
30 N - T=3a (ii) x-axis
From eq’s (i) & (ii) we get
v 5 2
40. (3) The retardation is a    1m / s
10 t 5
a   2m / s2
5
N  m g  a
36. (3)
N  60  9
N
R  54kg wt
10
41. (1) Given that V  29 t
dV
a=  29
dt
The force exerted by the vertical wall
2mg  T  2ma ...(i)
R  ma  29 N
T  mg sin30  ma ...(ii)
(i)+(ii) gives, Fnet  N 2  R2
mg g
2mg   3ma  a   (10)2  (29)  129 N
2 2
Force of friction = Weight of block

f W  N W
 N  W
W
F (N  F)

kx 15(0.2) As  1  F W .
42. (4) a   10 ms 2
m 0.3 55. (2) If applied force is less than limiting friction
force then frictional force is equal to applied
 F  F2  6 N  3N force.  f  F  0
43. (4) a 1   1m s 2
 m1  m2  3kg
56. (2)
F  3 N  1  1m s 2 Let f be the frictional force exerted by air on
the body
 F  4N
mg  f  ma
44. (3) 45. (1) 46. (2) 47. (2)
48. (4) As by an internal force momentum of the 25  9.8  f  25  9.2
system cannot be changed. f  25  0.6  1.5 N
49. (1)
57. (1) The forces that act on the car are shown
50. (3) Sand is used to increase the friction. in the figure. As the car has tendency to slip
51. (4) upwards the frictional force acts down the
incline.
52. (1) For moving on circular path without
slipping, centrifugal force must be equal to
frictional force.
That is,
 mr  2   m g  v  r 
 r2   g
At the verge of slipping
g 0.5  9.8 mg   N sin   N cos 
    0.7 rad / s.
r 10
mv 2
53. (1) Since,  mg cos  mg sin  N sin    N cos
R
 Force of friction is f  mg sin where f   N
1 By dividing the above two equations
f  10  2   10 N
2
v  gR
sin    cos   gR
   tan  
54. (3) Here applied horizontal force F acts as
normal reaction. cos   sin   1   tan 
For holding the block  38.2 m /s

58. (2) mg sin  mg cos
h
tan    
R
h  R
62. (4) As the lift is going down with g the
effective weight is zero.
3 N 0  f 0
N '  100cos30  100.  50 3N
2
Wf  0
 1 
Net driving force  100   30  63. (2) F  f  mg sin  mg ( cos  sin )
 2 
= 20 N upward 64. (1) Retarding force F  ma   mg
1  a  g
f max  50 3  50 N
 
3
Now from equation of motion v 2  v2  2 as
 Friction = 20 N(downward)
v2 v2
59. (3)  0  v2  2as  s   
2as 2 g
f smax   mg  0.54  2  10
v02
 s
 10.8N 2 g
As Fext  f smax  f  Fext  2.8 N 65. (1) f mass for A  1 ( mg cos45)
60. (2) 2 mg 2
  mg
Given that 3 2 3
3
sin   ,   0.6
5
The force required to move it up is
F   mg cos  mg sin 
4 3
 0.6  10   10   10.8 N
5 5
Also f mass for B  2 (2mg cos45)
61. (3) When the sand is piled like a cone the
particels start a particular angle of the cone. 1  2

 1 / 3  2mg   3 mg
 2
Total frictional force
2 2 2 2
 mg  mg  mg
3 3 3
But pulling force Applying equilibrium conditions,
 F2  F1 
2mg mg mg
  F x
0
2 2 2
 F1  1sin 45ο  2sin 45ο  0
 system can not acceleate.
or F1  2sin 45ο  1sin 45ο
So a A  aB  0
2 1 2 1 1
    N
2 2 2 2
1. (1) Net force on the particle is zero so the v
F2  2cos 45ο  1cos 45ο
remains unchanged.
2. (2) Let a is the acceleration of the rope 1 3
F2  2   N
2 2
F  Mg  Ma (i)
5. (2) Consider any one sphere
Let T be the tension in the rope at a distance x.
Let N 1 is the force between the spheres and

T  ( L  x)( g  a) (ii) N 2 is the force exerted by the cup on the sphere.
L
From eq’s (i) & (ii)
 x
T  F 1  
 L
3. (3) Let T be the tension in the string. From
the figure for the equilibrium of the system
As the sphere is in equilibrium N 2 sin 30  N1
N2
2
N1
6. (1) Let V be the velocity of the wedge. The

velocities of two surfaces in contact are shown
1
2T cos  2mg  cos     45 in the figure.
2
4. (2)
From the constraint relation along the normal

direction
V cos53  15cos37  5cos53  V  15
7. (1) The two surfaces which are in contact Vx   2 u (ii)

are shown below.
(Negative sign indicates that they are in opposite
v A cos(90  )  vB cos  direction)
By solving the above two equations
We get Vx  2u Vy  3 u
(The direction of V x should be towards right)

The velocity components must be same along
the normal. V  Vx 2  Vy 2  7 u
v A sin   vB cos  9. (1) The two surfaces which are in contact

8. (3) The surfaces which are in contact are are shown below.
shown below. Let vx & v y are the x and y
components of velocities of the sphere.
v2 component = net v1 component (along

normal direction)
Here V x direction can be taken towards left or
v2 sin   v1 y cos   v1 x sin 
right. The surfaces that are in contact are
shown below.
10. (3) Let Vx and Vy are the components of
velocity of the centre of the sphere. The angle
R/ 2 1
 is given by sin   R
 (According
2
to the figure given in question)
Constraint relation at A
Vx cos600  Vy cos300  u cos600
Vx 3Vy u
 
2 2 2
Vx  3 V y  u (i)
v x cos 45  v y cos 45   v A  0 ( vA is the
velocity
of tip A)
v x  v y is the net velocity component of the

spherical surface along the normal.
 v x  v y  (1) 12. (2) Let x be the distance of block B from

rigid support and y is the position of block A
Constraint relation at B w.r.to lower pulley connected to block B.
The length of the string is L=4x + y + c

 vx vy
( 2)cos 45  
2 2 By differentiating the equation w.r.to t is
vx vy dx dy d2x d2 y
1   (2) 04  & 04 2  2
2 2 dt dt dt dt
From (1) and (2) we get d2y
 4b  aAB
1 dt 2
vx  v y 
2 aA  4b (As the wedge B is moving away from
Velocity of centre of the sphere  vx2  v y 2 A and block A will lose contact with B hence
the horizontal component of acceleration of A is
=1 m/s zero. Since there is no horizontal force on the
11. (4) The positions of the wedge and block are block A at t=0)
expressed as shown in the figure. 13. (1) Express the position of moveable pulleys
and blocks with respect to fixed frame as
shown in the figure
The total length of the string is l  2x  y  c

dl dx dy dc dx dy
2   02 
dt dt dt dt dt dt
String – 1:
dx
Given, u l1  y A  yP1  y p1  c  y A  2 y P1  c
dt
0  VA  2Vp1
dy
  V Ay  2u
dt  0  a A  2 a p1  1
As the block A is fit in the wedge String – 2:
VAx  u , V Ax & V Ay are the x and y components of l2  ( yB  y p2 )  ( yC  y p2 )

velocities of block A, w.r.to the ground.
l2  yB  2 y p2  yC
VA  VAx2  VAy 2  (u)2  (2u)2  5u 0  VB  2Vp2  VC

0  a B  2 a p2  a C  (2) N cos  mg -(1)
From the diagram a p1  a p2 N sin   ma -(2)
Since two pulleys are connected by a vertical From (1) & (2)
string a dy
Tan     2x ( y  x2 )
From (1) and (2) we get g dx
aB  aC  a A  0 a  2 xg  3.9
We assume an arbitrary sign convention.Take 16. (3) Initially the block is at rest under action
downward direction as (+) ve of force 2T upward and mg downwards. When
the block is pulled downwards by x, the spring
Given a A  5 m / s 2 , a B  4 m / s 2
extends by 2x. Hence tension T increases by 2
 aC  1m / s 2 i.e., block C goes upward with kx. Thus the net unbalanced force on block of
1m / s 2
mass m is 4 kx.
14. (4) Let x be the separation of the blocks A

and B. y be the position of the block C w.r.to
the pulley connected to block A
The total length of the string is l = 2x + y + C

(C is a constant ) 4 kx
 Acceleration of the block 
m
By differentiating the equation w.r.to t.
17. (2) After string is cut, free body diagram of
d2x d2 y block A gives
02 2  2
dt dt
d2x
aAB  is the relative acceleration of the two
dt 2
blocks A and B, since x is the relative position
between A & B.
d2y
 2  a  b   aCA ,
dt 2
The tension in the string becomes zero.
15. (4) The force that act on the sphere are N,
mg and pseudo force ma mg g
2ma A  3mg  2mg or a A  
2m 2
Free body diagram of block B gives
maB  mg or aB  g
18. (4) If the tension in the string is T then for
insect
From equilibrium
a 3mg a0  g sin 
T  mg  m T  tan  
2 2 g cos 
3mg
T  N  3mg  N 
2
19. (1) Let T be the tension present in the block
connecting m 2 . The tension present in the string
22. (2)
connecting block m 2 is also T. Acceleration of
both the blocks are same and equal to a.
m2g  T  m2 a (i) F
F  2T cos  T 
2cos
T  m1a (ii)
From eq’s (i) & (ii) Magnitude of acceleration of the particle
m2g T sin  F tan   F x

a 
m1  m 2 m 2m 2m a  x2
2
20. (2) Let a be the acceleration of cubical block 

23. (4) Given that p (t )  A(iˆ cos kt  ˆj sin kt )
of mass M from ground frame. Then from the
frame of block of mass M, forces on m1 and  d 
m2 are as shown. Hence for m1 and m2 to F  ( p(t ))  Ak (iˆ sin kt  ˆj cos kt )
dt
remain at rest with respect to of M
 
F  p  A2 k ( cos kt sin kt  sin kt cos kt )  0
 The momentum and force are perpendicular

to each other at 90 .
2mg  mg
24. (2) Case I: a1  g
m1 m
m2 a  m1 g  a g
m2
(2m  m) g
m Case II: a2  (2m  m) g  3
 F  1 (m1  m2  M ) g
m2
(mg  mg  mg ) g
21. (3) Let T be the tension in the string Case III: a3  
2m 2
Hence a1  a3  a2
P
25. (2) Favg 
t
R
t 
v
T sin   ma0  mg sin 
T cos  mg cos
   29. (4) Let the effective mass of the entire

p i  mv sin iˆ  mv cos ˆj system be M. Then the effective mass of the
2 2
system below the first pulley is also equal to M.
  
p f  mv sin iˆ  mv cos ˆj Assume that the acceleration of the block is a.
2 2
    
 p  p f  p i  2mv sin   iˆ
2

2mv 2 sin
Favg  2
R
26. (2) The force of 100 N acts on both the boats
 250 a1  100 and 500 a2  100 From Newton’s second law

or a1  0.4 ms 2 Mg  T  Ma
and a2  0.2 ms 2 T  mg  ma
Then relative acceleration  M  m g 2Mmg

a &T
M m M m
 a1  a2  0.6 ms 2
The tension in the upper most string is 2T
Using S  ut  1 / 2 at 2 , we get 4Mmg
i.e., 2T   M eff g
100  (1 / 2)  0.6  t 2 M m
or t  18.3 s 4Mm
M eff   M  we have taken intially 
M m
27. (2) Consider the man and the child to be
simple blocks hung from two ends of a string  M  m g g
 M  3m  a  
passing over pulley. Being a system, the man M m 2
and child both have same magnitude of 30. (3)
acceleration but opposite directions.
31. (4)
28. (1) Let L1 and L2 be the portions (of length)
Pseudo force will act in noninertial frame.
of rope on left and right surface of wedge as
shown 32. (2) Rate of flow will be more when lift will
move in upward direction with some
 magnitude of acceleration of rope acceleration because pseudo force acts in
downward direction
Fupward  m ( g  a ) and Fdownward  m ( g  a)
33. (4)
M
[ L1 sin   L2 sin  ]g
a L 0
M
( L1 sin   L2 sin  )
When the whole system is accelerated towards as there is no friction. But work must be done
left then pseudo force (ma) acts on a block against gravity. So this statement (2) is incorrect.
towards right. (4) The normal reaction acting on the body
For the condition of equilibrium on an inclibned plane is given by,
g sin  N  mg cos
mg sin  ma cos  a 
cos Where  is the angle of inclination
 Force exerted by the wedge on the block As  increase, cos  decreases and hence N
R  mg cos  ma sin decreases. So this statement is incorrect
(3) The applied force needed to rub the
 g sin   mg (cos2   sin 2  ) duster upward,
R  mg cos  m   sin  
 cos  cos
mg
R
cos
34. (2)
Fapplied = mg+  R  0.05  10  0.05  11
 5  5.5  10.5N
 The work done in rubbing it upward through

a distance of 10 cm,
W  Fappiled  d  10.5  0.10  1.05J
Hence this statement is incorrect.
36. (1)
Net force acting along the tangent is
mg sin   mr 2 cos  0
 r 2  When the ANT is at the verge of slipping

or mg sin  1  cos   0 or
 g 
f  mg sin 
g 
  cos 1  2 
mg cos   N
 R 
35. (1)  mg cos   mg sin     tan 
(1) This statement is correct, because h  R  R cos 

moving vehicles are stopped by friction only.
 1 
(2) If a body is moved up an inclined plane, h  R 1  
 1 2
then the work done against friction will be zero  
37. (3) Let  is the coefficient of friction and x

is the length of the hanging part. At equilibrium
m m
(l  x) g  xg
l l
l As the particle is at rest w.r. to the bowl.

x
1  N cos  mg
38. (2) Horizontal acceleration of the system is N sin   mr 2
F F
a 
2m  m  2m 5m r 2
 tan  
g
sin  ( R sin  ) 2 g
  cos  
cos  g R 2
Let N be the normal reaction between B and C. 41. (4) f max  0.6  10 10 N  60 N
Free body diagram of C gives
Since the applied force is greater than f max
2
N  2ma  F therefore the block will be in motion. So, we
5
should consider f k .
Now B will not slide downward if  N  mB g
f k  0.4  10  10 N  40 N
2  5 This would accelerate the 40 kg block
   F   mg F  mg
5  2
Acceleration  40  1.0 ms 2
5 40 kg
 Fmin  mg
2
42. (3) As the lift is going down with g the
39. (3) F.B.D. of the box in the truck is shown in effective weight is zero.
the figure
N 0  f 0
2
mv
Here   mg 43. (1) Limiting friction between block and
R
slab   s mA g
 0.6  10  10  60N
But applied force on block A is 100 N. So the
block will slip over a slab.
Fk  k mA g  0.4  10  10  40 N
This kinetic friction helps to move the slab
2
v   g R  0.3  10  27 40 40 2
 Acceleration of slab  m  40  1 m/s
B
v  32.4 km / hr
44. (2) The frictional force on the box is
40. (1) Let m be the mass of the particle. The
centrifugal force acting on the particle is mr 2 Fk   k mg
u mg  ma  a  u g 2. (4) Force on the pulley by the clamp
The velocity of the box w.r.to ground is F  T 2  [( M  m) g ]2
vr2  ur2  2as F  (Mg )2  [(M  m) g ]2
vr & ur are relative velocities of the box w.r. to F  g M 2  ( M  m)2

belt
3. (2) Let T be the tension in the string and N
ur  2 m s , vr  0 is the normal force between the person and the
lift. The free body diagram is
a   u g
2 2
s  m
u g 5
45. (3)
Here mg acts downward and frictional forces
up the plane. Normal force direction is shown
in the figure
Then mg sin   2 k N  ma (1) Given that M = 30 kg & m = 50 kg

From the above eq’s we get
( there are 2 surfaces)
N = 100 N
also mg cos  2 N  0 (2)
4. (1) FBD of sphere :
W  T sin 
3
From (1) and (2) and cos      53
5
mg cos  W 5
mg sin   2k  ma  T  W
2 sin53 4
5. (4)
or a  g (sin   2 k cos )
1. (4) Net force applied by block on the inclined

plane is equal to the weight of the body
The FBD’s of A and B are:
Here b is the acceleration of A

From the figure
For the horizontal equlibrium of the system x  90    90    90  x  (   )  90

As the two surfaces are in contact
N AW  N BW
a cos[(   )  90]  b cos(90   )
For sphere A
a sin(   )  b sin
N AB sin   W  N AG
a
b [sin  cos  cos  sin  ]
N AB cos   N AW sin 
b  a[sin  cot   cos ]
For sphere B
8. (2) The particle and spherical surface which
N AB cos   N BW are in contact are shown in the figure below.
N AB sin   W
6
cos     53
10
From the above eq’s
3W
N AW  N BW 
4
6. (2) The acceleration of two surfaces that are We know that the
in contact are shown in figure.
Component of v2x = Net velocity of v1 y (along
normal)
v2 x sin   v1 y cos   v1 x sin 
9. (1) The top view of the three spheres present

on the horizontal surface is shown below.
a sin  g cos (As the block falls freely)
a  g cot 
7. (4) The block B slides along the fixed inclined Here C is the centroid of the triangle formed
surface. Acceleration of B makes an angle by the centre of the three spheres.
90   w.r.to vertical direction.The surfaces of The side view of the triangle connecting the
A and B which are in contact are shown below. centre of the top sphere ( O4 ) and centroid (3)
of the triangle and the centre of the sphere ( O2 ) 11. (1) The total length of the string is expressed
is shown below. as shown in the diagram
l  x  H  x  y  C  2x  y  H  C
y- position of block with respect to the wedge
As the spheres are in contact
x- position of wedge with respect to fixed frame
ao cos  a cos  90    (with respect to ground)
a  ao cot  By differentiating w.r.to t
From the triangle dx dy

02  0
dt dt
OP 2 R / 3
sin    dx 1  dy 
HY 2R    
dt 2  dt 
OP = Distance between C and O2
dx u
   velocity of wedge
HY = Distance between O4 and O2 dt 2
 cot   2 12. (4) The positions of the wedge and block are
expressed as shown in the figure.
 a  2ao
10. (4) The positions of block A and the pulley
are shown in the figure.
The total length of the string is l  2x  y  c

The total length of the string is l=x + y dl dx dy dc dx dy
2   02 
dt dt dt dt dt dt
On differentiating the equation
dx
d2 y d2x Given, u
 2 dt
dt 2 dt
dy
d2x   V Ay  2u
The acceleration of the wedge is b dt
dt 2
As the block A is fit in the wedge
The y-component of acceleration of the block
d2 y VAx  u , V Ax & V Ay are the x and y components of
is  2  b , The block has two components
dt velocities of block A, w.r.to the ground.
of accelerations. The net acceleration of block
A is VA  VAx2  VAy 2  (u)2  (2u)2  5u
b2  b2  2b 13. (1) The positions of the two blocks are shown

w.r.to the fixed reference frame.
(Since velocity is zero at the beginning)

 d 2x d 2 yC  d 2 yD
 sin  2  cos 
 dt dt 2  dt 2
d2 y
Total length of the string is l  4 y  3 x  c sin  a   cos  b    2D
dt
dl dy dx dc aD  a sin   b cos
4 3 
dt dt dt dt
15. (1)
dy dx
0 4 3 0
dt dt
4v2  3v1  4a2  3a1
14. (4) With respect to point O which is present
between the two pulleys, the positions of blocks
A and C are x and yC . The position of block D
w.r.to the pulley is y D . When fixed frame is not available then we
assume a fixed frame anywhere on the ground
as shown in the diagram.
Here neglect the sizes of the blocks.
The total length of the string is
l  xB  xA  xB  xA  xB  xC  C
The total length of the string is l  3 xB  2 xA  xC
L  x2  yC2  yD  3VB  2VA  VC  0
Differentiating the equation w.r.to t. 3aB  2a A  aC  0
dx dy 16. (2) As the amplitude is increased, the

2x  2 yC C maximum acceleration of the platform
0 dt dt  dyD
2
2 x  yC 2 dt increases.
x dx yC dyC dyD If we draw the FBD for coin at one of the
  0
2
x  yC2 dt 2
x  yC 2 dt dt extreme position as shown
dx dy dy
sin   cos C  D  0
dt dt dt
Again by differentiating with respect to t
d 2 x dx d From Newton’s law, mg  N  m 2 A
sin  2
 cos
dt dt dt
For loosing contact with the platform, N=0
d 2 yC dyC d d 2 yD
 cos    sin    0
dt 2 dt dt dt 2 So, A  g  2
At the beginning 17. (4) Before the string A is cut:
dx d Let x be elongation in the spring.
 0& 0
dt dt As system is in equilibrium.
Then for lower block,

kx  mg  20N
Driving force = 4500N
 5  4 10 4
(Resistance)   900 N
100
Resultant force  4500  900  3600 N
Mass of engine and coach  9 104 kg
So accleration of the train  0.04m / sec2

Now considering equilibum of the coach only
we have T  R   4  104  0.04
Just after the string A is cut:
For upper block, T  4 104  0.04  4  102  1600  400  2000 N
20. (2) Let T be the tension in thestring connected
ma  kx  mg
to block of mass 2kg. The tensions in other parts
 2a  20  20 are shown in the figure.
 a  20 m/s 2
18. (4)
If n balls each of mass m are hanging vertically
nmg
then, nmg  T  nma 
2
nmg
or T
2
also T  (16  n )mg sin 

 (16  n ) mg / 2
T  20  2 (1)
nmg 1 mg
on  (16  n )mg  (16  n ) a
2 3 2 60  2T  6   (2)
2
(testing (i))
From the above two eq’s we get
4 80
or n  20
3 6 a m / s2
7
80 3
or n   10 21. (4) From the equilibrium of pulley P1 which
6 4
is massless we can see that tension in upper
19. (2)
string is 50 N. Similarly, from equilibrium of P2
The engine, coach, coupling and resistance are we can see that tension in lower string will be
shown in the figure. 25 N.
N  mg cos30  F cos30
3 3 3 3mg
 mg  KR( 3  1) 
2 2 2
 (1  3)mg 
K  
 R 
24. (4)
Relative velocity of monkeys
 v  2v  3v
22. (3) (Force diagram in the frame of the car)
l l
Applying Newton’s law perpendicular to string Total distance covered    l
2 2
mg sin  ma cos l
 time taken by each monkey 
3v
25. (2) By considering the man and the platform
as two blocks the free body diagrams are
a
tan  
g
2F  N  Mg  Ma
Applying Newton’s law along string
2F  mg  N  ma
2 2
 T  m g  a  ma 4F  M  m g  M  m a
T  m g 2  a 2  ma 4F  M  m g
a
M m
23. (2)
26. (2) Method - I
The extension in the spring is x
x  AB  R  2 R cos30  R Consider a fixed reference frame w.r.to the
fixed frame. The positions of the three blocks
x  R( 3  1) are given as shown in the figure.
Forces -acting on the bead are shown in the
figure
As the strings are unstretchable the total lengths

of the strings are constant.
upper string l1  y1  C  y p The free-body diagrams of the blocks are
l2  y2  y3  2 y p
On differentiating the equations w.r.to time t. 2T  g  a1 (2)
dy1 dy p
0 0  V1  Vp  0
dt dt
a1  a p  0 (i)
T  2 g  2a2 (3)
dy dy dyp
0 2  3 2 V2  V3  2Vp  0
dt dt dt
a2  a3  2a p  0 (ii)
From the equations (i) & (ii)
T  3g  3a3 (4)
a2  a3  2a1  0
The constraint equation that we got cannot be From the equations (1) (2) (3) & (4) we get
used directly with Newton’s IInd law to find the a1 
19
g , a2 
17
g , a3 
21
g
29 29 29
accelerations. We have to make an assumption
The block of mass 1kg is moving upward and
that all the blocks are going up or all the blocks the other two blocks are going down.
are going down. Let us assume that all the blocks are going
Let us assume that all the blocks are going in downward with the accelerations a1 , a2 & a3 as
upward direction as shown in the figure. The
tensions present in the string are shown.
The free body diagrams of the three blocks are

w.r.to ground frame
2T  g  a (i)
w.r.to pulleys (P) frame
Fs  3a0
The free-body diagrams of the blocks are
T  Fs  2 g  2a ( Fs -Pseudo
force )
T  2a  2 g  2a (2)
g  2T  a1 2 g  T  2a2 3g  T  3a3
w.r.to pulley (p) frame
By solving the above three equations we get
19 g 17 g 21g
a1   , a2  , a3 
29 29 29
The 1 kg block is going up as it is (-)ve and 2kg 3g  Fs  T  3a
& 3kg blocks are going down.
Method II
3g  3a  T  3a (3)
In this method we assume that a is the
acceleration of the pulley that is going down. By solving the above three equations we get
Acceleration of the 2 kg block is a w.r.to the
2g 19 g
pulley as shown in the figure. a , ao 
29 29
19 g
a1kg    upward 
29
19 g 2 g 17 g
a 2 kg      downward 
29 29 29
19 g 2 g 21g
a 3kg      downward 
29 29 29
Method - III (Effective mass method)
Consider the pulley system given.
19 g 48g 24 g
we get a1  29 , 2T  T 
29 29
To find the accelerations of the remaining two
blocks we consider the free body diagrams.
24
2g  g  2a2
29
17
a2  g
29
4  (2kg  3kg )
M
2kg  3kg
24
M kg 24
5 3g  g  3a3
29
Let 2T be the tension in the string and a1 be the
acceleration of the lkg block. 21
a3  g
29
27. (3)
a
28. (3) tan  
g
So,  is directly proportional to a. If a decreases
then  decreases.
29. (4) For the man standing in the left , the
acceleration of the ball
  
a b  a b  a   ab  g  a
Where ' a ' is the acceleration of the lift

2T  g  a1 (1) (because the acceleration of the lift is ' a ' )
For the man standing on the ground, the
acceleration of the ball
  
a bg  a b  a g  abg  g  0  g
24 24 ag  0 ( acceleration of the ground)

g  2T  a1 (2)
5 5 10
30. (3) The deceleration a   2m / s 2
5
By solving the above two The normal force or apparent weight
is N  m  g  a 
N  60  8 From figure:
N N  ma and
R  48kg wt
10
F  mg   N  mg  ma  mg  a  g / 
31. (3) Let a is the acceleration of the lift. When
the lift is going up 35. (3) As car is moving in anticlockwise
direction and have tangential acceleration as well
N  mg  ma  N  ma  mg  W1 as radial acceleration. Friction component should
be along tangential and radial acceleration. The
When the lift is going down
correct option is (3).
mg  N  ma  N  mg  ma  W2 36. (3) Consider FBD of blocks A and B as
From the above two eq’s we get shown in diagram below.
W1  W2 W  W2
a  mg  1
2m 2
32. (3) The absolute accelerations of P and the
L
2 2
block are aP   , ab  L 
2
Acceleration of the block w. r.t. P is
As the blocks are in equilibrium
L
abP   2 For block A, f A =20N and for block B,
2
The forces acting on the block w.r.t P are f B  f A  100  120 N
37. (3) The system is at rest
For maximum M/m; Limiting friction will be
acting on both blocks
L L
T  FS  m  2  FS  T  m  2
 mg   Mg cos θ  Mg sin θ
2 2
L 2 Mg  sin θ   cos θ    mg
 FS  m 
2
T  ML2 
M 

33. (3) m  sin θ   cos θ 
l 38. (1) At the verge ofslipping mg cos  N
The length of the hanging part is x 
1 
l 1
Given that x    
3 2
34. (2) The friction force on the block will prevent
the block from falling it.
mg sin  f   N
From the above eq’s we get tan   
dy
 tan    in limiting case
dx
dy 3x2 1
   x  1
dx 6 2
1
So, y 
6
39. (4) The limiting friction on the lower block is
f   R   (mA  mB ) g
1. (22)
f  0.5 10 10N  50N
As, a force of 10 N is unable to start the motion
of the system. There is no relative motion Let the balls starts moving with velocity ‘u’ and
between A and B. it reaches up to maximum height
40. (1) For vertical equilibrium of the block: H max , then
R  F cos  mg u2
From H max  2g
While for horizontal:
u  2 g ( H max )
 2 10  2  2 10 m/s
This velocity is supplied to the ball by the hand
and initially the hand was at rest, it acquires this
velocity in distance of 0.2 metre
u2 40
 a   100 m/s 2
F sin   R 2s 2  0.2
From (i) and (ii). So upward force on the ball F  m ( g  a)
F sin   (F cos  mg )
 0.2(10  100)  0.2 110  22N
 mg
F  2. (23.3) Let T be the tension in the string
(sin    cos )
41. (2) As the block can slide down with
constant velocity
f k  mg sin 
when it is projected upward f k & mg sin  act

At equilibrium m1 g    m  m2  g
downward & net retardation is | a | 2 g sin 
5  0.15  m  10
1  u 
S  ut  | a | t2  0 t  
2  g sin  
m  23.3 kg
42. (2) The frictional force acting on M is  mg
So the correct option is (1).
 mg 3. (120) Normal force on block A due to B and
 Acceleration 
M between B and wall will be F.
As A is in equilibrium 1
  0.29
Fricition on A due to B=20 N 2 3
 Friction on B due to wall  100  20  120N 4. (4.9) mg sin   ma
4. (14) s  t3  5  a  g sin
ds where a is along the inclined plane
 velocity, v   3t 2
dt  vertical component of acceleration is
dv g sin 2 
Tangential acceleration at   6t
dt
 relative vertical acceleration of A with respect
v 2 9t 4 to B is
Radial acceleration ac  
R R
g
2
g (sin 2 60  sin 2 30)   4.9 m/s 2
At t  2s , at  6  2  12 m/s 2
in vertical direction.
9  16
ac   7.2 m/s 2
20
 Resultant acceleration 1. (4)
 at2  ac2 
2
12   7.2 
2
 14m / s 2 2. (2) When a metre scale is moving with
uniform velocity, the force acting on the scale
is zero and the torque acting about centre of
mass of the scale is also zero.
dP d  1  1 d 3. (1)
1. (1) F    2 1  t 
dt dt  1  t  1  t  dt
1
F 2
1  t 
At t = 2s
F = 1.
2. (180) Since action and reaction acts in opposite
direction on same line, hence angle between
them is 1800 .
3. (0.29) Since work done by friction on parts Given that
PQ and QR are equal
m1  8 kg , m2  12 kg
 mg cos  4    mgx
Let T be the tension in the string and masses
3 moves with an acceleration a when masses are
 mg   4   mg x released.
2
x  2 3m  3.5m
(m2  m1 )
 a g
(m2  m2 )
Applying work energy theorem from P to R
12  8
3 a 10  2 m/s 2
mg sin 30 4   mg  4   mgx  0 12  8
2
The tension in the string is When the body moves down the plane, f acts
up the plane.
T  m1 g  m1a
(m2 g  f )  m1 g sin 
 m1 ( g  a)  8(10  2)  90 N
m2 g  m1 g sin   f
m2 g  m1 g sin    m1 g cos
1. (3) In uniform translatory motion, all parts of
m2  m1 (sin    cos )
the ball have the same velocity in magnitude and
direction, and this velocity is constant. Choice (4) is correct.

2. (2) Here, m  5 kg , F  (3iˆ  4 ˆj) N
Initial velocity at t  0, u  (6iˆ  12 ˆj ) 1. (1) In region AB and CD, slope of the graph
is constant, i.e., velocity is constant. It means
F no force acting on the particle in these regions.
Retardation, a 
m 2. (4) The block will begin to move only when
the external force becomes equal to the limiting
 3iˆ 4iˆ  friction at the floor. The net force acting on the
     m/s 2
 5 5 block after it starts moving is
As final velocity is along Y-axis, its x-component dV
must be zero. ct  f K  ma  m
dt
From vx  ux  ax t , for X-component only,,
 V  f (t ) is a curve.
3iˆ
0  6iˆ  t mdV m
5 3. (1) F  t   ma  t  a
dt 
5 6
t  10 s
3 1 2
V t
3. (2,4) Let f is the force of friction. When the 2m
body moves up the plane, f acts down the plane. t  t 
V   
2 m 
at m 2
V  a
2 2
dv dv F  bt
4. (3) m  F e bt   e
dt dt m
v F t
 bt
  dv  e dt
0 m 0
t
F  ebt 
f max   R   m1 g cos v  b 
m  0
When m2 g  m1 g sin   f
F
v (1  e bt ).
m2 g  m1 g sin    m1 g cos mb
5. (1) The frictional force present between the
m2  m1 (sin    cos )
blocks is f  mg
Choice (2) is correct
As V2  V1 the friction accelerates the block. 1

 K
The velocity of the block at time t is V1  gt l
The velocity of the plank after time t is V2  at 2. (4) As relative velocity is increasing the
relative accleration should be non zero.
let t  is the time at which the relative velocity
between the block and Plank becomes zero. If F1  0 & F2  0 then a1  0 & a2  0
V1   gt  V2  at  ar  0
V1  V2 3. (1) Here force is constant then slope of
 t 
a  g surface should change from one constant value
(non zero) to another constant value (non zero)
For t  t 
1
Velocity of the block is original length i.e. x  1 K 
l
Vb  V1  gt 4. (3) The two blocks move together with the
same acceleration as long as the force of friction
slope is constant. between them is less than the limiting friction,
Here both the block and Plank move together as the only force on the lower block B is the
with common velocity as (a  g) force of friction. Limiting friction is reached, the
 f 
i.e., after t  t  the slope decreases. max
acceleration of B becomes constant.  m  ,
 B 
The correct answer is (1).
and the acceleration of A increases with faster
6. (1) When the applied force is P1 the friction rate.
 mg cos acts upward. When the applied force
is P2 the friction  mg cos acts downward. 5. (1) The force on the block as a function of
time is F  25t
7. (3) Since the graphs between x and t is a
straight line and passing through the origin. At t = 2s the block starts moving as F  f .
the net force acting the block is FNet  25t  50
At x  2, t  2
4
 xt The impulse mV  1 (25t  50)dt
At t  2, y  4 m  10kg  V  5m / s
6. (1) Extension in the spring is zero until
Since the graph between y and t is a parabola.
mgsin   mgcos 
 y  t2
Afterwards gradually x increases.
dx dv When   tan 1 ()
 vx   1 and ax  x  0
dt dt mg sin    mg cos
dy x
and v y   2t and a y  2 ms 2 K
dt
The force acting on the particle is dv
7. (2) F  at; m  at;  mdv   at dt
F  may  (0.5 kg )(2 ms 2 ) dt
 1N along y-axis at 2 ds at 2
mv  m 
2 dt 2
at 2
1. (4) We know that K 
C  mdS   2
dt
l
at 3 a  F 
3
 2  mgsin   mg cos  (1)
mS     ;
6 6 a  Case II When the force 10 N is acting on the
S  F 3 block, free body diagram of the block is given
8. (4) At 11th second lift is moving upward with as follow:
acceleration
0  3.6
a  1.8 m / s 2
2
T  m( g  a )
1500(9.8  1.8)  12000N
10 N  mg sin    mg cos (2)

1. (4) Limiting friction From eq’s 1 & 2
g sin   g cos  2 1   3 1 3
 ;  
g sin   g cos  10 1   3 5 2
3. (2) Time taken to slide along smooth surface
1
S g sin 450 t12
2
where S is thelrngth of the incline
f s  mg cos 450 t1 

2 2S
g
1
 0.6  10  10   42.43 N Time taken to slide along rough surface
2
when the block is at the verge of sliding down. 1
S  ( g sin 450   g cos 450 )t22
2
P  f  3N  mgsin 45  73.71N
2 2S
 P  73.71  42.43  31.28 N t2 
g (1   )
2. (2) Case I
Given that
When 2 N force is acting
t2  nt1
Resolve the forces along and perpendicular to
the plane. 2 2S 2 2S 1 1
 n2   1   2    1 2
g (1   ) g n n
4. (2) Let  be the minimum coefficient of
friction
 mg cos  mg sin   2 N
At equilibrium mass does not move so, N + Fsinθ - mg = 0 (1)

3mg sin  3mg cos
Fcosθ -μN = ma (2)
 min  tan 
Eliminating N from this two eqauations,
5. (1) Given that acceleration a can be expressed as:
m = 5 kg Fcos     mg  Fsin  
 a
at t  0s v1  6iˆ  2ˆj m
 F  Fsin  
at t  10s v 2  6iˆ a cos     g  
m  m 
 
 v 2  v1 6iˆ  6iˆ  2ˆj
a  
 
3iˆ  4jˆ 2. (30) Using conservation of linear momentum,
t 10 5  
pf  pi
F = ma

 3iˆ  4jˆ or, m  10ˆj  mv  m  10 3iˆ
6. (2) For pushing 
 v  10 3iˆ  10ˆj
N  mg  FA sin 30
 N  FA cos 30 
1

which makes an angle tan 1/ 3  30 with 
x-axis.
  mg  FA sin 30   FA cos30
 
3. (1) Acceleration of the particle

 F 2i  3j  5k
 mg a  
FA 
cos30   sin 30
 m 2
 Displacement of the particle
For pulling
   1
mg  r f  r i  ut  at 2
FB 
cos 60   sin 60
 2
FA 2
   1  2i  3j  5k  2
FB 3  rf  0i  0j  0k  0   
    4 
2  2 

or,   
r  8i  12 j  20k
f
 b  12.
4. (2) Statement - 1 :
On a horizontal ground,
v max  Rg  0.2  2  10  2 m/s

1. (1)
18
 2 km/hr
5
= 7.2 km/hr
Using  F  ma, we can write for vertical and
As speed of the cyclist (7 km/hr) is less than
horizontal directions,
7.2 km/hr, the first statement is correct.
Statement-2: On a banked road, v 2  v02  2 gh
 tan     1.2 Let ball travel distance ‘S’ before coming to rest
v max  gr    2  10   3.6 km/hr
 1   tan   0.8
S
v2 v 2  2 gh
 0
= 19.72 km/hr 2 g 2 g
v02 2 gh h v2
 tan     0.8     0
v min  rg    2  10   3.6 km/hr 2 g 2 g  2  g
 1   tan   1.2
8. (4) Acceleration produced in upward
= 13.15 km/hr direction
As V min < 18.5 < Vmax, the second statement is F
also correct. a
M1  M 2  Mass of metal rod
5. (3) Let T is the tension in the string
480
  12 ms 2
20  12  8
The tension at the mid point
 Mass of rod 
T   M2  a
 2 
 (12  4) 12  192N
mv 2 9. (2) Until F equals limiting friction a = 0 after
T sin   ,T cos   mg
r that a increases linearly with t. The correct
v2 graph is (2)
 tan  
rg 10. (4) At x  0    0 i.e., the body accelerates
down and v increases. When
  45  v  rg  0.4  10  2m / s
  tan  where
6. (4) Speed v  10m / sec
mg sin  mg cos
v2
a
R The body reaches maximum velocity.

| v | constant i.e., 0.3x  tan 45  x  3.33m
Beyond 3.33 m the friction dominates mg sin
1
a and the block decelerates and its velocity
R decreases.
11. (1) At equilibrium of the rotatig body
Kx  mr 2
Kx  m  l0  x   2
K 1cm  m  l0  1cm   2 (i)
7. (2) Initial speed at point A, u  v0 K  5cm  m  l0  5cm 4 2 (ii)
Speed at point B, v  ? From eq’s (i) & (ii)

l0  15cm
v 2  u 2  2 gh
Work, Power & Energy 109
8. (1)
5 5 5
W   Fdx    7  2 x  3x2 dx  7 x  x 2  x3  0
0 0
W=135 J
9. (3) Here, mass of rope m  l  d
1. (1)
For pulling the rope on the table,
2. (2)
Distance of centre of gravity moved = l/2
3. (1) Absolute displacement of the point of
l l dgl 2
application of force applied by the man through  Work done  F   ldg  
bars is zero. Hence work done is zero. 2 2 2
4. (2) Area of acceleration-displacement curves 10. (4) W net

 K
gives change in KE per unit mass As particle is moving slowly, this means
1 md
m  2  u 2   FS  s K  0
2 dt
WN  WF  Wmg  K
Changein KE d
  s  
Mass dt But WN  0 as N  dr
5. (1) When the block moves vertically 0  WF  mgh  0
g WF  mgh
downward with acceleration then tension in
4
11. (1) Work done in stretching a wire
the cord
1
 g 3 W  kx 2
T  M  g    Mg 2
 4 4
If both wires are stretched through same
  distance then W  k. As k2  2k1 so W2  2W1 .
Work done by the cord F .S  FS cos
12. (3) The forces that act on the mass are F,
 Td cos180 mg and tension (T). From work energy theorem
 3  d W   K
   Mg   d  3Mg .
 4  4
  WT  WF  Wg  0
6. (1) W  F  r  (5iˆ  3 ˆj  2kˆ)  (2iˆ  ˆj )
WT  0 since tension at a point is always
 10  3  7 perpendicular to the small displacement of the
mass.
1 F2 1
7. (2) W  kx2   W  (F = constant)
2 2k k  WF  Wg  U  mgL(1  cos )
110 Work, Power & Energy
13. (1) Mass of the chain hanging  4  3  12kg From work energy theorem W g  Wspr  KE
Shift in centre of gravity  4 2  2m The body velocity is zero both at initial and final
stages.
W  mgh  12  9.8  2  235.2 J
As KE  0
14. (3)
Wg  mg ( h  d )
c
15. (1) Given F  (c- is a constant)  Wspr   mg ( h  d )
v
dv c 1 2 kx 2
kx  mgh  h 
 ma  m dt  v 20. (3)
2 2mg
21. (2) Positive work done by a conservative

 mv dv   c dt force always decreases the potential energy
1 2 W  U
mv  ct  K .E  t
2
 U  (W)  (ve)  ve
16. (1) a  v / t1 , Displacement after time t:
22. (2)
1 1v 2 23. (1) Gravitational force is a conservative force
S  at 2  t
2 2 t1 and work done against it does not depend on the
path.
2
v 1 v 2 1 2 t  24. (2) The expressions for force components are
W  FS  maS  m t  mv  
t1 2 t1 2  t1 
U  xy U  x 2
Fx   , Fy  
x 2 y 4
F2 T2
17. (3) U  .
2k 2k
U
Fz  0
18. (1) Work done on the body = K.E. gained by z
the body
 xy ˆ  x 2 ˆ
1 1 F i j
Fs cos  1  F cos    2.5 N . 2 4
s 0.4
25. (3) Since the particle is moving in horizontal
19. (2) The situation is shown in figure. When circle, centripetal force,
mass m falls vertically on spring, the spring is
compressed by distance d. mv 2 k
F  2 (i)
r r
k
mv 2 
r
Kinetic energy of the particle,
1 k
K  mv 2  (Using (i))
2 2r
dU
As F 
dr
 Potential energy,, 1 2 1
Mgx  k  x  x0   kx02
r r
 k 
r
k 2 2
U    Fdr     2  dr  K  r 2 dr  .
   r   r
 Mg 
On solving x  2   x0 
k k k  k 
 Total energy  K  U    .
2r r 2r
31. (2)
26. (4) K.E = E-U, can be positive or zero
32. (2)
27. (3) Using conservation of energy principle
33. (1) In the case motion of stone is in vertical
v  2 gh  2  10  10  10 2m / s circle of radius L and centre at O.
The change in velocity is
 is independent of g

28. (3) From law of conservation of energy, we v  vjˆ  uiˆ
have the potential energy of vertically positioned
2 2
rod is converted into kinetic energy of rotation. v   v    u   v2  u 2
According to work-energy theorem
1 1
Wg  mv 2  mu 2 (i)
2 2
1 1
mgL  mv 2  mu 2
2 2
l 1  v 2  u 2  2 gL
Mg  I  2
2 2
  v  v 2  u 2  2  u 2  gL 
Moment of inertia of a thin rod is
34. (3) As the body moves on a circular track
1 the body can complete vertical circular motion
As, I  Ml 2
3
When the velocity at D is gR
2
M 1 Ml 2 3g
 gl     From conservation of energy
2 2 3 l
T  E A = T  ED
Given, l  1m   3 g
1 2
Also, v  r mgh  mg (2 R)  m
2
 gR 
 v  1  3  9.8  5.4ms 1
5R 2h
h R  2 cm
1 2 1 2 2 5
29. (2) kx  mv
2 2
From the given options Rmax  2 cm
600  9 104  15 103  v2
 d  P
 v  6m/s 35. (2) P  F  ma  m    dt   d
 dt  m
30. (4) Loss of gravitational potential energy of
1/ 2
block is equal to gain in elastic potential energy P 2  2P  1/ 2
of spring  t      t 
m 2  m 
2
 2P 
1/ 2
1 2 2 3   r3 
Now s    dt    t 
1/ 2
dt mv  0    t 2  dt or v 2   
 m  2 0
2   2 0
1/ 2
3/ 2  v = 2 m/s.
 2 P   2t 
 s     s t 3/ 2 .
 m  3 
P 1. (4) There is no displacement.

36. (3) P  Fv  mav  a 
mv
2. (3) As surface is smooth so work done against
dv P P friction is zero. Also the displacement and force
v   v 2 dv  ds of gravity are perpendicular so work done against
ds mv m
gravity is zero.
P s v2
3. (1) In explosion of a bomb or inelastic collision
  ds   v 2 dv
m 0 v1 between two bodies as force is internal,
momentum is conserved while KE changes.
P 1 m 3 3 Hence, KE of a system can be changed without
 s   v23  v13   s   v2  v1 
m 3 3P changing its momentum.
 4. (4)
37. (3) 
Here, Force, F  4iˆ  ˆj  2kˆ N 

5. (4)

Velocity,   2iˆ  2 ˆj  3kˆ ms
1
 6. (1) Work done  mgh  10  9  8 1  98J .
 

Power, P  F .  4iˆ  ˆj  2kˆ . 2iˆ  2 ˆj  3kˆ  7. (4) W  mg sin  s
 8  2  6 W  4W  2 103  sin15 10

38. (4) A truck is moving on an inclined plane = 5.17 kJ
therefore only component of weight 8. (3) Tension in the string
 mg sin  will oppose the upward motion g  Mg

T  M  g  a  M  g   
Power  force  velocity  mg sin   2 2
 1  30  5 W  Force  displacement
 30000  10     25kW .
 100  18
Mgh
39. (1) Work done in raising water = mgh 
2
W  (volume  density) gh   9 1000 10 10 9. (4)
 
5 W  F  s  (5iˆ  3 ˆj )  (2iˆ  ˆj )  10  3  7 J
 W  9 10 J

10. (2) Here F  iˆ  2 ˆj  3kˆ
work 9 105
Useful power    3kW 
time 5  60 S  4 kˆ
3  
 Efficiency  100  30%  W  F  S  (iˆ  2 ˆj  3kˆ)  4kˆ  12 J
10
1 2
40. (1) From work-energy theorem, 11. (3) W kx  W  k (x = constant)
2
 KE  Wnet or K f  K i   Pdt Opion (3) is correct.
1 W  F  S
12. (3) W  k  x22  x12 
2
F  mg sin    mg cos 
1
  800  152  52   104  8 J W  mg S sin    mg S cos 
2
W  mg y   mg x
2 1
13. (4) PE of spring  KE of mass  mv
2 yB xB
W   mg y  mg   x
1  150  2 yA xA
  0.2   3  10 J
3
 
2  1000 
W  mg (0)   mg ( xB  x A )
a
14. (2) W   Fy dy   ( Ay
2
 By  C )dy W   mg (10 m)
a
W  0.2  1  10(10)  20 J
3 a 2 a
y  y  a
 A    B    C  y  a 18. (1) Here, k  2 Ncm 1  2  102 Nm 1
 3 a  2 a
x1  5cm  5  10 2 m
2 Aa3 x2  15cm  15  10 2 m
W  2Ca
3
Work done = Increase in PE of spring
15. (3)
1
x1 x1  x2  1
x1
 k  x22  x12 
W   F .dx   Cxdx  C    Cx12 . 2
0 0
 0 2
2
1 2 2
  2  102 15  102    5  102  
L 2  
16. (1) The weight of hanging part   of chain
3    102  200  104  2J
1  19. (4)
is  Mg  . This weight acts at centre of gravity
3 
of the hanging part which is at a depth of L/6 20. (2) When a force of constant magnitude
from the table. which is perpendicular to the velocity of particle
acts on a particle, work done is zero and hence
As Work done  force  distance change in kinetic energy is zero.
Mg L MgL 21. (2)
W   
3 6 18 22. (4) Friction is a non-conservative force. Work
17. (3) Consider a small segment on the curve. done by a non-conservative force over a closed
Let  S is the length of the segment path is not zero.
2
p 2  Ft  F 2t 2
23. (2) Kinetic energy E   
2m 2m 2m
[as P = Ft]
F2 U k
24. (4) U  1  2 [ if forces are same]
2k U 2 k1
U1 3000 2
   .
Small work done by the external force is U 2 1500 1
25. (4) According to work-energy theorem 29. (3) Acceleration of the body : a  V T .
W=Change in kinetic energy
Velocity acquired in time t
1 1 V
FS cos  mv 2  mu 2 v  at  v  t
2 2 T
Substituting the given values, we get  K  E  v2
20  4  cos  40  0 Work done on the body is equal to gain in
the kinetic energy.
40 1
cos  
80 2 V 2t 2
So Work done 
T2
1
  cos 1    60 30. (1) By work -energy theorem
2
 
26. (3) Initial energy of body 1 2

W  K   10  10   0  500 J
1 1 2
2
 mv 2   1  20   200 J .
2 2
1 2 1 2
A part of this energy consumes in doing work 31. (1) mv  kx
2 2
against gravitational force and remaining part
consumes in doing work against air friction. 2
 mv 2  kx 2  0.5  1.5   50  x
2
i.e., K .E  W grav  Wairfriction

 x  0.15m
 200  1  10  18  Wair  Wair  20 J .
32. (4) Initial momentum  p1   p

27. (2)  
vi  6.00iˆ  1.00 ˆj m / s 2 Final momentum  p2   1.5 p
vi  vix2  viy2  37.0 m / s. p2

Kinetic energy  K    p2
 2m
v f  8.00iˆ  4.00 ˆj
2 2
K1  p1   p  1
v 2f  80.0m 2 / s 2      
K 2  p2   1.5 p  2.25
1
 K  K f  Ki  m  v 2f  vi2   K 2  2.25K
2
K  2.25K  K  1.25K or 125%
3.00
 80.0  37.0  64.5J . 33. (3) Among the given forces, force of friction
2
is a non-conservative force whereas all other
28. (1) The velocities of both the masses will be forces are conservative forces.
equal, i.e., v1  v2  v 34. (1) In case of non-conservative forces, the
work done is dissipated as heat, sound etc, i.e.,
1 it does not increase the potential energy.
m v2
K1 2 1 m 2 1
  1   . 35. (1)
K 2 1 m v 2 m2 4 2
2
2
36. (2) W  U
37. (2) 45. (2) By applying work energy theorem

38. (4) Any constant force is a conservative 1 v2 1 2 1
force. m  mv   kx 2
2 4 2 2
39. (1) The ice cube is in neutral equilibrium.
Since it has zero acceleration 3mv2 1 3mv2
   kx2 ; k 
8 2 4 x2
U ˆ U ˆ
40. (4) F  i j  7iˆ  24 ˆj 46. (3)
x y
1 2
Fx 7 0.75(12 J )  mgh  mv  v  18 m / s
 ax    1.4ms 2 along positive x-axis 2
m 5
47. (3) Acceleration due to gravity is
F 24 independent of mass of the body. Hence, the
ay  Y  
m 5 two spheres have the same acceleration.
 4.8ms 2 along negative y-axis 48. (1) Man possesses kinetic energy, because
of his velocity (v). When m is mass of man,
 v X  ax t  1.4  2  2.8ms 1 then
1
and vY  4.8  2  9.6ms 1 K  mv 2
2
Let v1  v, m1  m2  m
 v  vX2  vY2  10ms1
41. (1) As the total energy is constant v2   v  2 ms 1
T.Ei  T.Ef Then K 2  2 K1
K i  Ui  K f  U f K1 v12 K1 v2
Then, K v  
2
2
2 2 K1  v  2 2
1
Ki  m  02   0
2
 v 2  4v  4  0
U i  0 (At origin) 4  16  16
 v1 
2 2
U f   2   3  2   2J
v1  2  
2  1 ms 1
 K f  2J
49. (2) Let l is the length of the string. Tension is
42. (3) maximum at the point A and minimum at point
43. (2) B
1 2 1
44. (4) K E  mv  m(2 gh )
2 2
 K  E h
1 2 1
K E  mv  m( gt )  K  E t 2
2 2
At point A:
mu2
2 54. (4) T  mg 
mv l
T1  mg  (1)
l
 mg  5mg
At Point B:
 6mg
T2  mg cos (2)
55. (4) The velocity of the mass at point B is
Given that
vB  2 gl
T1  4T2 (3)
mvB2
TB  mg   3 mg
From the above eq’s l
mv 2 56. (1) The driving force on the car is
 mg   4mg cos
l
12
mv 2 F  1000  10 N  1200 N
 mg (4cos  1)  (4) 100
l
Conserving energy between points A and B P  Fv  1200 N  15ms 1  18kW .
1 2 57. (3) Constant power of a car P0  F .  ma.

mv  0  0  mgl (1  cos )
2
d m 2
2 P0  m .  P0 dt  m d  P0 .t 
mv dt 2
 2mg (1  cos  ) (5)
l
2 P0t
From equations (4) & (5)    t .
m
mg (4cos  1)  2mg (1  cos )
58. (2) Power: P = Fv
   60
1
50. (1) When the bottle is in rotation the pressure For power to be constant: F 
v
increases radially away so the bubbles come

towards the centre that is towards the neck. 59. (3) 
Here, Force, F  4iˆ  ˆj  2kˆ N 
51. (1) 
52. (2) Kinetic energy given to a sphere at lowest 
Velocity,   2iˆ  2 ˆj  3kˆ ms 
1
point = potential energy at the height of  

suspensions  
Power, P  F .  4iˆ  ˆj  2kˆ . 2iˆ  2 ˆj  3kˆ 
1
 mv 2  mgl
2  8  2  6 W  4W
 v  2 gl mgh P m t
60. (1) P  1  1 2
1 2 1 t P2 m2 t1
53. (3) mv  mgh  mu 2
2 2 [ as h = constant]
Given h = l
P1 60 11 11
    .
For umin , v  0. umin  2 gl P2 50 12 10
61. (2) 1 2 1
3 3. (1) Wmg  K f  Ki  m  v cos   mv 2
mgh p  t 2  10  60 2 2
P m   1200kg
t gh 10  10
1 1
 mv 2 cos 2   1   mv 2 sin 2 
mass 1200kg 3 2 2
As volume  density  V  103 kg / m3  1.2m
5,7  5,7 5,7
3 3
Volume  1.2m  1.2  10 litre  1200 litre. 4. (4) W  k   ydx  xdy    d ( x, y)  k  xy 3,5
 3,5  3,5
62. (3) Here, m = 1800 kg
 k  35  15  20k
Frictional force, f  4000N
1 1 2
v  2ms 1 5. (1) W  kx 2  k  x  y 
2 2
Downward force on elevator is
1 1
F  mg  f W  kx 2  k  x2  y 2  2 xy 
2 2
 1800kg  10ms 2   4000 N  22000 N 1

 ky  2 x  y 
2
The motor must supply enough power to balance
this force. Hence, The work done against elastic force is
P  Fv   22000 N   2ms 1   44 kW ky
Wext  W   2x  y 
2
63. (1) Here the car is under acceleration
x F y F
F  fr 6. (1) Here y  x
P  Fv  a 
m
That’s why force is conservative and particle is
F  1000
1   F  1500 moving on circle i.e., x 2  y 2  a 2
500
 P  1500  5  7500 Watt. Here work is independent of path
64. (1) Power consumed 0
Hence W  a Fx dx  Fy dy
80
  100W  80W  80 Js 1 K 0 K a
100  3 a
x dx  3  y dy  0
a a 0
7. (2) Let x be the length of the chain on the
table.
1. (3) We are applying force F therefore tension
in string is equal to F. Hence force acting on
weight is 2F upward, and displacement is h
upward.
The frictional force on the chain is
Hence W= 2Fh (work done on weight)
 M
2. (2) The ball rebounds with the same speed. f   ( x ) g   
 L
So change in its Kinetic energy will be zero, i.e.,
work done by the ball on the wall is zero. dW  fdx    g xdx
0
 Mg 0
 M  m v  M 2 gh
W   g  xdx   x2 
2 L /3
2 L   2 L /3
M 2 gh
v
2
M m
 Mg   2 L     Mg 4 L2
W       Let the resistance be F then
2 L   3   2L 9
Total work= change in K.E.
2  MgL
W 1 M 2 2 gh
9 i.e., F  M  m 2
  M  m g
2 d  M  m
 
8. (1)  
F  3x2iˆ  4 ˆj , r  xiˆ  yjˆ
M 2 gh
    M  m g
 d r  dxiˆ  dyjˆ  M  m d
Work done, 11. (1) By conservation of energy,
   3,0 2 1
2mgx  kx2  0  x 
4mg
W   F .d r  
 2,3 

3x iˆ  4 ˆj . dxiˆ  dyjˆ   2 k
12. (3) Let N be the normal force on the block
 3,0 3,0 

 2,3
 3x dx  4dy    x
2 3
 4 y  2,3
 
 (33  4  0)   23  4  3  7J
Change in the kinetic energy = Work done,

W = +7 J
9. (4) When level becomes equal then N sin   ma
h1  h2 N cos  mg
h
2 a
 tan    1    45
Work done by gravity = change in potential g
energy of the water column W  ( N sin )(vt )  150J
h  h gh  2 2 10
 2  Ah  g   Ah1  g 1  Ah2  2  13. (3) l1   4m, l2   m
2  2 2  sin 30 sin 37 3
10 2
 h2 A gh22  Displacement of point P : l1  l2  4   m
 A gh2   A g 1  3 3

 2 2 
 h  h 2 h 2  h22 
 A g  1 2   1 
 2  2 
2
h  h 
 A g  1 2 
 2 
10. (3) Using law of conservation of momentum, Work done, W  F  l1  l2   50  2  100 J
we get 3 3
14. (2) Let v be the final common velocity. 1

W  K f  K i  0  mv 2
2
1 2
   6  1024  3  104   2.7  1033 J
u 2
 m  2m  v  mu  v 
3
1
18. (3) sin  
x
From free body diagram of the body
 mg  mg g
a1    g , a2  
m 2m 2
Acceleration of B relative to A
g 3
a1  a2   g    g towards left
2 2
Linear momentum of the system is conserved
Pi F  mg sin   ma
so 1
Pf
g 
F  m  g sin   a   m   a  (1)
15. (1) Wspring  W100 N   k  onA  x 
 10  1 2
Displacement of the body till its velocity
Wspring  100      2  2  reaches v
 100  2
Wspring  4  10  6 J
v2
v 2  0  2as  s 
2a
dx d  t 3  2 m v2
16. (1) v   t
dt dt  3 

W  Fs cos 0   g  xa  
x 2a
When t = 0, then v = 0. When t = 2, then
mv 2
v = 4m/s   g  xa 
2ax
Work done in first two seconds = change in KE 
19. (4) If the force F  Fx iˆ  Fy ˆj is
1 2 2 1
W  m  4    0    2  16  16 conservative then
2   2
JFx JFy
17. (2) As T   2   , 
Jy Jx
So   2  3.15  10 7   1.99  10 7 rad s For the force given in option (4)
Now v  r  1.5 1011 Fx  x 2 y 3
1.99  10 7  3  10 4 m s
JFx
 3x 2 y 2
Now by work-energy theorem, Jy
Fy  y 2 x 3 21. (1) U  20  5sin(4 x)
dU
JFy 2 2
F  20 cos(4 x)
 3y x dx
Jx
At equilibrium F = 0
JFx JFy
  3 5

Jy Jx 4 x  , , 
2 2 2
For the remaining forces the above condition is 22. (4) The graph between U & x is
not fulfilled. 2  1
U  K1ex   K1 x2 
x4
x 2    e 
20. (1) Potential energy V  
4 2
A t x  0 U  K 1  1  0
For maximum kinetic energy, potential energy
of a particle should be minimum. As x   U  K 1  e   K
dV d 2V
For minimum value of V,  0 and 2  0
dx dx
3
 dV  4x 2x
Force F        0  x3  x  0
 dx  4 2
 x  x 2  1  0
It is an exponentially increasing graph of
i.e., at x  0, x  1, and x  1 force on the
potential energy U . From the graph it is clear
particle will be zero.
that at origin. Potential energy U is minimum
d 2V (therefore, kinetic energy will be maximum) and
Now,  3x 2  1 force acting on the particle is also zero because
dx2
dU
d 2V F  (slope of U  x graph) = 0
For x  1, and x  1 2  1 dx
dx
Therefore, origin is the stable equilibrium
It means the potential energy of the particle will
position. Hence, particle will oscillate simple
be minimum at x  1, x  1. harmonically about x  0 for small
Now substituting these values in expression of displacements. Therefore, option d is correct.
potential energy (1), (2) and (3) options are wrong due to the
following reasons:
 14 12  1
Vmin    J  J From the graph we can see that slope is zero at
 4 2  4
x  0 and x  .
 Kinetic energy max  Total energy Now among these equilibriums stable equilibrium
position is that where U is minimum (Here
 1 x  0 ).
  Potential energy min  2    
 4 Unstable equilibrium position is that where U
is maximum (Here none).
1 2 9 2 9 3
mmax   max   max  m / sec.
2 4 2 2 Neutral equilibrium position is that where U is
constant (here x   ). 26. (4) The potential energy stored in the spring
is converted into kinetic energy of the block.
Therefore, option (1) is wrong.
1 2 1 2
For any infinite non-zero value of x, force is kx  mv
directed towards the origin because origin is in 2 2
stable equilibrium position. Therefore, option (2) 1 2
is incorrect. As kx is constant for both blocks
2
At the origin, potential energy is minimum, hence
m1v12  m2 v22
kinetic energy will be maximum. Therefore,
option (3) is also wrong.
v1
2  dU mv12  4mv22  v2 
2
23. (2) F 3  2 
r r dr
v12 sin 2 
h1 
2  2g
dU  3
dr  2 dr
r r
v22 sin 2  h
U r
1 r
1 h2   h2  1
2g 4
 dU  2  r
0 
3
dr   
 r
2
dr
27. (4) From conservation of energy

 
U 2  1 1
r r m1v12  m2 v22  2.4
2 2
dU  x  or m1v12  m2 v22  4.8 (i)
24. (2) At equilibrium 0
dx
Now m1v1  m2 v2 or v1  48v2
1
12a 6b  2a  6
1 48 2
 11
 5 x  Using (i),
2
 48v2   v2  4.8
x x  b  1000 1000
a b b2 10 103
U at equilibrium    or v2  m/ s  cm / sec.
2
 2a   2a  4a 7 7
   b  28. (1) The initial and final position of the upper
b
 
block is shown in the figure.
U  x    0
 b 2  b2
D  0     
 4a  4a
25. (1) Work done   KE ,
 2r 
The C.M of the chain lies at a height of  r 
  
from the horizontal surface.
 2r  1
Mg  r    Mv 2
   2
 2 Here l0 is the natural length of the spring. The

v  2 gr 1  
  initial compression in the spring is x0 .
When the upper block reaches a height of x

above from the natural length of the spring then
the lower block gets lifted up.
kx = mg (i)
From conservation of energy.
1 2 1
kx0  mgh  kx 2  mg (h  x  x0 ) (ii)
2 2
From the above two equations. m  vb2  vt2 

Tb  Tt  2mg 
3mg R
x0  From conservation of energy
k
29. (3) Velocity of a ball to just reach the top of m  vb2  vt2  m  vb2  vt2 
the tube should be given by;   0  2mgR   0   4mg.
2 R
0  v 2  2 gh0 Substituting into the above equation gives
Where the velocity will be zero at topmost point. Tb  Tt  6mg
Here h0   2R  h   Tb  Tt  6mg
32. (1) The velocity of the bob at the lowest
Thus v  2 g  2R  h  position is
u  2 gL0
mv 2
30. (1) At point C: mg cos   N 
r If the bob has to make circular motion then
u  2 gL0  5g ( L0  L)
3
L  L0  9cm
5
33. (3) Power of motor initially  p0
(When block leaves the surface normal force Let, rate of flow of motor = (x)
v2
becomes zero, so putting N = 0)  g cos   work mgy  y
r Since, power, p0    mg  
time t  t
v 2  v02  2 gh, h  r  r cos
y
2  x  rate of flow of water

gr cos  0.5 gr   2 g  r  r cos  t
= mgx (i)
1 3
 cos    2  2cos  cos   if rate of flow of water is increased by n
4 4
times,i.e., (nx)
31. (4) Applying Newton’s Second Law at the
mgy '  y'
bottom (2) and top (t) of the circle gives Increased power, p1   mg   ,
t  t 
= nmgx
mv 2 mv 2
Tb  mg  b and Tt  mg  t The ratio of power
R R
p1 nmgx n
Adding these gives    p1 : p0  n :1
p0 mgx 1
34. (1) The force that accelerates the block is mass dm

39. (1) K 
mg sin  a  g sin  length dt
Velocity of the block after travelling distance l 1

KE  mv 2
is 2
v 2  2 g sin  l 1  dm  2 1  dm dx  2 1 1
P  v     v  kvv 2  kv3
The instantaneous power is 2  dt  2  dx dt  2 2
40. (1) Observer A is on the ground

P  Fv  mg sin  2 g sin  l
1
Wspring   K .E  0  mv02
2
P  2m 2 g 3 sin 3  l
1
Wspring   mv02
W nk 2
35. (3) P  P  nk
t 1s 41. (2) W.r.to the observer B the block is in
36. (4) Force is required to oppose the resistive dynamic equilibrium. The velocity and
force R and also to accelerate the body of mass acceleration of the block are equal to zero.
m with acceleration a. K .Eblock  0, ablock  0
Force F = R+ma The correct option is (2).
 Power = (R + ma) v 42. (4) As observer A is in ground frame all the
 options are correct.
37. (4) F  2tiˆ  3t 2 ˆj
43. (4) WN  NS

d N  Mg  Ma
m  2tiˆ  3t 2 ˆj (m = 1 kg)
dt
 
1 1
    S  aT 2  WN  M (a  g )aT 2
2 2
  d    
2tiˆ  3t 2 ˆj dt  v  t 2 iˆ  t 3 ˆj
0 0
 
Power  F .v   2t 3  3t 5 W .
1. (1) Normal force and displacement are
dv  perpendicular to each other.
38. (3) Using P  Fv  M  v
 dt  2. (4) Given
F  20kg  wt  20  9.8N and s  20m
P P
i.e., v 2 dv  vdt  dS
M M  Work done  Fs cos 
v S  20  9.8  20  cos60  1960J

P 2
Integrating  v dv  M  dS 3. (1) As shown in the figure
u 0
3PS
v3  u 3 
M
1/ 3
 3PS 
v  u3 
 M 
f  mg sin , vertical displacement in time at t  0s, v  2  0  6  6

t  t
at t  6s, v  2  6  6  6
Work done  f t sin
Initial and final KE are same hence no work is
2 done
 mg t sin 
4. (4) While moving from (0, 0) to (a, 0) 1

W  m  v12  v22   0
 2
Along positive x-axis, y = 0  F  kxjˆ
7. (2) The initial position of the chain is
i.e., force in negative y-direction while
displacement is in positive x - direction.
W1  0
Because force is perpendicular to displacement.

Then particle moves from (a, 0) to (a, a) along
a line parallel to y-axis (x = + a) during the

F  k  yiˆ  ajˆ 
W xt   U  U f  U i
The first component of force, kyiˆ will not by taking the reference line to measure P.E on
contribute any work because this component is the surface of the table
along negative x- direction  iˆ while U i  m1 g (0)  m2 g (0.3)
displacement is in positive
U f  mg (0)  0
y - direction (a, 0) to (a, a). The second
component of force i.e., kajˆ will perform 4kg 4kg
m1  1.6m, m2   0.6
negative work 2m 2m
W xt  m2 g (0.3)  3.6 J
  
W2  kajˆ ajˆ   ka  a   ka 2
x0 x0
So net work done on the particle W  W1  W2 cx03
8. (1) w   Fdx   cx 2 dx 
0 0
3
 0   ka 2   ka 2
9. (1) F  ax  bx 2
5. (4) Work done in stretching the rubber band
dW  Fdx
L
W    ax  bx  dx2 3
L
0
W   (ax  bx 2 )dx
aL3 bL4 0
 
3 4 aL2 bL3
W 
6. (4) Here t  x  3 2 3
2
10. (2) Work done in stretching a string to obtain
x   t  3  t 2  6t  9 an extension l is
dx 1
v  2t  6 W1  Kl
dt 2
Similarly, work done in stretching a string to 16. (4) Work done = Change in K.E.
obtain extension l1 is
1 2
i.e., Wg  Wair  mv
1 2 2
W2  Kl1
2 Where W g  mgs and Wair is the work done by
 Work done in second case air resistance
1 1
 W2  W1  K  l12  l 2  Wa  mgs  mv2
2 2
11. (1) From equation of motion m

2
 v2  2 gs    m2  2 gs  v2 
v  u  at  20  0  a 10
17. (3) K  W
20  a 10  a  2m / s 2
K W
Now, distance   Force
S S
1 1
s  ut  at 2  s  0   2  10  10 18. (4) Taking block as system the work energy
2 2 theorem is
or s = 100 m
W f   K  U  0  mgh
4
W  F .s  mas  50  2  100  10 J .
19. (1) We know that
12. (3) By work energy theorem, K .E sys  K .E of C . M  K .E w . r .to C . M .
1 2
W   K   10  10   500 J . 1 1
2 K .Esys   3  12   6  22  13.5 J
2 2
13. (1)
6  2  3 1
VCM   1m / s
4
t 9
14. (4) The position of the body is x  3
4 1
K .Eof CM   9  12  4.5 J
dx 4t 3 2
 0
dt 4  K .E w. r .to C . M  (13.5  4.5) J  9 J
v  t3 In the absence of external force on a system

W  k  k f  k i
K .Eof CM - always remains constant
1
 (2) (23 ) 2  03   64J So K .E w.r .to CM will be converted into P.E
2
15. (4) When body passes through the sand floor 1 2
K .Ew.r .to C .M  9 J  Kxmax
it comes to rest after travelling a distance x. Let 2
F be the resisting force acting on the body.
xmax  0.3m  30cm
Work done by all the forces is equal to change
in KE t3
20. (1) S  dS  t 2 dt
Fx=Mgh + Mgx (F-resistive force) 3
 h d 2S d 2 t3 
F  Mg  1   a   2tm / s 2
 x dt 2 dt 2  3 
Now work done by the force

1
  2
2  K .E  mv 2
2
W   F .dS   ma.dS
0 0
If mass is 40 kg then,
2 2
2 3 3 4 2 1 2
 3  2t  t dt   6t dt  2 t   24 J
0
K .E   40  10   2000 J
0 0 2
21. (2) The constraint relation between point P If mass is 100 kg then,
and block is 1 2
K .E   100  10   5000 J
2
24. (2) By work energy theorem w.r.to trolley
Wmg  WFs  K .E
Wmg   mgL (1  cos  )
WFs  maL sin 
The total length of the string is Where Fs is the pseudo force  K .E  0

l  yP  2 y B  C  a
 tan 
where C and l are constant 2 g
By differentiating the equation a

   2 tan 1  
0  y P  2 y B g
l
As (yP )  l  (yB ) 
2
mgl 
U  25. (1) If the force F  Fx iˆ  Fy ˆj is conservative
2
then
22. (4) The K.E at A and B are
Fx Fy

1 y x
K A   1 32  4  5 J
2 
F  xy 2iˆ  x2 y ˆj
1
K B   1  62  18J
2 Fx  xy 2 , Fy  x 2 y
K B  K A  13  5J
Fx 
  xy 2   2 xy
U BA  mgh  1  10  2  20 y y
W f  20  13  5  6.5J Fy  2
x

x
 x y   2xy
23. (1) The average speed of the athelete
Fx Fy
100  
v  10m / s y x
10

F  xyiˆ  xy ˆj x4 x2
27. (1) Potential energy V  
4 2
Fx  xy , Fy  xy
For maximum kinetic energy, potential energy
Fx F of a particle should be minimum.
 x& y  y
y x
dV d 2V
For minimum value of V,  0 and 0
dx dx2
Fx Fy
 
y x 3
 dV  4x 2x
Force F       0  x3  x  0
  dx  4 2
F  y 2iˆ  x2 ˆj
2 2
 x  x2  1  0
Fx  y , Fy  x
i.e., at x  0, x  1, and x  1 force on the
Fx  2 particle will be zero.
  y   2y
y y
d 2V
Now,  3x 2  1
Fy  2 dx2
x

x
 x   2x
d 2V
For x  1, and x  1 1
Fy Fx dx2

x y It means the potential energy of the particle will
 be minimum at x  1, x  1.
F  yiˆ  x2 ˆj
Now substituting these values in expression of
Fx  y , Fy  x 2
potential energy
Fx Fy
 1,  2x  14 12  1
y x Vmin   J  J
 4 2  4
Fx Fy
  Kinetic energy max  Total energy
y x
So option (1) is correct.  1

  Potential energy min  2    
 4
a
26. (1) Height of CG of mass m1 
2 1 2 9 2 9 3
mmax   max   max  m / sec.
2 4 2 2
b
Height of CG of mass m2  a 
2 28. (4) U  3x  4 y
 Gravitational potential energy of system
U
Fx    3
m  b x
  1  m2  ga  m2 g
2  2 U
Fy    4
y
 m  b
  1  m2  a  m2  g 
 2  2  F  (3iˆ  4 ˆj )
If the particle crosses y-axis then x displacement 1 K

will be equal to -6  K .E  mv 2  2
2 2r
1 Total energy  P.E.  K .E
x  vx t  ax t 2
2 K K
1  2
 2 0
6  0   3t 2  t  2sec 2r 2r
2
32. (1) When the net force is equal to mg then
The resultant velocity is the particle looses contact with the surface.
 
v  0  (3iˆ  4 ˆj )  2 | v | 10m / s
In one second the displacement of the particle
along x and y are
1
x  0   3  12  1.5
2
1
y  0   4  12  2
2
mv 2
Co-ordinate  (6 1.5,4  2)  (4.5,2) i.e., N   mg cos ( N  0) (i)
r
a b From conservation of energy
29. (4) U ( x)  
x10 x5
1
At equilibrium mgr (1  cos )  mv 2 (ii)
2
1
dU a 6b  2a  5 From (i) & (ii) we get
F   10 11  6  0  x   
dx x x  b 
cos  2 / 3
U  0
33. (1) Let the extension produced in the spring
a b b2 due to rotation be x.
U equilibrium  2
 
 2a  2
 a 4 a Then, for the body to be in equilibrium
   
 b   b 
Fspring  Fcentrifugal
 b2  b2
U ( x  )  Uequilibrium  0   .
 4a  4a  kx  m(l  x)2
30. (4) At x  5m
ml2
x
U  20  (5  2)2  29 J k  m2
Total energy  20  29  49J 34. (3) Both spheres will come at rest
simultaneously because net momentum of
v K system is zero. And when they come to rest,
31. (2) F  
r r 3 their KE converts into elastic potential energy
Since it is performing circular motion of spring.
mv 2 K 1 
F  3 U  2  mv02   mv02
r r 2 
K 35. (1) Equate loss of gravitational potential

mv 2  energy with gain of elastic potential energy.
r2
36. (1) Initial length of spring 1 2

2 2  Total energy, E2  mgh2  mv
  2R   1.5R   2.5R 2
So, E1  E2
x  2.5 R  2 R  0.5 R
From conservation of energy : 1
mgh1  mv 2  mgh2
2
1 2 1
mv  mgh  kx 2
2 2 v 2  2 g  h1  h2 
1 1 4mg 2
 mv 2  mg1.5R   0.5R  v  2  10   2.5  1.5   2 5ms
1
2 2 R
 v  2 gR 39. (2) By law of conservation of mechanical

energy
mv 2  K  U  A   K  U B
37. (2) T  mg cos  
R
Given T = mg 1
0  mgR cos   mvB2  mgR sin 
2
vB2  2 gR  cos   sin  
At B since particle leaves contact with the

surface hence normal reaction equals zero,
mvB2
mg sin  
R
m
mg sin    2 gR  cos   sin   
mv 2 R
mg  mg cos  
R sin   2cos  2sin 
v2  3sin   2cos
g 1  cos   
R
40. (1) Maximum tension in the string is in its
From conservation of energy
lower position.
1 2 1 2
 mv  mu   mg  R  R cos    0 From conservation of energy
2 2 
1 2
 v 2  u 2  2 gR 1  cos  mu  mgl (1  cos0 )
2
2
 v2   gl   2v 2 u 2  2 gl (1  cos0 ) (i)
Where u is the speed at the lowest position. l is
gl
3v 2  gl  v  the length of the string.
3
mv2
38. (1) At the highest point, v  0, h1  2.5m Tmax  mg  (ii)
l
 Toatal energy, E1  mgh1  0  mgh1 From eq’s (i) & (ii)
At the lowest point, v  ?, h2  1.5cm Tmax  mg  2mg 1  cos 

 mg  3  cos  44. (3) If the motor pumps water (density   )

continuously through a pipe of area of cross-
Block of mass 4m does not move. section A with velocity v , then mass flowing
So   4mg   Tmax out per second.
m  Av  (1)
Rate of increase of kinetic energy
1 dm 2 1 1
 v   Av   v 2  A v 3 (2)
2 dt 2 2
1
A v '3 3
P' 2 P'  v' 
   
P 1 A v 3 P v
 3  cos0  2
4 mg  mg  3  cos  or    
 4 
m ' A v ' v '
Now,  
41. (3) The power through a pipe m A v v
P  v3 As, m '  nm  v '  nv

where v - velocity of liquid P'
  n3  P '  n3 P
The rate of volume of the liquid that comes out P
dV
from a pipe is v 45. (1) p  Fv  K
dt
If rate is to be made three times then velocity is dv
m vK
also to be made three times dt
3 3
P1  v1   v1  K
   v dv  dt
 m
P2  v2   3v2 
P2  27 P1 v2 K 2K
 t v t
2 m m
42. (2) For instantaneous power
 
46. (1) P  F .v  ma  at  ma 2 t [ as u = 0]
dW F 2 2 F 2T
Pins    2t  Pins  2
dt 2m 2m v  mv 2
 m  1  t  21  t  a  v1 / t1 
2 2 2
 t1  t1
Pavg  W  F (T  0)  F T
 
T 2mT 2m F.s
47. (2) P
Pins t
2
Pavg
 2iˆ  3 ˆj  4kˆ  .  3iˆ  4 ˆj  5kˆ  38
43. (1) W  mgh  60  9.8   25 / 100  147 J    9.5W
4 4
Energy in one minute  147  20  2940J
48. (2) Volume of liquid flowing per second  Av
2490
Power   49W Mass of liquid flowing per second, m  A v
60
The rate at which kinetic energy is being 1

v   3t 2 0  3m / s
imparted to the liquid is
From work energy theorem
1 dm 2 1 1
 v   Av   v 2  A v3
2 dt 2 2 1
WF  K.E  m(v 2  u 2 )
49. (2) Kinetic energy of a body depends upon 2
the reference frame and so does the work done.
1
Since two observers are not accelerating w.r.t.  (1)(9  0)  4.5J
2
each other so they will observe same force acting
on mass and so they will observe same 4 . (12.89) Let fat used be ‘x’ kg
accelerations of block.
 Mechanical energy available
50. (1) W.r.to B the block is going down with
acceleration a. 7 20
= x  3.8 10 
100
1 
Wg  Fs  mg  at02 
2  Work done in lifting up  10  9.8  1000
51. (4) W.r.to B the pseudo force does not exist. 20
 x  3.8 107   9.8 104
WFs  0 100
52. (4)  x  12.89  103 kg
dU dx
1. (1) F  2 x+2 1. (5.28) v  3  8t  3t 2
dx dt
At equilibrium F  2 x+2  0  v1  3m / s and v2  19m / s
x 1 1
W m  v22  v12 
Just before and after x  1 the force and 2
displacement are in opposite direction. At
1
x  1 the particle is in stable equilibrium.   0.03  192  32   5.28 J
2
2. (150) As, force, 2. (15) Given that
 
F  2iˆ  15 ˆj  6kˆ , displacement,
10HP  10  735.5W
s  10 ˆjm The height to be rised from the ground is 30cm.

 Work done, W  F .s P.E  mgh  Vgh
  
 2iˆ  15 ˆj  6kˆ . 10 ˆj  150 J  103  22380 103 cm3 10  30
3. (4.5) F =6t =ma  P.E  22380  10  30
dv P.E
 a  6t   6t Power 
dt t
v 1
22380 10  30
 dv   6t dt
0 0
t
10  735.5
 15min
3. (60) Work  force  displacement 3. (4) Force between two protons = force
between a proton and a positron. As positron is
W   F cos  s much lighter then proton, it moves away through
much larger distance compared to proton. As
Given, W  25J , F  5N , s  10m work done = force  distance, therefore in the
same time t, work done in a case of positron is
W 25 1 more than that in case of proton.
 cos   
F .s 5 10 2 4. (4) Here, m = 5 kg, r = 1m
1 300
   cos 1    60  rps  5rps  5  2 rad s 1
2 60
4. (5.5) Let the block compress the spring by x
1 1 2 1 2
before stopping. Kinetic energy of the block = KE  mv 2  m  r    5 1  10   250 2 J
2 2 2
(P.E of compressed spring) + work done against
friction. 5. (2,4) When a man of mass m climbs up the
1 2 1 staircase of height L, work done by the
 2   4    10,000  x 2  15  x
2 2 gravitational force on man is (-) mgL, and work
5000 x2  15x  16  0 done by muscular force is mgL. The work done
by all forces on man is equal to -mgL + mgL =
2 zero. Further, force from a step does not do
15  15  4   5000  16 
x  5.5 cm work because the point of application of force
2  5000 does not move.
5. (2.5) Let u be the velocity with which the
particle is thrown and m be the mass of particle.
Then
1. (3) When the man gets straight up and stands,
1 reaction of ground on the man = mg. However,
K  mu 2
2 when the man is squatting on the ground, reaction
of ground is more than mg, as the man is to
At the highest point the velocity is u cos 60.
exert some extra force on the ground to stand
Therefore, kinetic energy at the highest point. up.
1 1 K 2. (4) As the road does not move at all,

K '  m(u cos 60)2  mu 2 cos 2 60 
2 2 4 therefore, work done by the cycle on the road
must be zero.
3. (2) As the shotput reaches the ground, its KE

1. (3) When an iron sphere falls freely in a lake,
its motion is accelerated due to gravity and =PE of shotput when it is thrown + KE given
retarded due to viscous froce. The overall effect
is increase in velocity and hence increase in 1 1 2
 mgh  mv 2  10  10  1.5   10 1
KE till the sphere acquires terminal velocity, 2 2
which is constant. Choice (3) is correct.
= 150 + 5 = 155 J
2. (1) As the magnetic forces due to motion of
electron and proton act in a direction 4. (3) Power P=Fv=mav
perpendicular to the direction of motion, no work If a is constant then v=at (u=0)
is done by these forces. That is way one ignores
the magnetic force of one particle on another. P  ma 2 t  P  t
5. (2) Length of the pendulum = 1.5 m

4. (2) Here, P   ML2T 3   constant
Potential energy of the bob at position A = mgh
As, mass M of body is fixed,
As bob moves from position A towards position
B its potential energy converted into kinetic L2T 3  constant
energy.
L2
 KE at positon B = 95% of its PE at position  constant or L  T 3/ 2 or
T3
A
displacement (S)  T 3/2 .
1 2 95
mv   mgh
2 100 5. (1) At stable equilibrium position U is
minimum, at unstable it is maximum and at neutral
2  95  gh
or v position it is constant.
100
6. (3) At stable equilibrium position U is
19 minimum. When U is (-) ve the force is attractive.
  9.8  15 = 5.28 m/s  5.3m / s
10
7. (4) F ( x)   kx  ax3 ;
F(x)  0;
1. (1) Work = Area under (F-x) graph
k
1 for x  0, x  
 (7  3)2   5  2  13J . a
2
2. (1) Work done W = Area under F - x graph. k
So slope is zero at x  0, x  
Area is (+)ve if F is (+)ve. Area is (-)ve if F is a
(-) ve
8. (2) At x   xm , the particle turns back.
1 
W    6  10  4  [5]  [4  5]  [2  (5)] Therefore, its velocity at this point is zero.
2  Therefore, kinetic energy K = 0. The total energy
 W  30  20  20  10  20 J E is in the form of potential energy, i.e., V  E .
According to work energy theorem,
K f  Ki  W dU
9. (1) From F  
K f  25 J  20 J  45 J dx
U ( x) x x
3. (4) Initial KE of the body  dU    Fdx   (kx)dx
0 0 0
1 1
 mv 2   25  4  50 J
2 2
kx 2
 U ( x)  
Work done against resistive force 2
= Area between F  x graph Therefore the correct option is (1).
1 10. (2) From the graph it is clear that force is
  4  20  40 J
2 acting on the particle in the region AB and
Final KE = initial KE - Work done against resistive due to this force kinetic energy (velocity) of
force the particle increases. So the work done by
 50  40  10J the force is positive.
6. (3) Slope of the graph is positive and constant

upto certain distance and then it becomes zero.
1. (1) Work done = Mass  Area covered in
between acceleration displacement curve and dU
F ,
displacement axis. dx
F = constant (negative) and become zero.
1
 10  (8  102  20  102 )  8  102 J .
2 7. (4) The z-component of the force and the x-
component of displacement are ineffective here.
2. (2) According to the graph the acceleration
a varies linearly with the coordinate x. We may dW  Fx dx  Fy dy  Fz dz  F y dy  3 xy.dy
write a   x, where  is the slope of the graph
Where Fz  0, dz  0 & Fx  0
20
   2.5ms 1 2
8 dW  6 x 4 dx  W   6 x 4 dx
0
The force on the brick is in the positive x-

direction and according to Newton’s second law, W  192 / 5.
its magnitude is given by 8. (1) For the system to be bounded one total
a  energy of the system is negative. The object
F  x having total energy E1 is bounded
m m
if x f is the final coordinate, the work done by 9. (4) The K.E decreases and it reaches
the force is minimum value at the top and then it. Starts
increasing.
xf  xf
W   Fdx  xdx
m
0 0
10. (2) From the graph it is clear that force is
acting on the particle in the region AB and
 2.5 due to this force kinetic energy (velocity) of
 x 2f   (8)2  8 J the particle increases. So the work done by
2m 2  10
the force is positive.

3. (4) Given F  ( xy 2 )iˆ  ( x2 y) ˆj
1. (2) P  fv  Mgv
W   Fx dx   Fy dy
Applied power  4000  v  20000v
W   xy 2 dx   x 2 ydy
60  746  4000v  20000v
(4,0)
1 2 2  x2 y 2  v  1 .9 m s
 d ( x y )  0
2  2 
  (0,0)
F
4. (3) When a pendulum oscillates in air, it loses 2. (1) Extension in spring, x  (At new
k
energy continuously in overcoming resistance due
equilibrium)
to air. Therefore, total mechanical energy of the
pendulum decreases continuously with time. The Using work energy theorem
variation of total mechanical energy E with time
t is shown correctly by curve (3). WF  Wspring  KE
5. (4) K + U = T.E = (-) ve at all points. So all Let v be the velocity of the block at new
points represent bound system. equilibrium.
 1  1 1 1
Fx    kx 2   mv 2  0 K .E  m2 v 2   r 3 (ii)
 2  2 2 2
2
Total energy=Potential energy + kinetic energy
 F 1  F 1
F    k    mv 2 Now, from eqn (i) and (ii)
k 2 k 2
Total energy = K.E + P.E.
F
v
mk  r3  r35
    r3
3 2 6
3. (1) Applying law of conservation of energy
1 2 1
mvA  mgh  mvB2
2 2 1. (2) Using linear momentum conservation
1 2 Pi  0.01  u  0  Pf   0.01  5.99   v

  20 103   5   20  103  10 10
2
(Where, u = speed of bullet and
1 v = intial speed of bullet-stick combination)
  20 103  vB2 
2
u
 vB  15 m / s  v
600
So, the angular momentum of the particle about
Using energy conservation
point O is
2
20 1  u 
mvB r   15  10  10  6   6  9.8  9.8  10
2
1000 2  600 
 6 kg  m 2 / s
 u  6  98  2  588 2 m / s
ds 2. (4) While going up
4. (3) va s 
dt
1
s
ds t (sin   tan   )
  a  dt 10
0 s 0
2 s  at
a 2t
v
2
1 1
W  mv 2  ma 4t 2
2 8
5. (2) As we know, dU = F.dr
 W
W sin   20  10  P
r
2ar 3
U    r dr  (i)  
0 3
W
2
W (sin  )10  10  P
mv 20
As,   r2
r
3W 4. (1) Consider a small element of length dx.

=P
2
While going down
1  v 
dK .E  dm v 2P  vp  x
2  l 
2
1 v 
KE   dm  x 
2  
1 m v2 2
 dx x
2   2
W P 3W
v  l
20 2 4 mv 2  x3 
= 3 × 
v  15m / s. 2l  3 0
3. (4)
mv2 l 3 mv2
  
2l 2 3 6
1
mgL  mgh  mv 20
2
h v20
When normal force becomes zero L 
 2g
mv 2
 mg sin  (i)
r
6. (4)
Where v is the velocity at Y.
Let  be the density of air and A be the cross-
1
mg r cos   mv 2  mrg sin  section area of the blades. Consider cylinder
2 with cross-sectional area A and length x
v2 1 1
 2cos  2sin  (ii)  K  mv 2   Axv 2 ,
rg 2 2
sin   2cos  2sin  x

Time taken t 
v
3sin   2cos
K 1
2  Power generated, P    A.v3
sin   cos  t 2
3
Centre of Mass, Linear Momentum & Collision 137
a 3a 
are (0, 0), (a, 0),  2 , 2  respectively..
 
The coordinates of centre of mass are

1. (3) Initially centre of mass is at the centre. m1 x1  m2 x2  m3 x3
X CM
When sand is poured it will fall and again after m1  m2  m3
a limit, centre of mass will rise.
2. (2) Refer figure,  3 a
1 0  2  a  2  2  5a
 
 3  9
1 
 2  2
The distance of CM from masses m1 and m2 m1 y1  m2 y2  m3 y3

are YCM
m1  m2  m3
m2 d
d1 
m1  m2  3 3a 
1  0   0  2  
2 2  2 3a 2a
d2 
m1d   
m1  m2  3  9 3 3
1
 2  2 
d1 m2
 
d 2 m1 6. (4) As boy walk from left to right on the
3. (2) block, the block will recoil towards left. The
friction acting on the block will act towards right.
r1  r2  r3 O  PQ  PR PQ  PR Hence the centre of mass of the boy and block
rCM   
3 3 3 will shift towards right.
m1Y1  m2Y2  .......

4. (1) Ycm 
m1  m2  .......
m  0  m  0  m  D   m  D   m  2D  4
 D
5m 5
3 7. (1) Let m1 , m2 are masses of two persons.

5. (2) Let the masses 1 kg, kg and 2 kg are
2 m3 is the mass of the boat.
located at the vertices A, B and C as shown in   
figure. The coordinates of points A, B and C m1 s1  m2 s2  m3 s3  0
138 Centre of Mass, Linear Momentum & Collision

s3 is the displacement of boat 17. (1) The gun has more mass than bullet but
  momentums are equal in manguitude. So
Where s1 & s2 are displacements of persons
K .EGUN  K .EBULLET
Given that
18. (2) Given, m1  m2  m1
  
s13  4iˆ  s1  s3
u1  v and u2  0
  
s23  4iˆ  s2  s3
v2  v1
e
From the above equations we get v0
  
60(4iˆ  s3 )  80(s3  4iˆ)  60s3  0 v2  v1  ev (i)

200s3  80iˆ mv  0  mv1  mv2

s3  0.4iˆ
v  v1  v2 (ii)
8. (2) If m1 , m2 ... m3 are the masses of the balls, From eq’s (i) & (ii)
the acceleration CM is.
v1 1  e

 m g  m2 g  ... v2 1  e
aCM  1  g.
m1  m2  ...
19. (2) If M  M ' , then bullet will transfer whole
9. (2) Here, m1  1kg and m2  2kg of its velocity (100% of its KE) to block and will
2
itself come to rest.
 m  m2  g
aCM  1  g 20. (1) As net horizontal force acting on the
 m2  m2  9
system is zero, hence momentum must remain
1 2 1 10 20 conserved.
sCM  at    4 
2 2 9 9 Let u is the velocity of m and V is the velocity of
10. (3) M after collision
11. (4) mu
V 
12. (3) M
13. (2) As per definition,

14. (1) (V2  V1 ) V  0 V m
e   
15. (4) In elastic collision the particles exchange (u1  u2 ) u  0 u M
their velocities after collision.
21. (1) u1   2 gh1 , v1  2 gh2
16. (4) By conservation of momentum,
mv  M  0   m  M V We know that v1  eu
 m 
Velocity of composite block V   v v1 h
mM  e  2
u1 h1
1
K.E of composite block   M  m V 2
2 22. (4) As all balls are identical in size, exchange
2
of velocities takes and finally the last two balls
1 m  2 1 2 m 
  M  m    v  mv  .
on right hand side will start moving with velocity
2 mM  2 mM  v.
23. (4) Let the initially particle x be moving in Let the mass B moves at angle  with the
anti-colckwise direction and y in clockwise horizontal with a velocity V.
direction.
Pi  mv (1)
As the ratio of velocities of x and y particles are
vx 1 Pf  mV cos (2)
 ,
v y 2 therefore ratio of their distance covered
 mv  mV cos   V cos  (3)
will be in the ratio of 2 : 1. It means they collide
at point B. From the conservation of vertical linear
momentum
mv v
 mV sin   0   V sin  (4)
3 3
By solving (3) and (4)
v2
v2   V 2 (sin 2   cos 2  )
3
After first collision at B velocities of particles 4v 2 2
get interchanged, i.e., x will move with 2v and  V 2 V  v.
3 3
particle y with v. Second collision will take place
at point C. Again velocities get interchanged 27. (1) The forces leading to the explosion are
and third collision take place at point A. So after internal forces. They contribute nothing to the
two collision these two particles will again reach motion of the centre of mass. Therefore, the
the point A. centre of mass of the fragments of the shell
continues to move along the same parabolic as
P12 P2 it would have followed, if there were no
24. (2)  2
2M 1 2 M 2 explosion as shown in figure.
M 2 P2 28. (1)

M 1 P1 29. (4) v1 & v2 are velocities of two parts after
25. (3) In case of glancing collision of two equal explosion
masses with one of them at rest. After collision
As the bomb initially was at rest therefore
two masses move at right angle to each other.
Initial momentum of bomb = 0
26. (2) Let mass A moves with a velocity v and
collides inelastically with mass B, which is at Final momentum of system  m1v1  m2 v2
rest. As there is no external force
 m1v1  m2 v2  0  3  1.6  6  v2  0
velocity of 6 kg mass v2  0.8m / s

Its kinetic energy
1 1
 m2 v22   6  (0.8)2  1.92 J
2 2
30. (2)  P  FΔt
2mv  FΔt
Δt  50  103 s  50 ms
31. (2) Pi  Pf 5. (3) Here, m1  1kg , m2  3kg

 
u r1  2iˆ  3 ˆj  4kˆ, r2  2iˆ  3 ˆj  4kˆ
mu  2mv  v 
2 The position vector of the centre of mass is
 
(u / 2)2 u 2  m r  m r
 h  r1  1 1 2 2
2g 8g m1  m2
32. (2) Fnet  FTh  mg

(1)(2iˆ  3 ˆj  4kˆ)  (3)(2iˆ  3 ˆj  4kˆ)

FTh  Fnet  mg 1 3
FTh  ma  mg 4iˆ  12 ˆj  8kˆ

  iˆ  3 ˆj  2kˆ
4
dm
vr  m(a  g ) 6. (1)
dt
7. (4)
dm m(a  g ) 4000(19.6  9.8)
   120kgs 1 8. (1)
dt vrel 980
9. (2) We may consider the entire mass of the
33. (3) The acceleration after 50 s is stick to be concentrated as a point mass at the
centre of mass of the stick. The center of mass
dm
vr moves as a projectile, it will move along a
FTh dt
a  parabolic path.
m 1500  10  50
10. (1) When the system is released, heavier
5000  10 mass move downward and the lighter one
  50m / s 2
1000 upward.
Acceleration of the blocks is
r1 m2
1. (4) m1r1  m2 r2  
r2 m1
1
r  .
m
2. (2)
3. (3)
 3m  m  g
4. (2) Centre of mass of particles system a  g 
 3m  m  2
m1 x1  m2 x2  m3 x3  
X cm   m a  m2 a 2 majˆ  3majˆ
m1  m2  m3 a cm  1 1 
m1  m2 4m
300  (0)  500(40)  400  70
X cm   a g ˆ
300  500  400 a cm   ˆj  j
2 4
20000  28000 48000 11. (4) To keep the centre of mass at the same
   40cm
1200 1200 position, velocity of centre of mass is zero, so
m1v1  m2 v2 22. (3)

0
m1  m2 23. (3) During inelastic collision of two bodies
kinetic energy after collision is not equal to kinetic
[where v1 and v2 are velocities of particles 1 energy before collision. Here, KE appears in
and 2 respectively.]
other forms. In some cases  KE  final   KE  initial
r1 r2 such has when initial KE is converted into

 m1  m2 0 internal energy of the product (as heat, elastic
dt dt
or excitation energy).
 m1r1  m2 r2  0
Note: In some cases  KE  final   KE  such
Let 2nd particle has been displaced by distance initial
x. as when internal energy stored in the colliding

particles is released. In this case the colliding
 m1 (d )  m2 ( x)  0 particles break into pieces.
m1d 24. (4) As in a perfectly inelastic collision, two
x bodies stick together after the collision and move
m2
with same velocity, so their relative velocity after
12. (1) Let m1 & m2 are the masses of the two the impact is zero.
balls. 25. (2) When two balls at the same temperature
collide, some fraction of the KE appears in other
m1 g  m2 g
aCM  g forms of energy, like heat energy, sound energy.
m1  m2
Hence, neither temperature, nor velocity or KE
13. (4) Velocity of centre of mass will remain will remain conserved. The only quantity which
unaffected as no external forcer acts on a will remain conserved is their momentum.
system. 26. (1)
14. (3) Net external force on system is zero so 27. (2) Let the direction in which ball is moving

v cm  zero be taken as positive, after collision the direction
of velocity is reversed. Hence, change in
15. (3) Here  Fint  0 So it can be added or momentum is
subtracted to the external force term. So options
1, 2 & 3 are correct.  mv  (mv)  2mv
16. (3) 28. (2) By the principle of conservation of linear

momentum,
17. (3) In an inelastic collision, the particle do not
regain their shape and size completely after Mv  mv1  mv2  Mv  0  ( M  m)v2
collision. Some fraction of mechanical energy
is lost in the form of heat energy. Thus, the Mv
 v2 
kinetic energy of particles is not conserved. M m
However, in the absence of external forces,
linear momentum is conserved. 29. (3) Mu1  Mu2  2Mv
18. (2) Conservation of linear momentum is Mu1  Mu2 u1  u2
equivalent to Newton’s second law of motion. v 
2M 2
19. (3)
2
20. (4) 1 1 u u 
K .ELoss  M u12  u22   (2M )  1 2 
2 2  2 
21. (2)
1 2
1  (u  u )2 
 M u12  u22  1 2 
K .E1  m 10 2
2
 
2  2 
Given that
1
 M [u1  u2 ]2 11 2
 1
4
4 2
  2

K .E2   m 10 2   mv22  v2  5 2
30. (1)
  P  0.05  5 2  0.35 N .s
31. (4) The velocities of two particles after
collision are 36. (4)
 m  m2   2m1  37. (2) When the sphere 1 is released from

v1   1  u1    u2 horizontal position, then from energy
 m1  m2   m1  m2  conservation,
2
m m 
1 1
 2m  Potential energy at top = kinetic energy at bottom
and v2   m  m  u2   m  m  u1
 1 2   1 2  1
 mgl0  mv 2 or v  2 gl0
2
u1 & u2 are initial velocities
Since, all collisions are elastic, so velocity of
 2m  m   2m 
sphere 1 is transferred to sphere 2, then from 2
 v2    ( 2v )    (v ) to 3 and finally from 3 to 4. Hence, just after
 2 m  m   2m  m 
collision, the sphere 4 attains a velocity to 2 gl0 .
2 2
 v v0 38. (1) Form conservation of energy
3 3
1 1
32. (2) The rebouncing height after nth collision Mv 2  kL2
2 2
is
2
hn  he 2 n  1 e2  1   0.6   0.36m. k
 v L
M
33. (4)
k
Momentum  M  v  M  L
M
Here, m1  m2  m = 0.25kg,  kM  L
39. (4) Momentum of skater
u1  3ms 1 , u2  1ms 1
A  30 1  30kg m / s
It is an inelastic collision.
Momentum of skater B  20  2  40kg m / s
According to conservation of momentum
They are at right angles to each other.
mu  mu2 u1  u2 3  1
v 1    1ms 1 Resultant momentum  p
2m 2 2
34. (4) By the law of conservation of momentum  p  (30)2  (40)2
mv  3mV  V  v / 3.  p  50kg m / s
35. (3) The velocity before collision is
p 50
Final velocity    1m / s
v1  2 gh  2  10  10  10 2 Total mass (30  20)

40. (4) The displacement vector  r between E 5
 100%  100%
the particles is E 9
  
 r  r2  r1  8iˆ  8 ˆj  55.6%
 42. (3) Because in horizontal direction no

 r  (8) 2  (8) 2  8 2 (1) external force acts
The relative velocity is given by 43. (3) If the velocities acquired due to
   explosions are in vertical direction, then options
v rel  v 2  v1  (  4)iˆ  4 ˆj (4) and (1) are possible. If one of them follows
parabolic path (or acquires velocity in horizontal

v rel  (  4) 2  16 (2) direction), then other also has to follow parabolic
path (or acquire velocity in horizontal direction
When they collide in time t to keep the momentum zero in horizontal
direction). Hence, option (3) can never be
 
r possible.
v rel     8.
t 44. (2)
41. (4) The resultant momentum is shown in the 45. (1) Let m is the mass of the particle.
figure.
  
I  P f  Pi
I iˆ  m 2uiˆ  muiˆ
1 1 I  3mu
E1  m(2v)2  2m.v 2
2 2
From work energy theorem
1
 m.4v 2  mv 2  3mv 2 1
2 W  K  m[4u 2  u 2 ]
2
From conservation of linear momentum
1
3mv '  2 2mv W  Iu
2
2 2v 46. (4) The charge in momentum
v' 
3
A
Mv  A  v 
1  2 2v 
2
3 8v 2 4 2 M
E2  3m.    m.  mv
2  3  2 9 3 47. (4) The velocity before collision is
v1  2 gh  2  10  10  10 2
4 5
E1  E2  3mv 2  v 2  mv 2
3 3
1 2

K .E1  m 10 2
2

E1  E2 5
 
E1 9 Given that
11 2
 1 2. (1) The (x, y, z) coordinate of masses 1g,
4 2  2
 
K .E2   m 10 2   mv22  v2  5 2
2g, 3g and 4g are
  P  0.05  5 2 ( x1  0, y1  0, z1  0), ( x2  0, y2  0, z2  0)
 0.35 N .s ( x3  0, y3  0, z3  0), ( x4   , y4  2 , z4  3 )
dm m1 x1  m2 x2  m3 x3  m4 x4
48. (2) Fr  Vret   X CM 
dt m1  m2  m3  m4
Fr  5  104  40 1 0  2  0  3  0  4  
X CM 
1 2  3  4
Fr  2  106 N
4 5
49. (1) Fnet  FThr  mg 1 or  
10 2
Here mg =0 The value of  can also be calculated by YCM
dm dm and Z CM also
Fnet  FThrust  vrel  210  300
dt dt 3. (4) The position of the rotating particle at
some arbitrary position is shown in the figure.
dm kg
  0.7
dt Sec
dv   dm 
50. (3) m  vrel  
dt  dt 
dm
dv  vrel
m
v m 1
 dv  vrel  dm  v  vrel log 2  2 log 2
0 m /2 m
The C.M of the two fixed particles 1 & 2 lies at
the origin. The resultant C.M lies at a distance
L
1. (2) Consider a small element of mass dm. from the centre which rotates in a circle of
3
K L
dm  dx radius
L 3
4. (2)
5. (3) Let the coordinates of 5 kg ( 3rd particle)
particle is at  x, y, z  . As the system C.M lies
at 1, 2,3
L
K 2
 L x dx  x 6 1  5  1  5  x 
0 3 1
 xcm  L
 L 655
K 2 4
0 L x dx  x3
6  2   5  3  5  y  Mx   M  m  x  m  l  x 
2  xCM 
655 2M
 y 1 mx  ml  mx ml
 
2M 2M
6  3  5  2   5  z 
3 9. (3) Let Q slides by a distance x towards left,
655 then net horizontal displacement of P w.r.t.
ground
 z 8
6. (2) After collision two bodies exchange their L cos  x
velocities. We can consider one body is at rest Now apply
and the other body moves with relative velocity m1 x1  m2 x2  0  Mx  m( L cos  x)
3v.
mL cos
2 R x
Time of meeting  M m
3v
7. (1) From conservation of linear momentum 10. (3) Let v1 is the velocity of the ball w.r.to
  the wedge at point B and v2 is the velocity of
Pi  Pf wedge w.r.to the ground.
   From conservation of LM
o  m1v1  m2 v2
m1  m, m2  M  v 
0  m  1  v2   mv2
    2 
v12  v1  v2  urel iˆ
v1  2 2v2 (i)
From the above equations we get
  murel iˆ
v2 
(m  M ) P.E  K1  K 2
8. (2) Mass of man = m 2 2
R 1   v1   v1   1 2
Mass of ladder = M - m mg  m   v2       mv2 (ii)
2 2   2   2   2
Let x is distance travelled by the ladder and the
weight M.
gR
The displacement of the man w.r.to the ground v2 
3 2
is l- x
11. (1) Let v1 & v2 are velocities after collision
From conservation of LM
Mv  0  Mv1  mv2 (i)

The coefficient of restitution is
v2  v1
e 1 
v0
v2  v1  v (ii)
  After 2nd collision velocity of 1st ball is v1 ' and

2Mv  2v 
v2    that of 2nd ball is v2
M  m 1 m 
 M 
mv1  v1 ' mv2
As M    v2  2v
v1  v1 ' v2 (3)
12. (2) As A and B has same masses so in first
collision A comes to rest and B starts to move v2  v1 '
e (4)
with v . Then B collides C and rebounds with v1
velocity
From eq’s (3) & (4)
 m  mC 
v1   B  v  0.6v. Now B collides with v1 1  e  u 1  e 
2
 mB  mC  v2  
2 22
A and B comes to rest and A moves with
 0.6v. u 1  e 
N 1
So velocity of Nth ball is vN 

2 N 1
u2
13. (1) Let h 
2g 15. (4) Let m & M are masses of small and large
spheres.
After first collision velocity becomes
Before collision the two sphere have equal
eu  h'  e2 h.
velocity 2 gh . Let V1 & V2 are velocities of
After second collision velocity becomes
the spheres just after collision.
e2u  h''  e4 h.
From conservation of linear momentum
' ''
The total distance travelled  h  2h  2h  ...
M 2 gh  m 2 gh  MV1  mV2 (i)
 e2   e2   1  e2  V1  V2
 h  2h  h 1 
2
  1  e2   h  1  e2 
1 
2 2 gh
(ii)
1  e     
14. (4) After 1st collision velocity of incoming
ball is v and that of 1st ball is v1 m
3 2 gh  2 gh
V2  M
m
1
M
From conservation of LM V2  3 2 gh (M > > m)
mu  mv  mv1
V22
h'   9h
u  v  v1 (1) 2g
v1  v 16. (1) From conservation of LM

e (2)
u 2
 v 2
From eq’s (1) & (2) mv   m    mv2 
 3
u 1  e 
v1  2 2
2 v2  v.
3
17. (2) Let v is the velocity of the particle after 20. (1) For two particles to collide, the direction
collision. Consider particle as first body and wall of the relative velocity of one with respect to
as second body. other should be directed towards the relative
v2  v1 2u  v
position of the other particle
e 1   
u1  u2 u  2u r1  r2
i.e. r  r  direction of relative position of 1
 v  5u 1 2
 
Charge in momentum of the particle is v1  v2
w.r.t.2. & v  v  direction of velocity of 2
P  m(5u  u)  6mu 1 2
w.r..t.1
vA 1  e so for collision of A & B
18. (2) Given that v  1  e
B
   
r1  r2 v1  v2
     (Both are unit vectors)
vA 1 e r1  r2 v1  v2
1  1
vB 1 e 21. (3) After each collision vertical velocity
becomes e times whereas horizontal velocity
2e
 vB  vA   vB (1) remain same. So after each collision time of
1 e flight becomes e times of previous one. So that
vB  v A horizontal displacement
But e (2)
u  R  eR  e2 R
2vB u 2 sin 2
u   u  eu  2vB  (1  e  e2 )
(1  e) g
22. (3) Velocity along common normal  v cos 

 u   vB  vA   2vB  u  v A  vB
19. (1) After elastic collision first block comes As m1  m2 they exchange their velocities along
to rest and second block moves with speed v0 . common normal. Velocity of the disc
3v
 v cos  v cos30 
2
The maximum compression in the spring is 23. (1) Let v1 & v2 are the velocities of ring and
block
1 2 1
Kxmax   vr2
2 2 1 1
 mv12  (2m)v22 (i)
2 2
m(2m) 2m
 
m  2m 3 From conservation of momentum along
horizontal direction
Relative velocity vr  v0  0  v0
mv1  2mv2 (ii)
2  2m  2
Kx max   v0
 3  From the eq’s (i) & (ii) we get
2m 8 gl
xmax  v0 v1 
3K 3
24. (2) Shell is fired with velocity v at an angle 27. (1) Let the speed of the body before
 with the horizontal. explosion be u. After explosion, if the two parts
Velocity at the highest point  v cos  move with velocities v1 and v2 in the same
direction, then according to conservation of
Momentum of shell before explosion  mv cos momentum,
 Mv1  1    Mv2  Mu
The kinetic energy T liberated during
explosion is given by
1 1 1
T   Mv12  1    Mv22  Mu 2
2 2 2
When it breakes into two equal pieces one piece 1 1 1

  Mv12  1    Mv22 
retraces its path to the canon, then other part 2 2 2M
moves with velocity V.
2
 Mv1  1    Mv2 
1
 M  1    v12  v22  2u1u2 
2
2 2T
v1
 v2  
 1    M
So momentum of two pieces after explosion
2T
m m
(v cos )  V  v1
 v2  
  1    M
2 2
By the law of conservation of momentum 28. (2) Let u is the velocity of A before collision.
v1 & v2 are the velocities after collision.
m m
mv cos   v cos  V  V  3v cos  .
2 2
25. (3) As momentum is conserved the
momentum of both the particles are equal to P.
Energy of explosion  K .E1  K .E2

Coefficient of restitution
2 2 2
P P P (m1  m2 ) v2  v1
   e (i)
2m1 2m2 2m1m2 u 0
26. (2) The y position of CM must be zero.  Jiˆ  mv1iˆ  muiˆ
 m  3m J  mv1  mu (ii)
15    y
4 4
ycm    0 Jiˆ  mv2iˆ
m
y = -5 cm J  mv2 (iii)
From the above equations 31. (2) The impulse is resolved into two
components. The component of J along the string
J  mu  J 
 will produce an impulsive tension in the string.
v1  v2 m  m 
e 
u u
J J
u 
e m m  2J  1
u mu
2J
e  1 ( P  mu )
P J cos60  (mA  mB )vx  2mvx
29. (1) From adjoining figure the component of
J
momentum along x-axis (parallel to the wall of  vx 
4m
container) remains unchanged even after the
J sin 60  m B v y  mv y
collision.
3J
 vy 
2m
J2 3J 2
vB  vx2  v 2y  2

16m 4m 2
13J
vB 
 Impluse = change in momentum of gas 4m
molecule along y-axis,
32. (4) When the bullet gets embedded then the
 2mv cos two blocks will move with common velocity v.
30. (1) Linear momentum of water striking per mu  5mv
second to the wall Pi  m  A  A 2  ,
u
similarly linear momentum of reflected water v
5
per second
The impulsive tension
Pr  A 2  .
 u  3mu
 3m   
5 5
33. (1) The Newton’s II nd law for variable
mass system is given by
  
dv
Fext  FTh  m
dt

Change in momentum of water per second Fext  Net external force
 Pi cos  Pr cos  dm
FTh  ur (Thrust force)
dt
 2 A 2  cos
dm
 5kg / s
Force exerted on the wall  2 A 2  cos  . dt
 3. (4) The CM has both x and y coordinates.

ur  0.75m / s

dv
 0 (Trolley is moving with constant velocity)
dt

Fext  5  0.75  0

Fext  3.75 N
By using papus theorem, rotate the material

about x-axis.
1. (1) Linear density of the rod varies with
distance as Volume covered= distance moved by CM  Area
of the material
 2 x
2 3  R2
dm  R  2 yCM 
  [given] dm   dx 3 4
dx
4R
yCM 
3
4R
Similarly xCM 
3
2 2 4 2R
rCM  xCM  yCM 
3
Position of centre of mass
4. (4) If mass of removed disc is m then mass
 dm  x of big disc will be 16 m.
xcm 
 dm Here, m1  16m, m2  m3  m
3
3 3
 2 x3  16m(0)  m(0)  m(3R) 3
(  dx )  x (2  x )  xdx x  3  xcm   R
0 0  0 16m  m  m 14
 3  3 
2 3
 x  Because of symmetry we can say ycm will also
0  dx 0 (2  x) dx 2x  2 
 0 have same magnitude
9  9 12
  m. 3
9 7 ycm   R
6 14
2
m1 x1  m2 x2  .......  3
2. (2) X CM  rcm   R (iˆ  ˆj )
m1  m2  ...... 14
ml  2m.2l  3m.3l  .... 5. (2) Take m1 as mass of vertical portion and

m  2m  3m  ....
m2 as mass horizontal portion
l n(n  1)(2n  1)
m1  1.5kg , m2  1.5kg
ml (1  4  9  ......) 6 l (2n  1)
   .
m(1  2  3  ....) n(n  1) 3 x1  0, x2  1m, y1  m, y2  0
2
 
0  1.5(1) vcm 10(3 ˆj  4iˆ)
xCM   0.5m 10. (2) F m   5(3 ˆj  4iˆ)
3 t 2
(1.5)(1)  0 
yCM   0.5m F  5 32  42  25 N
3
11. (2) Due to net force in downward direction
 m2iˆ  m2 ˆj ˆ ˆ
6. (4) vcm  i  j and towards left centre of mass will follow the
mm
path as shown in figure (2).
 m(3iˆ  3 ˆj )  m  0 3 ˆ 3 ˆ 12. (4) Let m is the mass of the block and the
acm   i j
mm 2 2 tube. As there is no external force along
As both vectors make 45 with x-axis so both horizontal direction then the C.M should not
move along a straight line. move along horizontal direction.
 
7. (1) External force acting on the system is m1 s1  m2 s2  0
zero. The displacements of man and boat are
  m1  m2  m
related as m1 s1  m2 s 2  0 (1)
 
 s1  s2  0 (i)
m1  m, m2  m 3
Given that
 
s1 & s 2 are displacements w.r.to ground.   
s12  Riˆ  s1  s2 (ii)
  
Given that s12  s1  s 2  Liˆ (2)
From the above two equations we get
 L
 R ˆ
From eq’s (1) & (2) s1  iˆ s2  i
4 2
8. (3) The CM of the system should be at rest. 13. (1) For wedge (1)
From conservation of angular momentum.
0  0  mV1  MV2
Where V1 & V2 are velocities of block and the

From the figure it can be observed that the person wedge
L mV1
displaces by V2 
4 M
m1v1  m2 v2 2  0  5  14 1 1
9. (4) Vcm    10m / s mgh  MV22  mV12
m1  m2 25 2 2
2Mgh
V12  (1)
M m
For wedge (2)

mV1  (m  M )V0
v1cm  v1  Vcm  14  10  4m / s
1 1
mV12  (m  M )V02  mgx
v2cm  v2  Vcm  0  10  10m / s 2 2
V12  M  2  10  3  0
2  m  M   gx v   4m/s
  23
ghM 2 Loss in kinetic energy is K = K i  K f
 gx (From equation 1)
(m  M ) 2
1 1  1 
hM 2   m1u12  m2 u2 2    (m1  m2 )v 2 
x 2 2  2 
(m  M ) 2
14. (2) 1 1  1 
   2  (10) 2   3  (0) 2     (2  3)(4) 2 
 2 2   2 
15. (3)
 100  40  60J
17. (2) The velocity after second rebound is
v  e2u
From conservation CM
2 gh
 e 2 2 gh
mu  mv1  2mv2 2
u  v1  2v2 (i) 1
e
From coefficient of restitution 2
v2  v1  u (ii) 18. (1) Let initial velocity of the bullet be v.
From eq’s (i) & (ii) By linear momentum conservation
2 1 m m 
v2  v, v1  v v    m  v1 ( v1  combined velocity)
3 3 2  2 
1 8 2 v
m u
K2 v1  (1)
% K  100  2 9  100  88.8% 3
Ki 1
mu 2 Retardation   g
2
When the block comes to rest
16. (4) After collision the metal ball sticks with
2
stationary ball, then the kinetic energy is lost in v
0     2  gd  v  3 2  gd
the form of heat, 3
From law of conservation of momentum, 19. (2) p after 1st impact = ep- (-p) =
m1u1  m2u2  (m1  m2 )v p(1+e)
Similarly p after 2 nd impact =ep (1+e)
m u  m2 u2
v 1 1
(m1  m2 ) So, pnet  p 1  e)[1  e  e2  .........
Given,
p(1  e)
Pnet 
5 (1  e)
m1  2kg , u1  36   10m/s ,
18
20. (3) Let v1 and v2 are the velocities of the
m2  3kg , u2  0 ball and wall after collision.
(5)  (v1 )
1  v1  20m / s
(10)  (5)
Hence ball will start moving towards upward

direction with velocity 20 m/s.
Let impulse imparted by plate force on the ball
is J in upward direction.
Here the wall mass is assumed to be very large  
ˆj  P f  P i  1 20  ˆj  110 ˆj
and hence its velocity remain constant.
v2  v  J  30kg m / s
The coefficient of restitution 22. (2) Applying the law of conservation of

momentum,
v2  v1
e 1 m1v1  ( m1  m2 )V
u1  u2
u1  u , u2  v
where v1  2 gd is the velocity with which m1
collides with m2 . We get :
v1  v1 , v2  v
m1
v  v1 V 2 gd
1  v1  2v  u ( m1  m2 )
uv
The force on the ball is Now, let the centre of mass rise through a height
h after collision. In this case, the kinetic energy
p Mv1  Mu
F  of ( m1  m2 ) system is converted into potential
t t
energy at maximum height h.
M [2v  2u ]
F
t 1
(m1  m2 )V 2  (m1  m2 ) gh
2M [v  u] 2
F 
t 2
1  m1 
21. (3) (m1  m2 )    2 gd  (m1  m2 ) gh
2  m1  m2 
2
 m1 
hd 
 m1  m2 
23. (3) The reduced mass of the system
mM

mM
Let is the velocity of the ball after collision. The maximu compression in the spring is given
Velocity of the platform remains constant. by
v 2
 v1  1 2 1
  vr2
As we know e  2
K .xmax
2
(u1  u2 )
mM If t is the time of imapct.

Where   , vr  v1  v2
mM For upper ball
mM (v1  v2 ) 2 2 N cos30t  Mv0 (i)

xmax  Any one of the lower ball
K (m  M )
 N cos 60t  Mv (ii)
24. (3) The displacement vector  r between
From both eq’s
particles p1 and p2 is
v0
   v
 r  r2  r1  8iˆ  8 ˆj 3
 27. (1) P  P2  P1
 r  (8)2  (8)2  8 2 (i)
The relative velocity is given by P2  2mv cos 45 iˆ

  
v rel  v2  v1  (  4)iˆ  4 ˆj 28. (3) The velocity of the canon at the heightest
point is v cos . Let m is the mass of the canon.

vrel  (  4)2  16 (ii) Pi  Pf
When they collide in time t m m

mvcos  (0)  v'
  2 2
r
v rel 
t v'  2vcos
  p2
Substituting the values of v rel and r from 29. (4) E ( Linear momentum is
equation (i) and 2m
(ii) t = 2s, then on solving we get   8. conserved )
25. (4) A constant external force (m1  m2 ) g acts 1 E m

E  1  2 .
on the system. The impluse given by this force, m E2 m1
in time t = 0 to t  2t0 is (m1  m2 ) g  2t0
2 1
30. (1) Let E0  KE  mv0
 Change in montemum in this interval 2
'  '1  
m1 v1  m2 v 2  (m1 v1  m2 v2 )  2(m1  m2 ) gt0 . Where v0 is the velocity of one of three
fragments
26. (2)
From conservation of LM
Let N is the force exerted by the moving ball on
both balls. 
0  mv  m  v0iˆ  v0 ˆj  v0 kˆ 
v  3v0
The energy of explosion is
1 2
1 
KE 
2
m  3v0   3  mv02 
2 
1
E  6 mv02  6E0
2
31. (1) Form conservation of linear momentum Change in velocity

206v '  4v 1
 2v sin 30  2  12   12 m s
1 2
2
KE of recoiling nucleus  (206)(v ')
2 mv 0.5  12
Force    24 N
1  4v 
2
4 t 0.25
 (206)    K' K
2  206  206 35. (2) When the string becomes tight an impulse
tension is generated which will effect the
32. (1) Let m1 , m2 & m3 are the masses
velocity component of B along the string.
Given that m1 : m2 : m3  1: 3 : 3 Conserving momentum in direction of AB
m1  k : m2  3k & m3  3k (Where k is constant)
m
m  m1  m2  m3  7k  k 
7
m 3m 3m
m1  , m2  m3 
7 7 7
m u cos 30  mv A  mvB  alongAB 
The resultant momentum of m2 & m3 is
vB  alongAB   v A since string is tight
 3m 
  (15) 2
 7 
u 3 u 3
vB alongAB    vA 
From the conservation of linear momentum 4 4
m  3m 
 v    (15) 2 Component of vB perpendicular to the string is
7
   7 
v  45 2m / s u
u sin 30 
2
33. (2) F t  P  F t  m  v1  v2 
36. (2) Collision takes place with A only so
v1  2 gh1 vB  0 (just after collision). For collision with
A
v2  2 gh2
v0
mv0  (m  m)v  v 
F

m 2gh1  2gh2   0.1 2 10  5  2 10 1.25  2
Δt 0.01 37. (2) The velocity of exhaust gases with
 150N respect to rocket 100ms 1
The minimum force on the rocket to lift it
Fmin  mg  1000  10  10000 N
Hence, minimum rate of burning of fuel is given
by
34. (2)
dm Fmin 10000
  1
dt ur 100 = 100kgs
38. (4) Conserving the linear momentum of the

(water + wagon) system
3. (2)
 5  10 3
 1.2   103  0   6  103 v
 v  1m / s By conservation of linear momentum
1 m
KEf   6  103  12  3000 J mv  mv1  v2
2 2
1 2v  2v1  v2 (i)
KE   5  103  (1.2) 2  3600 J
2
v2  v1
e
Change in KE  600 J u1  u2
v  v2  v1 (ii)
1. (25) U1  mgh1 and U 2  mgh2 From equations (i) and (ii)
U1  U 2 v 4v
% energy lost  U1
 100 v1  ; v2 
3 3
mgh1  mgh2 h h  h h
 100   1 2   100 A  ; B 
mgh1  h1  PA PB
2  1.5 m
  100  25% PA  mv1 & PB  v2
2 2
2. (2.35) The force exerted by the molecules
on the wall is A 2

B 1
dp
F  2n mv cos 45
dt
1. (3) Consider a small element of thickness dy

present at a distance y from the vertex
where n is the number of molecules strikir per

unit time.
F 2n mv cos 45
Pressure  
A A
 1 
2 1023  3.3 1027 103    r
 2 tan  
 y
2 104
1 1
yCM  ydm   (  r 2 dy) y
 2.35 103 N / m2 m m
h From momentum conservation

 h
3 2   y4  2
  y tan  dy  tan   
m 0 m  4 0 mV0  m  V1  V2  (2)
2
 R2 h43h  1 2 
 V1  V2   V0 2
yCM  2
   z0  m    R h 
m h 4 4  3  V02
2V1V2  
2
 Z 0  3m
2 3 V2
2. (0.33) Let the mass per unit area be  .  V1  V2   V12  V22  2V1V2  V02  0
2 2
Then the mass of the complete disc
V1  V2  2V0  2 m / s (V0  2)
2
    2R    4 R2 4. (5) It is case of elastic collision. As masses
 
of two balls are equal and 2 nd ball is rest before
collision, hence after the collision the speeds are
just exchanged,
1. (4) The position of the centre of mass of the

system shown in figure is likely to be at C. This
is because lower part of the sphere containing
2
The mass of the removed disc    R    R 2 sand is heavier than upper part of the sphere
containing air.
Let us consider the situation to be a complete
disc of radius 2R on which a disc of radius R of 2. (2) Given, separation between the nuclei of
negative mass is superimposed. Let O be the H and Cl is 1.27Å  1.27 1010 m
origin.
Let mass of hydrogen atom = m
xc.m 
 4 R   0    R  R
2 2
 Mass of chlorine atom = 35.5 m
2 2
4 R   R
 R2  R
 xc.m 
3 R2
R 1 Let mass of hydrogen atom be at origin, i.e.,

 xc.m    
3 3 position vector of it, r1  0
 Position vector of chlorine atom
3. (2) Let V1 and V2 are the velocities of the
two particles after collision. r2  1.27  1010 m
m1r1  m2 r2
rCM 
m1  m2
m  0  35.5 m  1.27  1010

1 1 31  
mV12  mV22   mV02  m  35.5 m
2 2 2 2 
3 35.5 1.27 1010

 V  V  V02
2 2
(1)   1.24Å
1
2
2
36.5
3. (1) The given system can be asumed as a M R



combination of nm at the centre of the hexagon MR 4
x 4 2  
and -m at the place where valancy is present. M 8 3M
M
4
(nm)0  (m)a a
rCM   R
nm  m (n  1) x
6
4. (1) In inelastic collision between two bodies,
total linear momentum remains conserved. 3. (3) m1v1  m2 v2  (m1  m2 )v
 2  3 1 4  (2  1)v
6  4  3v
2
1. (4) 2  3v  v  m / s
3
4. (3) Centre of mass of a bangle lies at the
centre of the bangle, which is outside the body.
5. (1) Initial mass of the rocket m  20000 kg
Initial acceleration a  5.0 m/s 2 in upwards

direction.
Let initial thrust of the blast be F.
From figure,  F  mg  ma
L L L
x1  , x2    L  F  mg  ma
2 2 2
 m( g  a)
L L L 5L
x3    
2 4 2 4  20000  (9.8  5.0)  2.96  105 N
M 1 x1  M 2 x2  M 3 x3
 xCM 
M1  M 2  M 3
d  KE  d 1 2 dv
L 5L 1. (2)   mv   mv
M M LM  dt dt  2  dt
 2 4
M M M K increases continuously. When the ball hits the
ground it decreases rapidly to zero and then it
11
ML increases again to maximum. Graphs (1) & (2)
11L
 4  dK dv dK
3M 12 satisfies this  m v  mgv,  0 initial
dt dt dt
2. (3) Let mass per unit area of the disc be  . dK
and v also increases and hence slope also
2
dt
 Mass of the disc (M)   R  . increases so graph (2) is correct.
Mass of the portion removed from the disc ( M ' ) 2. (3) After giving the impulse the body will gain
some speed and because of friction it will be
2
R M under deceleration.
    
2 4 3. (3) Initially particle was at rest.
The position of centre of mass of the remaining Final momentum of the particle = Area of F - t
portion is graph
 mV  Area of semi circle 4. (2) Velocity between t  0 and t  2sec

 r2  r1r2  ( F )(T / 2) dx 4
 mV     v1    2m / s
2 2 2 dt 2
 F T
u  . Velocity after t  2sec , v f  0
4m
4. (2) The impulse given to the particle is equal Impulse = Change in momentum  m (v f  vi )
to the area under the F - t graph.
1 1  0.1(0  2)  0.2kg m sec 1 .
P   4 103  10  10  4 103   4 103  10
2 2 5. (2) As the given graph is periodic the
P  8 10 2 velocities of the body just before and just after
2s are
The initial momentum of the particle is
Vi  1m / s & V f  1m / s
Pi  0.07  50  102  3.5  102
P  mV  0.8 Ns.
Hence, the particle will reverse in direction.
1
Its average acceleration is 6. (4) Velocity after 10 sec 
m
Favg Area 8  102 Area enclosed between F  t graph and time
aavg     1m / s 2
m m  ttotal 0.07  12  103 axis
Area =
5. (2) After head-on collision with the similar
particle at rest the first particle comes to rest. 1 1 1 
 2  2  10  2  10  2  10  2  2  10  2  4  20 
1
| (0  mu ) |  T  F
2  50m / s
2mu 7. (1) The x-position of CM of the rod is
F 
T
L L L n
 x
 xdm  x dx
 k   .xdx
L
xCM  0L  0L  0L n
dp dm  dx x
1. (3) As F  , so the force is maximum 0 0 0  L  dx
k
dt
when slope of graph is maximum. L
 Kx n 2 
(4) From O to T, area is (+) ve and from T to 
2.
  n  2  Ln  0 L  n  1
2T area is (-)ve, net area is zero, hence, no  L

change in momentum.  kx n 1  n2
 n 
3. (3) The impulses are   n  1 L  0
P1  0.25  1  103  2.5  104 L

For n  0 xCM 
2
1
P2   2  103  0.3  3  104
2 n  1,
1
P3   1 103  1  5  104
2 2L
xCM 
1 3
P4   1 103  0.5  2.5  104
2
3L
The impulse is maximum for (III) n  2, xCM 
4
 1 Substituting (2) and (3) in eqn. (1)

L 1  
n
xCM   3  8 m
 2 mv    v  m  M  v 
1   5   10 M 
 n
As n   8 8
 mv  v mM
5 10
xCM  L
 M  4m
The correct graph is (1)
1 2
1. (4) Initial K.E is K i  mu
2
3. (3)
Using conservation of linear momentum
mu   2m  M  v
mu
v 2.5  0  1.5  3  1 0
 2m  M  x cm 
2.5  1  1.5
1
Also, K f  K i x cm = 0.9
6
2.5  4  1 0  1.5  0
1 11 2  ycm 
  2m  M  v   mu 
2
5
2 6 2 
m2u 2 1 y cm  2.0 cm
 2m  M   2
 mu 2
 2m  M  6 4. (4) Final K.E is 50% of initial K.E
M 1 11
 4 mv  mu 2
m 2 22
2. (1) Let M be the mass of the nucleus. u
v
Applying conservation of linear momentum, 2
mv  mv1  Mv2 (1) 1

v  eu  e 
2
1 2 36 1 2 3
Also, mv1  mv  v1= v (2)
2 100 2 5  1
1
Applying conservation of kinetic energy,  1  e   h  2   3h
2
H  h 2 
 
 1 e  1 1 
1 2 1 2 1  2
mv  mv1  Mv22
2 2 2
5. (2) Centre of mass of the rod is given by
1 64 1 2
 Mv22  mv L
2 100 2 bx 2
 (ax  L
)dx
xcm  0 L
8 m bx
 v2  v (3)
10 M 0 (a  L )dx
aL2 bL2 L  a  b  2m  2u cos 60 î  2u sin 60 ˆj

  
 2 3   2 3
aL 
bL
a
b 
 (m  2m  3m)v
2 2
a b  u ˆ
L  
7L 2 3 v 
12
i  3 ˆj  
Now 12  b
a
2 4. (1) Mass per unit length  0  kx
On solving we get, b = 2a
L
M   ( 0  kx)dx
0
1. (4) For particle C

K  L2
M  0 L 
From conservation of momentum 2
2mv' sin   mv cos30  mv sin 45

2  M  0 L 
K
' 3v v L2
2v sin    (i)
2 2
2M 2 0
Along horizontal direction  K
L2 L
L
2mv' cos  mv sin30  mv cos 45 2
 dm(r)  (dx)x  ( x  kx )dx

0
0
v v rcm   
2v' cos   (ii) M M
2 2  dm
From eq’s (i) & (ii) 1   0 L2 KL3 
   
M 2 3 
3 1
 substitute ‘k’
tan  
2 2  2 3
1 1 1 2
 2L 0 L2
2 2 rcm  
3 6M
2. (2) Space between the supports for motion
5. (4) From conservation of linear momentum.
of beads is L - 2nr
4
After each collision the change in momentum  300  0.8v0
1000
of any bead is mv.
Where v0 is the velocity of the block after
Average force experienced by each support,
collision
p 2mv mv 2 3
F   v0   1.5m / s
t 2(L  2nr) L  2nr 2
v
v 2  v02  2as
3. (1) From conservation of linear momentum
  0  v02  2 gs
pi  p f
v02
m3uiˆ  3m   u cos 60 î  u sin 60 ˆj s  0.375m
2 g
162 Rotational Dynamics
 Moment of inertia of the system
2
 L
  ML2
 M  2
 2  
 2  3 12
1. (4) Moment of inertia of a body depends on
the mass of the body, its shape and size,
distribution of mass about the axis of rotation, ml 2 ml 2
9. (4) I   ml 2
and the position and orientation of the axis of 3 3
rotation.
5
2. (4) I  ml 2
3
3. (4)
2  2 
mR 2 mR 2 5mR 2 10. (3) I XX '  2  mR 2   2  mR 2  mR 2 
(I )  ,( II )  ,( III )  2mR 2 ,( IV )  5  5 
2 4 4
18 2 
I ML2 / 12 L  mR 2  9  mR 2 
4. (4) K   5  5 
M M 12
2
ml 2 17ml 2 Given that mR 2  I
5. (4)  mx 2  5
12 12
I XX '  9 I
2 16l 2  x  2l
x    
12 3 11. (3) Apply right hand thumb rule for   r  F
6. (3) I  I1  I 2 12. (1)   I , if   0 then   0 (since I  0 )
m 2 m 2 2m 2 13. (2) Let R1 & R2 are the reactions at the
  
3 3 3 supports.
7. (3) Moment of Inertia of sphere about its R1  R2  10
tangent
 A  0  R2  1  10  0.3  0
2 7 7
 MR2  MR2 = MR 2  MK 2  K  R  R2  3 N ,R1  10  R2  7 N
5 5 5
8. (1) Moment of inertia of a uniform rod about 14. (1) Let T is the tension in the string. As the
rod is in equilibrium the net torque about the hinge
Ml 2
one end  s is zero.
3
Rotational Dynamics 163
Tangential acceleration is
20
at  r   2  40m / s 2 .

21. (4) As, I  1.2kg  m2 , E  1500 J

 s   mg  T  0
  25rads 2 , 0  0, t  ?
l
mg sin 60  Tl sin 60 1
2 Now, Er  I 
2
2
mg
T 2 Er 2  1500
2     50 rads 1
I 1.2
 
15. (2) Here, F   Fkˆ, rabout P  iˆ  ˆj
From,   0   t
  

As,   r  F  iˆ  ˆj   Fkˆ    50 = 0 +25 t  t = 2 s
22. (4) As the wheels are in pure rolling the
  F  iˆ  kˆ  ˆj  kˆ  velocity of the top point of the wheel is twice
that of centre of mass and the speed of centre of
  F   ˆj  iˆ  mass is same for both the wheels (Angular
speeds are different).
16. (2)
So vF  vr
17. (1) Weight of the rod will produce the torque.
23. (3) The displacement of P is shown in the
The torque about the given point A is
figure
l ml 2
  I   mg   
2 3
3g
Angular acceleration   .
2l
18. (3)
19. (4)  2  02  2  0  4 2 n 2  2

Displacement of point P after half revolution
2
 1200 
4 2   S  R 
2
 (2R)2  R  2  4  5  2  4.
  60   200  2 rad
2 4 24. (3) Consider two points P1 and P2 on the string
2
 2 n  200  n  100  314 shown in the figure. Let S is a point on the pulley.
Where the string looses contact
20. (3)
 80
   4 , 0  0,  2 (n)  4
r 20 / 
( As n  2)
 2  02 16 2
 2  02  2      2
2 2  4 As the string is unstretchable
aP1  aP2 cases. But as acceleration a  sin  , therefore

acceleration and time to descent will be different.
As there is no slipping between string and the
pulley g sin 
28. (2) a
I
as  aP1 1  CM2
mR
(at  aCM )  aP1  aP2  0 29. (1) According to conservation of mechanical
at  aCM  R  a energy
25. (3) 1  k2 
mgh  K .Etra  K .Erot  mv 2  1  2 
2  R 
 
2
 2 gh 
v  2 
 1  k 
 R2 
Writing torque equation about an axis passing It is independent of the mass of the rolling body.
through A, perpendicular to the plane of paper.
For a ring, k 2  R 2
2 
 F  2 R   mR 2  mR 2    (i) 2 gh
5  vring   gh
1 1
a  R (ii)
2 R2
From eq’s (i) & (ii) For a solid cylinder, k 
2
7
F  ma 2 gh 4 gh
10 vcylinder  
1 3
1
26. (2) Let us assume friction is in backward 2
direction and a is the acceleration of the disc.
2 2 2
For a solid sphere, k  R
Then F  f  ma 5
The torque about centre is 2 gh 10 gh

vsphere  
2 7
mR 2 1
(F  f )R   5
2
Among the given three bodies the solid sphere
As the body is in pure rolling has the greatest velocity and the ring has the
a  R least velocity at the bottom of the inclined plane.
on solving the above eq’s we get 30. (1) The loss in potential energy = gain in
kinetic energy
 ma
f 
4
As f is (-)ve so our assumed direction for friction
is opposite to the original direction.
That means ‘f’ should act in forward direction.
27. (2) In pure rolling, mechanical energy remain
conserved. Therefore, when heights of inclines
are equal, speed of ring will be same in both the
mgL 1 2
 I 
 iˆ  0  0  ˆj  0  0  kˆ mxvy  myvx 
2 2

2
mL 3g

L  m xvy  yvx kˆ 
I  
3 L 40. (2) the angular momentum of a rigid body is
31. (1) Sphere possesses both translational and    
L  r  PCM  ICM 
rotational kinetic energy.
l ml 2
1 1 2 L  mv  
32. (1) K  I  2   2.4   25  t  2 12
2 2
41. (2) From law of conservation of angular
1
1500   2.4  625  t2 momentum if no external torque is acting upon
2
body
t  2 s
L  I   constant
2 gh 2 gh 10 When liquid is dropped, mass increases hence I

33. (4) v 2
  gh .
K 2 7 increases  I  mr 2  . So,  decreases, but as soon
1 2 1 
r 5
as the liquid starts flowing out from the disc
34. (1) Power    increases again.
 P  I  42. (2) L=I  =constant

As population migrates to the poles then I
d decreases and then  increases and T decreases.
   K    K
d
43. (1) Since net external torque is equal to the
rate of change of angular momentum.
  3
  2 d  K 0    K   3  3K .2 n 
3 As speed changes L  r  mv
0
44. (2)
  n1/3 
The direction of L remains same whereas its
35. (3) magnitude changes.
36. (2) Angular momentum is an axial vector, so 45. (4) Angular momentum of system cannot
its direction is perpendicular to plane of motion remain conserved as some external unbalanced
which is not going to change because of change torque is present due to forces at axles. Kinetic
in speed. So only its direction remains same but energy is not conserved, because slipping is there
its magnitude will vary. and work is done against friction.
37. (3) 46. (3)
2
38. (1) L2 K L2 K  100  100
K  1  12  1    
L  2KI  2 10  8 107  4 103 kgm2 s1 2I K 2 L2 K 2  110  121
  
39. (3) We use L  r  p K 2  1.21 K1
iˆ ˆj kˆ % change in K .E is K2  K1 100  21%


L x y 0 K
mvx mv y 0
Increase in kinetic energy = 21%
47. (1) From conservation of angular momentum 5. (1) The moment of inertia of the system about
the given axis is
Li  L f , I11  ( I1  I 2 )2
2 2
 L  L ML2 ML2 ML2
2 IM  M    
1 1  I  2 2 4 4 2
K .Eloss  I112  ( I1  I 2 )  1 1 
2 2  I1  I 2 
2
Ml 2 l Ml 2
2 6. (4) I M  
I1 I 2 1
12 2 3
K .Eloss 
2( I1  I 2 )
7 7
48. (3) Given, mass of bullet (m) = 10g = 0.01kg 7. (2) I  MR2  10   0.52   3.5 kgm2 .
5 5
Speed of bullet (v) = 500 m/s
1
Width of the door (l) = 1.0 m 8. (2) I MR 2  MR 2  4 I
4
Mass of the door (M) = 12 Kg
2 2 3
As bullet gets embedded exactly at the centre We have to find: I '  I CM  MR  MR
2
of the door, therefore its distance from the hinged
1 3
end of the door is  m  4I  6I
2 2
Angular momentum transferred by the bullet to 9. (4) Radius of gyration

the door,
I
K
1 m
(L)  mv  r  0.01 500  = 2.5 J-s
2
Moment of inertia of the door about the vertical 1
mR 2  mR 2
2 3
axis at one of its end, K disc   R
m 2
2
Ml 2 12  1
(I )    4kg  m2 mR 2  mR 2
3 3 K ring   2R
m
Angular momentum, (L)  I
3
 2.5  4  
K disc 2  3

  0.625 rad / s K ring 2 2
10. (3) The M.I. of disc about its diameter
1. (3) Analogue of mass in rotational motion is MR 2 R

moment of inertia.   Mk 2  k 
4 2
2. (4) 11. (2) Length of each side of hexagon = 2L
3. (2) Mass is present at large distance in the Mass of each side = m
ring.
Let O be centre of mass of hexagon.
4. (3) As mass decreases the moment of inertia
also decreases and hence radius of gyration also Therefore, perpendicular distance of O from
decreases. each side
r  L tan 60  L 3   r F sin 
F sin  is the transverse component of force

correct option is (2)
  
16. (1) T  rF
    
T is  to both r and F so r.T  0 and F .T  0
17. (1) Net force on centre of mass is zero, i.e.,
I 0  6  I oneside 
the centre mass cannot move at all. Hence, the
2 body may rotate about centre of mass.
 m  2L    mL2 2

 6
 12
 mr 2   6 
  3
 
m L 3 

18. (4) Let m gram be the mass of the metre stick
concentrated at C,
 mL2  In equilibrium, taking moments of forces about
 6  3mL2   20mL2 C, we get
 3 
10g (45 - 12) = mg (50 - 45)
12. (1) I  I ring  I spokes
10 g  33  mg  5
 mR 2 
 MR 2  3   m
10  33
 66 g
 3  5
  M  m R2 19. (2) As both forces produce same torque
13. (3) As, m1  m2  1.6

  1.6  1  1.6 Nm, F    4N .
d 0.4
  R12 d1   R22 d 2
20. (3) As the net torque about the hinge is zero
2
R 1d
 2
2 l l
R 2d1 T    mg  
 3 2
1
mR 2 3mg
I1 2 1 R2 d T
  12  2 2
Now, I 2 1 2 R2 d1
mR2
2 
21. (3) Here, r  î  ˆj  kˆ
14. (3) In case of raw egg, liquid inside it is 
thrown towards the outer surface. So moment F  7î  3ˆj  5kˆ
of inertia is more for raw egg.   
Torque,   r  F
15. (2) Consider a force F that acts at point P.
The torque produced by the force about O is
î ˆj kˆ

  1 1 1
7 3 5
 î  5  3  ˆj  7   5   kˆ  3   7  

or   2î  12 ˆj  10kˆ

22. (2) Given that F   FKˆ 27. (2) Given, r = 0.4 m,   8rad s 2
 m = 4kg, I = ?

r  iˆ  2 ˆj 
Torque,   I  mgr  4  10  0.4  I  8
  

  r  F  iˆ  2 ˆj  FKˆ   16
I  2kg  m 2 .
8
  
 F iˆ  kˆ  F 2 ˆj  kˆ  28. (2)
29. (4)
   
 F  ˆj  F 2iˆ   F 2iˆ  ˆj   30. (3)
23. (4) Taking torque about hinge A.  4500  1200 
2  
2 (n2  n1 ) 60
    rad / s 2
Wl t 10
PL  Wl  0  P 
L
3300
24. (3) The bar is in equilibrium, the total torques 2
 60  360 degree   1980 degree / s 2 .
must sum to zero. Let TA and TB are tensions in 10 2 s 2
the strings. 31. (4)
1 1 2
θ  0 t   t 2   2  2    3 2   10 rad
2 2
32. (1) Here, moment of inertia, I  3  102 kgm 2
Torque,   6.9 102 Nm
Initial angular speed, 0  4.6 rad s 1

The net torque about point O is
Final angular speed,   0 rad s 1
 3L  L
  T  A   Mg    0
4  2
As   0   t
  0 0  4.6 4.6
     rad s 2
Therefore TA  2Mg 3 t t t
 Now, negative sign is for deceleration torque,
25. (1) Since net force is zero, so a cm  0
  l
26. (3) The torque experienced by rod about the
pivot given 4.6
6.9 102  3 102 
t
3 102  4.6
t  2s
6.9 102
33. (3) Angular velocity becomes zero when

  0    t  t 
  g (2mb  m(3b))  5mgb [  force × distance] 
and I  (2mb 2  m(3b2 ))  11mb 2
d
Also    d   dt
Since,   I  , we have  =5g/11b
1b dt

t   41. (1) The two forces produce torque in
  d 
0     t  dt clockwise direction. In the absence of friction
0 the body slips backward. So friction acts in
forward direction. So C.M moves towards right

1 
2 1 2 1 2 side
   t  t 2   
2 0  2  2  42. (3) When a body is in pure rolling on an
inclined plane then the friction always act upward.
 4 2
34. (2)    4 rad s g sin  g sin 30 g
t 1 43. (2) a   .
K2 1 3
1 2 1
  I R 2
 2  103  4  8  103 Nm g sin  g sin 

44. (2) a 
35. (3) As the body is rigid therefore angular ICM K2
1 2 1 2
velocity of all particles will be same, i.e.,  = mR R
constant
For a given  , K is same for both the rings. So
From v  r , v  r (  = constant) a is same and hence both will reach at the same
36. (4) vt  velocity due to translatory motion time.
45. (4) Let T is the tension in each string.
R  velocity due to rotational motion
As the body is in pure rolling vt  R  v
vnet  v2  v2  2v  2 2 m / s.
37. (1)
38. (1)
39. (3) Consider a point P on the string that lies Force equation mg-2T=ma (1)
at a position where the string gets detached from
mr 2
the pulley Torque about C.M 2Tr   (2)
2
For no slipping a = r (3)
2g
From the three eq’s a 
3
As there is no slipping between string and the 46. (1)

pulley the acceleration of the point P is equal to 2
the tangential acceleration of the pulley. As the 1 2 1 v
47. (2) K .Ering  mv  (mr 2 )    mv 2
string is unstretchable 2 2 r
a P  ablock 1 2 1  2 2  v  7
2
K .Esphere  mv   mr    mv 2
2 2 5 r
  5
R  a  aP  at 
40. (3) As there is no force along horizontal K .Ering 5

direction. COM of rod falls vertically down K .Esphere 7
1 1 mr 2  v2  mv 2 1 2  k2 
48. (4) KErot  ICM  2    mv 1  2   mgh
2 2 2  r2  4 2  R 
mv2 1 2 3mv2 where h is the height of the inclined plane

KEtotal   mv 
4 2 4
2 gh
v
KErot 1 k2
  1 2
REtotal 3 R
49. (3) Rotational kinetic energy.
k2 2
For a solid sphere 
1 2 R2 5
K I
2 Substituting the given values, we get
1
Here,   1rad s 2  10  7 2 10  7  5
v   10ms 1
 2 7
1 1  
Hence, kinetic energy, K  I 1  5
2
53. (2) Power  
 I  2K
2  1800
50. (2) 54. (2)   2 n   60 rad / s
60
1  K2  1 K2
K R  KT  mv 2  2   mv 2  2  1 P 100  103
2 R  2 R P       531N  m
 60
i.e., the body is ring
L2
51. (4) Here, 55. (3) Kinetic energy K 
2I
Mass of the coin, M  10 g  10  103 kg
1
If angular momenta are equal then K 
2
Velocity, v  6 cm / s  6  10 m / s I
The total kinetic energy of the coin is If I A  I B and K A  K B
1 1 56. (4) As the block rotates with constant speed

 Mv 2  I  2 F=f
2 2
1 11  R 
 Mv 2   MR 2   2 Wext  W f  mg s   mg    1Joule
2 2 2   2 
1 57. (2)
[I  MR 2 ]
2 58. (2) L  I
1 1 59. (1) He decreases his moment of inertia by
 Mv 2  Mv 2 [ v  R ]
2 4 this act and therefore increases his angular
velocity.
3
 Mv 2  27 J . 60. (2) L  I
4
52. (2) According to the law of conservation of L2
61. (2) E  L  2 EI .
energy, we get 2I
62. (3)
2
I  mr 2  10   0.2   0.4kgm2 73. (4) I11  I 22
2100 MR 2   MR 2  2mR 2  2
  2 n  2  rad / s
60 M
2 
0.4  2  210 M  2m
 L  I   88kg  m2 / s
6 74. (2) As no torque is applied, angular
momentum
iˆ ˆj kˆ
   2  2 
63. (1) L  r  p  1 2 1   ˆj  2kˆ L  l  constant=  MR 2    constant i.e.,
5  T 
3 4 2
 R2 R12 R22
L lies in yz plane   constant  
T T1 T2
64. (3) Angular momentum about
2
origin  I translation  I rotation R2  xR 
 T2  22 T1   1   24h  24 x 2 h
R1  R1 
1 3
 MvR  I c  M  R  R  MR 2  MR 2
2 2 75. (4) Consider the two disc as a system. As
there is no external torque the angular
L
65. (2)  momentum is conserved about the central axis.
t
2
Given, L  4 A0  A0  3 A0 , t  4sec
 2  0.2 
I11   50 
2
3  4  0.1 200
2
  A0
4 I 22   
2
66. (4)
Li  L j
67. (1) L=I  =constant
As ice melts then it reaches the equator and I11  I 22   I1  I 2  
hence I increases and then  decreases and T 2 2
increases. 2  0.2  4  0.1
 50    200 
 2 2
68. (2) 2 2
2  0.2  4  0.1
69. (4) Since there is no resultant external force 
2 2
linear momentum of the system remains constant
  100 rad sec
1 1 L2
70. (1) K= Iω 2 
2 2 I
L=I  =constant
As I increases  and K decreases. 1. (4) The situation is shown in figure.
71. (2) L  I   constant
L2 1
Kinetic energy K =  K
2I I
K1 2
72. (3) L  I   MK 2  constant  
K2 1
 l 
2
2 5. (1)
I xx'  2m  
 2
  m  2l  3ml 2
2. (1) We can imagine the given triangular plate

as a half part of a square of side length l.
Ml 2
I AC 
12 dI  dmx 2 sin 2 
I   x 2 sin 2  dx (dm   dx )
x
I  sin 2   x 2 dx
0
3. (3) Treat each half separately. For the wood, sin 2 l 3
the moment of inertia about the given axis is I
3
sin 2  l 3 (m)  m
I   
6l  2 l
ml 2 sin 2 
I Net  2I 
3
1
2 2 MR 2
L  L 1 6. (4) ITotal disc 
I w  mw    mw    mw L2 2
12  2  4  12
Similarly, for the brass, the moment of inertia
about the given axis is
2 2
1  L  L 1
Ib  mb    mb    mb L2
12  2   4  12
 The moment of inertia of the given rod about M  R M

2
the given axis is I  I w  I b M Removed  2

   
R  2 4
1 1 1
 mw L2  mb L2   mw  mb  L2 . I Removed (about same perpendicular axis)
12 12 12
L M  R / 2
2
M  R  3MR 2
2
4. (1) L  2 r  r  ,m   L     
2 4 2 4  2 32
I Remaining disc  I Total  I Removed
MR 2 3 13
  MR 2  MR 2
2 32 32
3 2 7. (3) Momentum of inertia about perpendicular
Lxy  mr
2 bisector is
2
3  L  3  L3 m 2 mR 2
 L   I 
2  2  8 2 12 4
m  2 8 Mπ
2 = MR 2 -10mR 2 [here m= ]
I  R  (i) 3 16
4 3 
Also m   R 2   8 5  2
   MR
3 8 
m 10. (2) Moment of inertia is more when mass is
 R2  Put in equation (i)
  farther from the axis. In case of axis BC, mass
distribution is closest to it and in case of axis
m  2 m  AB mass distribution is farthest. For the axis
I   
4  3    passing through C and perpendicular to the plane
the moment of inertia is maximum.
For maxima & minima
 IC  I P  I B  I H
dI m  2 m 
   0  
d 4  3   2   11. (3)  av .t   L
2 m 2  R 2  2u sin
  2   Here t = time of flight = 
3   3  2  g
Change in angular momentum about point of

2 R 2 2 3
or   2  projection (initially it is zero)
3  R 2
  
L  L f  Li  ( mu sin )Range  0
 3
 
R 2  mu sin   u 2 sin 2  mu 3 sin sin 2
 
8. (4) MOI of removed part about axis passing g g
through COM &  to plane of disc 
  L mu 3 sin  sin 2 g
 I cm  Md 2 Now  av 
t

g

2u sin 
2
( M )  R / 3  4 R 2  MR 2
M mu 2 sin 2
2  2 
 9  2
So MOI of remaining portion 12. (1) Let true weight of the object be W and
=[MOI of whole disc-MOI of removed part] weight of right pan is x more than left pan.
Now suppose the object is placed in left pan:
R 2 MR 2
 (9 M )   4 MR 2
2 2 W  W1  x (i)
9. (4) Mass of each disc Now if the object is placed in right pan.
M  W2  W  x (ii)
m 2
  R2  M
16 R 16
From (i) and (ii)
Moment of inertia of the remaining part is
W1  W2
I = I square  4 I disc W
2
M(4R)2  mR 2  13. (2) As the block remains stationary therefore

= -4  +m( 2R)2 
6  2  for translatory equilibrium
l
 mgl sin   mg cos
2
1  1 
tan   or   tan 1  
2  2 
16. (1) For the equilibrium of the entire structure,

 Fx  0 F  N
N1  N 2  2Mg
and  Fy  0 f  mg
For rotational equilibrium   0
By taking the torque of different forces about
point O
   
 F   f   N   mg  0
As F and mg passing through point O

 
 f   N  0 N 2  Mg , N1  Mg (since N1  N 2 )
  For individual boards, T1  f1 , N1  Mg
As  f  0  N  0 and torque by friction and
normal reaction will be in opposite direction. l
and T1  l sin   Mg cos
14. (1) If the hanging light is in equilibrium then 2
the torque about the hinge S is zero. As the
horizontal string approaches down then lever Mg
f1  T1 
decreases and hence tension increases. Possibility 2 tan 
of breaking of horizontal string is more in case -
For safe equilibrium, f1  s Mg
(C) and least in case - (A)
15. (4) The forces on the ladder are shown in Mg 1
 s Mg  tan      45
the figure. Net force along y and x directions 2 tan  2s
are zero.
17. (1) During toppling normal force passes
mg  N1 through A
 N1  N 2 or N 2   mg
Taking moment about O, we get
a F 3a 2mg
mg   F
2 4 3
18. (2) The sphere is on the verge of toppling
when a line of action of weight passes through
the edge.
21. (2) From the centre of the hemisphere the

R
C.M lies at a distance . Frictional force acting
2
on the sphere is zero. The normal force passes
through the C.M
 R  h
cos    h  R  R cos 
R
19. (2) From the figure The torque is measured w.r.t. C.M
y = 50 - 25 =25 cm  N   mg  0
x  50 2  25 2  25 3
R
c  F  Ic 
Fy  5 gx 2
From parallel axis theorem
2
R
IO  Ic  m  
2
2 2 mR2
mR  I c 
3 4

 F  25   5  10 25 3 
5mR 2
 F  50 3 N Ic 
12
20. (3) The tendency of rotation will be about
point C. For the minimum force the torque of F R 5mR 2
F  
about C has to be equal to the torque of mg 2 12
about C.
6F
 
5mR
22. (3) The gravitational force on the rod is
shown in the figure
 3 a mg
F  a   mg    F 
 2   2 3
Taking torque about pivot   I    

(ii) V C  V B  2viˆ  viˆ;V B  V A  viˆ  0  viˆ ,
 m2 correct
mgsin   
2 3    
(iii) V C  V A  2v, 2 V B  V C  2v ,correct
3g option is (1).
 sin 
2
27. (4) Let  be the angle made by the rod with
23. (1) If distance between them remains horizontal
constant, then
v cos 30  u cos 60
3 1 10
v  10  v 
2 2 3
  (v sin 30  u sin 60 ) / d
 10  1 10  3
   2 The coordinates of C.M are
 3 2 5
  rad / s.
2 3 l l
x  l cos  cos  cos (i)
2 2
24. (4) As cotton reel roll point of contact with
ground will act as instantaneous axis of rotation. l
y  sin  (ii)
2
By squaring the two equations
l2
x2  y2 
4
i.e., the C.M moves in circular path having radius
l
2
28. (1) As sphere rolls, the lower point of the
vP    R  r 
sphere should have the same acceleration as
vp the plank.
   20rad / s
Rr Hence, a1   R  a2
Velocity of centre of reel
2  2  4    3rad / s 2
vP R 6  20
vC   R    4m / s 29. (1) C1 and C2 are IC (instantaneous centre
 R  r   20  10
of rotation) of the two cylinders. The cylinder
  
25. (4) For a rotating body d,  &  are can be considered as rotating about C1 and C2 .
perpendicular to the plane of rotation. so In the absence of slipping between the plank
  and the cylinders, points A1 and A2 have the
d  0
same velocity. Angular velocity of the larger
26. (1) We know
   2v v
cylinder is 
V C  2viˆ;V B  viˆ,V A  0 4R 2R
Acceleration of point B = acceleration of point

A
 v 
vCM   2 R   v r1  acm  r (iv)
 2R 
30. (2)  acm  2r
32. (1) For no slipping, vcm   R (i)

The velocity of point B is
 r
vB   r  vcm  vcm  1   (ii)
 R
 B vB  r  r
  1     B   1  
 vcm  R  R
Angular velocity of the upper cylinder is
Where lB & l are the distance travelled by B
2v    v  3v and C.M.
up   ; for the lower cylinder
2R 2R
33. (2) 2vCM  vA So when cylinder moves a
v0 v distance l, length of the string that passes through
lower  
2R 2R the person is 2l.
up
Hence, ratio  3
lower
31. (3) Let T be the tension in the string. Torque

on the upper disc is
mr 2
Tr  1 (i) 34. (1) Torque equation: The internal forces are
2
the reaction forces at O. Then the net torque
Torque on the lower disc is about O is
mr 2
Tr   (ii)
2
1   (iii)
From (i) and (ii)

 0  mgl  mg  2l   3mgl mgL / 2 3g

 
mL2 / 3 2 L
Which produces an angular acceleration  .
Tangential acceleration of the centre of mass,
Using Newton’s 2nd law of rotation   I 
L 3g
We have 3mgl  I 0 where at    . Let R1 & R2 are the forces
2 4
2
I 0  ml 2  m  2l   5ml 2 exerted by (P). Use Newton’s second law of
equation.
3g 3  10
This gives     3rad s 1
5l 5 2
Force equation: Referring FBD, we have
 mg  mg   R  ma1  ma2
3mg
R1  mar 
4
3mg
Substituting a1  l and a2  2l , we have mg  R2  m  at 
4
R  m  2 g  3l  mg
R2 
4
 2  2 10  3  2  3  2  20  18  4 N .
So, reaction force by the pivot on the rod,
35. (4) The angular velocity of the rod about the pivot R  R12  R22  10mg / 4 at angle of
when it passes through the horizontal position is tan 1  R2 / R1    tan 1 1 / 3   with the
given by
horizontal.
L mL2  2
mg  sin 30   36. (1) Friction force acts on the ball in forward
2 3 2
direction that makes it to move towards right
side.
3g

2L 37. (3) The distance of CM from the ring centre
O
3m  0   m  r  r
x 
3m  m 4
We can apply torque equation about point of
contact as the ring is in pure rolling.
Radial acceleration of the centre of mass (as
centre of mass is moving in a circle of radius L/
2) is given by
L 3g
ar   2 
2 4
Torque about (P), in the horizontal position, is
L
  mg  I  P  I P
2
40. (2) In pure rolling mechanical energy remain

r 2
4mg      3mr 2  3mr 2   m  AP    conserved. Therefore, when heights of inclines
 4  
are equal, speed of sphere will be same in both
2 the cases. But as acceleration, a  sin ,
 mgr  6mr 2  m  2r   
 therefore acceleration and time to descent will
be different.
g
 mgr  8mr 2    41. (4) To maintain the cylinder in rotational
8r equilibrium the friction should act upward. Let
38. (3) After breaking the tension let a is the us assume cylinder is not moving, then for
acceleration of C.M.  is the angular translation equilibrium
acceleration. From force equation mg - T = ma T  f s  mg sin 
and
For rotational equilibrium
mg 3
T .R  f s R  0  f s 
4
But  f s max   N   mg cos 
1 mg
 0.4.mg  
2 5
Torque equation about C.M Here, f S  f Smax which is not possible so the
body can’t be at rest. So, friction existing must
mgl ml 2 a
 I   be kinetic
2 3 l
  1 mg
2 fk   mg cos  0.4  mg  
2 5
 l 
If the string is tight  a    42. (4) From normal reactions of roller, we can
 2 
conclude it moves towards left.
3g
a
4 43. (4) Rolling can be considered as pure rotation
mg about point of contact
So, T
4 1 1  mR 2  3
Krolling  I P 2    mR 2   2  mR2 2 (i)
39. (4) When a body is in pure rolling on an 2 2 2  4
inclined plane the friction is static and it always
acts up the plane. The rod translates with the velocity v hence
velocity of centre of disc will also be v
So the forces on the plane are
Normal force (Into the plane) v  R (ii)
2 3
Friction (Down the plane) From (i) and (ii); K rolling  mv
4
Kinetic energy of the rod
1 m mv 2
K rod    v 2 
2 2  4
K total  k rolling  k rod  mv 2
44. (4)
v  u cos iˆ
mu 3 sin 2  cos  ˆ

L  m( r  v )  k
2g
47. (3)
1 2 1 2 1 2 1 2
mbott  I bott  mg 2  R  r   mtop  I top
2 2 2 2
7 2 7 2
mbott  4mg  R  r   mtop  4mg  R  r 
5 5
7
  g  R  r   L0  mv cos 45  4
5
1
27  53 2   4  60 g cm 2 s
or bott  gR  r 2
7
45. (3) As there is no slipping then mechanical 48. (1) Since surface is frictionless, the body does
not rotate about its centre of mass. Only it slides
energy is conserved and we can state that the
increase in kinetic energy of the system equals parallel to the surface, i.e.,   0 and
the decrease in potential energy. Since v   g sin   t
K i  0 (the system is initially at rest), we have
The angular momentum after time t:
K  K f  Ki
L  mvr  m  g sin  t  r  mgrt sin 
1 1 1
 m1v 2  m2 v 2  I  2 49. (2) In non-inertial frame torque produced by
2 2 2
the pseudo force should be considered. So eq-
Where m1 and m2 have a common speed. But
B will not be correct w.r.t. non-inertial frame.
1 I    
v  R so that  K   m1  m2  2  v 2 50. (1) We know that L  r  mv
2 R 
We see that the system loses potential energy

because of the motion of m1 . Applying the law
of conservation of energy.
 K  U  0, gives
1 I 
 m1  m2  2  v 2  m2 gh  m1 gh  0
2 R 
2  m1  m2  gh
v  R  ^
I
m1  m2  2 LO  mv   k   Dto A
R 2 
Rˆ 
46. (2) r i  hmax ˆj  R ^
2 LO  mv   a  k C to D 
 2 
u 2 sin 2 ˆ u 2 sin 2  ˆ
 i j So the false option is (1)
2g 2g
51. (3) Li  I   mvR the time of stopping is zero. We know the angular
momentum of the disc about P remains constant
L f   I  mR 2   ' because frictional force f, pass through point P
and net torque produced by mg and N must be
I  mvR zero.
L f  Li   ' 
I  mR2
52. (1) The angular momentum of the disc about
P, just before and after the impact, remains the
same
 L2  L1
 Linitial  LFinal  mv0 r  I 00  0
1
 mv0 r  mr 20  2v0  0 r
2
55. (3) Just after releasing the bola we consider
where L1 = angular momentum of the disc L about C.M then L will be conserved since
about P just before the impact tension in the string produce zero torque about
C.M. The angular velocity decreases since the
1 2 system gains some translational motion.
L1  I 0  mr  (i)
2 56. (3) By conservation of angular momentum
L2 = angular momentum of the disc about P just
after the impact
1 
L2   mr 2  mr 2   ' (ii)
 2 

Just before the impact, the disc rotates about  M 9R2 Md 2  8
MR2   MR2   
O. But just after the impact, the disc rotates  8 25 8  9
about P.
 200R2  9R2  25d 2  8
1 3 1 R2   
 mr 2  mr 2 '   '    8  25 9
2 2 3
53. (2) Conserving angular momentum about 225R 2  209R 2  25d 2
C.M Li = Lf 4R
d  .
5
3mR 2 ω
mvR   3mR 2 ω ' 57. (2) Conserving angular momentum about O
2
(just before and just after) is
2v - 3Rω a  ma2 ma2  3v
 ω'= mv      
6R
2  6 2  4a
54. (4) Since the disc comes to rest, then v = 0 58. (1) Let the impulse due to the particle be J
and   0. That means, the angular momentum
about the instantaneous point of contact just after J  mvCM
J Angular impulse = Change in angular

 vCM  ( vCM -velocity of C.M of the rod) momentum
m
Angular impulse about centre of mass is l ml 2
P  
2 12
ml 2 12 Jx
Jx     2
12 ml 6P

For point A to be at rest instantaneously, ml
  Time taken by the rod to turn by 90 is
vtrans  v rot  0
l T 1 2   ml
or vcm  t    
6 4 4  2 12 P
61. (1) Let v be the velocity of the centre of mass
of the sphere and  be the angular velocity of
the body about an axis passing through the centre
of mass.
The linear impulse is
J = Mv
The angular impulse is
2
J  h  R   MR2  
J l 12 Jx 5
Thus   2
m 6 ml
2 2
From the above two equations, v  h  R   R 
l 5
which gives, x 
2
From the condition of pure rolling, v  R
59. (2) Let  is the initial angular velocity and
R is the radius. 2R 7R
hR h
5 5
By conservation of angular momentum
62. (3) Impulse provided by the edge in the
m0 R 2  m R2  m horizontal direction:
   0  tR 2   t  0
2  2  2 2 
 Ndt  mv    mv 
'
(i)
60. (4)
Angular impulse by friction in the vertical
direction
2 v
fR   R  Ndt  mR 2   (ii)
5 R
Angular impulse about center of mass
pl

2
From eqs. (i) and (ii), we get

a I1 0  1 
 (iii)
' b I 22
 Ndt  2 mV and v v
63. (4) From linear impluse When slipping ceases between the discs, the
contact points of the two discs have the same
J= mv (v-velocity of CM) linear velocity, i.e.,
From angular impluse
a1  b2 (iv)
2
mL
Jx   On substituting 2 in eq’s (iii), we get
12
 I10 
1 
 I1   a 2 I 2 / b 2  
 
ml 2
As the point P velocity is zero. 1. (2) I yy ' 
12
l Using parallel axis theorem:(about AD)
vP  v    0
2 ml 2 ml 2 ml 2
I AD   
12 4 3
l
v 
2
L
From the above three eq’s x 
6
64. (1) The two discs exert equal and opposite
forces on each other when in contact. The torque
due to these forces changes the angular
momentum of each disc. Let 1 and 2 are the
angular velocities of the two discs.
2. (2) Moment of inertia of each of rod AC and

BC about the given axis OO’ is
m 2 m 2
I AC  I BC  sin 2 60 
3 4
The angular impulse on the two discs are
fat  I1 0  1  (i)
and fbt  I 22 (ii)

From eqns. (i) and (ii), we get and M.I of rod AB about the given axis OO’ is
 3
2
3 2
The square can be assumed as a combination
I AB  m 
2   4 m of four small squares each having a dimension
 
l
Hence,
2
m 2 m 2 3 2 5 2 By symmetry for remaining portion it must be
I  I AC  I BC  I AB    m  m
4 4 4 4
 2  R 2  21 3  M 2  M 2
I  2  M     M (2 R )2  MR 2 
3. (1) 4  6  8
 5  2   5
4. (1) The moment of inertia, 7. (2) M.I. of complete disc about ‘O’ point
1
ITotal   9M  R 2
2
dmr 2
dI   x 2 dm
4
r2 L/2
I  dm    ( r 2 ) x 2 dx
4  L/2
 L2 r 2   L2 D2  R
I  m    m   Radius of removed disc 
 12 4   12 16  3
5. (4) 2
9M R
 Mass of removed disc  2
    M
R 3
M.I. of removed disc about its own centre

2
1 R MR 2
 M  
2 3 18
Moment of inertia of removed disc about ‘O’

2
MR 2  2R  MR 2
2 I removed disc  I cm  mx 2  M  
2 1  c  18  3  2
I PNQ  I AOB  M . ON  ,I PNQ  MR 2  M . 
4  2
M.I. of complete disc can also be written as
1 R I total  I removed disc  I Remaining disc
But I PNQ  MR 2  c  
2 2
9MR 2 MR 2 8MR 2
Hence (4) is correct.  I Remaining disc     4MR 2
2 2 2
6. (1) For full square about an axis passing 8. (2) For maximum possible volume of cube
through ' O '
2 R  3a , a is side of the cube. Moment of inertia
M 2 ma2 a2
 about the required axis I   a3 ,
6 6 6
M 7 MR 2 55MR 2
where   4 
2
 2
 6 M   2R  
2

R 3
3
5 I P  I o  md 2
3M 1  2 R  3M 1 32 R5 4MR 2
I  
4R3 6  3  4R3 6 9 3 9 3 55MR 2 181
  7 M  3R 2   MR 2
2 2
I
9. (2) I AC  I BD  (From  axis theorem)
2 13. The angular momentum of the particle with
respect to P is
I
Similarly, I EF 
2
 I AC  I EF .
10. (3) Because distribution of mass is closest
around the axis DB. Lp  mv y R
11. (2) As shown in the figure
v y  2 gH
2H
R  uT  u
g
2H
 Lp  m 2 gH u  2muH
g
14. (1) Torque due to weight of left m:

 1  mgl1kˆ
I CD  I C ' D '
Torque due to weight of right m:

I  2   mgl2 kˆ
So, 2 I CD  I  I CD 
2   
So net torque:    1   2  (mgl1  mgl2 )kˆ
I
Also, 2 I AB  I  I AB  15. (2) Net force and net torque should be zero.
2
In case (2) the net torque can be zero if it is
So I CD  I AB measured about the right most end. In other
12. (3) options the torque is not zero w.r.to the right most
end.
16. (3) Position vector of the point at which force
is acting
 
r1  iˆ  2 ˆj  3kˆ, r 2  3iˆ  2 ˆj  3kˆ
  
r12  r1  r2  (iˆ  2 ˆj  3kˆ)  (3iˆ  2 ˆj  3kˆ)
 2iˆ  4 ˆj  6kˆ
  
I o  I cm  md 2 Now   r12  F  (2iˆ  4 ˆj  6kˆ)  (4iˆ  5 ˆj  3kˆ)
iˆ ˆj kˆ 21. (2) Taking moment of forces about point P,

 we get
  2 4 6
4 5 3
 iˆ(12  30)  ˆj(6  24)  kˆ(10  16)
 (42iˆ  30 ˆj  6kˆ) N  m
17. (3) As f  0 1,2 & 4 are not possible since
tension has non zero horizontal component of
force. Only option 3 is possible as net torque At the toppling condition
can be zero about any point. F  0.6  100  0.8
18. (2) As ladder is at rest so net torque
about A is zero 100  0.8 800
or F  N N  133.3N
0.6 6
22. (1) About C, for toppling  F   mg
l
mg  cos  N 2 6  0
3 a mg
Fa cos30  mg F
l4 2 3
 500  N 2 6  N 2  111N
3l At the verge of slipping
19. (4) The block shall topple about its edge
fs  F
through O. The torque FL of the applied force
is clockwise. The torque MgL/2 of the weight but f s   mg
is anticlockwise.
Applying condition for rotational equilibrium. mg 1
   mg   
3 3
mg
 FL  mgL / 2  0 or F 
2 23. (2) As   I
20. (2) The cube will topple when the line of 1
action of CM passes through the point A.  (  is constant)
I
Moment of inertia of figure (ii) is smaller hence
acceleration is greater.
24. (4) Torque exerted on the disc   TR
Now   I
 TR 2TR 2T
AX a / 2    
That is, tan       45 I 1 2 MR 2
MR
OX a / 2 MR
2
28. (2) v p  vt2  vCM

2
 2vt vCM cos(80   )
As the body is in pure rolling vt  vCM  v
25. (2) For rigid body, separation between two

points remain same
v1 cos 60  v2 cos30
v1 3v2 v p  v2  v2  2vv cos 180   

  v1  3v2
2 2
 vp  2v2 1  cos   2v sin  / 2

29. (4) Here,   3iˆ  4 ˆj  kˆ

r  5iˆ  6 ˆj  6kˆ
  
As v    r
iˆ ˆj kˆ
v2 3v1  3 4 1
v2 sin 30  v1 sin 60 
disc   2 2
5 6 6
d d
 iˆ  24   6    ˆj 18  5   kˆ  18   20  

v2  3  3v2 2v2 v2
disc     18iˆ  13 ˆj  2kˆ
2d 2d d
30. (3) From the given figure
26. (2) Angular velocity,
Let OA = x, OB = y
20  10
  20rad / sec.
0.5 x 2  y 2  l 2 and x  l cos
Differentiate x 2  y 2  l 2 with respect to time.
dx dy
2x  2y  0
dt dt
v  dy x dx 4
27. (1) rad   po int   rel  . vB     vA cot   v0 
 r  dt y dt 3
vrel . being the velocity of one point w.r.t other.. 31. (2) For pure rolling velocity of the point of
contact has to be equal to the velocity of the
‘r’ being the distance between them.
surface.
3v  v 2v Let us say cylinder rolls with angular velocity
 
r r .
x  l cos   b cos  (l  b)cos (1)
y  b sin  (2)
From equations 1 & 2
x2 y2
sin 2   cos2    1
At point B, (l  b) 2 b 2
vC  r  vP  r  vC  vP The value of b depends on point P. The path

followed by the point P is ellipse.
At point A,
34. (3) Let ω is the angular velocity of the wheel
v A  vC  r  2vC  vP vA  Rω & vB  2Rω
vB  2 v A  y = 2 x
35. (2)
32. (1)
Since, there is no relative sliding between the

cylinder and the planks 1 and 2, the points A and
B of the disc will move with velocities equal to
the velocities of the plank’s.
 
v A  viˆ and v B  2viˆ If the spool rolls without slipping then
Joining A and B and A ' and B ' we find the point vP  2vCM  xP  2xCM
P as IC. Then we have the similar triangles l
Given that xP  l  xCM  .
PAA' and PBB ' . Using the properties of similar 2
triangles we have If the point P (person) moves a distance l then
2v v BP l l
  AP  CM travels by . So the additional length
BP AP 2 2 2
has to come out from the spool.
4R
AP  BP  2R  BP 
3 36. (4) The forces acting on the rod are
Hence, IC is located at a distance of 4R/3 below mv 2
the top of the disc. N  mg 
r
33. (1) Let l be the length of the rod and  be From conservation of energy
the angle made by the rod with respect to the
horizontal. The coordinates of the point P are. 1 2 l
I   mg
2 2
1 ml 2 2 l
  mg
2 3 2
mv 2 3mg

r 2
3 5
N  mg  mg  mg
2 2
7
For no sliding, f   mg  F   mg
2
39. (1) Net torque about point O is  O  I O
R a
4 mg    fR  I CM   I CM  
37. (4) KE of the rod is maximum when it is in 2 R
the vertical position.
From conservation of energy, we get
2
 m Rα + f =2mg
5
mgl 1  ml 2 ml 2  2 & f =ma=mαR

   
4 2  12 16 
10 10
 f= mg  mg  μN
24 g 7 7
2 
7l
10  2
 mg   (5mg )     
and Ry  mg  m  l / 4   2 7 7
40. (3) Let T be the tension in each string.

 6 g  13
Ry  mg  m   = mg
 7  7
38. (4) As sphere is rolling we can apply torque

equation about point of contact (about P)
FR  I P
7 
 FR   MR 2   Force equation mg-2T=ma (i)
5 
5F mr 2
 R   a (acceleration of cm) Torque about C.M 2Tr   (ii)
7M 2
Equation of translatory motion as there is no slipping a = r (iii)
2 2g
F  f  Ma  f  F mg From the three eq’s a 
7 3

2s
41. (2) t
a 2 13
2 
MgR   MR  
22 
g sin 
a 4g
 K2  
1  R 2  3R
 
Rg
2s  k 2  At the lowest point on the rim, v  2 R  4
t 3
1  
gsin   R 2 
45. (3) From the conservation of energy
 k2   k2 
 1  2 
  1  2  1 1
 R ring  R disc mgh= mv 2  Iω2
2 2
tCoin  t ring As the cord does not slip v=r
42. (1) In four wheel drive, all wheels are
2mgh 2mgh 2mgh
powered wheels. In the absence of friction the   2 2
  .
wheels slip backward so friction must act in mr  mk mr 2  I I  mr 2
forward direction. 46. (3) Point D is instantaneous axis of rotation.
43. (3) When the ring is at the maximum height, Take pure rotation about D, the total kinetic
the wedge and the ring have the same horizontal energy of the rod is equal to the rotational
component of velocity. As all the surfaces are kinetic energy about instantaneous centre
smooth, angular velocity of the ring remains
constant. 47. (1) Angular momentum of earth about its axis
of rotation,
From conservation of mechanical energy, we
get 2 2 4 MR2
L  I  MR2  
1 2 1 2 1 '2 1 2 1 '2 5 T 5T
mv  I   mv  I   mv  mgh
2 2 2 2 2 48. (3) t  L
'
where v is final common velocity
Fr t  L
From conservation of momentum
As F & t are same
2
v v
v'  h
4g
L2  L1  r2  r1 
2
44. (4) From the parallel-axis theorem, L2 F 2 r 2 t 2 F 2 t 2

K  
1 3 2I mr 2 m
I  I CM  MD 2  MR 2  MR 2  MR 2 2
2 2 2
The pivot point is fixed, so the kinetic energy is
entirely rotational around the pivot.  K1  K 2
49. (2) 4 y  3x  2
3 1
y x
4 2
  37
52. (2) Let  is the angular velocity of the rod.
Angular momentum about A will be conserved,

The angular momentum about origin is
i.e.,
1 1
L  mv cos37  3  5cos37 Li  L f
2 2
mva  I 
 6kgm 2 s 1
2
    m  2a  3v
50. (3) Since L P  Lcm  r  pcm  mva  .   
3 4a
53. (4) Angular momentum about point of
contact will be conserved
Li  L f
2 2  v0 2 2 v
 5 Mr  r +Mrv0 =  5 Mr  r +Mrv
   
 I cm0  kˆ  Mv0 R  kˆ    
Since sphere is in pure rolling motion hence 6

 v  v0
7
  v0 / R
54. (4) The initial and final motion of the
 2
ball is shown in the figure
v  
  R 
   
 L P   MR 2  0   kˆ  Mv0 R  kˆ 
5 
7
 Mv0 R kˆ
5
 
51. (3) The initial angular momentum of the By conserving angular momentum about ‘O’
system is zero. The final angular momentum of
L1  L2
the girl-plus-merrygo-round is  I  MR2   ,
which we will take to be positive. The final 2 v
angular momentum we associate with the throw mv0 R  mvR  mR 2  
5 R
rock is negative: -mRv, where v is the speed of
the rock relative to the ground. 5v0
v 
From angular momentum conservation 7
mRv
0   I  MR 2    mRv   
I  MR 2
55. (3)
mvR 2
The girl’s linear speed is R 
I  MR 2
By conserving angular momentum about P By conservation of angular momentum about

pivot
Li  L f
L  I
2 v
2
mr 0  mvr  mr  0 2
r mvd  Md 2 d 
  m  
2  12  2  
0 r
v
2
 md 2 md 2 
56. (3) Angular momentum of the pendulum   
about the suspension point O is  2 4 
(M=6m)
2v
 
3d
60. (2) Here we assume that the centre of the
ring is fixed.
Using conversation of angular momentum about
O, we get
 mvR  (mR 2  mR 2 );  

v
is along the direction shown. r is defined from 2R

O to the body and v is along tangential direction.
As the bob is in rotation the angular momentum 61. (3) Let l be the length of the rod.
changes its direction.
57. (3) Let the angular speed of the disc when
the balls
reach the end be  . From conservation of
angular
momentum, J
v
1 1 m m 2m
mR 20  mR 2  R 2  R 2
2 2 2 2 2 2
l  l l 
0
J    m    m   
   2    2  2 
3
58. (3) Apply conservation of angular momentum J

about the hinge we get ml
m  l J J J
mvR   R 2    m  R  R  vA  v     
2  2 2m 2m m
62. (4) Let v be the velocity of COM of ring just
2v 2  5 10
R    m / s. after the impulse is applied and v’ its velocity
3 3 3 when pure rolling starts. Angular velocity  of
59. (2) Let  is the angular velocity of the v'
system after collision. the ring at this instant will be   .
r
1. (0.9)
From impulse = change in linear momentum, we

have
J = mv  v = J/m
Between the two positions shown in the figure,
force of friction on the ring acts backwards.
Angular momentum of the ring about bottom
most point will remain conserved 1 2 l
I   Mg  cos37  cos 60 
2 2
 Li  L f
Ml 2 2
mvr  mv ' r  I    Mgl  0.8  0.5 
3
mvr  mv ' r   mr 2   v '/ r   2mv ' r mv 2 v2
 0.9mg   0.9 g
l l
v J
v'   v2
2 2m Frad  dm  0.9  dm  g
l
63. (1) Given system of two particles will rotate
2. (10)
about its centre of mass.
L
Initial angular momentum  Mv  
2
2
L
Final angular momentum  2 M   
2
From conservation of angular momentum

 contact po int  0   mg   ma
2
L L v
Mv    2 M       g
2 2 L
 R mg sin 300  R ma sin 600  a 
3
64. (4)
3. (20) If the belt does not slip, the linear speeds
From linear impulse at the two rims must be equal. A point on the
rim of wheel C has the same tangential
J  mv
acceleration as a point on the rim of wheel A.
From angular impulse This means that  A rA   C rC , where  A is the
2 v angular acceleration of wheel A and  C is the
Jh  I   mR 2   
5 R angular acceleration of wheel C. Thus
h 2  rA   10cm 
 1.6rad / s   0.64rad / s
2 2
 C   C  
R 5 r
 C  25cm 
With the angular speed of wheel C given by but as  g sin  (ii)

C   C t ,
Dividing the two equations
the time for it to reach an angular speed of
ar 3
12.8rad s 1 starting from rest is 
as 5
C 12.8rad / s
t   20s
 C 0.64rad / s 2
1. (1) Linear momentum of particle before

colliding  mv  mve y
1. (0.25)
Linear momentum of particle after it bounces
The representation of velocities are shown in 
  mv   mve y
the figure
Change in linear momentum,
  
p   mv  ( mv )  2 mve y
 
Change in angular momentum L  r  p,
where

r  ( ye y  aez )
v1 v v1
 2 x l  ( ye y  ae z )  ( 2 mve y )  2 mv aex
x lx v1  v2
2. (3) 2. (1) If we keep the small piece at the centre

then its moment of inertia about the central axis
Acceleration required: is zero. The mass of the remaining portion
g  2R    g / R decreases and hence moment of inertia also
decreases.
The three rockets provide:
3. (3) When a body of mass m slides down an
3FR  MR 2    3F / MR
inclined plane, then v = 2 gh
g MR
time needed: t   /   When it is in the form of ring, then
R 3F
2 gh 2 gh 2 gh v
3. (0.6) vring  2
  
 K  11 2 2
For rolling 1  R 2 
 
g sin  4. (1, 2, 3, 4) When net external torque on a system
ar 
1 K 2 / r2 of particles about an axis is zero, i.e.,
 
K 2
2   r  F  rF sin 
For shell 
r2 3 For   0
3 r  0 or   0 (since F  0 )
ar  g sin  (i)
5
If force acts parallel to the axis then torque
For sliding produced by the force is zero.
l 3 Fx
a   
2 2 ml
1. (3) When a disc rotates with uniform angular
or a  x
velocity, angular acceleration of the disc is zero.
2. (2) When the person jumps off radially away i.e., a  x graph is a straight line passing through
from the centre, no torque is exerted i.e.,   0. origin.
According to the principle of conservation of ml 2
angular momentum, I   constant. As mass 4. (4) I sin 2 
reduces to half (from 2M to M), moment of 12
inertia I becomes half. therefore,  must 5. (4) No of revolutions = Area of trapezium
become twice
1 1
1 3 103  2.5 103  3  1.5  3 103
3. (1, 4) For general rotational motion, angular 2 2
momentum L and angular velocity  need not
be parallel. (example is a compound pendulum)  11250 rev
Again, for a general translational motion, linear 6. (3) The perpendicular distance from point O
momentum p and linear velocity v are always to the extended line of the linear momentum
parallel. This is because p is directed along v remains constant. Angular momentum also
only. remains constant.
4. (1, 3) Choice (2) is false, as theorem of 7. (2) Angular momentum about rotational axis
perpendicular axes only to a plane lamina.
Lt   I  m(vt )2  
Now, Z axis parallel to Z ' axis and distance
a dLt
 2mv 2t;
between them  . Therefore, according to dt
2
the theorem of parallel axes,
2
 a  ma 2
I z'  I z  m    I z 
 2 2
Choice (1) is true
Again, choice (4) is false as Z'' axis is not parallel
to z-axis. choice (3) is true from symmetry,
Torque   (2mv 2 )t.
2
1. (4) I  MR 2  I  R2
5
1. (3) The perpendicular distance from point O
This relation shows that graph between I and R to the extended line of the linear momentum
will be a parabola. remains constant. Angular momentum also
remains constant.
2. (1) L  I  L
So graph between L and  will be a straight 2. (1) I  Ic  mx2
line. (m = mass of the sphere)
3. (2) The rod will rotate about point A with
I  I c at x  0
angular acceleration :
Therefore, I versus x graph is a parabola with
 F .x 3Fx
   minimum value of I  I c at x  0 .
I ml 2 ml 2
3 Therefore, the correct graph is (1).
3. (2) As, L  rp MR 2
I1  mR 2 
2
 log e L  loge P  log e r
and moment of inertia when the tortoise is at B
If graph is drawn between log e L and log e P
MR 2
then, it will be a straight line which will not pass I 2  mr 2 
2
through the origin.
After time t the tortoise reaches point B
4. (2) Angular momentum
AB  vt
L  m[r  v]  mvr sin
2
Here r 2  a 2   R 2  a 2  vt 
From conservation of angular momentum

 I 1   ( t ) I 2
Substituting the values we can find that the

variation of  ( t ) is non linear..
Given y  x  4 6. (3) The moment of inertia about the axis

shown is independent of the angle  . The three
Comparing with general equation y  x  4 , axes are present in the plane of the square.
we get m  1
tan   1    45
By considering the angular momentum
at (-4, 0) or (0,4)
1
mvr sin   5  3 2  4   60 unit.
2
5. (2) Since there is no external torque, angular ml 2
I x  I y  I 
momentum will remain conserved. The moment 12
of inertia will first decrease till the tortoise moves
from A to C and then increase as it moves from I  constant
C to D. Therefore  will initially increase and
then decrease.  x dm , x
7. (3) xcm  cm
 dm
n
L x
0 xk  L  dx  n  1 
 n
 L
L  x   n2
0 k  L  dx
L
at n0 xCM 
2
 1
1 n 
xCM  L
Let R be the radius of platform m the mass of 1 2 
tortoise and M is the mass of platform.  n
Moment of inertia when the tortoise is at A As n   xCM  L
8. (1) The positions of the two points p & q are

shown at different times. 1 4 gh

r 3
2
M 2 
2. (3) I M 
12 4
7
I M 2
48
7
Mk 2  M 2
48
7
k 
48
3. (4)
Assume m be the mass of one rod

At equilibrium the torque about o is zero.
i.e.,  rod1   rod 2  0
L 
1. (2) mg  sin  
2 
L 
mg  cos   Lsin    0
2 
3 L 1
Lsin   cos   tan  
2 2 3
4. (2) The moment of inertia I(x) at distance x
is
2
Assume that the bob travels down by h. I  x   MR 2  Mx 2
5
From energy conservation
1 2 2 
1 1 x2   I  x   MR 
mgh  mv 2  I  2 (1) M 5 
2 2
This equation resembles the standard equation
v of parabola. Hence the correct graph is (2)
 (2)
r
5. (3)
Solving (1) & (2), we get
 
The angle between r and v is 90
Angular momentum, LP  mvr sin 90
 2  0.6 12 11

Balancing torque w.r.t. point of suspension [As v  R ,sin 90  1 ]
l  So, LP  14.4 kgm 2 / s
mg x  Mg   x 
2 
 l 1
mM  M
 2 x 2
   2
Y 
1
 C equation of a straight line
1. (2) I  2m 
 2
 m  2 
x
2m 2
1   2m 2
The graph between m and is a straight line 2
x
6. (3) Moment of force will be maximum when  3m 2
line of action of force is perpendicular to line
AB. 2. (2) 1 revolution  2π radian
Mass of disc = 2 kg
Radius = 0.2 m
Torque about the centre:
  I
2 1 or, F.R  I
tan   
4 2
2  0.2  0.2  
7. (1) From the concept of dimensions  20  0.2 
2
I 0  Kml 2 ( BC  l )   100 rad/s 2
where K is a constant
Now, 2  02  2
2
m l  K 2
I DEF  K    ml  50 
2
 2  100    ω0  0 
4 2 16
I0
I DEF     12.5radian
16
I 0 15I 0 12.5
I remain  I 0    N  2 turns
16 16 2
8. (1) The point P is shown in the figure. 3. (1) N  m2 R
 4 2 
 m  2  R 
T 
4  9.8 
  0.2   0.2 
1600
 9.8 104 N
4.    RT = I 
  rF
MR 2 a
i j k RT  .
2 R

  2 3 5 Ma
T
4 3 1 2
 From the above equations.
or,   i  3  15  j 2  20   k  6  12 
Ma
or,   12i  18j  6k mg 
2
 ma
 2 2 2 2mg
   12   18   6   504 a
M  2m
= 22.4 8. (3) Angular momentum of the block is
conserved about point O. The moment of inertia
of the cube about the coner is
2
 ma 2  a   2ma 2
I0    m   3
5. (3)  6  2  
a 2 3v
mv  ma 2 ω     5 rad/s
2 3 4a
Using  F  ma along the incline, 9. (1) Let a and  are the linear and angular
accelerations of the cylinder. f is the static
mg sin   f  ma cm (1) friction acting in backward direction.
Using   I , F  f  ma and fR  I
2
ma
f .a  . (2)
2
As the disc is rolling without slipping
a am   a (3)
For pure rolling a   R
Eliminating f, m and a from these three
equations: F  f  ma .....(i)
2 mR 2 a
a cm  gsin  fR  .
3 2 R
b3
ma 3
2 f   F  ma
u 2 2
6. (4) S
2a 10. (1) Using perpendicular axis theorem.
2
2
S1 u u  I0  I x  I y
  S2   2  S1  (2)2 (40)  160m
1
2
S2 u 2  u1  I 0  2 I ( I x =I y =I )
7. (4) Let a is the acceleration of the block and
 is the angular acceleration of the disc. Force
on the block is
mg - T = ma
Torque on the disc is
I 0 ma 2 L  (mu cos60)H
I 
2 12 2 2
 mu  u sin 60
11. (1) The acceleration of rolling body on an  
 2  2g
incline plane is
gsin  mu 3 sin 2 60
a   3.0 kg m 2 / s
 K2  4g
1  R 2 
 
14. (3) The angular momentum of the particle is
Given that ac =as
LO  rP ( P -component of momentum which
g sin c g sin s is perpendicular to the radial distance)

1 2
1 1
2 5
sin c 3 / 2 15
 
sin s 7 / 5 14
12. (2) Let f be the friction present between the

plank and cylinder. Let a1 and  are linear and
angular accelerations of the cylinder.
LO  rmv cos  (2a cos )mv cos
LO  2mva cos 2   mva(1  cos 2 )
LO  va(1  cos 2 ) (If m=1)
15. (3) As there is no external force angular

momentum of the system is conserved.
f  Ma1 (i)
Li = L f
MR 2
fR   (ii)
2 Let  ' is the new angular velocity of the system
constraint relation MR 2  (MR 2  2mR 2 ) '
a  a1  R (iii)
M
' 
From the above three eq’s we get (M  2m)
Ma The loss is
f 
3
13. (1) The angular momentum of the particle at 1 1 (MR 2  2mR 2 )M 2 2
(K.E)  MR 2 2 
the heighest point is 2 2 (M  2m) 2
mM
 ( 2 R 2 )
(M  2m)
Gravitation 201
GMmx
F
( R 2  x 2 )3/ 2
Here x  3r , R  r
GMm
1. (3) When earth stops suddenly, centrifugal  F 3
force on the man becomes zero so its effective 8r 2
weight increases. 11. (2) The apparent acceleration of a body at
the equator is
2. (2)
g '  g  R2
3. (2)
4. (4) As  increases g ' decreases and hence
apparent weight decreases.
5. (2) Gravitational force is independent of the
medium. Thus, gravitational force will be same 12. (1)
i.e., F. 13. (2) Acceleration due to gravity at a height h
6. (2) is
2 2
7. (1)  R  1  h
g1  g   g 2
 g 1  
 R  h   h  R
8. (2) F  xm2 1  x  1  
 R
dF  2h 
For maximum force 0 g1  g 1   (h  R )
dx  R
dF Acceleration due to gravity at a depth h is
  m2  2 xm2  0
dx
 h
g 2  g 1  
 x  1/ 2.  R
4 
2
 h
G   R3  1
g 2  R   h  2h 
1
Gmm  3    1  1  
9. (3) F 2
 2 g1  2h   R  R
 2R  4R  1  
 R
4
  2 2 R 4  F  R 4 g2  h  2h   h 2h2 
9  1  1    1   2 
g1  R  R  R R 
10. (4) From Newton’s 3rd law the magnitude
g2  h
of force exerted by ring on sphere is equal to  1  
g1  R
the force exerted by the sphere on the ring.
g2
Force by the ring on the sphere g1 increases linearly with h.
202 Gravitation
14. (3) Acceleration due to gravity at an altitude The acceleration above the surface is
h above the earth’s surface is
GM GM GM  2h 
gRE2 g2  2
 2
 1  R 
gb   R  h  h R2
 RE  h 
2 R 2 1  
 R
Where g is the acceleration due to gravity on
the earth’s surface and RE is the radius of the  2h 
 g 1  
earth.  R
So acceleration due to gravity decreases with Given that g1  g2

increasing altitude.
d  2h
Acceleration due to gravity at a depth d below
the earth’s surface is 18. (3) Gravitational Potential energy
 d  GMm GMm
g d  g 1   U 
 RE  r Rh
So acceleration due to gravity decreases with GMm GMm
increasing depth. U initial   and U final 
3R 2R
Acceleration due to gravity at latitude 
GMm GMm GMm
g   g  RE 2 cos 2  Loss in PE  gain inKE   
2R 3R 6R
So acceleration due to gravity increases with 19. (4) Work done by the external agent is
increasing latitude () .
W  U f Ui
Acceleration due to gravity of body of mass m
is placed on the earth’s surface is
 Gm2 
GM E Ui  3  
g  a 
RE2
So acceleration due to gravity is independent  Gm2 
of the mass of the body but it depends upon the U f  3 
 2a 
mass of the earth.
4 Gm2 3Gm2 3Gm2
15. (4) g  GR W  3  
3 2a a 2a
g1 R11 20. (1) From conservation of
 
g 2 R22
1
2 1 1 
energy mV  GM e m    (i)
GM 2  R R h
16. (3) We know g  2
R
GM 6.67 1011  7.34 1022 GM e
R  Also g  (ii)
g 1.4 R2
 1.87  106 m On solving (i) and (ii)
17. (3) The acceleration below the surface is V 2 R2 R

h 
GM GM  d 2GM  V 2 R  2 gR 
g1  r  3  R  d   g 1    2  1
R 3
R  R  V 
Gravitation 203
30. (3) From kepler’s law

Gm1 Gm2
21. (3) Vg  Vg 1  Vg 2   
r1 r2 T2
T 2 R 3   constant
2 3
R3
10 10 
 6.67  1011     1.47  10 7 Joule/kg 31. (4) Angular momentum of the earth around
 0.5 0.5 
the sun is
1 2 1 L  M e v0 r
22. (1) mve  m2 gR  mgR.
2 2
23. (1) From conservation of mechanical energy. GM s  GM s 
 Me r  v0  
r  r 
GMm 3GMm 1 2
 0  mv
R 2R 2 12
 L  M e2 GM s r 
GM ve
v  Where, M e  Mass of the earth
R 2
24. (1) Potential energy of the particle is M s  Mass of the sun
r  Distance between the sun and the earth

 
 GMm 
U  4    L r
 a  32. (2) Time period of a revolution of a planet
 
 2 
2 r 2 r 2 r 3 2
T  
v GM s GM s
 GMm  1 2 8 2GM
4    mv  0  v  r
a 2 2 a
Where M s is the mass of the sun
vp Mp Re 1 Squaring both sides, we get
25. (4)    6  3
ve Me Rp 2
4 2 r 3
T2 
 v p  3ve . GM s
26. (1) Earth rotates about its axis from west to The graph between T 2 and r 3 is a straight line
east. As earth rotates about sun also that is why 4 2
whose slope is GM .
sun rises in the east. s
27. (1)
vB r 4R
28. (4) 33. (1)  A  2
vA rB R
29. (4) When a person is in satellite the net force
on the person is zero in the satellite frame. In  vB  2  vA  2  3v  6v.
satellite frame force exerted by earth and
34. (1) When a satellite is in stable orbit then its
centrifugal force (pseudo force ) gets balanced.
time period is
The force exterted on the person who is in the
GMm r3 2
satellite is F  ( R  h)2 T  2 
GM 
204 Gravitation
GM 1
 
r3 r3
1. (1) Comparable to other planets moon is
close to the earth so moon is responsible for
GM 1
v0   v0  tides in the sea.
r r
2. (4)
GMm 3. (3)
U As r increases U also increases
r
4. (4)
The centripetal acceleration is
5. (3)
2
v0 GM 1 6. (1) Work done by the gravitational force is
ar   2  ar 
r r r independent of path followed. It is conservative
35. (4) As body covers equal angles in equal in nature.
time intervals of time, so magnitude of linear 7. (1)
velocity is constant.
8. (1) The gravitational force obeys Newton’s
36. (3)
third law of motion. Thus, it form action-reaction
GM pair. The gravitational force is independent of
37. (3) v if r1  r2 then v1  v2
r inter-vening medium. In order words, the force
Orbital speed of satellite does not depend upon between two masses remains the same whether
the mass of the satellite. they are in air, vacuum, or separated by a wall.
38. (3) Consider a planet of mass m that displaces The gravitational force is a conservative force.

s in time t as shown in the figure.
The gravitational force is a central force.
9. (2)
10. (3)
11. (2) Time period is related with radius as
    
L  r  P  r  mv
T 2 r 3
  
s m  
L  r m  r  s 2
 T1   r1 
3
t t
   
 T2   r2 
 m
L  2A 2 3
t  T1   r1  3
    4
 r  s  Area of the parallelogram  T2   (r1 / 4) 
A is the area sweep by the planet in time t. T1 T

 43/ 2  8  T2  1
T2 8
A L

t 2m 12. (2) Let m is the mass of each body
39. (1) From conservation of angular momentum Gm2
of the planet of mass m. F
d2
mv1r1  mv2 r2 . [angular momentum is constant] After transferring unit mass
Gravitation 205
20. (4) An astronaut experiences

G(m  1)(m  1) G(m2  1) Gm2 G
F'   2  2
d2 d2 d d weightlessness in a space satellite. It is because
the astronaut experiences no gravity.
G
So the force decreases by 21. (2) At the centre of the earth g = 0
d2
l
GM1 M 2 G  T  2 
13. (3) F  2  100  100 g
R2 R
22. (1) Gravitational attraction force on the
G G
F '  2 125  75  2  9375 particle is
R R
GM p m
F 15 F 2
F'
1000
 9375 F '  F
16
 D / 2
14. (4) The force exerted by the ring on the point Acceleration of particle due to gravity
GMmx Fg 4GM p
mass is F  ( R 2  x 2 )3/ 2 a  .
m Dp2
When the force F is maximum
2
 Mp   Re   1  2
dF d 
 GMm  2
x 
0 23. (2) g p  ge      9.8    2 
dx 2 3/ 2 
dx  ( R  x )   Me   Rp   80 
d d  9.8 / 20  0.49m / s 2 .
x ( R 2  x 2 )3/ 2  ( R 2  x 2 )3/ 2 ( x)
 dx dx 0 24. (3) For the condition of weightlessness at
( R 2  x 2 )3/ 2 equator
3 g  R 2
x ( R 2  x 2 )1/2 (2 x)  ( R 2  x 2 )3/ 2  0
2
3x 2  ( R 2  x 2 )  0 g 1 1
   3
 rad / s.
R 640  10 800
R
 x 25. (2)
2
4 R p  g p   e  1
15. (4) g   GR       1   
3 Re  g e    p   2
16. (3)
Re R
dV  Rp   .
17. (3) Gravitational field is E   2 2
dr
If E = 0 then V is constant. GM
26. (1) g
R2
18. (4) Conceptual
19. (1) The acceleartion due to gravity is Me R
According to problem M p  and R p  e
2 2
GM M1 M 2
g  
R2 R12 R22 2
gp  Mp   Re  1 2
          2   2
Given that M 2  4 M 1  R2  2 R1 ge  M e   Rp  2
206 Gravitation
gm 1  Gm 2  mv 2
 Fr  2  2  cos 30 
27. (2) g e 6 (Given)    r
Where subscripts m and e refer to moon and 2 

Where r  
earth respectively. cos 30 2 cos 30

GM m G 4 Gm
 m  Rm3 v
gm Rm2 Rm2 3  R 
   m m
g e GM e G 4
 e  Re3 e Re
2 2 30. (2) The gravitational potential of a spherical
Re Re 3
shell of radius R and mass M is
Rm GM
 V
Re ( m  e ) R
After compressing to half radius its new
R g 1 potential is
 m  m 
Re g e 6 GM 2GM
V' 
28. (4) Resultant force on particle ‘1’ ( R / 2) R
So gravitational potential decreases.
Gm Gm 3Gm
31. (1) V  
R 2R 2R
32. (3) The work done by the external agent is
Fr  2 F1  F2 W  U f Ui
Gm2 Gm 2 mv 2 Ui 
GMm
0
Fr  2 2
 2

 2r  
 2r  r
3GMm
Uf 
2R
Gm  2 2  1 
v   W  ()ve
r  4 
33. (2) As gravitational force is a conservative
29. (1) The centripetal force required for a planet force, work done is independent of path.
is provided by the resultant gravitational force
W1  W2  W3
of other two planets
34. (2) Gravitational potential at the centre is
 GM  GM
Gm 2 U  4     32
F 2 L
 L 2
Gravitation 207
35. (1) When reference point potential is changed

GM
then potential difference between any two points v0  (ii)
remains constant. R
VOld  VNew From (i) and (ii), we get ve  2v0

41. (4) Escape velocity from the earth
V 
 VP Old  V  VP  New
2GM
ve 
0   5   10  VP New
  VP New
 5unit R
For a given planet,
36. (1) T .Ei  T .E f
2G  6 M 
GMm 1 2 GMm 1 2 vP   3ve  3v
  m 0   mv 2R
 2 R 2
42. (1) As minimum energy is required so the
2GM K .E f is zero.
v  2 gR
R
T .E i  T . E f
 Gm1m2 
Gm1m2 r
37. (1) F ˆ
r 
r2 r2 r U i  K . Ei  U f  K . E f
Gm1m2 
 r GMm 1 2 GMm
r3  mv 
r1 2 2r2
38. (2) The escape velocity of a body is
r1  R, r2  3R
2GM e
 ve 
Re 1 2
where mv - is the energy given to the satellite.
2
The above formula shows that escape velocity
is independent of the mass of the projectile. 1 GMm GMm 5GMm
 mv 2   
2 r1 2r2 6R
39. (2) When a body is projected with escape
velocity then T.E is zero at infinity.
43. (2) The P.E. of the mass at d/2 distance from
GMm 1 2 the earth and moon is
 mve  0
R 2
If the initial velocity is less than ve then the body

is in bound with earth and hence T.E = (-) ve
40. (1) Escape velocity,
GM 1 m GM 2 m
U  2 2
2GM d d
ve  (i)
R
2Gm
Where M and R be the mass and radius of the U  ( M 1  M 2 ) (Numerically)
d
earth respectively.
The orbital velocity of a satellite close to the 1 2 G
mve  U  ve  2 (M1  M 2 )
earth’s surface is 2 d
208 Gravitation
44. (4) 58. (4)

45. (4) 59. (3)
46. (2) 60. (3) From Kepler’s 2 nd law
47. (3)
A
48. (4) As the satellite come close towards the  constant
t
earth then its P.E decreases and K.E increases.
A1 t1
49. (4) 
A2 t2
50. (2)
A 7 days
51. (3) As, the escape velocity on the surface of 
A ' 21 days
moon is small i.e., 2.4 km/s in comparision to
11.2 km/s that of earth and also this velocity is A'  3A
smaller than rms velocity of gas molecules, 61. (1)
therefore gas molecules escape moon,s
gravitation. Thus, moon cannot have 62. (4) The areal velocity is
atmosphere. dA L dA
  r 2
52. (4) The gravitational force of attraction dt 2m dt
between earth and satellite provides centripetal 63. (2) Given that
force for the circular motion of earth. Hence,
no fuel is required for the motion of satellite. k mv 2
F 
53. (1) The time period of an artificial satellite is r3 r
given by k c 2r
v2  v 
mr 2 r T
( R  h)3
T  2
gR2  r 2 T
where, R is radius of earth and h the height of So Kepler’s law of period does not hold.
satellite. From the above formula, it is clear that Kepler’s law of area is derived from
time period of revolution does not depend upon conservation of angular momentum. Angular
the mass of the satellite. moment is conserved for central force .So
54. (2) Orbital velocity is independent of mass kepler’s law of area still holds.
of the object.
k Mv 2
55. (1) F  . Hence v R 0 1. (3) Let particle A lies at origin, particle B and
R R C on y and x-axis respectively
56. (1) The kinetic energy of the satellite is
1 2 1  GM  GMm
K mv0  m  
2 2  r  2r
1 1
 K   2/3  T 2/3
r T
57. (2) Angular momentum
L  mvr  2mE .r  2mEr 2 .
 GmA mC ˆ
F AC  2
i
rAC
Gravitation 209
3. (1)
6.67 1011 11 ˆ
 2
i  1.67 109 Niˆ
 0.2

Similarly F AB  1.67  109 Njˆ
 Net force on particle A

  

F  F AC  F AB  1.67  109 N iˆ  ˆj 
2. (1) During normal weighing method we
Centripetal force = net gravitational force
maintain equal lengths for both pans. In this case
the acceleration due to gravity is same for both mv02 2Gm2 1 Gm2
bodies. In this case pan heights are different   2 F cos 45  F1   2
r 2r 2 2 4r
and hence g is different at both locations. As
with height g varies as
mv02 Gm2
  2 2  1
g'
g  2h 
 g 1   r 4r 2  
2
1  h / R   R
1

 Gm 2 2  1
 v0  
  2
 4r 
 
The angular velocity of a particle is given by
1
v  Gm(2 2  1)  2
 0   
r  4r 3 
and in according with figure h1  h2 ,W1 will be
4. (3) The gravitational field intensity at a point
lesser than W2 and
inside the spherical shell is zero.
5. (1) When reference point potential is
 h1 h2  changed then potential difference between any
i.e., W2  W1  mg2  mg1  2mg   
R R two points remains constant.
GM h VOld  VNew
or W2  W1  2m
R2 R
V
 VP Old  V  VP  New
 GM 
 As g  R 2 and  h1  h2   h  0   5   10  VP  V  5unit
  New P New
2mhG  4 2  6. (4) The sphere with cavity can be assumed

or W2  W1   R   as a combination of two spheres.
R3  3 
8
 Gmh
3
 4 3  Potential at internal point of solid sphere at a

 as M  3 R   distance ‘r’
 
210 Gravitation
GM  3 r 2  1 4
V   h g   RG  g  R
R  2 2R2  9. (1) g here 3
R h1 g 2 R2  2 1
At r   
2 h2 g1 R1 1 6
GM  3 R2  11 GM
V1    2  h2  3.0m
R  2 8R  8 R
Potential by the removed sphere is 2h h
t t 
g g
 M 
G
3  8  3 GM t1 hg h 1
V2  .   1 2  1 
2 R 8 R (Sphere mass is (-)ve) t2 g1h2 h2 6
2
11 GM 3 GM 10. (3) Here the particle exicutes SHM. Time
Net potential, V  V1  V2    period of oscillation is independent of the distance
8 R 8 R
of the point from the centre of earth. In both
GM cases

R
7. (2) Acceleration due to gravity at a height h Re
T  2
g
GM
above the earth’s surface is g = ( R  h)2
11. (2) Acceleration due to gravity at latitude 
where, G = Universal gravitational constant is given by g '  g  R2 cos 2 
R= Radius of the earth 3
 2 2  2
At 30 , g30  g  R cos 30  g  R
GM g 4
g' 2
 2
 h
2  h
R 1   1   3
 R  R  g  g30  2 R
4
g
According to given problem, g ' 
100 12. (4)
g g M
  h  9R. Mass of small element of length, dM  dx
100  h 2 l
1  
 R
8. (3) Let g & g ' are the accelerations before The gravitational potential energy between this
expansion & after expansion. element and point mass is
M 
GM GM ' Gm  dx 
g0  2 & g 2 GmdM  l 
R R dU  
x x
M 4 3 
M '   R  GMm a dx
4
  2 R 
3  3  U   
3 l a x
M g GMm  a   
M ' g 0 U  1n  
8 8 l  a 
Gravitation 211
16. (4) First we find a point where the resultant

field due to both is zero. Let the point P be at a
distance x from centre of bigger star.
13. (1)
VA  (Potential at A due to m1 ) + (Potential at A G(16M ) GM

 2

x (10a  x)2
due to m2 )
 x  8a (from O1 )
Gm Gm2
 VA   1  i.e., once the body reaches P, the gravitational
R 2R
pull of attraction due to M takes the lead to make
Similarly, m move towards it automatically. A minimum
KE has to be imparted to m from surface of
VB  (Potential at B due to m1 ) + (Potential at
16M such that it is just able to reach point P. By
B due to m2 ) law of conservation of energy.
Gm2 Gm1 (Total mechanical energy at A)

 VB   
R 2R = (Total mechanical energy at P)
1 2  G (16M )m G ( M )m 
Since, WAB  m VB  VA   mvmin   
2  2a 8a 

Gm  m1  m2   
2 1  GMm G (16 M )m 
0  
 2a 8a 
2R
1 2 GMm
14. (2) The potential energy of the system  mvmin  (45)
2 8a
GMM GM 2 3 5GM
U    vmin 
R R 2 a
15. (1) Assume that the total mass M is present 
at infinity. We bring the total mass and form 17. (3) 
Gravitational field, g  5iˆ  12 ˆj N / kg 
like a sphere. Let m is the mass of the shell
having radius R. To bring further mass dm from dV
g
infinity the work done by the external agent is dr
x y
dWext  dU  U f  U i  U R  U  V     g x dx   g x dy     g x .x  g x . y 
 0 0 
 Gm Gmdm 
dWext   dm     5  7  0   12  3  0  
 R  
  35   36    1J / kg
G M GM 2
Wext  mdm 
R 0 2R i.e., change in gravitational potential 1J/kg.
GM 2 Hence change in gravitational potential energy
Wext  U Self 
2R 1 J.
212 Gravitation
18. (1) Escape velocity for that body

22. (3) T
2r
v 
 
2r (2) 10 (10 )
11
2GM
ve  v T 1.6  107
r
5
ve should be more than or equal to speed of  10  104 m / s
4
2GM 23. (2) The orbital velocity of a planet is
light i.e., c
r
GM r
19. (3) The total energy of a satellite present in V 
an orbit of radius r is r t
GMm GM
T. E  r  t
2r r
The difference in the energies of two orbits is
r1 1 r
equal to the additional K.E to be supplied  2
r2  2 r1
GMm  1 1 
K.E     3/2
2  R1 R2   r2 
1   2  
20. (2) Here both satellites rotate in opposite  r1 
direction. W.r.to an observer standing on earth
the relative velocities for west and east motions 2 3r
Given that r  9
are v0  R and v0  R. The time periods are 1
2R 2R
 2  90  1  270
Twest  and Teast 
v0  R v0  R 24. (1) The energy required to remove the satllite
from its orbit around the earth to infinity is called
 T  Teast  Twest binding energy of a satellite. It is equal to
negative of total mechanical energy of satellite
 2R  4R 2 in its orbit.
 Teast  Twest  2R  2 2 2 
 2 2 2
 v0  R   v0  R 
GMm
21. (2) Conserving angular momentum of the Thus, binding energy   E 
2r
particle w.r.to the centre of the earth.
GM
but, g   GM  gR 2
v2 R2
m(v1 cos60)4 R  mv2 R   2.
v1
gmR 2
 BE 
Conserving energy of the system: 2r
25. (4) As in case of elliptical orbit of a satellite,
GMm 1 2 GMm 1 2
  mv1    mv2  GMm 
4R 2 R 2 mechanical energy E     remains
 2a 
1 1 3 GM 1 GM constant, at any position of satellite in the orbit.
 v22  v12   v12 
2 2 4 R 2 R
GMm
1 6 8000 i.e., KE  PE   (1)
 v1  64 10  m/s 2a
2 2
At a distance r
Gravitation 213
GMm GMm
PE   (2) PE   (ii)
r r
So, from eqs. (i) and (ii), we have
So, from eqs. (1) and (2), we have
1 2 GMm GMm
1 2 GMm GMm mv   ,
mv   , 2 r 2a
2 r 2a
2 1
i.e., v 2  GM   
2 2 1  r a
i.e., v  GM   
r a
29. (2) During path DAB planet is nearer to sun
26. (3) The gravitational force between the planet as comparing with path BCD. From
and the star is conservation of angular momentum
vr=constant. So time taken in travelling DAB
GMm is less than that for BCD because velocity of
F 
R5/ 2 planet will be more in region DAB.
Where M and m be mass of star and planet 30. (3) From conservation of angular momentum
respectively.
mv1 r1  mv2 r2
For motion of a planet in a circular orbit;
GMm
2
mR  5/ 2
R GmM S mv12 GmM S mv22
  
r1 2 r2 2
2
 2  GMm  2 
mR    5/ 2    2
T  R  T  GM S GM S v22 v22  r2 
     
r1 r2 2 2  r1 
42 GMm 2 42 7/2
  T  R
T2 R7/2 GM
2GM S r1
 v2 
T  R 7/4 r2  r1  r2 
27. (1) Applying conservation of angular

2GM S r1r2
momentum at positions A and B and L  mv2 r2  m
r1  r2
mvA  OA  mvB  OB
vB OA
Hence,  x
v A OB GM
1. (1) F1  F2 
a2
28. (4) As in case of elliptic orbit of a satellite,
 GMm  GM
mechanical energy E     remains Resultant of F1 and F2 is 2
 2a  a2
constant, at any position of satellite in the orbit.
GMm
i.e., KE  PE   (i)
2a
At a distance r
214 Gravitation
GM GM  2h 
F3  2
 5. (3) g1  g 1  
Now, 2a 2
 2a   R
 h
2 GM GM g 2  g 1  
Now, and act in the same  R
a 2
2a2
direction. g1  g 2h
For height  100%   100%  1%;
g R
GM  1
Their resultant is 2 
a 2  2 g2  g h 1
For depth  100%   100  %  0.5%
2. (3) g R 2
GMm  42  1 6. (4) The acceleration due to gravity on earth

m2 R   m  2  R  n  T 2  Rn1
R n
T  R is
 n 1 
GM E
  g (i)
T  R  2 
RE2
3. (1) Centripetal force is provided by the
gravitational force of attraction between two ME 4
  M E   RE3
particles As 4 3 3
RE
3
Substituting this values in Eq. (i), we get
 4 
G   RE3 
g  3
3   4 GR or   3g
E
RE 3 4GRE
mv2 Gm  m
i.e.,  2
R  2R   x  gx
7. (1) g depth  g surface   
R R
1 Gm
v where, x = distance from centre of earth.
2 R
 Total force on wire
4. (3) Differentiating the equation of the curve
w.r.t. t we get, 2 R
R  x  g  x 
F    dx g     
dx dy 4 R /5
 R  R  2  4 R /5
2x  500
dt dt
9Rg
2
F  10.8KN
d x  dx  dx  dy 50
2x 2
 2     500
dt  dt  dt  dt
GM
2 8. (4) g
 dx  d y 2 2
d x  R2
2    500  2  dt 2  0 
 dt  dt   As M is constant
d 2 y 2  104 dg dR
   4 g. The effective g '  5 g .  2
dt 2 500 g R
Gravitation 215
13. (3) The four particles are shown in the figure.

dg dR
 100  2  100
g R
dg
 100  2%
g
g will increase if R decreases.

9. (4) The acceleration due to gravity inside the From figure
GM
earth is g '  E  r AB  BC  CD  AD  l
R3
 AC  BD  l 2  l 2  l 2
g GM g R
 3 r  r r 
3 R R 3 Total potential energy of the system of four
particles each of mass m placed at the vertices
10. (1) The escape velocity is A,B,C and D of a square is
v  2 gR
 G  m m   G mm 
U     
 AB   AC 
vA g A RA
    k1  k2  k1k2
vB g B RB  G  m m   G  m m 
     
 AD   BC 
T .E i  T . E f  G  m m   G  m m 
     
 BD   CD 
GMm GMm 1 2
 0   mv
2R R 2 4Gm2 2Gm2 2Gm2  1 
   2 
l l 2 l  2
GM
v
R 14. (4) W1  U Surface  U centre
12. (4) Let gravitational field is zero at a distance W2  U   U Surface

x from the mass M.
GM  3 GM 
GM G  4 M     1
 W1 R  2 R 
x2 r  x
2 
W2  GM  2
0 
2
 R 
 4x 2   r  x 
15. (2) The height (h) of the particle is very small
r so acceleration of the particle can be treated as
2x  r  x  x 
3 constant.
The potential at that point is
GM G  4M 
Vp   
x rx
3GM 6GM 9GM

  
r r r
216 Gravitation
The acceleration of the small particle is 2  tan 1  4   7556' (Anti clockwise)

GM GM GM
a 2
 2
 so 1  2  90
(2 R) (2 R) 2R 2
The time taken to reach A is i.e., the line y  4 x  6 is perpendicular to I.

18. (1) Potential energy of the body at a distance
2 h 2  h  2R2 hR 2
 t  2 4 Re from the surface of earth
a GM GM
16. (1) Apply the principle of conservation of GMm mgRe

U 
linear momentum and conservation of energy. 5Re 5
Let velocities of these bodies at r distance from Minimum energy required to escape the body
each other be v1 and v2 respectively.. will be
By conservation of momentum mgRe

5
m1v1  m2 v2  0
19. (2) A geostationary satellite goes around the
 m1v1  m2 v2 (i) earth in west-east direction. The time period a
From conservation of energy geostationary satellite is 24 hours. The angle
between the equatorial plane and the orbital
KEi  U i  KE f  U f plane of geostationary satellite is 0.
20. (1) Gravitational force provides the required
Gm1m2
0  KE centripetal force
r
m2 R  mg
Gm1m2 1 1
 m1v12  m2 v22 (ii)
r 2 2 GMm 42 GM
m2 R    4  T  R2 .
From (i) & (ii) R3 T2 R
21. (4) Potential energy = 2 (total energy)  2E0

2Gm22 2Gm12
v1  v
, 2 
r (m1  m2 ) r (m1  m2 ) 22. (4) Air resistance F  v 2 , where v is the
GM
2G orbital velocity v 
 vrel  v1  v2  (m1  m2 ) r
r
r- distance of satellite from planet’s centre.
17. (1) Work done by the gravitational field is
zero, when displacement is perpendicular to GM
gravitational field. Here, gravitational field, F 
r
I  4iˆ  ˆj. If 1 is the angle which field makes
The work done by the resistance force
with positive x-axis, then
3/ 2
GM  GM  GM   dt
1 1  1   dW  Fdx  Fvdt  dt  (i)
tan 1   1  tan    14 6 ' (Clockwise). r r r 3/ 2
4 4
The loss of energy of the satellite = dE
If 2 is the angle which the line
dU d  GMm  GMm
y  4 x  6 makes with positive x-axis, then   
dr dr  2r  2r 2
Gravitation 217
GMm T1 T
 dE  dr (ii)  T2   Here T2 is complete time
2r 2 2 2 2 2
From work energy theorem the loss in T.E of period for elliptical path so time of fall is
satellite is equal to work done by the resistance
force. T2 T

2 4 2
dE  dW
24. (2) Linear velocity of earth,
GMm (GM )3/ 2
dr   dt
2r 2
r 3/ 2 2 Re 6.28  6.4  106
ve    463 m s
Te 24  3600
t
m
R
dr

m R  
n 1
2 GM nR
 r  GM
Orbital velocity, v0  Re g  7.9  103 m s
m
t  
n 1
 gR According to question,
23. (4) When the satellite is revolving around vKP  v0  ve  7900  463  7437 m s
earth
vKQ  v0  ve  7900  463  8363 m s
2 2
EKQ vKQ  8363 
  2
 
EKP vKP  7437 
25. (2)
T12  r 3 (i)
Where T1 is the time period of the satellite

When the satellite is stopped it follows the
straight line path but we can approximately take
the path as a narrow ellipse whose focii present
at the ends.
From Kepler’s Law, T 2 r 3

2 3 2 3
T  r   1   10 
4
 1    1       
 T2   r2   8   r2 
Where a is the length of semni major axis of the

 r2  4  10 4 km
ellipse.
ar /2 2 r
v  r 
From eq’s (i) & (ii) T
3
T12  r  2 T12 r r 
    8  T2   v2  v1  2  1  2     104 km / hr
T22  a  8
 T1 T2 
218 Gravitation
26. (1) Here, 31. (1) The earth moves around the sun in
8
TS  29.5 TE , RE  1.5  10 km, Rs  ? elliptical path as shown in the figure, So by using
the properties of ellipse
According to Kepler’s third law, T 2  R 3
TS2 RS3
 
TE2 RE3
2/3
T 
RS  RE  S 
 TE 
r1  1  e  a and r2  1  e  a
2/3
 29.5TE 
RS  1.5  108    1.4 109 km where, a = length of semi-major axis
 TE 
b = length of semi-minor axis
27. (2) Centripetal force
e = eccentricity
1 Given that
F  m 2 R
Rn
r1  r2
a
2
1   1
 2  n 1
R n 1 R2 and r1r2  1  e2  a 2
n 1 Now, required distance = semi latusrectum

Time period T  2  R 2
 b2
F1 P 
a
28. (3) At height h escape velocity a 2 1  e2  r1r2 2r r

   12
2GM a  r1  r2  / 2 r1  r2
ve 
hR
32. (3) According to Kepler’s first law, every
GM planet moves in an elliptical orbit with the sun
Orbital velocity v0  situated at one of the foci of the ellipse.
Rh
In options (1) and (2) sun is not at a focus while
 Increase in orbital velocity required to escape
in (4) The planet is not in orbit around the sun.
gravitational field Only option (3) Represents the possible orbit for
planet.
 ve  v0  gR  2 1
29. (1) Let K is the kinetic energy to be given to
the satellite when it is present on the surface.
1. (0.267) Let m1 is mass of core and m2 is of
outer portion
GMm GMm 5GMm
K  K 
R 2(3R) 6R 4 4
m1   R 3  , m2   [(2 R)3  R 3 ]2 
30. (3) The areal velocity is 3 3
dA L dA Gm1 4
   vr  r 2 g1    RG,
dt 2m dt R2 3
Gravitation 219
3. (9.8) Since gravitational acceleration on earth

G  m1  m2  G 4 3
g2  2
  R 15  is defined as
 2R  4R2 3
GM e
ge  (1)
g 4 Re2
 1 
g 2 15 Me Re
MP  & RP 
2. (11.2) Escape velocity is independent of mass 80 4
of the object. It depends only on the mass of the GM P
planet as. So gP  (2)
RP2
2GM
v From (1) &(2), we get
R
ge
GM 1 2GM   2 m sec 2
3. (6400) v  5
Rh 2 R
4. (160) According to question,
 2 R   R  h   h  R  6400 km.
G MP GM
g' 2 on the planet and g  2 e on
RP Re
the earth
1. (12800) If g be the acceleration due to gravity
at the surface of the earth, then its value at a  RP  Re and M P  4 M e
height h above the earth’s surface will be -
g' 2
g Now, g  4  g '  4 g  40m / sec
g'
 h
2
g' 1
Here 
1   g 9 Energy needed to lift 2 kg mass through 2m
 Re 
distance  mg ' h  2  40  2  160 J
1 1
  2
 h  2 Re  12800km
9  h 
1   1. (2) Moon is revolving around earth in almost
 Re  circular orbit. Sun exerts gravitational pull on
both, earth and moon. When observed from sun,
2. (9.2) Here, G  6.67  1011 Nm2 / kg 2
the orbit of the moon will not be strictly elliptical
Mass of the pulsar, M  1.98  1030 kg because the total gravitational force (i.e., force
due to earth on moon and fource due to sun on
Radius of the pulsar, R  12 103 m moon) is not central.
Acceleration due to gravity on the surface of 2. (4) Weight of the body at earth’s surface
GM W  mg  250 N
the pulsar is g 
R2 Acceleration due to gravity at depth h from
Substituting the given numerical values, we get earth’s surface
g
 6.67 10 11
Nm2 kg 2 1.98 1030 kg   h
g '  g 1  
3  R
12 10 m3
R
 0.092 1013 m / s 2  9.2 1011 m / s 2 . Here, h 
2
220 Gravitation
 R  (12  103  4  9.87  1.44)

 
g '  g 1  2  g
 6.82 105 m / s 2
R g'
 2
  As g  r 2 , therfore the object will remain stuck
to the star.
Weight of the body at depth h
6. (1, 2, 4) If the law gravitation becomes an
mg inverse cube law, then
W '  mg   125 N
2
GMm mv 2 GM
3. (4) If the earth is an approximate sphere of F 3
 v
non-uniform density, then the centre of gravity r r r
of earth will not be situated at the centre of earth.  Time period of revolution of a planet,
The distance of different points on earth will be
at different distances from the centre of gravity 2r 2r 2
T  or T 2  r 4 . It means a planet
1 v GM
of earth. As, g  , so g is different for
will not have an elliptical orbit. The circular orbit
r2
different points on the surface of earth but not of a planet may not be possible as the
zero. gravitational attractive force obeys inverse
square law and not inverse cube law. There will
4. (3) The viscous force acting on satellite be some gravitational field inside a spherical
decreases the energy of satellite. As a result of shell of uniform density. A stone thrown by hand
it, the value of r gradually decreases, on the surface of the earth will follow nearly
consequently the height of satellite decreases. parabolic path, under the gravitational force.
5. (2) If the gravitational force acting on an
7. (2, 3, 4) When G '  10G , then
object is equal or greater than the centripetal
force required for the orbital motion, then will (10G ) M
g'  10 g .
remain stuck to the surface of the star due to R2
gravity
 Weight persons  mg '  m  10 g  10mg i.e.,
mv 2 v2 gravity pull on person will increse. Due to it,
mg  or g 
r r walking on ground would be very difficult.
Critical velocity of rain drop,
Given, mass of the star M  2.5  2  1030 kg
2r 2 (  ) g
 5.0  1030 kg vc 
9
Radius R  12km  12 103 m
 vc  g . Since g increases, hence vc
GM increases.
Acceleration due to gravity g 
R2 To overcome the increased gravitational pull of
earth, the areoplanes will have to travel much
6.67 1011  5.0 1030 faster.

3 2
 2.3 1012 m / s 2
12 10 
Now, centripetal acceleration
1. (1) The earth is revolving in circular orbit
 r 2  r (2n)2 (   2n) around sun due to gravitational force (F) which
act along the radius of circular path, towards
 12  103  (2  3.14 1.2)2 the sun. As torque
Gravitation 221
2
   r  F  rF sin 0  0  2  GM
    3s
 2r 
v  
T  r  T 
Therefore, torque is zero
2. (3) Resultant force on mass m due to masses 42 r 3
 Ms 
GT 2
G2M  m GMm
at A and C is F  
 AB2   BC 2  . Therfore, 4  (3.14) 2  (1.5  1011 )3
 30
m will move towards 2M. 6.67 1011  (365  24  60  60)2  2  10 kg
3. (4) Given, height of the satellite above the 6. (3, 4) When G decreases with time, the
earth’s surface GMm
gravitational force F  will become
6 r2
(h)  400km  0.4  10 m weaker with time. Then radius of the orbit (r)
increses with time. Due to it, earth will be going
Mass of the satellite  m   200kg
around the sun not strictly in closed orbit and
after long time it will leave the solar system.
Radius of earth  Re   6.4 106 m
7. (2, 3, 4) Due to large amounts of opposite
Mass of earth  M e   6.0 10 kg 24
charges on sun and earth there will be a large
force of electrostatic attraction as well as
Gravitational constant gravitational attraction. Both the forces obey
 G   6.67 1011 N.m2 / kg 2 inverse square law and are central forces. Due
to it, the distance between sun and earth will
Energy required to send a satellite out of earth’s decreses. In this situation, all the three Kepler’s
gravitational influence is called its binding energy. laws will be valid.
GM m
e
Binding energy  2  R  h 
e
1. (1) Let M is the mass of the sphere and R is

6.67 1011  6.0 1024  200 its radius

2  6.4 106  0.4 106 
GM
For r > R, E 
6.67  6.0 1015  2 r2
  5.9  109 J
2  6.8 106
GM
4. (4) Asteroids move in circular orbits like For r = R E 
R2
planets and they obey Keplar’s laws
5. (4) Mean orbital radius of the earth around For r < R, E = 0.
the sun
2. (4) Kepler’s law T 2  R 3
8 11
r  1.5 10 km  1.5 10 m
3. (4) When the particle is present at r  R
Time period of earth around the sun =1 yr
mv 2 GM G 4 
 365 days  365  24  60  60s  3 r  3    R3  r
r R R  3 
The orbital velocity is
v r
2GM s
v  When the particle is present at r  R
r
222 Gravitation
mv 2 GM
 2
r r
mv02 GMm GM
1 1. (4)  5/ 2  v0  3/ 4
v r r r
r
3
4. (1) Let m is the mass of the body and M is  ln v0  ln GM  ln r
4
the mass of the earth. The orbital velocity is
GM The slope of the graph between ln v0 and ln r
v 
r
3
is
1 GMm 4
K .E  mv2 
2 2r 2. (4) For hollow sphere Potential remains
constant inside the sphere and it is equal to
GMm
P.E  potential at the surface and increase when the
r point moves away from the surface of sphere.
GMm GM GM GM
T .E  Vin  ,Vsurface 
2r and Vout 
R R r
5. (1) F  0 when 0  r  R1 3. (4) U  mV
because intensity is zero inside the cavity GMm
U  ( r  R)
F increase when R1  r  R2 R
1 GMm
F when r  R2 . U  ( r  R)
r2 r
6. (1) The force on a particle at a distance x GMr

4. (2) g= inside the Earth (straight line)
From the centre is R3
where r is the distance from the centre.
GM
g outside the Earth
R2
Where M is Mass of Earth
5. (3) Intially the weight of the passenger
 60  10  600N
Finally the weight of the passenger
GMm
F x (M - Mass of the earth)  60  4  240N
R3
and during the flight in between some where its
The component of force perpendicular to the
weight will be zero because at that point
GMm gravitational pull of earth and mars will be equal
tunnel is F sin   .
2 R2 in the magnitude and opposite in direction.
Gravitation 223
6. (4) GM
u2   v12
7. (2) The time period of a revolving planet is R
 4M 2  3 Again after ejecting rocket
T2   R
 GM   
Pi  P f
 GM  2
R3   T
 4 2 
 GM 
3log R  log    2logT
 4 2 
1  GM  2
log R  log  2   logT
3  4  3
Where v0 is the orbital velocity
2
Scope 
3 GM GM
v0  
r 2R
1 GM
y  intercept  log 2  6
3 4 9m GM m 
 v2 ( p conservation
GM 10 2 R 10
 1018
4 2 along tangent)
M  4 2  1018
GM
G v2  9
2R
M  6  1029 kg Conservation of momentum along radial
direction
m
1. (1) mv  v1  10v
10
1m 2
K.E of rocket =
2 10
 v1  v22 
1 m  2 GM  GM 
  100  u    81 
2 10   R  2R 
m 2 100GM 81GM 
1 2 GMm GMm 1 2  100u   
mu    mv1 20  R 2R 
2 R 2R 2
2GM GM  119 GM 
u2    v12 K .E  5m  u 2  
R R  200 R 
224 Gravitation
GM Gm
. g  2GK
2. (1) The velocity of the satellite is r2
r
The acceleartion due to gravity at a distance r
1 (r>R) from the centre is
m v2
TA 2 A A 1 2R
    1 GM
TB 1 m v 2 2 R g Where M is the total mass distributed
2
B B r2
until r = R.
3. (2) The angular momentum of the planet The correct graph is (1)

about the sun is L  r  mr
GMm
s m 6. (3) Gravitational force F  R  h 2
L  rm   rs   
t t
1
m A L 7. (3) The gravitational field varies as from
L   2A    r
t t 2m the centre of the cylinder. The force on the star
Where A is the area sweep by the planet in is
time t. k
F   m
r
GMm
4. (3) E1   k is some constant
2R
mv 2 km
GM (m / 2) GM (m / 2) 2GMm so   v  constant
Ef     r r
R
   3 R  3R
2  2 
 2  2  2 r
T  T r
v
4GMm 2 Mm
  8. (4) When an object turns around an internal
6R 3R
axis (like the earth turns around its axis) it is
GMm  2 1  GMm called a rotation. When an object circles an
E f  Ei      
R  3 2 6R external axis (like the earth circles the sun) it is
5. (1) Given that the density of the body is called a revolution.
K Rotation period of earth  1 day  24  3600s


r
Revolution period of earth  365 days
The mass of a sphere of radius r(r < R) is
 365  24  3600s
K
dm  dV  4r 2 dr
r
 dm  4K  r dr 1. (2) From conservation of angular momentum
r L  m v max rmin  m v min  rmax

2Kr 2
m  4K  rdr 
2
0
8 104  6 1010  vmin 1.6 1012
The acceleration due to gravity at a distance r
(r<R) from the centre is  v min  3 103 m / s
Gravitation 225
2. (3) E  U f  Ui 5. (3) The gravitational potential is
GM
 3 GMe2  3 GM 2e V (3R 2  r 2 )
E  0  2R 3
 5 Re  5 Re
Graph (3) closely depicts the correct variation
So x = 3 of V(r)
3. (1) The apparent acceleration due to gravity 6. (3) When the mangalyan travels from earth
is to mars we assume that it travels in elliptical
path and it covers exactly half of the new ellipse.
g   g  2 R cos 2  Let r is the length of semi major axis of the
ellipse
3g g
 g  2 R , 2 R 
4 4
g 1
 
4R 2  8 100 am  ae 1.5  2.28
r   1.89
2 2
1 1
  102  0.6 103
1600 16 We know that the time period T is T 2  r 3
4. (1) 3/ 2
2 3
 TE   rE  T  T  r 

T  r  , E  
     rE 
Time taken to travel from E to M is

3/ 2 3/ 2
T T r 365  1.89 
For the given elliptical path. t EM   E    
2 2  rE  2  1.5 
3A
Area of sabc = t EM  257.3 days.
4
7. (4) From M.E conservation
A
Area of sadc = (Given in the problem) Gm1m2
4 O  KE
d
Area of sabc = Area of abc + Area of csa
Gm1m2 1 1
 m1v12  m2 v22 (i)
A A 3A d 2 2
= + =
2 4 4 Since momentum is conserved
From Kepler’s 2 nd law m1v1  m2 v2 (ii)
ΔA From eq’s (i) & (ii) we get

=constant
Δt
2G
v1  m2
3A A d (m1  m2 )
  t1  3t 2
4t1 4t 2
2G
v2  m1
d (m1  m2 )
226 Gravitation
8. (4) Let P is the centre of the smaller sphere V = Ed

as shown in the figure.
 V(12 m ,0)  E x  12 J / kg
and V(0,5 m )  E y  5 J / kg
(Given : potential at the origin is zero)
V(12m,0) E x  12 5  12
   1
V(0,2m) E y  5 12  5
10. (3) Let mass of smaller sphere (which has
The field at P by the big sphere having cavity is to be removed ) is m
E P . We can assume the big sphere having cavity R
as a combination of solid sphere and (-)ve mass Radius  (from figure)
2
sphere of radius R/2
  
As densities are same
E P  ESolid sphere  E negative mass
M m

4 3 4  R 3
 GM ˆ G(M / 8) ˆ R  
EP  i 2
i 3 3 2
(3R)2  5R 
 
 2  M
 m
8
 GM ˆ GM ˆ 41GM ˆ
EP  i i i Mass of the left over part of the sphere
9R 2 50R 2 450R 2
So force on the smaller sphere present at P is M 7
M'M   M
8 8
  M 41GM 2
F  EP  iˆ A finite mass behaves like a point mass at large
8 3600R2
distances. The gravitational field due to the left
9. (2) From question, over part of the sphere
E x  5N / kg and E y  12N / kg
7 GM

8 x2
227
Mechanical Properties of Solids (Elasticity) 227
K is the spring constant
T1
l1  l  (i)
K
1. (4) Young’s modulus of wire depends only T2

l2  l  (ii)
on the nature of the material of the wire. K
By eliminating K from eq’s (i) & (ii)
2. (3) YAl
F T l Tl
l 2 1
we get l  T  T
1 2
The graph between F & l is a straight line. 2 1
3. (4) 11. (1)
4. (3) 12. (3)

13. (3) Energy stored per unit volume
Fl
5. (4) L  1
AY   Stress  Strain
2
2
l1 A2  d 2 
     16 1 2 1
l2 A1  d1    Young's modulus   Strain    Y  x2 .
2 2
l1 1
l2   mm 14. (3) Stress  Y  Strain
16 16
6. (2) As the two different rods are connected  2 1011  0.15 Nm2  3 1010 Nm2
in series and they have same cross-section so 2
stress must be same. Where as strain is different u1  (strain)1  x 2 4 L2

15. (3)     1:1.
u2  (strain)2  L2 4 x 2
7. (4)
16. (1) Energy stored per unit volume is
F
8. (4) l  2 2
r 1 1  stress  S2
u   stress  strain  
2 2
2 2 Y 2Y
l1 F2  r1  1
    (4)     1 l2  l1  1mm. 17. (2) The energy stored per unit volume is
l2 F1  r2  2
1
9. (1) The stress in both strings shall be same u stress  strain
2
so the extensions in them also same.
1 stress
 stress 
10. (4) l1  l  l 2 Y
F ( stress ) 2 P 2
u 
where l  2Y 2Y
K
228 Mechanical Properties of Solids (Elasticity)
F V l
22. (2)  (1  2)
106  10 V l
18. (4)  A   2  108
x 0.05
L 23. (2) Y  3K (1  2), Y  2(1  )
19. (1) The vcross-section AB is shown in the
figure For Y  0, we get 1  2  0, also 1    0
1
  lies between to  1.
2
24. (3)
F cos x
The shearing stress on AB  1. (2) Elongation of a wire of density  and
A' length L is
A ' is the area of section AB
L2  g
From the figure it can be observed that the L 
2Y
projected area is A '  AB  on the vertical plane is
2. (4) Due to tension, intermolecular distance
A 'cos  90  x  between atoms is increased and therefore
potential energy of the wire is increased and with
A 'cos  90  x   A the removal of force interatomic distance is
reduced. This change in potential energy
A produces heat in the wire that will increase its
A' 
sin x temperature.
F sin x cos x 3. (3)
Shearing stress 
A 4. (1) Elastic nature for a plastic body is zero
P
20. (1) Bulk modulus, B   (V / V ) so Y  0
5. (2)
1 V
Compressibility, K    6. (1)
B PV
7. (2)
Decrease in volume, V  PVK
8. (3) Metals have larger values of Young’s
7 10
 4 10 1 6 10 modulus than alloys and elastomers.
 24  103 litre 9. (3)

10. (1)
=24  103 103 cm3  24cc
11. (1)
21. (2) At constant temperature
PV = constant 12. (2) When a spring undergoes stretching the
length and shape both changes and the weight
PdV  VdP  0
of the load behaves as a deforming force. The
PdV  VdP change in length corresponds to logitudinal strain
dP and change in shape corresponds to shearing
PB
 dV / V  strain.
229
2
Fl 1 W1 r12 L2 L2  r1 
13. (1) l  l  (F,L and Y are     
AY A W2 L1 r22 L1  r2 
constant) 2
W1 1  1  1
A2 l1  0.1     
  A2  A1  2
  2 A1  2  4  8mm . W2 2  2  8
A1 l2  0.05 
W2  8W1  16 J
F l F
14. (2) Given that l  l , Y   Y
24. (4) V= Al
A l A
15. (3) Stress
YAL 2 0.2 Young's modulus Y 
 0.9  1011     0.3  103    51N Strain
L 100
16. (2) Given that 1
Work done, W   Stress  Strain  volume
2
 L  104 m , F  20 N , A  106 m 2 , L  1 m
1 2
FL 20 1 W=  Y   Strain   Al
Y   2
A L 106  104
2
1  y 1  YA 
10 11 2   Y     Al    y 2  W  y 2 .
 20 10  2 10 N / m 2 l 2 l 
Fl YAl 25. (2). Elastic potential energy per unit volume
17. (1) Y F 
Al l is
F  YA t  1
u   stress  strain
18. (1) 2
19. (2) stress

Y=  stress  Y  strain
strain
20. (2)
21. (4) Breaking force  Cross-sectional area. Given, strain  x
As cross-section remain same so the same load 2 1 2
of 800 N will break the table. Therefore, u  Y x  u  0.5Y x
2
22. (1) Breaking force  Cross-sectional area.
Yx 2 Al
max stress mg / A U  uV  uAl 
max strain   2
Y Y
26. (3) There will be both shear stress and normal
6 2
A  3  10 m stress. So shape and size both will change
27. (4) Here, F  5000 N
11 3 6
Y strain A 2  10  10  3  10
m    60kg. L  25 cm  25  102 m
g 10
A  (25  102 m)2  625  104 m 2
23. (3) Work done in stretching the wire is
  80GPa  80  109 Pa
1  YA 
W  Kx 2 K  
2  L 
Angle of shear,   F  5000
A 625  104  80  109
1 YAx 2
W  106 rad
2 L
viscous force
28. (3) shearing stress 
area
1. (2)
dv
A
 dx  103  5  0.5  103 N / m2 Y
stress

L stress
  2.5  104
A 10 L L Y
29. (1) Solids are least compressible where as L
gases are most compressible.
Volume of a cylinder is V   r 2 L
30. (3)
V r L
31. (3) According to definition of bulk modulus, 2 
V r L
P
B
V / V 2r V L
 
r V L
Here, P  100atm  100  105 N m2  107 N m 2
r
V  (99.5 100)litre  0.5litre  0.25  104
r
V  100 litre 2. (2) A rod of length L and density  , is fixed
to rigid support. Consider a small element of
107 N m2 length dx present at a distance x from the lower
 B  2  109 N m2
(0.5 / 100) end of the rod. Let  is the linear density of the
rod.
V l
32. (2)  (1  2)
V l
x 2 L
33. (1) Volume of cylindrical wire, V 
4
where x is diameter of wire
Differentiating both sides
The small extension produced in the element of
dV   dL  length dx is dl
  2 xL  x 2 . 
dx 4  dx 
Fdx
dl 
dV AY
Also volume remains constant  0
dx
where F   Axg
dL dL
 2 xL  x 2  0  2 xL   x2 ,  gdx
dx dx dl 
Y
dx L
1 g  gL2
 x  l   xdx 
dL 2 Y 0 2Y
L
2
1500  10  10 
1 l   15  104 m.
 Poisson’s ratio   2  5  108
2
3. (3) The tension present in the string in the
34. (4) first case is
231
2 6. (1) Consider an element of length dx at

 2 
F1  m  2l    m  2l  
2
 distance x as shown in the figure. The extension
 T  produced in the element dx is dl
8ml 2 YA  l 
F1   (i)
T2 l
Similarly
2
 2 
F2  m(3l )  
 T1 
12ml 2 YA  2l  r2  r1 r  r1
F2   (ii) tan   
T12 l l x
r  r1  x tan 
3
T1  T dr  dx tan 
2
4. (2) F (dx) Fdx
dl   2
AY r Y
Elongation of total wire is

F dr
l 
 Y  r 2 tan 
2 r
F dr
 
tan  Y r1 r 2
T T T T 1
strees   1  2  1 
A A1 A2 T2 2
Fl 1 1 
   
and for equilibrium,  Y (r2  r1 )  r1 r2 
3T2 Fl
mg  T1  T2  
2  r1r2Y
and mgx  T2 .(2)  0 7. (2) Consider a circular slice having radius x
and width dx. The mass of the slice is dm and
3T2 4 its areal density is
x  2T2  0  x  m
2 3
M
FL 4 FL 
5. (2) Y  ; F  mg  r  r12 
2
2
Al d 2l
Substituting the values, we get Y  2 1011 N / m2
Y d l  0.01  0.05 9
 2   2  
Y d l  0.4  0.8 80
9 9
 Y   Y   2 1011  0.2 1011 N / m2 .
80 80
The toque required to rotate the slice with angular  2F 

T  F   xa   x  
acceleration  is  m 
d    dmat  x   dm x  x 2  xF  m
TF   
m  L
d   dm x 2    2xdx  x 2 2Fx
T F
3
L
 d   2x   x dx Consider a small element of length dx. The
r extension produced in the element is dl.
2M 2 3 2M  r24  r14 
 x dx  2 2
 r22  r12 r1 r2  r1  4 
  dl 
Fdx 1  2 Fx 
 F   dx
AY AY  L 
M  r24  r14   L
F  2x 
 l   1   dx
2  r22  r12  AY 0  L 
8. (1) Initial length (circumference) of the  L 2  x2  L 

F
ring  2r   x 0    
AY  L  2  0 
Final length (circumference) of the ring  2R
Change in length  2R  2r. F  2 L2  2 FL
  L  
AY  L 2  AY
change in length 2( R  r ) R  r
strain   
original length 2r r 12. (3)
Force mg volume  density  g
F/A F/A
Stress   
Now Young’s modulus E   area A Area
l / l ( R  r ) / r
L3  g
Stress 
Rr L2
 F  AE  .
 r 
Stress  L
K 7 Stress will become 9 times the initial value
9. (3) Y   2.33  1010 N / m2 .
r0 3  1010 13. (2)
YA
10. (4) The force constant of a wire is K 
l
for series combination
k1k2 Y1Y2 A
ke  
k1  k2 Y1 L2  Y2 L1
2F
11. (2) The acceleration of the rod is , where
m
m is the mass of the rod. The tension present in Consider a small element of length dx at a
the rod at a distance x from the left end is T. distance x from the axis of rotation.
m
Mass of the element, dm  dx  dx
L
233
The centripetal force acting on the element is Mass of the element (dm)  dV  ( AR)
dT  dm2 x  2 xdx 2T sin  / 2  (dm)2 R
The tension in the rod at a distance x from the  T  A2 R 2

axis of rotation will be due to centripetal force The stress at any section of the ring
due to all elements between x  x to x  L
T A2 R 2
L
 f    2 R 2
2 2 A A
 T   2 xdx  [L  x2 ] (i)
x 2 Rupture takes place when f  
Let dl be increses in length of the element. Then

   2 R 2   
T/A R 2
Y
dl / dx
1  1 
and n 
Tdx 2 2 2 R 2
2R 
dl   [ L  x 2 ]dx
YA 2YA
15. (2) When a wire is stretched through a length,
[using (i)] then work has to be done, this work is stored in
So the total elongation of the whole rod is the wire in as elastic potential energy.
1 1 1
L
2 2 U = Kx 2   Kx  x  Fx
dl   [ L  x 2 ]dx 2 2 2
0 2YA
where K is the effective spring constant of the
L rod.
2  2 x3 
l  L x  16. (4) The tension T in the string at a distance x
2YA  3  0
from its free end is given as:
1 2 L3 1 m2 L3 F
  T x;
3 YA 3 YA l
Breaking Stress & Breaking Strain T F

Stress   x
A Al
14. (2) Let us consider small part of ring, which 2
subtend an angle  at the centre. Due to rotation, 1  stress 
Potential energy U   dV
each part of ring experience an outward force. 2 Y
1 l F2 2
U  x dv,
2Y 0 A2l 2
F 2l
where dV  Adx; This gives U 
6 AY
17. (1) Consider two points P and q as shown in
the figure. After rotating the rod the point q gets
shifted to q '. Let  is the angle of shear..
From the figure qq '  a  L V P V

V '  V  V  V   ( B  P)
Consider a circular slice of radius r and width B B
dr as shown in the figure.
M V B B
'    
V ' V ' B  P  B   n  1 p0 
20. (4) Change in pressure P  gh

Bulk modulus
P m dV d 
B ; V  
V / V  V 
V 

V 
dF / dA
The small shear force on the slice is  
tan  P (gh) 2 gh
   
B B B
dF  dA tan 
 r  21. (2) In a static fluid the pressure variation is

dF   2 rdr   
 L  given by
2 r 3 dr
dP
d  dFr   g  dp  gdh (i)
L dh
The bulk modulus is defined as
2 a 3
 r dr
L 0 dP
B
dV / V
2  a4   a4
  dV d 
L  4  2L where 
V 
18. (4) The pressure exerted by a 3000 m column
of water on the bottom layer is  BdV Bd  (ii)
dp  
V 
P  hg  3000m  100 kgm3  10ms 2
Combining equations (i) and (ii) we get
V Bd 
Fractional compression is  g dh
V 
V P 3 107 Nm2 
d  h gdh 1 1 gh
  9 2
 1.36 102  1.36%  2  0 B or 0    B (iii)
V B 2.2 10 Nm 0
19. (2) Increase in pressure

BdV d
(P)  nP0  P0  (n  1) P0 as dP   B
V 
The decrease in the volume due to increase in p 
d 
pressure P is V  V P Hence  dP   B  or P  P0  B ln  (iv)
p0 0 0
B
235
On multiplying equation (iii) by 0 , we get A r

 2
A r
0 0 gh
1 
 B  l  2 T
 2   
   gh   l  AY
so that ln   ln 1  0 
0  B 
Substituting this in equation (iv), we get
  gh   F 
P  P0  B ln 1  0   
 B     2 R 2 
1. (4) Y   (i)
volumetric stress l1
22. (2) [Bulk modules  volumetric strain ] 2L
 F 
mg  2
R 
K and Y   (ii)
 dV  l2
a 
 V  L
dV mg l2
 . (i)  2.
V Ka l1
4 3 LT LT  T  L
Volume of sphere  V   R 2. (1) L   2   2
3 AY  r Y   Y  r
Fractional change in volume
L
dV 3dr  max for L = 100 cm & r = 0.2 mm
 (ii) r2
V r
3. (1) Let L is the original length of the wire
3dr mg and K is force constant of wire.
Using eq. (i) & (ii) 
r Ka
Final length = initial length + elongation
dr mg
 F
r 3Ka L'  L 
K
 r  4
 
r Fl For first condition a  L  (i)
23. (2)    ; Y  K
 l  Al
 
 l 5
F For second condition b  L  (ii)
Y  1.6  108 N / m 2 K
 r 
A  By solving (i) and (ii) equations we get
 r 
1
r L  5a  4b and K 
ba
24. (1) Poissons ratio    r
l
Now when the longitudinal tension is 9N, length
l
of the string
l T 9
(Longitudinal strain)   L  5a  4b  9(b  a)  5b  4a.
l AY K
4. (4) The bar is at equilibrium. The net force 1 2 TL TL

from right or left of a section of BC is 70KN. As Young’s modulus, Y  AL  AL
1 2
We know that the extension due to external forces
F is given M1 g M2g
 [ Using (i) and (ii)]
L1  L L2  L
l 
Fl
; lBC 
 70 10  1  3.5 107 m
3
AY 1 2 1011 L1 M 2  L2 M 1
L .
M 2  M1
YAL
5. (2) T
2L 9. (1) The system can be considered as two
springs connected in series.
Increase in total length of the wire is
K1 K 2
L  2 L2  d 2  2L K
K1  K 2
1/2
 d2  Y A Y A
L  2L 1  2   2L K  1 , K2  2
1 l l2
 L  1
Y1Y2 A YA
 d2  d2 K 
L  2L 1  2   2L  Y1l2  Y2 l1  l1  l2 
 2L  L
 Y  2  4  1011 N / m 2
Y  r 2 .d 2
T .
2 L2 F0
10. (4) Acceleartion  , where   mass per
6. (2) F> (shear stress)  area L
unit length , tension at a distance x from front
 Fmin  (3.45  108 )(2rl ) end
 (3.45  108 )(2  3.14  0.73  102 1.27  102 )  ( L  x) F0  x

T   ( L  x) a   F0 1  
L  L
 200kN
Consider a small element of length dx at a
7. (1) K  Yr0  20  1010  3  1010  60 N / m distance x from the front end.
N
 6 109
A
8. (4) L be original length of the wire.
The change in the length of the element is dl.
T1 and T2 are tensions in the wire when masses
Tdx
M 1 and M 2 are hung. When a mass M 1 is dl 
SY
suspended from the wire, change in length of
wire is L1  L1  L L F 
0 x
 Total change in length  0 SY 1  L  .dx
 
When a mass M 2 is suspended from it, change
L
in length of wire is L2  L2  L F x2 F  L F L
l  0 x   0 L  0
SY 2L 0 SY  2  2SY
From figure (2), T1  M1 g (i)
F0
From figure (3), T2  M 2 g (ii)  Total strain 
2SY
237
11. (1) As the weight of wire acts at centre of 1 F 2L L

gravity. 16. (1) U Fx  . U  2 [F and Y are
2 2 AY r
 Only half the length of wire gets extended.
constant]
F ( L / 2) Mg ( L / 2)
Now Y  .  U A  LA   rB 
2
1 3
2
A l Al         (3)     .
U B  LB   rA   2 4
MgL ALgL
 l   l 
2 AY 2 AY 17. (1) Angle of shear
r 4 101
L g 2
   30  0.12
 l  L 100
2Y
18. (1) Y  2(1  )  3  2(1  )
So the correct choice is (2).
12. (3) Stress at B is maximum, so the wire will 3 1
 1 
break first at B if stress here becomes 105 Pa : 2 2
Now substituting the value of  in the following
expression.
Y
Y  3K (1  2)  K  
3(1  2)
1 V
19. (4) Compressibility, k  
B PV
mg ALg 105 Decrease in volume, V  PVk

 105   105  L   5m
A A g
 4 107 1 6 1010  24 103 litre
mg Ahg
13. (4) Stress    gh  24 103  103 cm3  24cc.
A A
P
h g  breaking stress 20. (4) Bulk modulus, B   ,
V
V
30 107
h  10Km
3 103 10 -ve sign shows, with increase in pressure the
volume decreases.
14. (4) We can treat the rod as a spring and its
spring constant is
P 100 103 pa
B   2000  106 N / m2
YA V 5  105
K
L V
Work done in stretching the spring (rod) is The speed of the sound
1 YAx2 B 2000 106 N / m2

W  Kx2  v   1400m / s.
2 2L  103 kg / m3
1 Mgl 21. (2) If coefficient of volume expansion is 

15. (4) Work done  Fl  .
and rise in temperature is  then
2 2
P P P Volume of the rod is

Volume elasticity B     
V / V  B V  r 2l
22. (4) Isothermal bulk modulus is V l 2r
 
V l r
P  1atm  1.013 105 N / m2 .
As length increases radius decreases.
dV P  2 103  2  103  0
23. (2) The volumetric strain  ,
V B
V L dr F  F 
26. (4)  2   2  
dV gh V L r AY  AY 
where p  gh Then,   (i)
V B
4 3 1. (3) The young’s modulus is

Since, the volume of the sphere is V  r ,
3
F/A F/A
Y 
dV 3dr l / l T
  (ii)
V r F
 Y T
Using eqs (i) and (ii), A
dr gh F
    Pr essure  1.1  105  100  2  1011
r 3B A
24. (3) Due to thermal expansion  2.2 108 Pa
The volume strain is
V 1. (3) As both have same values

 3T
V
Due to external pressure
V P

V K
Equating both we get
P As shown in the figure, the wires will have the

3T 
K same Young’s modulus (same material) and the
P length of the wire of area of cross-section 3A
T  will be l/3 (same volume as wire 1).
3 K
For wire 1,
r
F/A
25. (3) Here Poisson’s ratio   r Y (i)
l x / l
l F '/ 3 A
l For wire 2, Y  x / (l / 3) (ii)
In this case  2  103 and   0.5
l
r l F l F' l
  3 3 From (i) and (ii),     F '  9F
r l  2 10  0.5  10 A x 3 A 3x
239
 Fractional change in volume

 V  5
1. (2) As,    2.74  10
 V 
mgL
Y (Y= Young’s modulus) 4. (2) Given, for steel wire
 r 2 l
Length (l1 )  4.7 m
1
 Y
d2 Area of cross-section ( A1 )  3.0  105 m2
1 For copper wire

 d
Y
Length (l2 )  3.5 m
d copper Yiron
  Area of cross-section ( A2 )  4.0  105 m2
diron Ycopper
Let F be the given load under which steel and
2. (3) Given, radius of steel cable copper wires be stretched by the same amount
(r )  1.5cm  1.5  102 cm l.
Maximum stress  108 N / m2 F / A F l

Young’s modulus (Y )  
l / l A  l
Area of cross-section of steel cable ( A)   r 2
F l
1
 3.14  (1.5  102 )2 m2 For steel, Ys  A  l (i)
1
 3.14  2.25 104 m2 F l

2
For copper, Yc  A  l (ii)
Maximum force 2
Maximum stress =
Area of cross-section Dividing Eq. (i) by Eq. (ii), we get
Maximum force  Maximum stress  Ys F  l1 A2  l
 
Area of cross-section Yc A1  l F  l2
 108  (3.14  2.25 104 )N  7.1104 N l1 A2 4.7 4.0 105

   
l2 A1 3.5 3.0 105  1.8
3. (1) Pressure( P)  10atm  10  1.013  105 pa
5. (3) Given, each side of cube
( 1atm  1.013  10 pa)
5
(l )  10cm  0.1m
 1.013  106 pa Hydraulic pressure ( p)  7  106 Pa
Bulk modulus for glass (K )=37  109 N / m2
Bulk modulus for copper (k )  140  109 Pa
 V 
Fractional change in volume  ? Volume of the cube (V )  l 3  (0.1)3  1  103 m3
 V 
P p pV
Bulk modulus ( K )  Bulk modulus for copper ( K )  
V / V V / V V
V P 1.013 106 pV 7 106 1103

   V    5  108 m3
 2.74  105 K 140 109
V K 37 109
6. (3, 4) From graphs it is clear that ultimate 5. (1, 4) The given wire is massless. So at any
strength of material (ii) is greater than that of point
material (i). Therefore, the elastic behaviour of
material (ii) is over a large region of strain as force F
Tensile stress  
compared to material (i). If the fracture point area A
of a material is closer to ultimate strength point,
Tension = applied force =F
then the material is a brittle material. Therefore,
the material (ii) is more brittle than material (i). 6. (2, 4) Let the mass m be suspended at distance
x from left end of the rod for equal stress in the
wires. Let F1 and F2 be the tensions in the wire.
Then,
1. (1) According to the figure,
Increases in length  BO  OC  BC
L  2BO  2 L  2( L2  x 2 )1/2  2 L
 x2  x2
or L  2L 1   2 L 
 2L2  L The net torque about point O is zero
L x2 / L x 2 F1 l  x
 Strain    2 F1 x  F2 (l  x) or F  x (i)
2L 2L 2L 2
2. (4)
F1 F F F2
3. (2) Because a liquid at rest begins to move S1   16 and S 2  2 
A1 10 A2 2  106
under the effect to tangential force.
4. (2) Young’s modulus of copper For equal stress, S1  S2
stress F / A
(Y )   F F F1 1
strain strain 1 2
or 106  2  106 or 
F2 2 (ii)
1 1
 Y  2 or d 2 
d Y
lx 1
2
From Eqs. (i) and (ii), we get 
d1 Y x 2
  2
2
d Y1
2 On solving, x  2l / 3
d1 Y 190  109
It means mass m is suspended close to wire B.
or  2 
d2 Y1 110  109 stress
As, Strain 
Y
19
  1.73  1.31
11 F1 / 106 F2 / (2  106 )
For equal strain, 
YS YAl
 d1 : d 2  1.31:1
241
F1 Y (200  109 ) 10 stress applied. Rubber is an example of

 S   (iii) elastomer the correct. Graph is (1)
F2 2YAl 2  70  109 7
7. (4) As, Y  tan 
l  x 10
From Eqs. (i) and (iii),  YA tan 300 1 / 3 1
x 7    
YB tan 600 3 3
7
On solving, x  l
8. (2) From the graph l  104 m, F  20 N
17
It means mass m is suspended close to wire A. A  106 m, L 1m
FL 20 1
   20 1010  2  1011 N / m2 .
l 106 104
1. (3) Region I is called as elastic region where
9. (2) Young’s modulus of the material
as region II is called as plastic region. The wire
decreases with increase in temperature.
regains its shape only in the elastic region. the
hooke’s law is obeyed between O  P Y2  Y1  T1  T2
2. (4) As stress is shown on x - axis and strain
on y - axis
Stress  1  1 1. (4) Graph between applied force and

Y 
Strain tan  slope extension will be straight line because in elastic
range,
So the elasticity of wire P is minimum and of
wire R is maximum. F x
3. (1) In ductile materials the plastic region EC The graph between extension and stored elastic
is large and hence the materials can be easily energy will be parabolic in nature
changed into different shapes. Whereas for
brittle materials the plastic region between EC As U  1 / 2kx 2 or U  x 2 .
is small and they will break soon after the elastic
2. (2) Area of hysteresis loop gives the energy
limit is crossed. Graph B represents brittle
loss in the process of stretching and unstretching
material and graph A represents ductile material
of rubber band and this loss will appear in the
4. (3) form of heating. When the extension becomes
zero the force becomes zero. That is the rubber
 dU 
5. (2) F    band reaches to initial state after removing the
 dx 
force.
dU 3. (3) For Hooke’s law, stress  strain i.e., the
In the region BC , is positive.
dx graph between stress and strain is a straight line.
 F  negative, i.e., force is attractive in 4. (2) The relation between force and potential
nature.
 dU
dU energy is F    
 dx 
In the region AB, is negative.
dx
In the region between T and R
 F  positive, i.e., force is repulsive in nature.
dU dU
6. (1) Within the elastic limit the stress-strain  ()ve  F    ()ve
dx dx
graph for elastomers is not a straight line. Strain
produced in the wire is much larger than the In the region between P and R
dU dU 9. (2) The rubber which has less area for

 ()ve  F    ()ve
dx dx hysteresis graph is more suitable for car tyres.
Since heat generation should be minimum for
The region between T and R is attractive and
car tyres during compression of the tyres, so
between P and R is repulsive.
that the tyres last longer.
At-R
10. (4) Let  is the density of the rod and A is
dU the area of cross section.
R  0  F  0 and the molecule is in
dx
equilibrium.
At-S
dU
S  ()ve
dx
dU
F   (ve)
dx
The tension at a distance y from the lower end
dU is
The slope is maximum at S and minimum
dx T   Ayg
at T. Attractive force is maximum at A and
minimum at T. T
stress    yg
A
5. (4)
1
6. (2) The slope of the graph is
Y
F/A
Given that 1. (2) Y
l / l
1 1
 AY l
YA YB Y
l
 YA  YB F  AY  T
 1 2. (delete) Young’s modulus
 T1  T2  Y  
 T
Fl 50  103 l
Y   5
7. (2) 2 3 2
 r l  (5  10 ) 10
FL 1
L  L  2 [ Y, L and F are constant ] Given, radius r = 5mm, force F  50 kN , l  1
AY r
i.e., for a given load, thickest wire will show l  0.01mm
minimum elongation. So graph D represents the
F l
thickest wire. Y   2 1014 N / m2 .
 r 2 l
8. (3)
3. (1)
2
Fl Fl Fl F
Y  l    l  l 2 .  stress  3.5  108 N / m2
Al AY VY A
Graph between l and l 2 is a straight line. A  2 rt (t- hole thickness)
243
1 r  R 3R  R  2x 
 2   0.3  104 m2 tan     r  R 1  ;
2 x L  L
F = A  stress
Mg
4
Y
4
 1.05 10  3.3 10 N dL
r 2
dx
lateral strain
4. (3) Poisson’s ratio,   longitudinal strain
L
Mg dx MgL
For material like copper,   0.33 L 
Y R2  
2

3 R 2Y
0 1  2 x 
 L 
And, Y  3k (1  2 )
9 1 3  1 Mg 
  L '  L  L  L 1 
 3  R 2Y 
Also,
Y k 
k Y 2. (4) The bulk modulus is
Y  2(1   )
P NV
B 
 Y  V / V A V
Hence,   Y  k where V   a 2b  V  2 aba
5. (4) Stress is same in both wires. A  2 ab (Across which the pressure is

exerted)
Yc  (Lc / Lc )  Ys  ( Ls / Ls )
BAV
 1103 
11 11  Ls 
N
 110     2 10    V
 1   0.5 
 B AV
0.5  103 The frictional force f   N 
  Ls   0.25mm V
2
f  4 Bba
Total extension of composite wire
 Lc  Ls  1.25mm 3. (3) Compressibility = 1
Bulk modulus
As bulk modulus is least for ethanol (0.9) and
maximum for mercury (25) among ethanol,
1. (3) Consider a thin disc of radius r and width mercury and water. Hence compression in
dx. The small extension produced in the disc is V
dL. volume 
V
Ethanol > Water > Mercury
4. (2) The acceleartion of the blocks is
(M - m)g
a
M+m
 2mM 
Tension in the wire, T   g
mM
Tension 2mM dr 1  dV dl 
Stress   g     (ii)
Area A m  M r 2  V l 
dr
2(m  2m)g  0.25  104
 (M  2mgiven) r
A(m  2m)
From eq’s (i) and (ii) putting the value of
4m2 4mg l , l0 and V and sloving we get
 g
3mA 3A  r  4
   0.25  10
r
 
5. (3) Given, Y  2 1011 Nm2
ls r Y l
6. (3)  a, s  b, s  c, s  ?
F 7 2 lb rb Yb lb
Stress    5  10 Nm
A
 
Fl Fl
As, Y   l 
V Al AY
 0.02%  2  104 m3
V
3Mgls
ls  [ Fs  ( M  2M ) g ]
stress dl stress  rs2 .Ys
Y    2.5  104 (i)
strain l Y 2Mglb
lb  [ Fb  2 Mg ]
 rb2 .Yb
dV dr dl
V   r 2l  2 
V r l ls 3a
  2
lb 2b c
245
Fluid Mechanics 245
1. (2) Pressure depth

10. (1)
 P0  hg  3P0  hg  2 P0
11. (3)
Now (after removed water ) the pressure at depth
12. (2) When ice having lead ball melts, the
 h 4h 13P0 volume of water formed and lead balls left will
 P0   h   g  P0  g 
 5 5 5 be smaller than the volume of displaced water.
Due to which level of water will go down.
 h 2
s   g
2. (2)  g  (2Rh)  (gh)  R  h  R 13. (2)
 2
If m s  m g then Fb  Fg (since  g   s ) and
3. (4) Pressure  h g
both can show same weight.
Pressure at the bottom is independent of the area
of the base of the vessel. It depends on the height 14. (1) Upthrust  V liquid ( g  a ) ; where,
of water upto which the vessel is filled with water. a  downward acceleration, V  Volume of liquid
As in all the three vessels, level of water are the displaced. But for free fall a  g Upthrust  0
vessels is also same.
15. (1) Level will fall if initially the impurity pieces
Hence, PA  PB  PC . were floating along with ice and later they sink.
4. (3) Level will remain unchanged if initially they were
floating and later also they keep floating.
5. (1) Force acting on the base
mass
F  P  A  hdg A  0.4  900  10  2  103  7.2 N 16. (4) Volume of raft  V 
density
6. (2) As g eff  0 . The mercury will not fall
down and shows 100 cm. 120 1 3
V  m
600 5
7. (2) As the container is accelerating towards
right side pressure increases in backward When the raft is totally immersed in water, it
direction because of pseudo force. Pressure 1 3
displaces m water, so the
increases in downward direction also because of 5
gravitational force.
1
8. (3) upthrust  1000  g newton
5
9. (3) In the tilted position FB and W produce If x kg is the extra weight put on the raft to
restoring couple. fully immerse it in water, then
246 Fluid Mechanics
21. (2) When the cylinder is floating in liquid its

1
( x  120) g  1000  g or  x  120  200 downward weight is balanced by buoyancy force.
5
When the cylinder is depressed further then
x  80kg buoyancy force increases. The increase in
buoyancy force will produce acceleration.
17. (1) The weight is balanced by upward
buoyancy force and tension. Fb  ma
T  Fb  mg l Ayg  2a
T  lVg  mg 103   (0.1)2  0.04  10

a m / s 2  2m / s 2
2
m
T  l g  mg 22. (4) Pull on the

rope  Buoyant force  weight of He
m
T  (1.0)W 10  4  10 T  (1500 1.3 10 1650 10) N
10W
T  3000 N
T  36 N
18. (4) The weight is balanced by the tension and Fb1  Fb2
23. (4) m
buoyancy force g
Fg1  Fg2  T  Fb m  102  6  0.6  102  4  1  760 gm
24. (4) The effective weight of the block in liquid

4  3 4 will becomes less than 2 kg due to buoyancy of
  2R   R3  6.g  R3w g
3   3 liquid. As a result of which A will read less than
2 kg. As the body also exerts force on the liquid
4
 T  (2R)3  w  g in downward direction (Newton’s 3rd law) so
3
the reading of B will be more than 5 kg.
4 28 25. (1)
 T   7  (6  w ) gR3  R3 (6  w ) g.
3 3
26. (2) Reynold’s number is small
M 27. (4) R is dimensionless.
19. (4) Volume of ice  , volume of water

1
M 28. (2) p  v2  gz  constant.
 2

Av  constant
M M 1 1 At narrow cross-section velocity is high and
 Change in volume   M  
     hence pressure is low. So height in the tube is
low at narrow cross-section.
20. (3) Given that V    50; V ( 1)  40
29. (1)
 50  V  40  V  10cc; 30. (3) From the equation of continuity
   5g / cc , Av = constant
Weight in a liquid of specific gravity A  3  A  1  5  (1  5 A)v
 V ( 1.5) 10(5 1.5)  35gm.  v  1m / s
247
Fluid Mechanics 247
31. (4) If the liquid is incompressible, then mass 1

of liquid entering through left end should be equal   1.3  1202  902 
2
to mass of liquid coming out from the right end.
 4.095  103 Nm2
M  M1  M 2
Gross lift on the wing   p1  p2   area
 Av1  Av2  1.5 A.v (   is constant)
A  3  A1.5 v 1.5v =4.095 103 10  2
v  1m / s  81.9  103 N
  
32. (1) Due to increased velocity of air, the 41. (4) Power  F .v  PAv   ghAv
pressure between the two balls decreases.
 13.6 103  10  150 103 / 60watt
33. (1)
34. (4) 102
 watt  1.70 watt
60
35. (1) P represents pressure energy per unit
volume 42. (3) Pressure is minimum at the highest point
of the tube. Pressures are same at B and C
E Pm
P  E  PV 
V 
36. (1) 1. (1) Increased thickness at the bottom

37. (4) Conceptual increases the strength of the wall against bending
and also increases the restoring moment of the
38. (1) Pressure at the bottom of tank wall.
N 2. (1) Pressure at bottom of the lake  P0  hg
P  h g  3 105 and velocity of water is v
m2
h
1
 P  v 2 Pressure at half the depth of a lake  P0  g
2
2
According to given condition
1 3 2
 (3  1)105  10 v
2 1 2 1 1
P0  hg   P0  hg   P0  hg
2 3 3 6
v  400 m / s
2 P0 2  105
h   20m
39. (2) The velocity of efflux is v  2 gy g 103  10
3. (4) P  pa   gh
2h
 R  2 gy
g Pa
4. (1) h
g
R2
So, y As the atmospheric pressure Pa is constant.
4h
1 1 h g
40. (2)
2 2
As, p1   v1  2   v2  
2 2 h g
1 5. (4) Pressure  h g i.e., pressure at the

or p1  p2    v22  v12  bottom is independent of the area of the bottom
2
248 Fluid Mechanics
of the tank. As in both the tanks, the levels of 14. (2)

water are the same, pressure at the bottom is
15. (2) For a body immersed in liquid, when the
also same.
weight of the body is less than the upthrust, then
6. (3) the body will float partially immersed.
Pressure force is same on both dams, so they 16. (4) Buoyant force due to air will be absent
can be built with same strength.
17. (2)
 D2 4mg 18. (4)
7. (4) P  mg or P  pascal
4  D2 19. (4)
20. (2)
8. (2) 21. (3) Since, up thrust (F )  V g i.e., F  V
where V is the volume of the solid and  is the
If the level in narrow tube goes down by h1 then density of liquid.
in wider tube goes up to h2 , 22. (3) Pseudo force acts on the cork in the
downward direction. So spring undergoes
Now,  r 2 h1   ( nr )2 h2  h1  n 2 h2 (i) compression.
Now, pressure at point A = pressure at point B 23. (2) Weight of man = Change in buyoancy
force
h  g  (h1  h2 )  ' g (ii)
From the above two eq’s 1
 volume  density  3  2   103  60kg
100
h '
h2  2 where s 24. (3) In the absence of gravity liquid particles
(n  1) s 
will be still moving and hence pressure exists.
9. (1) P '  ( g  a)h Archimedes’ upward thrust will be absent for a
fluid, if there was no gravity.
10. (2)
25. (4) As the cube is in equilibrium
11. (1) Upthrust  103  1  2   2  1 N  2kN Fb  Fg
 2
lV ' g  sVg
12. (1)
13. (1) According to law of flotation l  103 Kg / m3 , s  102 Kg / m3
mice g  V1 g (  = density of liquid) V ' - volume of the cube present inside water
V - volume of the cube
mice
Volume of liquid displaced V1   (1) 103V '  102 V

When ice melts volume of water formed V' 1

  0.1
V 10
mice
V2  (2)
Water The fractional volume present outside is 0.9
26. (2) Here, mass of block, m = 1kg
As   Water in the given situation  V2  V1
Volume of a block, V  3.6 104 m3
i.e., liquid level will fall.
249
Fluid Mechanics 249
Tension in the string, T  mg  V  water g 32. (4) Conceptual

33. (2) New reading = weight of water (X) +
 Decrease in the tension of string
weight of water displaced by the object (Z)
 mg  [mg  V water g ]  V water g
 New reading = X + Z (Z - buoyancy force)
  3.6  10 m   10 kgm
4 3 3 3
  10ms   3.6 N .
2
34. (1)
27. (1) Let R be external radius of sphere and r 35. (4)
be radius of cavity. 36. (1)
4 3  3 3 4
3 3 3 37. (4)
Given,  r    R  ; r  R
3 4 3  4
38. (1)
4 3 3 1 4 3 3
Further,  ( R  r )  g    R  w  g dV dy
3 2 3 2 39. (1)  A  Av
 dt dt
Using (i),   3 40. (4) If velocities of water at entry and exit
w
points are v1 and v2 , then from equation of

28. (3) As the object just sinks mg  FB
continuity,
4 4  4
 8    R3   r 3  g  1  R 3  g v A r  2 4
2 2
3 3  3 A1v1  A2 v2  1  2   2      .
v2 A1  r1   3  9
1/3
 r3   r  7  71/3 
 8 1  3   1          41. (2) According to equation of continuity
 R   R 8  2 
a1v1  a2 v2 or
29. (1) Water will take spherical shape.
2
4 4  3 
13
v1 a2  d22 / 4  d2   3.75  9
V  R 3  1cm3  R 3  R    cm       .
3 3  4  v2 a1  d12 / 4  d1   2.50  4
23
2  3  2 42. (4) Equation of continuity av  constant
Surface  4R  4   cm
4
  2
13 13 13 v2 a1  r1 
  4    9  cm   36  cm2 2
a1v1  a2 v2    
v1 a2  r2 
30. (2) Given that mg =15
 v2  3  (2)2  12 m / s.
mg  wVg  12  wVg  3
43. (3)
mg  lVg  13  lVg  2
44. (2)
l 2
 45. (3) Siphon works without atmospheric
w 3
pressure. Siphon works because of gravity.
2V
31. (3) In first case w g  W 46. (3) Pressure below the pan decreases on
3
account of increased velocity.
V
second case l g  W 47. (1) When we blow air through the opening
2
between the balloons then kinetic energy of air
l 4 between balloons increases and pressure

w 3 decreases hence balloons move towards each
250 Fluid Mechanics
other due to pressure difference outside and

108 103
between the balloons. v2  m / sec  30m / sec
3600
48. (1)
v1  0 ; h2  h1
49. (2) Velocity of the water that comes out w.r.
to the cup is v  2ar h p1  Atmosphere pressure  105 N / m2
ar  0 (For free - fall)   density of air  1.26kg / m3
v=0
1 1
50. (2) Pressure energy per unit volume of a liquid  p1  p2   v22   1.26  302 N / m2
2 2
equals pressure.
 567 N / m2
51. (2) Let A1  Cross-section of the tank
 Aerodynamic lift on the roof
A2  Cross-sectional hole
 A( p1  p2 )  20  567 N  11.34 KN
v1  Velocity with which level decreases
1
v2  Velocity of efflux 54. (3) p1  p2    v22  v12 
2
From equation of continuity, A1v1  A2 v2
1
By using Bernoullis theorem =  1.3  (1202  902 )  4095 N / m2
2
1 1 55. (1) Velocity of efflux when the hole is at depth
p   gh   v12  p  0   v22
2 2
h, v  2 gh
2 gh 2 10  (3) Rate of flow of water from square hole
 v22  2

A  1  (0.1)2
1 2 2  Q1  a1v1  L2 2 gy
 60 m / s  A1 
2
Rate flow of water from circular hole

52. (2) Horizontal range will be maximum when
H 90 Q2  a2 v2  R 2 2 g  4 y 
h   45cm , i.e. hole 3
2 2
Given that Q1  Q2
L
 L2 2 gy  R 2 2 g  4 y   R 
2
56. (3) Point (1) is on the free surface of water

in the supply reservoir. and point (2) is just inside
the pipe.
53. (2) Here,
1 2 1 Applying Bernoulli’s equation between (1) and
 v1   gh1  p1   v22   gh2  p2
2 2 (2),(Taking the reference line at (2))
1 1
p1   (02 )   gh  p2  V22   g (0)
2 2
p1  p2  p
251
Fluid Mechanics 251
V2  2 gh 4. (4) Bulk modulus,

p p
The volume rate of flow through the pipe is B  V0  V  V0
V B
Q  AV2
 p 
d2  V  V0 1 
A V2  2 gh  B 
4
1
gh  p   p 
2  Density,    0 1    0 1 
Q d  B   B 
8
where, p  p  p0  h 0 g = pressure
difference between depth and surface of ocean
1. (2) The torque and force due to liquid force
and torque and force due to F must be balanced   gy 
   0 1  0 
then N=0  B 
5. (4)
due to liquid   yg  dy b( H  y )  F .h

0
where b is the breadth of the plate

H
 Hy 2 y 3  gbH 3
gb      Fh (i)
 2 3 0 6
H Let R be the radius of the tube

 dF  F   yg dyb  F
0 EF=R sin  ; CI=R cos ; BG=R-R cos
H2 BH  R  R sin 
gb  F (ii)
2 Pressure at point B must be same on both sides
From (i) & (ii)
d1 g ( BH )  d1 g ( BG )  d 2 g (CI )  d 2 g ( EF )
2F H 3 H
2
  Fh  h 
H 6 3
d1 R(1  sin  )  d1 R(1  cos )  d2 ( R cos )
2. (2) Weight of liquid  A1hg  A2 hg
 d 2 ( R sin  )
Walls of the vessel at level x exert a force
d1 sin   cos  1  tan 
 gh( A2  A1 ).  
d 2 cos   sin  1  tan 
3. (2) The total centrifugal force on the liquid is
6. (4) The center gravity of liquid in the first
balanced by the normal force exterted by the
right end of the tube. h1
column is at a height and that in the second
2
ML 2
Fc  MrCM  2  h2
2 column is
2
252 Fluid Mechanics
Let initial Potential Energy be U, 10. (4) The force mg and pseudo force ma act
on a liquid particle on the surface as shown in
h  h  the figure
 U1  ( Ah1  ) g  1   ( Ah2  ) g  2 
2
  2
1
 U1  A g (h12  h22 )
2
When the water column in the two equalises
h1  h2
then equivalent height is and the center
2
As the trolley moves down with acceleration
h h
of gravity in both the columns is 1 2 . g sin  on the inclined plane so a  g sin . The
4
pseudo force ma( mg sin ) is balanced by
If U 2 be the final potential energy in both the
mg sin  (gravitational force component). The
columns then
only unbalanced force is mg cos . A liquid
  h  h2   h1  h2   surface is always present perpendicular to net
U2  2  A 1  g 
  2   4  force so surface of the liquid is parallel to the
inclined plane. So   
1 2
 U2  A g  h1  h2  11. (2)
4
Since work done is equal to the decrease in
potential energy.
2
 h1  h2 
So, W  U1  U 2  W  A g  
 2 
7. (4) Pressure difference created =10 mm of
Hg. This must be equal to the pressure of water
column being created in the straw. If height of
water column be h.
hg  1cm  13.6  g
h  1  13.6  h  13.6cm. For the given situation, liquid of density 2
8. (3) As a  a0 (iˆ  ˆj  kˆ) the pseudo force should be behind that of . Let A and C are
present very closely on both sides of B.
components act along -x, -y & -z directions. So
pressure increases along l
From the right limb, PA  Patm  gh  a
D  A, D  C & D  H so maximum pressure 2
is possible at point F. l
PC  Patm  2gh  2a
9. (1) If the liquid does not come out then the 2
pressure exerted by vertical and horizontal From above two equations
columns must be balanced at the bending point.
As the liquid section at B is in equilibrium PA  PC
(aL) A  (gH ) A
3al
gH  h
a 2g
L
253
Fluid Mechanics 253
12. (3) As the cylinder is in equilibrium Fg  Fb  Volume of the body outside the liquid is
0.8(h  1.2  0.8)  1.2  0.7  0.8 1.2 b

Vout  V  Vin  V  V
l
 h  0.25 cm [Using (i)]
13. (3) Let l be the length of the cylinder, present
  
inside water initially. Let A be the cross-sectional  V 1  b 
area of the cylinder. Equating weight of the  l 
cylinder with the upthrust, we get
Volume of the body out side the liquid
mg  Al  g  m  Al   Volume of the body
  
 1  b 
 1 
As this ratio is independent of the effective value
of acceleration due to gravity, so the percentage
When the cylinder is tilted through an angle  , of the volume of the body outside the liquid will
remain unchanged.
l
length of cylinder in water  16. (3) The weight of the rod is
cos
W  LAg
l
Weight of water displaced  A g
cos where L is length of the rod and  is its density
lA g L
Restoring force   lA g OT 
cos 2sin 
 L 
 1   1  Fb    AW g
 lA g   1  mg   1  2sin  
 cos    cos 
14. (2) When brakes are applied then pseudo

force acts forward and pressure increases in
forward direction and hence the bubbles move
back ward (towards the less pressure side)
15. (2) Let V be volume of a body.
b and l be the density of the body and liquid

respectively. When the body is floating in the
liquid, then
OT l
OP  
Weight of the body = Weight of the liquid 2 4sin 
displaced At equilibrium about O
V  b g  Vin  l g mg  Fb
Vin b L b  L 
 (i) LAg cos    AW g 
V l 2 4sin   2sin  
254 Fluid Mechanics
W 1 20. (1)
sin 2    sin 2  
4 2 21. (4) Let T be the tension in the string.
   45 T  mg  Fb  mg  m g   m  V  g
17. (4) At equilibrium the pressure force on the   density of water..

block is balanced by its weight
As the system is in equilibrium
T1  T2
m
1
 V1  g   m2  V2  g
36 48
36  1   48  1
9 2
 2  3 g cm3
22. (1) When the liquid is in horizontal part of
Here the effect of atmospheric pressure is not
the tube it has velocity v. When it starts moving
considered since Pa effect is there both on top vertically its velocity decreases because of
& base of the block. FP1 and FP2 are pressure gravitational force. The total height travelled by
forces. the liquid calculated from conservation of
mechanical energy. For a liquid particle of mass
FP1  Fg = FP2 m
1
 mg (h  H )  mv 2
gh1 (2r )  gH (2r )2
2
2
3
v2
2 2 hH 
 g (h1  H ) (2r )  r  2g
5H 23. (2) v22  v12  2 gh

h1 
3
 (0.4)2  2  10  0.2
18. (3) Let Fb ( Vg ) is the buoyancy force on
the cylinder. Let F is the pressure force exerted
by the liquid on the base of the cylinder.
From the definition of buoyancy force
Fb  F  R 2gh
Vg  F  R 2 gh
 v2  2.04 m / s.
 F  gV  R 2 gh
19. (2) Buoyancy force=Net pressure force d 22 d 2

Again,  v2  1  v1
2 2
4 4
1 H H
   H g   P0  gH      Fv
3  2  2 v1 0.4
 d 2  d1   8 103  m
v2 2.04
H 2 gH 3 
 P0 
4 6  d 2  3.54  103 m  3.6  103 m
255
Fluid Mechanics 255
24. (2) From equation continuity

dx 2h 1 dy
 vb    2g 
A 1v1 = A2v2 dt g 2 y dt
2 2
8 2
   (0.25)      v2  v2  4m / s h dy
 2 2 vb  . (i)
y dt
The horizontal displacement is
Using equation of continuity
x  v2t  4t
Ady
2h 2 1.25 1  a 2 gy (ii)
 t   s dt
g 10 2
h a
1
 x  4   2m
Equation (i) and (ii) vb  
y
2 gy
y
2
dV a 1
25. (3) =(Q-av) vb  2 gh   20   1ms 1 .
dt y 20
h2 t 29. (2) If h is the initial height of liquid in drum

Adh Adh
=(Q-av)  h Q-K h =0 dt above the small opening, then velocity of efflux,
dt 1
v  2 gh . As the water drains out, h decreases,
2A   Q  K h2   hence v decreases. This reduces the rate of

t K
K 2 
 
h1  h2  Q ln  
 Q  K h  discharge, a longer time is required to drain out
 1 
the same volume of water. So, clearly t1  t2  t3 .
26. (3) Effective value of acceleration due to
30. (1) Let  be density of liquid flowing in the
gravity becomes ( g  a0 ) .
tube.
The velocity v= 2(g+a0 )h A1v1  A2 v2
27. (4) As the vessel is accelerating down then v1 A2
pseudo force acts on the block in the upward 
v2 A1
direction and hence the spring length increases.
According to Bernoulli’s equation for horizontal
28. (1) Velocity of efflux v  2 gy flow of liquid,
1 1
2h P1   v12  P2   v22
Range x  2 gy  2 2
g
1
P1  P2  (  v22   v12 )
2
1
h g   (v22  v12 ) ( P1  P2  h g )
2
v22  v12  2hg
31. (1) Using equation of continuity, we have
 dx  A1
The velocity of the block is  . v2  v1
 dt  A2
256 Fluid Mechanics
From Bernoulli’s theorem, 34. (2)   due to F1  due to F2

1 1  F1 ( R  R)  F2 ( R  R)
P0  gh1  v12  P0  gh2  v22
2 2
 ( F1  F2 )2 R F1  F2 (Both torques are
1 additive, forms a couple)
 g  h1  h2    v22  v12 
2
 h
 A2    2 F  2 R  F  av 2 , v  2 g   
 60   12  1 v12   2  

 A2 
h
A 4  4a 2 g   R  4aghR
 1  2
A2 1
dm  
32. (4) As the momentum of the liquid gets 35. (2) The thrust force is F  v2  v1
dt
reversed the net force on the tube is
 
1   Av v2  v1 v2  v1
F  2  Av 2  x   av 2 
2 
angle between v2 , v1 is 120
 x4
 Av ( 3v)
33. (4) Here, v1  2g  h  x  ; v2  2gx
36. (3) Pressure is minimum at the highest point
Let, a  area of cross-section of each hole of the tube. Pressures are same at B and C
  density of the liquid 37. (2) From the equation of continuity
AV=constant the velocity of the liquid is same
everywhere inside the tube as cross sectional area
is constant.
VA  VB  VC  VD  VE
Take the reference line at E. By applying
Bernoulli’s equation at D and E we get
1 1
pD  V 2   g (h1  h2 )  p  V 2   g (0)
The force exerted on the lower hole towards 2 2
left
pD  p   g (h1  h2 )
2
 av 1
38. (4) Take the reference line at point 1. By
Similarly, the force exerted on the upper hole applying Bernoulli’s equation at the points 6
towards right and 1 we get
1 1
 a  v22 pa   (0)2   gh2  pa  V12   g (0)
2 2
Net force on the tank, F  a   v12  v22 
v1  2 gh2
F  a   2 g  h  x   2 gx   2a  gh From the equation of continuity
 F h V1  V2  V3  V4  V5
257
Fluid Mechanics 257
By applying Bernoulli’s equation at 1 and 2 H
due to liquid   yg  dy b( H  y )

1 1
pa  V12   g (0)  p2  V22   gh2 0
2 2
Where b is the breadth of the plate
p2  pa   gh2 H
 Hy 2 y 3  gbH 3
similarly we can prove   gb    
 2 3 0 6
p3  p4  pa   g (h1  h2 )
3. (1) At a depth y from the surface of the fluid,
p5  pa   gh2 the net force acting on the gate element of width
dy is
p3  p4  pa   g (h1  h2 )
dF  ( P0  gy  P0 )  1dy  gydy
p5  pa   gh2
39. (3) Take the reference line at point 3
By applying Bernoulli’s equation at 1 and 3 we
get
1
p   (02 )   g (3)
2
1
 p  V32   g (0)
2
V3  6 g  7.76m / s
Torque of this force about the hinge is
By applying equation of continuity at 2 and 3
we get d   gydy y
V3 A3 Torque experienced by the gate due to liquid is

V2   3.03m / s
A2 1
   gy 2 dy
0
By applying Bernoulli’s equation at 2 and 3
1
 gy 3  g
1 1   
p2  V22   g (0)  p  V32   g (0)  3 0 3
2 2
As the torque is balanced by the external force
p 2 =125.5kp a
F
g g
  F (1)  F 
3 3
1. (2) Force  area of base  pressure due to
4. (2) Thrust on
liquid column of height h   r 2  h g
lamina  pressure at centroid  Area
2. (2) The small force exerted on a strip of width
dy and length b is h g 1
  A  A gh
3 3
dF  ygdyb
5. (2) Gauge pressure  h g
H
h
2
Hg
The torque about O is  Gauge pressure is
2
258 Fluid Mechanics
6. (2) If F is additional force applied then

The mass of the entire liquid  (h1  h2  h) A
F
pressure increases at every piston by which If this moves with a velocity v,
A
will be carried out through out the liquid since
liquid is in static state. 1 2
Kinetic energy  (h1  h2  h) A υ (ii).
2
7. (4)
g
v   h1  h2 
2(h1  h2  h)
10. (3) gd   p  p0 
g
p p  0
m
d 
V 
GM  p  p0 V
 
Vertical height of the liquid in portion AC R2 dm
h1  DO  OE  R sin   R cos  R(sin   cos )  p  p VR 0

2
M 
Vertical height of the liquid in portion CG Gdm
h2  R  R cos  R(1  cos ) 11. (1) Pressure due to pseudo force causes the
rise in level in both the limbs.
Vertical height of the liquid in portion GB
12. (1) The pseudo force components ma cos 30
h3  R  R sin   R(1  sin  ) & ma sin30 are shown for a liquid element on
In equilibrium, the pressure due to liquid on the the surface.
both sides must be equal at the point G
 gh1   gh2   gh3 [As pressure  h g ]
 gR (sin  cos )   gR(1  cos )   gR(1  sin )

 (sin  cos )  (1  cos )   (1  sin )
(   )sin  (   )cos
(   )     The surface of the liquid is always perpendicular

tan      tan 1  
(   )     to resultant force
8. (1) F  P  A  glA0 3 3
ax a cos30
9. (3) From the above problem we can write tan     2  0.23
a y  g a sin 30  g 3  9.8
the decrease in gravitational potential energy as 2
2
h h  a
U   1 2  A g (i) 13. (3) We know, tan  
 2  g
259
Fluid Mechanics 259
h Note that volume of water in the tank remains

tan   constant.
l
Vi  V f
h a al
So,  or h 1
l g g AH   Ah   h  2H
2
3b  b a g
14. (2) tan    a h a
4b g 2  tan   
L g
15. (4) Consider an element of the liquid of width
dx and area of cross-section  , at a distance x  2H 
a g
from the front of the tank.  L 
17. (1) OP is the portion of the rod immersed in
water.
Mass of the element  dm  ( dx) .

Net force to the right on the element
 ( p  dp)  p   dp
40
cos 60 
From Newton’s 2 nd law OP
  dp  ( dx)a
40 40
 OP  cm  cm  80cm
dp  adx cos 60
12
C C
dp    adx or pC  pA   al The centre of buoyancy is at the centre of the
 A A
immersed part of the rod. So, the required
(Refer to the figure in the question to identify A, distance is 40 cm.
B, C) 18. (2) Let x is the length of the candle above
the water surface at equilibrium.
Also pB  pC   gh
Given that h  2L, x  L
pB  ( pA   al )   gh
pB  p A  h g  l  a  h  2x
16. (3) The liquid will not come out of the tank at dh dx
the point A if the free surface passes through 2
dt dt
the point itself.
dh cm
2
dt hr
dx cm
 1
dt hr
So the top of the candle falls at 1cm/hr.
260 Fluid Mechanics
19. (3) Let V be the volume of wooden ball. The 1

mass of ball is m  V  . water. This means that 20   2m of ice layer
10
Upward acceleration, will be above the water surface.
upward thrust  weight of ball

a
mass of ball
V 0 g  V  g  0    g
 
V 
If v is the velocity of ball on reaching the surface
In other words, in the hole, the water level will
after being released at depth h is
rise up to 18 m of the ice layer. So minimum
   
1/ 2
length of rope required  2m
v  2as   2  0  gh 
     22. (2) Oil will float on water so, (2) or (4) is the
correct option. But density of ball is more than
If h ' is the vertical distance reached by ball that of oil, hence it will sink in oil.
above the surface of water, then
23. (3) From figure, kx0  FB  Mg
v 2 2  0    1
h'   gh 
2g  2g
   0 
 0 h    1 h
     
20. (4) Acceleration when block B is in the liquid
mA g  (mB g  upthrust)
a1 
(mA  3m)
 3m 
mA g   3mg  g
 2  kx0  
L
Ag  Mg

(mA  3m)
 2
L
Accelaration when block B is outside of the liquid  kx0  Mg   Ag
2
mA g   3mg  LAg
a2    Mg 
(mA  3m)
 x0  2  Mg 1  LA 
 
k k  2M 
Given a1  a2 , we get
Hence, extension of the spring when it is in
3 Mg  LA 
mA g  mg  3mg  mA g equilibrium is, x0 
2 1  
k  2M 
9 9 24. (3) According to equation of continuity,
 2mA g  mg  mA    m
2  4 av  constant. As v increases, a decreases.
9 25. (1) According to equation of continuity
21. (1) Given that density of ice   density
10
A1v1  A2 v2  A3 v3
1
of water, so only th of ice will be outside the  4  0.2  2  0.2  0.4  v3  v3  1m /s
10
261
Fluid Mechanics 261
26. (3) From equation of continuity and similarly time taken for the level to fall from
A1v1  NA2 v2 h A 2 h 
to zero t2    0
2 A0 g 2 
m
2.4cm2 1.5  20  2 102 cm2  v2
s 1
1
t1 2  2 1
v2  9m / s  
t2 1
0
27. (4) By equation of continuity 2
Av=av' 31. (2) Let A and a be the cross-sectional areas
2
of the vessel and hole respectively. Let h be the
 A   πR  height of water in the vessel at time t.
 v'=   v=  2 ×v
 a   πr 
 dh 
 v '  400cm/sec Let    represent the rate of fall of level.
 dt 
28. (1) From equation of continuity
dh 
2 Then, A     av  a 2 gh
20  10  v  1 1  dt 
1 dh a 2g
v cms 1  5cms 1 or   dt
20  10 2
h A
29. (1) If v  2 gh is the velocity of the liquid 0 1 a 2g t

 dh  dt
when the free surface height is h then h
h A 0
dh 0 dh t 2H a 2g
v  2 gh     2 g  dt  t  (2 h )  t
A
dt H
h 0
g
A 1 A 2h
t 2 h  t 
2H a 2g a g
t
g
Now, t  h
t1 H1 H1 When h is quadrupled, t is doubled.
  
t2 H2 H1 2
t1 10 min
 t2    7 min
2 2 32. (2)
30. (3) Time taken for the level to fall from H to
H'
From Bernoulli’s equation
A 2 
t H  H '  1
A0 g  2 gh2 +1 gh1= 1v 2
2
According to problem - the time taken for the
A 2  h 2 g ( 1h1   2 h2 )   
h t1   v2   v  2 g  h1  2 h2 
level to fall from h to  h  1  1 
2 A0 g  2
262 Fluid Mechanics
33. (3) Rate of inflow = rate of out flow P

or v2  v1 
 vav
Q  av  a 2 gh
v1  v2
Q2 Here, vav   720 km /h
 h 2
a2  2g
34. (4) Water rushing from a hole made in the 5

 720  m/s  200 m/s
tank below the free surface follows a parabolic 18
path. 4
 104
Let h is is the depth of the hole from the free v2  v1 P 5
  2 
surface vav  vav 1.2  (200) 2
The velocity of efflux of liquid, v  2 gh 4 104

  0.17
5 1.2  4 104
2( H  h)
Time, t  , Horizontal range, R  vt 36. (1) By applying Bernoulli’s equation and
g equation of continuity at 1 and 2
1/ 2
 2( H  h)  1 1
R   2 gh   P1   v12  P2   v22   gH
 g  2 2
The range is maximum if dR /dh  0 v1 A1  v2 A2
 R 2  4h( H  h)  4( Hh  h 2 ) Given A1  A2  v2  v1
dR  P1  P2
or 2R  4( H  2h) or 0  (H  2h)
dh
37. (1) If a ball is moving from left to right and
or h  H/2 also spinning about a horizontal axis in clockwise
35. (3) The weight of the aircraft is balanced by direction about a horizontal axis perpendicular
the upward force due to the pressure difference. to the direction of motion then relative to the
ball air will be moving from right to left.
i.e., P  A  mg
mg (4  105 kg )(10m /s 2 ) 4
P     104 N /m2
A 500 m2 5
Let v1 , v2 are the speeds of air on the lower and

upper surface of the wings of the aircraft and
The resultant velocity of air above the ball will
P1 , P2 are the pressures there.
be (v  r) while below it (v  r) . So in
Using Bernoulli’s theorem, we get accordance with Bernoulli’s principle, pressure
above the ball will be less than below it. Due to
1 1 this difference of pressure an upward force will
P1   v12  P2   v22
2 2 act on the ball as shown in the figure.
1 38. (2) Let A = The area of cross section of the
P1  P2    v22  v12  hole
2
v = Initial velocity of efflux

P  (v2  v1 )(v2  v1 )   vav (v2  v1 ) d = Density of water,
2
263
Fluid Mechanics 263
Initial mass of water flowing out per second 42. (2)

 Avd dk d  1 2  v2 dM v 2  dM dl 
  Mv       
Rate of change of momentum  Adv 2 dt dt  2  2 dt 2  dl dt 
 Initial upward reaction  Adv 2 dk 1 2 dl 1 3 dM

  mv   mv (Given that =m )
dt 2 dt 2 dt
 Initial decrease in weight  Ad  2 gh 
43. (3) Point (1) is on the free surface of water
1 in the supply reservoir. and point (2) is just
 2 Adgh  2     1 980  25  12.5 gm  wt. inside the pipe. Applying Bernoulli’s equation
4
between (1) and (2),(Taking the reference line
39. (1) The thrust force exerted by a liquid of at (2))
density  flowing through a cross section of
area A is  Av 2 (v - velocity of the liquid). 1 1
p1   (02 )   gh  p2  V22   g (0)
2 2
As the momentum of the liquid gets reversed
p1  p2  p
the net force on the left tube is F1  2  Av 2
V2  2 gh
 A
Force on the right tube is F2  2    v '2
2 The volume rate of flow through the pipe is
Q  AV2
 F1  F2  v '  2v
d2
40. (3) Velocity of efflux  2gh A V2  2 gh
4
dm gh
Thrust force  v   av 2 Q  d2
dt 8
Frictional force  mg 44. (3) By applying Bernoulli’s equation at B and
C we get (reference line is taken at C)
 av 2   Ahg
1 1
2
p   (02 )   g (3.6)  p  V 2   g (0)
a  2 gh    (  Ah) g 2 2
2a V  6 2 m/s
 
A By applying Bernoulli’s equation at B and A we
2a get
 min 
A
1
41. (3) Velocity of a liquid that comes out from a p   (02 )   g (3.6)
2
hole which is present at a depth H is v  2 gH 1
 pA   (V 2 )   g (3.6  1.8)
2
Let H1 & H 2 are depths of holes from the free
surface. The thrust forces exerted by the liquid pA  4.6  104 N / m2
on the container are
2 d
F1   av12  a  (2 gH1 ), F2   av22  a  (2 gH 2 ) Discharge rate =AV   ( )  V
2
Fnet  a  2 g ( H1  H 2 )  a 2gh  96 2  104 m3 / s
264 Fluid Mechanics
45. (1) By applying Bernoulli’s equation at A and

B
1 1 1. (3.55) From Bernoulli’s theorem,
p   (02 )   gh2  p  V 2   g (0)
2 2 1 1
P0  v12 gh  P0  v22  0
V  2 gh2 2 2
v2  v12  2gh  0.16  2 10  0.2  2.03m / s

46. (3) h1  h2  3.0m
From equation of continuity
By applying Bernoulli’s equation at A and C
are A2 v2  A1v1
1 D22 D2
p   gh2  pc  V 2   g (h1  h2 )   v2   1  v1
2 4 4
v1
The gauge pressure =  g (h1  h2 )  D2  D1  3.55  103 m
v2
= 6 104 N / m2 2. (0.25) Let  is the density of water. When
So absolute pressure is the cylinder is in equilibrium initially
Fg  Fb
 1105  0.6 105  0.4 10 N / m2
V
47. (3) h1  h2  3.0m mg   g (V- volume of the cylinder)
2
By applying Bernoulli’s equation at A and C are V
m (i)
1 2
p   gh2  pc  V 2   g (h1  h2 ) After pouring liquid
2
The gauge pressure =  g (h1  h2 ) = 6 104 N / m2  2V   2V 

mg     g   g (ii)
 3   3 
48. (4) The velocity of the liquid that comes out
at point (1) is V1  2 g (6m)  10.95m / s . 
From the above two eq’s  
4
By applying Bernoulli’s equation at (1) & (2)
3. (560) T  F  mg
1 1
P2  V22  g (h  6m)  P1  V12  g (0) mg  0.65  103  10  900  5600 N
2 2
 m  560kg
We know that V1  V2
P1  P  105 N / m 2
When water bubbles form at (2) 1. (1) Buoyant force = weight of the body in air
- weight of the body in liquid
P2  0
2. (10) Let a is the acceleration of the tank
5 2
g (h  6m)  10 ( g  10m / s )
(h  6m)  10m
h  4m
i.e., when h  4m the liquid flow will become
discontinuous.
265
Fluid Mechanics 265
aL
2. (2, 3) Reynold’s number
a h
The angle  is tan     h
g L g vD
R
From conservation of volume of the liquid 
Vi  V f  - Density of the liquid
1 D - Pipe diameter (inside)

( Lb)1.5  (2  h)bL  [ Lbh]
2  - Coefficient of viscosity
Where b is the base width If R is small then the liquid is streamlined. So 
gh is low and  is high.
 h 1  a   1m / s 2
L 3. (1, 2) When the coin falls into water the block
The time to reach velocity 10m/s is gets lifted up and hence h decreases. As some
portion of block goes out the liquid height l
v  at  t  10s decreases.
3. (10) According to Bernoulli’s theorem.
1 1
P1  v12  P2  v22 1. (3) According to the equation of continuity,
2 2 when cross-section of duct decreases, the
velocity of flow of liquid increases Av =
For closed tap v1  0
constant.
 v1  0 From Bernoulli’s theorem,
2  P1  P2  1 2
 v2  p+  v +  gh = constant
 2
2   3.5  105  3  105  When velocity increases pressure decreases.

 v2  3
10 2. (4) From the equation of continuity and
Bernoulli’s theorem Av = constant
 v2  10ms 1
1 2
p+  v +  gh = constant
2
1. (4) In a streamline flow the two streams When A increases velocity decreases and the
cannot cross each other. pressure increases non linearly.
2. (1) According to equation of continuity 3. (1)
a1v1  a2 v2 dh
4000 104   0.5 104 2 10  h  104  2
2 2
dt
v1 a2  d22 / 4  d2   3.75  9
      
v2 a1  d12 / 4  d1   2.50  4 dh  5  20  10 4 
  5  10 4     h
dt  4 
when h=costant,
1. (2) Along a streamline, the velocity of every 2
fluid particle while crossing a given position is dh  5 4 
 0, h     0.8m
the same. dt  5  20 
266 Fluid Mechanics
4. (3) From Archimedes Principle, the amount 2
of water displaced is equal to the amount of  dy   x g dx
water added, so the rate the water level

increases is the same before and after the cup 2 x
 2 x2
is submerged. When the water level is at the
y
g  x dx  y 
0 2g
edge of the cup, and as water is added, the water
fills the cup, displacing air in the cup which When y  H  x  R
results in a small decrease in the water level of
the tub.  2 R2
H
2g
5. (4) When the liquid surface takes a curved
path take the origin at the base centre of the 1
container and Consider a tangent on the surface.  2Hg
R
6. (3) The shape of the surface of the liquid is
The tangent is considered at point P whose

coordinates are (x, y). The volume of the empty space in the container
is given as follows.
dy
The slope of the tangent is tan   (i) The small volume of the space having thickness
dx
Consider a small liquid element at point P. The dy is dV   x2 dy
forces that act on the element are shown in the
figure. From the property of a liquid surface, Let the equation of the surface is y  Kx 2
the tangent to the surface at a point is always
perpendicular to the net force at that point. 2
Where K 
2g
dy  2Kx dx
dV   x2 (2Kx dx)
R
V  2 K  x 3 dx
0
mx 2  Pseudo force or centrifugal force
2 KR 4  KR 4
mg - Weight of the element V 
4 2
Fr  Resultant force we know that
2 2
mx x
tan    (ii)  2 x2  2 
mg g y  H    R2  KR2
2g  2g 
From Eq’s (i) & (ii)
dy x 2 H
K
we get dx  g R2
267
Fluid Mechanics 267
1 F  mg  Fb
V   R 2 H  Vempty space
2
when the cylinder is coming out from the liquid
VTotal  Vliq  Vempty space
then Fb decreases. When the cylinder comes
VTotal   R 2 H out finally Fb  0 and F = mg.
1 4. (2) Initially the cube is floating so buoyancy
Vempty   R 2 H
2 force is equal to weight and tension in the string
As the volume of the liquid remains constant is zero. When the block is pulled inside the water
the buoyancy force increases linearly and hence
V f  Vi
tension also increases. When the block
1 completely submerges in water the tension in
 Vliq   R 2 H   R 2 h the string remain constant.
2
h - initial level of the liquid 5. (1) Let A is the cross sectional area of the
H hole. The area of the surface of water is  x2 .

h
2 From the equation of continuity
Note : When we whirl a container the initial dy
 x 2  A 2 gy
level of the liquid in the container, is dt
hmax  hmin dy A 2g
h as shown in the figure Given that  constant  x 2  y
2 dt 
2
Kx4  y where K 
2gA2
6. (1) Specific gravity of A is more than that of
B.
In the given problem hmin  0 & hmax  H  h1  h2

When the ice starts melting the level of water
H
hinitial  remain constant in both containers. After some
2
time the level decreases in both containers and
finally levels will be equal when both the
materials A and B sink.
1. (4) When we stir the tea the cup is at rest
and the particles of the tea which are in contact
with the walls of the cup also be at rest. When 1. (2) The linear speed of the liquid at the sides
we move from center to circumference, the is r. At the centre of the liquid and on the top
velocity of liquid goes on decreasing and finally
surface the pressure is equal to atmospheric
becomes zero.
r 2 2
2. (1) As the liquid is non-viscous the rotation pressure.    gh
of container will not have any effect on the 2
liquid. 2gh  2 r 2
3. (3) Consider the cylinder when it is inside the 2 2
liquid. Both weight and buoyancy forces are  2  2   5  10 2
h  0.02m
constant. 2  10
268 Fluid Mechanics
r4  H 4
2 gH  2gh  4 2gH  x  r  
x H  h
dh
2. (2) Let the rate of falling water level be 
dt
2. (2)
Initially at
t  0; h  h
t  t; h  0
Equating the pressure at G from both sides
 dh  2 2
1 g ( R  R sin  )  1 g ( R  R cos )  2 gR sin  Then, A      a .v   a 2 gh
 dt 
 2 gR cos
A
dt   2
dh
1 gR(cos  sin  )  2 gR(cos  sin  )  a 2 gh
1 (1  tan  )  2 (1  tan  ) Integrating both sides
   2 
t A 0
   tan 1  1  dt    h 1/ 2 dh
 0
2 g a 2 h
 1  2 
3. (Delete) The given problem is wrong as area  h1/2 

A 2A h
0
t
of cross-section is less at A then velocity of fluid [t ]  
0 .
2   t  2
2 g a 1/ 2  h a g
is more and pressure is less. So the liquid height
in A should be smaller than height in B. 3. (1) By applying the bernoulli’s equation at the
4. (2) The thrust force exerted by the liquid on top and the base
the hand is
F  Av 2  103  102  (1.5)2 F  22.5 N
1. (1) From Bernoulli’s Principle,

The velocity that comes out from the hole is
v1  2 gh 1 1
P0  0.8(v1 )2  1 g  3  0.8  g  2  P0  0.8v22
2 2
Let v2 is the velocity after falling height h
Here v1  0
a1v1  a2 v2
1
4.6 g  (0.8)v22  v22  115  v2  10.7m/s
r 2
2 gH   x v22 2
1
r2 4. (3)
2 2
Pressure difference P2  P1   (v2  v1 )
2 gH  v2 2
x2
1
v22  v12 =2 gh  v22  2gH  2gh   1.2  (150)2  (100)2   7500Nm2
2
Viscosity 269
6. (1)
2 R v 1
7. (3) The terminal velocity is v r  v  9 .
3R
mg
1. (3) 8. (1) has the dimensions as that of
r
2. (2) velocity.
3. (1) The viscous force on a sphere is 9. (2) Viscous force = Weight - Force of
buoyancy
f  6rv
 f v  mg  FB  mg  d 2 vg
f

6rv  d vg   d 
 mg 1  2   mg 1  2 
 mg   d1 
MLT 2
  ML1T 1
LLT 1 10. (4) Relative to liquid, the velocity of sphere
4. (3) Viscosity of a liquid depends on is 3v0 upwards.
v  Viscous force on sphere  6r 3v0
temperature. Viscous force f  A
d downward
Viscous force depends on area, separation
between two layers and relative velocity between  18rv0 downward
two layers.
11. (3) The rate of flow of liquid is
Fricition force between two solids f K   K N
depends on only normal force.  Pr 4
V
8l
5. (2) Here,   103 Nm 2 s, v  5 ms 1 ; l  10 m
Pr 4 1
12. (3) Q ; Q
v 8l l
Strain rate 
l Q2 l1 l 1
    Q2  20 106 m3 / s
Q1 l2 3l 3 3
Shearing stress
Coefficient of viscosity,   Strain rate 13. (3) When a viscous liuid flows through a
cylindrical tube the velocity of the liquid particles
 Shearing stress    Strain rate
touching the surface of the tube is zero and it
(103 Nm2 s ) (5 ms 1 ) increases as we move towards the centre of
  0.5  103 Nm 2
(10 m) the tube.
270 Viscosity
12. (2) The terminal velocity of a sphere of radius

 Q  Q1 
4  r24  r14 
14. (4) Q  r , 2 100   4 100 R moving in a liquid of coefficient of viscosity
 Q1   r1  2 R     g
2
 is   
9 
 1.1r 4  r 4 
  1 4 1  100  46% Where ρ - density of the sphere
 r1 
 - density of the liquid
   R2
1. (4) 13. (1) As the steel ball is small therefore
2. (2) When water flows in a river the velocity upthrust can be neglected.
is maximum at the surface and it decreses as we
go down. mg
Now, 6 rv0  mg or v0 
r
3. (1)
14. (4)
4. (2)
2 gr     
2
2
5. (2) Cohesive forces are the intermolecular vt   vt  r 2   V1/3   V 2/3
forces like hydrogen bonding, Vander waals 9 
forces etc. Cohesive forces present between 2/3
liquid layers resist the motion thus are responsible vt1 v 
 1 
for viscosity. vt2  v2 
6. (1)
2/3
7. (1) 9  v1  v1 27
   
4  v2  v2 8
8. (1) The velocity gradient is
15. (4) When the rain drop reaches terminal
dv F
  1.5 s 1 velocity then
dx A
4
F  2  105 v   r 3  g
9. (1) A  (0.1)2  0.01 m 2 , 3
  0.01 Poise  0.001 decapoise (M.K.S.unit) 4 22 3

 2 105 v    1.5 103  103 10
3 7
dv  0.1m/s and F  0.002 N
On solving, v  7.07ms1  7ms 1
dv 16. (3) When two drops of radius r each combine
F  A
dx to form a big drop, the radius of big drop will be
given by
 Adv 0.001 0.01 0.1
 dx    0.0005 m
F 0.002 4 3 4 2 4 3
R  r  r
3 3 3
10. (4) Viscosity force on a moving sphere in a
liquid is f  6rv or R3  2r 3 or R  21/3 r
11. (1) Viscous force 2 2 1
vR  R 
Now     2 3  43
 6rv  6  18  105  0.03  100 vr  r 
 101.73  104 dyn (dyn  1g cm / s 2 )  vR  5  41/3 cm / s

Viscosity 271
17. (4) 2. (1) As the liquid is present on both sides of

the plate the viscous force on the plate by both
18. (4) As the poiseulle’s equation has r 4 so the the liquids is
radius is to be measured with more accuracy.
19. (4) Av Av
f  
d1 d2
Pr 4
20. (3) V  V   Pr 4
8l 3 Ns
Given that   0.01 Poise  10
m2
[  and l are constants]
4 4
A  2 m 2 , v  2m / s
V  P  r  1 1
 2   2  2   2    
V1  P1  r1   2 8 d1  0.5 m, d 2  0.1 m
Q f  12  103 N
 V2 
8 3. (4) Let v is the velocity of a spherical shell
21. (3) The rate of liquid flow is or radius r
The viscous force between two adjacent layers
 Pr 4 P having radius r is
Q 
8l  8l 
 4
 r 
By comparing the above equation with ohm’s

V
law i  then the resistance offered to the
R
8
fluid is R 
r 4
22. (2) Given that
dv
df   2rL
PD 4  D dr
V r  
64L  2
where L is the length of the cylinder. dv is the
 PD4  V
relative velocity between two layers which are
V '  4  separated by dr.
 8  642L  4
As we move towards right v decreases and r
increases.
1. (3) As the block moves with constant velocity dv  ()ve & dr  ()ve
Viscous force  mg sin dv

f  2rL
v dr
 A  mg sin 
t r v
dr 2L
v R r  f v dv
a 2  a3g sin  1 0
t
tg sin a r 2L(v0  v)

 ln  (i)
v R1 f
272 Viscosity
R2
dr 2L 0 2 r 2 (  ) g
  f v dv 5. (1) Terminal speed, v 
9 
R1 r 0
Where  and  are the densities of spherical

R 2Lv0 body and medium respectively, r is the radius of
ln 2  (ii)
R1 f the spherical body and  is the coefficient of
viscosity of the medium.
Here, r  1 mm  103
ln(r / R1 )  v0  v 
 
ln( R2 / R1 )  v0  Specific gravity of Al  2.7
Density of Al ,   2.7  103 kg /m3

 ln(r / R1 )  v0 ln( R2 / r )
v  v0 1  
 ln( R2 / R1 )  ln( R2 / R1 ) Density of water,   103 kg /m3
4. (1) Consider a thin cylindrical shell of radius   8  104 , g  9.8 m /s 2

r. Let L is the length of the cylinder. The viscous
force present between two adjacent cylinders 2(103 )2  (2.7 103  103 )  9.8
separated by dr and having radius r is  v  4.6m / s
9  8 104
d 6. (1) A stream lined body has less resistance
f   2rL r
dr due to air. Here cigar shaped body will have least
resistance.
3 d
The torque    2Lr 7. (4) Retardation
dr
Fb  mg VDg  Vdg Dg
r 
   g
dr 2L m Vd d
r  d
R2
3
 2 (i)
 D  g(D  d )
a  g   1 
d  d
R1 0
1 2L
R r 3 dr    d  (ii) The time at which the body comes to rest is
2 2
u
t
Dividing the two equations we get a
(u - initial value)
1 1

r 2 R22   2 2 gh d 2h  d 
 t   
1 1 2 g (D  d ) g  Dd 

R12 R22
g
8. (1) Given that a 
2
1 1
2
 2
 r R2 ma  mg  Fb  f
1
2 1 1
2
 2 g
R1 R2 m  mg  Fb  f
2
1 1  R2 R2 g
  2  2  2  2 1 2 2 m  Fb  f
 R1 r  R2  R1 2
Viscosity 273
g 12. (4) The rate of liquid flow is

V  Vg  6rv
2
 Pr 4
Q
  8l
Vg     2
v 
2   r g (  2)
6r 9 Let P1 & P2 are the pressure difference across
the tubes.
9. (1) KE = 0, since v = constant
P  P1  P2 (i)
1 2
KE = mv
2 As the flow of liquid is same in both tubes
Mass is falling down so potential energy
Q1  Q2
decrease at rate
 dh  r
4
 mg    mgv [decreases] P2  
 dt  Pr 4
 2   P  P2
1
 1 (ii)
10. (4) After shaking different particles take 8l l 2
8
different times to settle down. Here all the 8
particles have to travel same distance. We assume
that the particles reach terminal velocity From eq’s (i) & (ii)
immediately but they have different values
P 2P
because of different radii. P1  & P2 
3 3
s 13. (3) Consider the cross-section view of the
We know the terminal velocity v 
t tube.
s 2 r 2 (   )g 9s 
   r2 
t 9  2t (    ) g
Given s  2  102 m, t  3600s
   1  102 poise  1  103 kg m1 s 1
Substituting given values, we get

9 2  102 1 103 The small rate of flow of liquid across the
r2   
2 3600 (1.8  103  1 103 )  10 circular of radius r and width dr is
100
 r  106 m  1.77  106 m dQ  (dA)v
32
Diameter D  2r  2 1.77 m  3.54 m  r3 
dQ  (2r )drv  2v0  r  2  dr
 R 
11. (1) The condition for terminal speed (vt ) is
Weight= Buoyant force +Viscous force R 1 R 
Q  2v0   rdr  2  r 3 dr 
0 R 
 v1 g  v2 g  kvt2 0
Vg (1  2 )  R2 1 R4  v0 R2
 vt   2v0   2  2
k 2 R 4
274 Viscosity
3. (1) The viscous force on the cylinder is
1. (1) The profile of the velocity of fluid is as dv v0

F  A  (2Rh)
shown in the figure. dx R  R'
F v
  (2R)
h R ' R
4. (4) The top view of the disc is shown in the
figure. Consider a circular slice having radius r
and with dr. The linear velocity of any point on
the slice is r.
The velocity of the layer of fluid, which is in
contact with metal plates (fixed), is zero. As we
move towards the centre from either plate, the
velocity of the layer of fluid increases and it
becomes maximum at the
location of moving plate.
2. (2) Case I:
The small viscous force on the circular slice is
v r
df  dA  (2rdr )
D D
Here viscous force acts on the plate on both The torque

sides 2r 3 dr 2 R 3
d   dfr   r dr
Drag force = F1  F2 D D 0
R 4
 v v  4 vA 
 0  0 A 0 2D
 h/2 h / 2  h
5. (2) From Stoke’s law
Case II:
Viscous force F  6Rv
hence F is directly proportional to radius &
velocity.
6. (4) Let the radius of bigger drop is R and
Drag force on the plate that of smaller drop is r then
4 3 4
 v v vhA R  8   r 3
   A 3 3
 h y y y (h  y )
or R  2r (i)
As drag forces are equal
Terminal velocity, v  r 2
40 vA vhA
  2
h y (h  y ) v ' R 2  2r 
    4
v r2  r 
40 y  y
  1  
h  h  v '  4v  4  8  32cm s 1
Viscosity 275
7. (3) If v is the terminal velocity, then the Where  is the coefficient of viscosity of the
force, liquid
xg  yg  6rv 2 R12 (1  ) g 2 R 2 (  ) g
 vt1  and vt2  2 2
9 9
( x  y) g
 v
r 6 According to the given problem, vt1  vt2
( x  y) R12  2  
 v
r R ( 1   )  R ( 2   ) or R 2    
1
2 2
2
2 1
8. (1) Initially the terminal velocity V of sphere Substituting the given values, we get
of radius a is
R12 (11 103  2  103 ) 3
Weff  6 aV  
(i) R22 (8  103  2  103 ) 2
(Weff  weight  buoyant force) R1 3


As the radius is doubled, mass is increased to 8 R2 2
times and new terminal velocity will be
11. (1) In both cases the sphere travels same
8Weff  6 2 aV ' (ii) distance. The viscous force in the first part
gradually increases from zero and reaches
From (i) and (ii) V '  4V maximum. In the second part the the maximum
9. (1) When the ball is just released, the net viscous force acts throughout the motion. So
work done in second part is more than the work
force on ball is ma  mg  Fb
done in the first part.
The terminal velocity ' vt ' of the ball is attained W2  W1
when net force on the ball is zero.
12. (1) Viscous force on a falling sphere in a liquid
mg  Fb  6rvt
(2)r 2 (  ) g
F  6rvt where vt  is terminal
9
2
When the ball acquires rd of its maximum velocity,   density of sphere,   density of
3
liquid.
velocity v f
Rate of production of heat  Fvt
 2v 
ma '  mg  Fb  6r  t  2
 3   2 r 2 (   ) g  8  g 2 (    )2 5
 6r  
 27 r
9   
2
ma '  ma  (ma)
3 dQ
  r5
dt
a
a' 
3 r
1 1
13. (2) Given, l1  l2  l and r  2
10. (2) The terminal velocity of the spherical body 2
of radius R, density  falling through a liquid As the tubes are connected in series.
of density  is given by
4
2 R2 (  ) g Pr 4 P r 4 P r 
vt  Q  1 1  2 2  1   2   16
9  8l 8l P2  r1 
276 Viscosity
 P1  16 P2
Since both tubes are connected in series, hence 1. (6.25) The condition for terminal velocity is
pressure difference across cmbination,
V 1 g  6rv  V 2lg
P 32
P  P1  P2  2  P1  1  P1   1.88m
16 17  Vg (1  2 )  6rv
14. (2) The rate of flow of liquid is
Vg (  l1 )  61rv1
Q  Q1  Q2  Q3
Vg (  l2 )  62 rv2
4 4 4 4
Pr Pr Pr Pr
  
8L 8L1 8L2 8L3 1v1    l1 
 
2 v2    l2 
1 1 1 1
  
L L1 L2 L3    l2   1v1 
v2     
   l
L1 L2 L3  1   2 
L
L1 L2  L2 L3  L3 L1
(7.8  1.2) 10  8.5 104
 
(7.8  1) 13.2
 v2  6.25 104 cm / s
2r 2 (d1  d2 ) g
1. (0.1) Terminal velocity, vt 
9
2. (2) Given that vt1  vt2
vt2 (10.5  1.5) 9
  vt2  0.2  2 r2 2 r2
0.2 (19.5  1.5) 18 g [1  w ]  g [ 2  l ]
9 w 9 l
 vt2  0.1 m / s
w 1  w 3.2  1 1
   
2. (0.2) Terminal speed of spherical body in a l 2  l 6  1.6 2
viscous liquid is given by
3. (4)
2
2r (  ) g
vt  4
9 Mass  Volume  Density  M   r 3  
3
where   density of substance of body and As the density remains constant
  density of liquid.
 M r3
From given data
1/3 1/ 2
r M   M  1
vt ( Ag )  Ag    1  1     (i)
 r2  M 2   8M  2
vt (Gold ) Gold  
10.5  1.5 2 r 2 (  ) g
 vt ( Ag )   0.4  0.2m / s Terminal velocity, vr 
19.5  1.5 9 
Viscosity 277
2
vt1 r 
 vr  r 3 ,   1 
vt2  r2 
 pr 4
1. (4) Q
2 2
8 L
v  r1  1 1
   or    [From (i)] The rate of flow is same in both tubes
nv  r2  n 2
p1r14 p2 r24
 n4 
L1 L2
p1r14 4 p1r24 r4
  r24  1
1. (3) When the pebble is dropped from the top L1 L1 16
of a tall cylinder, filled with viscous oil the pebble 4
acquire terminal velocity (i.e., constant velocity)
after some time. r1
r2 
2
b c
1. (3, 4) In gases the molecules widely spaced as Q  S    1 

2. (4) Let a    
compared to liquid. Hence, the intermolecular A  h   g 
forces are weak. The increase in temperature Using dimensions
causes the rise of the energy of gas molecules, a
MT 3  [ML1T 1 ] [LT 2 ]b [M1L2 T 2 ]c
which increase the randomness of molecules.
Which results retard the motion of gases, so or, MT 3  [M a-c L a  b  2c T  a- 2b  2c ]
viscosity increases. In liquids, the molecules are Equating powers and solving
loosely packed as compared to gases. The we get, a  1,b  1,c  0
intermolecular attraction is strong.
Q S
When temperature increses the energy of  
A h
molecules increases. It causes a decrease in
inter molecular attraction between them, which
reduce viscosity.
1. (4) The acceleration of the body is
1. (4) The acceleration of the body is maximum  K

a  g  v    
initially and decreases to zero when the body  m
reaches terminal velocity.
dv
 g v
dt
v t
1. (4) As the ball ascends upwards in addition dv  g v 
to weight, air friction also acts downwards.
   dt  ln 
g v 0  v 
 t
0
Hence the initial magnitude of acceleration will

be greater than magnitude of acceleration due dv
v = v0 (1  et ), a   v0  et
to gravity. So the only possible option is (4) dt
278 Viscosity
g 3 N
where v 0  3. (3)   102 poise  10 s
 m2
2. (3) The forces exerted on the steel ball are 18000
shown in the figure. v  18km / h   5m / s
3600
l  5m
v
Strain rate 
l
shearing stress
Coefficient of viscosity,  
where Fg  Fb is the effective force on the strain rate
sphere.  Shearing strees    strain rate
6 rv  Fg  Fb (i) 5
 103   103 Nm2
(ii) 5
6 rv'+Fg  Fb  2( Fg  Fb )
From the above two equations

v'  v
Surface Tension 279
10. (3) As the pressure in smaller bubble is greater

than bigger bubble so air flows from smaller to
 1
larger bubble  P  l 
 
1. (1) Weight of spiders or insects can be 11. (3) As both bubbles have same radii the
balanced by vertical component of force due to pressure difference across them is same. So after
surface tension. removing the valve there will not be any change.
2. (2) 12. (2) The pressure difference across the air
3. (1) Because film tries to cover minimum 2T
surface area. bubble is P 
r
4. (4) 2 rT
1 cos  2 r2T cos   mg P  Pin  Pout , Pout  P  hdg
mg  The pressure inside the soap bubble is

 T ,
2 (r1  r2 )cos 2T
Pin  P  hdg 
r
here   0, mg  7.48  103  10, r1  8  102 ,
m
13. (2) PV  RT i.e., m  PV
r2  9  102 M
Substituing T  70  103 N / m m1 PV
 1 1
m2 PV
2 2
5. (1)
6. (2) Both liquids water and alcohol have same 4T 4T
p1  p  , & p2  p 
nature (i.e., wet the solid). Hence angle of R1 R2
contact for both is acute.

 4T  4 3  4T  3
7. (3)  P    R1  P   R1
 R1  3  R1 
 
8. (2) W  2  4 R2  T , R is increased by a  4T  4 3  4T  3
P
     R2  P   R2
factor of 2. So, W is increased by a factor of 4.  R2  3  R2 
9. (4) 14. (2) If angle of contact is 0, then meniscus

4 4 3 4 4 3 shape is hemispherical.
V   R3 ;2V   R' ;2   R3   R' ;
3 3 3 3 15. (1) As net rised remain constant so the slanting
length of the liquid column in the capillary tube
R '  21/3 R
is more than the height H.
2
W '  2(4 R ' T )  22/3  2  4 R 2T  41/3W 16. (1)
280 Surface Tension
1 and (3) are not possible. In case (1) attraction

17. (2) h . between solid molecules and liquid molecules is
R
more than the attraction between the liquid
2T cos molecules.
18. (2) h
rdg Adhesive > cohesive
Where as in case (4) attraction between liquid
1
 h molecules is more than attraction between liquid
g molecules & solid molecules
ge Cohesive > adhesive.

As g m 
6
 hm  6he .
1. (2) Surface tension tends to make the area of
2T cos rain drop minimum.
19. (3) Height, h  r  g 2. (4) Soap helps to lower the surface tension
of solution, thus soap get stick to the dust particles
2  Tw  cos0 and grease and these are removed by action of
 For water, hw  r  1 g water.
3. (1)
And, for mercury,
4. (4)
2  Tm  cos135
hm   5. (3) Force on each side  2TL [due to two
r 13.6  g
surfaces]
hw 2  Tw 1 r 13.6  g  2  Force on the frame  4(2TL)  8TL
  
hm r 1 g 2  Tm 1
6. (4) Let R be the radius of the biggest
[ cos135  1/ 2] aluminium coin which will be supported on the
surface of water due to surface tension.
10 Tw Then, mg  S  2R
   13.6  2
3.42 Tm
 R 2 tg  S  2R
Tw 10 1
    R  2S / gt
Tm 3.42  13.6  1.414 6.5
7. (3)
Tw : Tm  1: 6.5. 8. (3) Tangent drawn at point of contact makes
20. (3) The pressure difference across the 90 with wall of container..
surface is 9. (3)
2S 10. (2)
Patm  PA 
R
11. (4)
2S 12. (2) Surface energy = surface tension 
PA  Patm 
r increment in area  T  A
21. (4) Liquid having positive capillary action 13. (4) Effective area  2  0.02m2  0.04m2
have concave meniscus while having negative
capillary action have convex meniscus. So (2) Surface energy  5Nm1  0.04m2  2 101 J
Surface Tension 281
20. (1)
14. (2) P  4T
r
21. (3) When no gravity or insufficient length the
15. (3) Because energy is liberated. liquid level rise upto top. When capillary tube
16. (2) Soap bubble has two surfaces kept in a place no gravity or artificial satellite, or
freely falling lift or insufficient length the liquid
Total length 2(2 r) rises upto top.
F  (4 r)  T 22. (4)
17. (2) Difference of pressure between inside 23. (2) When no gravity or freely falling lift then
and outside of the spherical drop is given by liquid rises to top. When capillary tube kept in a
place no gravity or artificial satellite, or freely
2T falling lift or insufficient length the liquid rises
P 
R upto top.
Given, R  1 mm  1  103 m
2T cos
24. (2) h
T  70  10 N m 3 r g
2  70 103 25. (2)

P   140 N m
1103 26. (4) Because of large force of cohesion or
When capillary tube kept in a place no gravity
18. (3) Given that P1  0.01P ; P2  1.02 P
or artificial satellite, or freely falling lift or
4T insufficient length the liquid rises upto top.
P1  P  0.01P  (i)
r1 27. (1) In the condition of weightlessness, water
4T rises to the whole of the available length.
P2  P  0.02 P  (ii)
r2 28. (2) The pressure difference across the surface
2T
r2 1 is PA  Patm 
From (i) & (ii) r  2 R
1
R - radius of the meniscus
3
V1  r1 
The ratio of volumes is   8 R  r (For   0)
V2  r2 
m 2T
19. (4) From, PV  RT i.e., m  PV PA  Patm 
M r
4T 4T 29. (4) The pressure difference across the

P1  P  0 ;
R1 R1 meniscus is
4T 4T 2T
P2  P  0 PA  Patm 
R2 R2 r
 4T 4 3 2T
PA  Patm 
   R1 r
m1 PV R 3
 1 1  1
m2 PV
2 2  4T 4 3
   R2
 R2 3
1. (2) The ring is in contact with water along its
m R2 inner and outer circumference. Total force on it
 1  12
m2 R2 due to surface tension will be
282 Surface Tension
F  T  2 r1  2 r2   105  1.5 105  4 105  15 103

5
 105 (1  1.5  4  0.15)  6.65  10 Pa
mg
 T
2  r1  r2  5. (2) Net force = Average pressure  Area
T  2 R
3.97  980 h
 
2  3.14   8.5  8.7    P0   g  (2Rh)  T 2R
 2
 36.02 dyne/cm  2 P0 Rh  R  gh 2  2 RT
2. (2) F  mg  2Tl 6. (3) From conservation of volume of the liquid
F  l g 2Tl 4 3 R
R  N (4R 2 )
3 1000
F   gl
T  0.041N / m  N  1000
2l
i.e., 1000 bubbles can be blown from the drop
F = 5mN,   1.75  103 kg/m, l  50 mm
W  T (A)
3. (2) The free-body diagram of wire is showing
in figure. A f  1000(2  4R 2 )
Ai  4R 2
W  T (2000  1)4R 2  8000T R 2
7. (4) As surface energy = Surface tension 

surface area
y i.e., E  S  2A
sin   tan  
a
New surface energy, E1  S  2( A / 2)  S  A
Let l be the length of the wire. If  is mass per
unit length of wire, the weight of wire E  E1
% decreases in surface energy  100
lg  2Tl sin  (acting vertically downward) E
2SA  SA
 g  a g  100  50%
 T 2SA
2sin  2y
8. (3) The radius of the bubble is given by
4. (3) In side the air bubble the pressure =
Pressure due to atmosphere + Pressure due to
upper liquid layer + Pressure due to lower liquid
layer +Pressure due to surface tension.
2T
 P0  h1 1 g  (h  h1 )  2 g 
r
 105  15  103  10  (20  15)8000  10

From the figure R cos60  r
2  7.5 102
 The radius of the bubble is R  2r  2cm
0.01103
Surface Tension 283
The pressure difference is q R

V  (iv)
4T 4  0.05 4 0 R 0
P    10Pa
R 2 102
9. (2) Although not given in the equation, but From Eq’s (i), (ii), (iii) & (iv)
we have to assume that temperatures of A and
B are same.  0V 2 R
we get P0 ( R 3  r 3 )  4T ( R 2  r 2 )  0
2
11. (4) The force exerted by the blown air on
the bubble is F   AV 2   ( b2 )V 2
when the bubble detaches
nB pBVB / RT pBVB
 
nA p AVA / RT p AVA
( p  4T / rB )  4 / 3 (rB )3
 (T = surface tension)
( p  4T / rA )  4 / 3 (rA )3
Substituting the values, we get The surface tension force is

nB 2
 b  2 b T
 3.2 Fx  F sin   (2 b)T   
nA R R
10. (1) The pressures on the bubble are shown
2T
in figure The radius of the bubble is R 
V 2
12. (3) Let P1 and P2 be the pressures inside

the two bubbles as shown in figure
Let S is the coefficient of surface tension

pressure difference across small bubble
4T
P1  P0  (i) 4S 4S
r P2  P   P2 
2R 2R
 2 4T Pressure difference across large bubble
P2  P0   (ii)
2 0 R 4S 6S
P1  P2   P1 
From isothermal condition, R R
3
Let n1 and n2 are the number of moles in the
PV
1 1  PV
2 2 Pr
1  P2 R 3 (iii)
two bubbles n is the number of moles in the bigger
The potential of the sphere is bubble after the first bubble bursts.
284 Surface Tension
n1  n2  n 2  5.06 102
1105  x  (0.11  x)
PV PV PV 1105
1 1
 2 2 (T = constant)
RT RT RT Solving for x, we get x  0.01m
1 1  PV
PV 2 2  PV Length of capillary to be immersed = 0.01 m
6 S  4  4S  4 
  (2 R ) 3  4  R 3   4 S 4  r 3 15. (1) The capillary raise
  R  
3
R  3  2 R  3 3  r 3
2T cos 2T cos 2T cos
r  5r h , x , y
 rg  rg  rg '
13. (3) Resultant of weight of liquid and surface
tension makes the orientation of liquid in the pipe x g'

as in option (3). y g
14. (3) Let P0 be atmospheric pressure and P GM  d

g' ( R  d )  g 1  
the pressure of air within the sealed tube. Then R 3
 R
for equilibrium pressure just below the meniscus
x  d
should be equal to atmospheric pressure because  1
y  R 
levels of water inside and outside the tube is
same. 16. (2) The pressure of the water changes
linearly with the increase in height. At the bottom
of the meniscus it is equal to the external
atmospheric pressure P0 , and at the top to
P0   gh. The average pressure exerted on the
wall is Paverage  P0   gh / 2.
 2T  2T
i.e.,  P    P0 or P  P0 
 r  r
If L  0.11m is the length of tube and x is the
length of immersed part, then from Boyle’s law
PV
1 1  PV
2 2
Consider the horizontal forces acting on the
 2T  volume of water enclosed by the dashed lines in
P0 LA   P0   ( L  x) A
 r  the Figure.
where A is the cross-sectional area of tube, The wall pushes it to the right with force F1 ,
 2T  the external air pushes it to the left with force
i.e., P0  L   P0   ( L  x)
 r  F2  lP0 h, and the surface tension of the rest of
2T the water pulls it to the right with a force F3  lT .
P0  L  P0 ( L  x)  ( L  x) The resultant of these forces has to be zero,
r
2T  1 
i.e., P0 x  ( L  x) That is,  P0   gh  lh  lT  lP0 h
r  2 
Surface Tension 285
2T 2  0.073 T (2l )   (dlh) g

h   0.0038m.
g 1000 10
2T
h (ii)
Water rises by approximately 4 mm up the wall  gd
of the aquarium.
2T 2l
17. (4) The plates attract due to pressure From (i) & (ii) F   gd 2
difference between water lying in between the
plates and outside the plates. The force on a 18. (4) Let R be the radius of the meniscus
plate is F   P0  Pavg  lh formed with a contact angle  . By geometry

P0  atmospheric pressure this radius makes an angle   with horizontal
2
Pavg  average pressure that acts on the plate by
the liquid
P1  P2
Pavg 
2
P2  Pressure on the surface of the water
P1  Pressure at the highest point of the liquid
T
P0  P1   (cylindrical surface)
r
2T
 P1  P0 
d
  b
cos     
P2  P0  2 R
( P0 )  ( P0  2T / d ) T 2S
Pavg   P0  P0  P1  & P2  P1   gh
2 d R
From the above equations
  T  Tlh
F   P0   P0    lh 
  d   d  
2S cos    
(i) 2S  2
h 
 gR  gb
The height of the liquid is given by
1. (3)
FST  Fg
286 Surface Tension
Weight of metal disc = total upward force  Surface energy of all droplets
= upthrust force + force due to surface tension  S  1000  4r 2  S  1000  4( R / 10)2
= weight of displaced water  T cos [2 r ]  10( S  4R 2 )  10U
 W  2 rT cos 9. (3) W  2T A
2. (3) Effective length  2 r  2 R . W  2T 4 (5)2  (3)2   104
The surface tension force is F  T 2(r  R)
 2  0.03  4 25  9 104 J
3. (4) 2TL  mg (Since film has two surfaces)
 0.4103 J  0.4mJ
2
mg 1.5 10 1.5
T    0.025 N /m 10. (1) The pressure difference across the soap
2L 2  30 102 60 bubble is
4. (3) The force exerted by film on wire or
4
thread depends only on the nature of material of Initially Pi  P1  (i)
the film and not on its surface area. Hence the r
radius of circle formed by elastic thread does After pushing the piston
not change.
4
5. (4) Suppose tension in thread is F, then for Pf  P2  (ii)
(r / 2)
small part l of thread
As the bubble is compressed isothermally
3
4 4 r
Pi  r 3  Pf   
3 3 2
Pf  8 Pi (iii)
From the Eq’s (i), (ii) & (iii) we get
24
P2  8P1 
r
11. (2) As number of moles remains constant
l  R and 2F sin  / 2  2T l  2TR
n  n1  n2
   PV PV PV
 sin    1 1 2 2
 2 2 RT RT RT
TR TR PV  PV
1 1  PV
2 2
 F   2TR
sin / 2  / 2
 4T  4 3  4T  4 3  4T  4 3
6. (1) Cohesive force decreases so angle of  P    R   P    R1   P    R2
 R 3  R1  3  R2  3
contact decreases.
P  R13  R23  R3   4T  R2  R12  R22 
7. (2) W  [2  4 (3r ) 2  2  4 r 2 ]T  64 r 2T
8. (1) Given, U  S  4R 2 ; when droplet is 12. (1) The angles between any two tangents
splitted into 1000 droplets each of radius r, then which are drawn to the interfaces are equal
4 3 4       120
R  1000  r 3 or r  R /10
3 3 13. (2)
Surface Tension 287
14. (3) The drop that is formed is shown in the 16. (4) When radius is decreased by dr
figure
Decrease in surface energy = Heat required for
vaporisation
2dAT  dV L
2T
(4rdr )  T  2  4r 2 drL  r 
L
17. (2) The given soap bubble is
The free body diagram of the drop is
Let P1 , P2 & P0 are the pressures shown in the

figure
4T
Fy  F sin   (2 rT ) sin 
P2  P0  (i)
3cm
2
 r  2 r T 4T
  2 rT     P1  P2  (ii)
R R 1cm
15. (1) The bubble will detach if the upward From Eq’s (i) & (ii) we get
buoyancy force is balanced by the downward
surface tension force.  1 1  16T
P1  P0  4T     3cm
 3cm 1cm 
The radius of the single soap bubble that
maintains the same pressure difference is
16T 4T 3
  r  cm
3cm r 4
18. (1) The free body diagram of the water that
raises is given as
4 
(W )  R 3  g  (T )(2r )sin 
3 
r
 sin  
R
2W R 4 g 2W g
Solving, r   R2
3T 3T F1  F2  Fg
288 Surface Tension
F1  2 r1S & F2  2 r2 S
Fg   (r22  r12 ) hg
2 (r1  r2 )s   (r22  r12 )hg
2S
we get h   g (r  r )
2 1
mg  4aT
h
2S 2s a2  g
19. (1) PA  P0   h  g and  h g
r r 23. (1) Let P0 is the atmospheric pressure and
The pressure inside a concave meniscus is less P1 and P2 are the pressures just below the
than the pressure outside (atmospheric). surfaces. As the liquid is in equilibrium the
Assuming the meniscus to be spherical (as for pressure is same at everywhere in the horizontal
thin capillaries), excess pressure is 2S/r, where r tube.
is the radius of the hemispherical surface.
20. (2) When the capillary tube is tilted by an angle
 with the vertical, the capillary rise to distance
l in the tube will be such that the meniscus will
remain at the same vertical height above the
level of water in the container.
2T cos
P0  P1  (i)
r1
2T cos
P0  P2  (ii)
r2
Hence, h  l cos PL  PR &   0
h 10 P1   gh1  P2   gh2
l   10 2cm
cos cos45 2T 2T
P0    gh1  P0    gh2
r1 r2
2T cos
21. (2) h , Height in capillary depends
rdg 2T  1 1 
h2  h1    
upon the surface tension of the liquid.  g  r2
r1 
Surface tension of soap water solution is less
than water. So, height will be less in second case. 1.0 1.5
r1  mm & r2  mm
2 2
22. (2) At equilibrium mg  FST  Fb
T  0.0736N / m
mg  4aT  a 2 h  g  h2  h1  10mm
Surface Tension 289
24. (1) The top surface of the liquid is curved

r
and the pressure difference across the surface R
cos
is
T 2S 2S cos
P0  P  We know that P0  P1  
r R r
where r  d / 2 P2  P0  P2  P1   gh
2T 2S cos
 P0  P    gh
d r
Note : Pressure difference (P) When the tube is long enough
2T 2S cos1
P  (spherical surface)   gh1 (i)
r r1
When the tube is having smaller length
T
P  (Cylindrical surface) 2S cos2
r   gh2
r2 (ii)
25. (2) Here S  7.0 10 N / m;
2
cos 1 h1
3 3

3
r  0.2  10 m,   10 kg / m cos 2 h2
We know that the height of liquid rise in the tube 1  0 h1  7cm & h2  5cm
is h  2S cos / rg
h2 5
where r is the radius of the tube and  is the cos  2    2  cos 1  
h1 7
angle of contact.
2(7.0  102 )
h   7 cm
(103 )(0.2  103 )(10)
1. (3) Sum of volumes of 2 smaller drops
When the length of the capillary tube above the = Volume of the the bigger drop
free surface of the liquid is less than the height
of liquid that rise in the tube, radius of the free 4 4
2  r 3  R3  R  21/3 r
surface is not equal to the radius of the tube. 3 3
It is greater than r as the surface tends to be Surface energy
flatter.
 T  4R 2  T 422/3 r 2  T  28/3 r 2
2. (4) Water wets glass and so the angle of
contact is zero. Neglecting the small mass the
capillary rise,
2 rT   r 2 h g
2T 2  0.07
or h  r  g  0.25 103 1000  9.8  5.7cm
But here the tube is only 2 cm above the water

and so water will rise by 2 cm and meet the
From the figure r  R cos  tube at an angle such that
290 Surface Tension
2 rT cos    r 2 h '  g 2T 2T
 h '  g or R 
R h'g
 2T cos  h ' r  g
Here h '  25mm  25  103 m
h 'r g
 cos 
2T
2  73 103
2 3
2 10  0.25 10 1000  9.8  R  0.6 103 m  0.6mm
cos  25 103 103  9.8
2  0.07
3. (1) Let A be the circular area over which the
   70 liquid wets the plate and d be the distance
3. (3) We know that capillary raise between two plates. Mass of liquid drop,
2T cos 
m  Ad . If S is the force of surface tension of
h water, then excess of pressure inside the liquid
 rg
film is given by
h1 T1 cos1 2
 
h2 1 T2 cos 2
1
140  S S 2S
h2 2 1 p  
or  or r d/2 d
h1 2 70  1
Force of attraction between the plates,
h 6
h2  1  cm  3cm
2 2 2S  F
F A  p  A 
d  
2S 2Sm
1. (3) Water fills the tube entirely in gravity less F  Ad  2
d 2 d
condition i.e., 20cm.
2. (1) In the capillary tube, the water should rise 2  0.07  (80  106 )
  0.28 N
2T 103  (4  108 )
to a height h  r  g
Here T  73 103 N / m
1. (3) The menicus of liquid in a capillary tube
0.50mm will be convex upwards if angle of contact is
r  0.25  103 m ; obtuse. It is so when one end of glass capillary
2
tube is immersed in trough of mercury.
  103 kg / m3
2  73 103 1. (1) Surface tension decreases with increase

 h  59  103 m  59mm
0.25 103 103  9.8 in temperature. The approximate relation with
temperature is T  C1  C2 t
Now, h  h ', So the contact angle and radius of
the surface changes. Where T- Coefficient of surface tension
If R is the radius of meniscus, then we have t - temperature.
Surface Tension 291
The pressure difference between inside and

outside the liquid is
4T 1
1. (2) P  P  1 1 
r r P  T   
r R
Where r is the radius of the bubble.
1 1 1
P is the pressure difference between inside As r  R   
r R R
and outside pressures.
As radius of soap bubble increases with time T
P 
1 R
 P  .
r 3. (1) Let N be the number of drops each having
2. (3) The capillary rise h is given by, radius ‘a’. When they coalesce the total volume
remains constant
2S cos
h
 gr 4 4
N  a 3   b3
hr  constant for a given glass and liquid. 3 3
The correct graph is hyperbolic in shape. b  aN 1/3

The change in surface area is
A  N 4 a 2  4 b2
1. (Delete )
4 
d  r 3 
dV 3 A  4  Na 2  a 2 N 2/3   4 a 2  N  N 2/3 
 const.    K
dt dt
The change in K.E  T A  4 a 2T  N  N 2/3 
4 dr dr
  3r 2   K  4r 2  K
3 dt dt 1 4 3 2 1 4 3 2
   b V   N   a V
13 2 3  2  3 
3K  3Kt 
 r3  tCr   C
4  4   4 a 2T  N  N 2/3 
The pressure inside the bubble is
6T  1  6T  a 
V  1  N 2/3    a 1  b 
4S a    
P  P0 
r
6T  1 1 
4S V  
 P0  13   a b 
 3Kt 
 4  C  (b  aN 1/3 )
* None of the given options is correct. 4. (1) Let h is the depth of the bubble. Pa is the
2. (4) Let r be the radius of the liquid coverage atmospheric pressure. The pressure inside the
bubble.
P2  Pa  1100 (Given) (i)
We know that
292 Surface Tension
2T 2. (1) As the number of moles remain constant

P2  P1  (ii)
r
n  n1  n2
P1  Pa  gh (iii)
PV PV PV
 1 1 2 2
Where P2 & P1 are the pressures outside & RT RT RT
inside the bubbles
PV  PV
1 1  PV
2 2
 4T  4 3  4T  4 3  4T  4 3
2T  P0    R   P0    r   P0    r
gh   1100  R 3  r 3  r 3
r
8 4  4T
h 103  4.8  1100 
2  0.06 P0   r 3   R3    4 R 2  8 r 2 
 980 3 3  3
0.1102
4T
980 PV
0  S
h  0.1m 3
9.8  103
3PV
0  4TS  0
3. (3) The capillary rise
1. (2) The height rised by the liquid is
2T cos
h
2T  rg
h
rg
When water is on a waxy or oily surface
2T cos is negative i.e., 90    180
On the surface of the earth x  rg
i.e., angle of contact  increases
At a depth d below the surface of earth the And for   90 liquid level in capillary tube fall
height rised is
i.e., h decreases.
2T
y 4. (2) Surface tension of a liquid decreases with
rg '
the rise in temperature. At the boiling point of
liquid, surface tension is zero.
x g'

y g 2T cos 
Capillary rise h 
rdg
GM  d
where g '  [ R  d ]  g 1   As surface tension T decreases with rise in
R3  R
temperature hence capillary rise also decreases.
x  d
 1
y  R 
293
Thermal Expansion of Solids & Liquids 293
40  0 T  10

100  0 130  10
40 T  10
 or
1. (4) 100 120
2. (1) 200
5T  50  240  T   58.
3. (2) 5
4. (4) The temperature is measured by the value 8. (2)

of the thermodynamic property of a substance 9. (4)
i.e., the property which varis linearly with the
temperature. 10. (3) On heating the system; x, r and d all
increase, since the expansion of isotropic solids
5. (2) The temperature scales are shown in the
is similar to true photographic enlargement.
figure
11. (1) The strip on cooling bends in the form of
an arc with less contractable metal (K) on outer
side
12. (2) If the junction does not not shift then
force exerted by both rods must be same.
l
From the scale it can be written that F  AY  AY T
l
100C  0C 90cm  50cm YAB  AB
 Y11  Y2  2   BC   1.2  105 / C
x0 60cm  50cm YBC
 x  25C
FL
C F  32 13. (1) Y where Y is young’smodulus, A
6. (4)  AL
5 9
is area
C F 9
   F  C
5 9 5 YAL
F  (i)
F  54 F L
7. (4) The given two scales are From the formula for linear expansion
L
 (ii)
L 100
According to the condition the bar should not

bend or expand
294 Thermal Expansion of Solids & Liquids
Now from equations (i) and (ii) 18. (3) Thermal expansion of isotropic bodies is
independent of shape, size and availability of
F  YA  100 hole/cavity.
Hence, force is independent of length L. 19. (2) The change in volume is
l pV pV p
14. (2) Strain   αθ dV    V T   T 
l B B B
Stress=Y  Strain  Yα θ 20. (1) Water has maximum density at 4C , so
 Force or tension, if the water is heated above 4C density
T  Stress  Area  YAα θ decreases, i.e., volume increases. In other words,
it expands so it overflows in both the cases.
πYαd 2 θ  πd 2 
  A 
4  4 
π  2  1011  105  104  25

T
4
 3926 N  4000 N
15. (1) The relation between fractional change
in area and length of a body is A  L2 21. (1)
22. (3) Given that
A L
 2
A L  r  1.5  1.5(2)
A  r  3
  2  2  4%
A So volume of liquid does not change.
16. (3) From conservation of angular momentum, 23. (2)
I 00  I 
24. (3) Body is in equilibrium, buoyancy force will
2 2 2 2 balance the weight of the body.
Now I 0  ma0 and I  ma . Hence
5 5 mg  FB
2 The gravitational force does not change with
 I0  a 
   0   0  0 where a is the radius of change in temperature. So Buoyancy force
I  a remains constant. By increasing temperature
the sphere at T. density of liquid decreases so volume of body
inside the liquid increases
17. (3) Density of substance is d  d 0 1  γθ 
25. (4)
9.7  10 1  γ100 
 γ  3  10 4 1. (4)
Coefficient of linear expansion is 2. (3)
γ 3  104 3. (1)
α   104
3 3 4. (4)
295
5. (1) Fl Al
6. (4) Y  F Y  YAT
Al l
C F  32 C 140  32
    C  60C.
5 9 5 9 F l
7. (3) The densest layer of water will be at 20. (2) AS the coefficient of thermal expansion
bottom. The density of water is maximum at of the brass is greater than steel. Hence, the
4C. So the temperature of bottom of lake will lengthof brass strip will be more than steel strip.
be 4C. Therefore, brass strip will be on convex side.
F  32 C K  273 x  32 x  273 l
8. (2)     21. (4) Thermal strain is
9 5 5 9 5 l
 x  574.25 l
   T
C F  32 25 F  32 l
9. (1)     F  77F .
5 9 5 9 Strees  Y  T
10. (2)
So stress is independent of length of the rod.
11. (1)
22. (2) Increase in length L  L0  T
12. (3)
13. (1) If we heat water present at 0C then it  10 10 106  100  0   102 m  1cm.
contracts until it reaches 4C. 23. (2) Moment of inertia of a rod,
14. (4) When hot water is poured into the glass,
the inner portion of the glass comes into the 1
I ML2 (i)
contact of hot water immediately. As a result 12
the inside of the glass expands faster then the Where M is the mass of the rod and L is the
outside of the glass and hence causes the glass length of the rod
to break.
1
15. (1) A bimetallic strip on being heated bends  I  2MLL (  M is a constant) (ii)
in the form of an arc with more expandable 12
metal (1) outside (as shown). Divide (2) by (1), we get
I L
2 (iii)
I L
L
As L  L  T or   t (iv)
L
16. (3) The strip will bend with metal of greater I
 2t.
 on outer side i.e., convex side. So I
 brass   Iron
24. (4) Initially (at 20 C ) length of composite
17. (3) Wood contraction is less than the metallic system L  50  100  150cm
tape. So lwood  30 cm.
Length of iron rod at 100 C
18. (3)
19. (2) Here the rod must be fixed between rigid  50[1  12  106  100  20 ]  50.048cm
supports on all sides. The strain developed is
thermal Length of aluminium rod at 100 C
 100[1  24  10 6  100  20 ]  100.192cm unchanged while due to thermal expansion

volume will increase and becomes V '.
Finally (at 100 C ) length of composite system
V '  V 1  T 
L '  50.048  100.192  150.24cm
where  is the coefficient of volume expansion
25. (2) When the ball is heated, expansion of ball of liquid
and cavity both occurs, hence volume of cavity
increases. m m 
  '  V '  V 1  T   1  T
26. (4) As, V  L3
V  100 L 100  '

 3  3  0.2%  0.6% Fractional change in density 
V L 
27. (4) The hole also expands.  '  1 

 1    1 
28. (3) Percentage change in volume  γt  100     1  T 
 3αt  100  300αt  300  2  105  200  1.2 T 49  105  30

 
29. (3) 1  T 1  49 105  30
30. (4) Conceptual 0.0147

  0.0145  1.5  102
31. (4) γ  α1  α2  α3 1.0147
40. (2) Apparent coefficient of volume expansion
13 107  231 107  231 107  475  107
32. (2)  app .   L   s  7 s   s  6 s (given  L  7 s )
33. (1) Water density is maximum at 4C. If we ratio of absolute and apparent expansion of liquid
heat or cool water present at 4C it expands.
 L 7 s 7
 
34. (1) When water is heated from 0C to 4C  app. 6 s 6
then its density increases and hence buyoancy
force also increases and tension in the thread 41. (3) It is given that the volume of air in the
decreases. flask remains the same. This means that the
expansion in volume of the container is exactly
35. (1) equal to the volume expansion of mercury,
36. (1)
i.e. Vv  Vm
37. (1) On heating density of liquid decreases and
hence buoyancy force decreases and the tension
c  Vc γc T  Vm γm T
in the spring increases.
38. (2) Since the vessel is partly filled, volume of
the vessel is greater than that of the liquid. Hence Vc γc Vc  3α
Vc    γ  3α 
on heating expansion of vessel will be greater γm γm c
than that of liquid, since  V   L . It means

unoccupied volume will necessarily increase. So, 2000  3  9  106
 Vm 
option (2) is correct. 1.8  104
39. (4) When the temperature of a liquid is
increased by T C the mass will remain Vm  300cc
297
4. (4) As, F  (T ) AY
1. (1) As the different in their lengths is same  1.1105  (50  30)  2 106  2 1011
Lcu  Ls  88N
LCu  Cu  LS   S  5 LC  2 LS (i) 5. (4) Here, l  80.3  80.0  0.3cm
5 l  80cm,  1.2  106 C 1

LS  LC
2
l
LS  LC  5cm (ii) Rise in temperature T 
l
2.5LC  LC  5 0.3
T   312.5C
80 1.2 105
5  2 10 6. (2) Fractional loss of time per second
LC    3.33 cm
3 3 1
  T
2. (1) The coefficient of linear expansion of the 2
rod as a function of x is Therefore
   1  1
  1   2  x;  (T0  15)  (24 hrs)  5
 L  (i)
2
Let dL is the small change in the length of the
1
small part of length dx, for a rise in temperature and  (30  T0 )  (24 hrs)  10 (ii)
2
of T
from eqn (i) and (ii)
dL  dxT
L
on solving T0  20C
L    dxT
0
7. (4) As the resistors are connected in parallel
   2   1  2  1 1 1
L 1  
 LT ; eff    Req R1 R2
 2   2 
3. (4) Due to heating the length of the wire 1 1 1
increases.  
Req R(1  T ) R(1  T )
 Longitudinal strain is produced
1 1
  2  (2)T 
L Req R
    T
L
R
Req 
1 1
E   Stress  Strain   Y  (Strain)2
 2  (2)T 
2 2 1
Req  R  2  (2 ) T   R  2  (2 ) T 
2
1  L  1 2 2
 E  Y      Y    T
2  L  2 Req  2 R 1  T 
[As   3 and T  T (given)]
 2R 1   eq T 
2
1   1
 E  Y     T 2   2YT 2
2 3
  18   eq  
8. (3) For distance between A and F to remain Similarly for tungsten plate
constant, extension in CD = extension in AB +
extension in EF  ( R   / 2)  l0 (1  tT ) (ii)
 l2  2l1  l2 2   2l11 To eliminate the unknown quantities, and l0 we

divide the equation (i) by (ii)
l1  2
  ( R   / 2) (1  CT )
l2 21  
( R   / 2) (1  tT )
9. (1) Let l11 & l21 are new lengths of the rods [2  (c  t )T ]
 R 
after increasing temperature T as shown in [2( c  t )T ]
the figure. 11. (4) After expansion as 15 cm & 12 cm
coincide then
15 cm(1  1T )  12 cm(1  2 T ) (i)
Let x is the mark on the first scale that coincides
with 10 cm mark on second scale
x(1  1T )  10 cm(1  2 T ) (ii)
As h remains constant
Eq (i) / Eq(ii)
2
l12 2 l1 15 12
l22   l21  1   x  12.5 cm
4 4 x 10
square on both sides. 12. (1) Fractional loss of time per second
2 12
l 2 l 1
l22  1
 l21  1   T
4 4 2
l12 l 2 (1  1T ) 2 Therefore

l22   l22 (1   2 T ) 2  1
4 4 1
 (T0  15)  (24 hrs)  5 (1)
l  2
 1 2 2
l2 1 1
and  (30  T0 )  (24 hrs)  10 (2)
10. (1) The length of copper plate at a 2
temperature T is lC  l0 (1   CT ) , where l0 is the from eqn (1) and (2)
length of copper plate at 0C . The length of the on solving T0  20C
tungsten plate is lt  l0 (1   t T )
13. (2) The power loss is given by
l
I 2 R   (T  T0 ); Strain    (T  T0 )
l
I 2 R (T  T0 )

 l  (T  T0 )
 
 l 
Length of copper plate lC   ( R   / 2)
I 2R  l I 2 R
Consequently,   
 l   l 
 
 ( R   / 2)  l0 (1  C T ) (i)  l 
299
14. (2) If the three rods are free to expand then 15. (4) For isotropic material separation between
their extensions are two points varies as
ls  l0  s T l  l0 (1  T )
la  l0  a T After heating, all the dimensions x, y & z

increase.
16. (2) After increase in temperature both
spheres volume increases. The C.M of sphere
A moves up and hence its gravitational P.E
increases. The increase in gravitational energy
comes from heat supplied. So a part of heat will
be utilised to rise in temperature in A and hence
its temperature change is low. In case of B
potential energy decreases. The decrease in P.E
should be utilised to increase temperature of B.
pV pV p
17. (2) dV    V T   T 
B B B
18. (4) Due to volume expansion of both mercury
fig (1)- Initial position of the combination before and flask, the change in volume of mercury
rising temperature. relative to flask is given by
fig (2)- Expected natural extensions in the rods V  V0  L   g    V0  m  3 g  
after rising temperature.
fig (3)- Final position of the combination after  50 180  106  3  9  106   38  18  0.153cc
rising temperature.
19. (1) Let Vco and Vlo are the initial volumes of
As the three rods are rigidly fixed they will have
a common extension l. As  s   a the steel container and liquid Vco  Vlo  constant (Given)
rods try to exert force in forward direction on
the aluminium rod. Vc  Vl  Vc0 (1   c T )  Vl0 (1  l T )
As the system is in equilibrium Vc  Vl  Vc0  Vl0  Vc0  c T  Vl0 l T
Fa  2 Fs (i) Vc0  c  Vl0  l
The strain in steel rods Given Vc0  1000cc, l  20  c

(l0  s T  l )  Vl0  50cc
strain 
l0
Ys A(l0  s T  l ) 20. (1) As with the rise in temperature, the liquid
 Fs  (ii)
l0 undergoes volume expansion therefore the
Similarly fraction of solid submerged in liquid increases.
Ya A(l  l0 a T ) Initially when the block is floating at temperature

Fa  (iii) t1
l0
From the three eq’s we get Fb  Fg  V1  l g  V0  s g
l0 T (2Ys  s  Ya  a ) V1  s s0 (1   l t1 )
l   f1    (i)
Ya  2Ys  V0 l l0
Here V0 is total volume of block which is h- total height of the cylinder

constant and s is also constant that is h ' - height of the cylinder present inside the
s  s0 . V0 is the volume of the solid present
liquid.
inside the liquid. h ' (1   l 50)

  (1  l 50)(1   s 50)1
h (1   s 50)
Similarly at temperature t 2
 s0 h'
 (1   l 50)(1   s 50)
f2  (1   l t2 ) (ii) h
l0
h'
f1 1   t1  1  (l   s )50
From Eqs. (i) and (ii), f  1   t h
2 2
f1  f 2 h'
 1  7.7 105  50
 
f 2t1  f1t2 h 
21. (3) The relation between apparent and real h'

 0.99615
expansion of a liquid is h
The percentage of the cylinder present outside
 real   app   vessel ;  vessel  3
is (1  0.99615) 100  0.38%
For vessel ‘A’   real   1  3 A
For vessel ‘B’   real   2  3 B

1. (2) Let l1 be the length of Cu and l2 that of
1   2 steel
Hence,  1  3 A   2  3 B   B  A
3
(l21  l11 )  (l2  l1 )  (l2 2  l11 )T
1   2
B    l2 2  l11  0 (l21  l11  l2  l1 )
3
22. (4) As the cylinder is completely submerged l2 1
that is, l  
in the liquid the densities of cylinder and liquid 1 2
are same at 50C . Let 0 is the initial density..

10 1.6
After decreasing the temperature 
l1 1.2
0 0
s  & l  or l1  7.5 cm
(1   s 50) (1   l 50)
At equilibrium of the cylinder 2. (1) Here, (steel)  1.1 105C 1
Fg  Fb (copper)  1.7  105C 1
mg  lV ' g As the difference in the lengths remain constant
sVg  lV ' g l0 ( s) ( s)  l0 (c) (c)
s hAg  l h ' Ag l0 ( s) (c) 1.7  105

As,    1.545
l0 (c) (s ) 1.1 105
h ' s

h l  l0 ( s)  1.545 l0 (c)
301
The coefficient of linear expansion is given as

Also l0 ( s )  l0 (c)  5
R

0.545 l0 (c)  5 RT
R R 1
5    .t   (ii)
 l0 (c)   9.17cm R R  .T
0.545
From equations (i) and (ii)
and l0 (s)  1.545  9.17cm  14.17cm
F
Y  F  Y .S.T
3. (1) Let l0 be the length of the copper at 0c. S.T
The length of iron at 0c is l0  10cm. After a  The ring presses the wheel from both sides,
change of temperature T the difference in Thus Fnet  2 F  2YST
their lengths must remain same
7. (2) Here, T  20  15  5C
 l0  10 cm 1   Fe T   l0 1   cu T   10 cm   0.000012C 1  12 106C 1
Given that 1
Time lost per day    T   86400s
2
 cu  17  106c 1
1
6 1
=  12  106  5  86400s  2.560s
 Fe  11 10 c 2
8. (1) Loss of time due to heating a pendulum
 l0  18.3 cm is given as
1
4. (2) As dl   l dT T   T
2
2l
dl T
1
 12.5     (25  0)C  86400
 l l 0 T dT   aT 
 a 2
1/2 1
T2  ln 4    C
ln 2  a  T  86400

2  a 
9. (4) Fractional change in period
5. (3) Given l1  l2 or l1 a t  l2 s t T 1 l

T 2 l
l1  s l1 s
   
l2  a l1  l2  a   s T 1 1
      2  106  10  105
T 2 2
6. (4) The Youngs modulus is given as
T
F/A % change   100  105  100  103 %
Y T
L / L
10. (3) Effective linear coefficient
Here it is given as
L  L1  L2
F
Y  2 R [ L  2 R] L  T  1 L1 T   2 L2 T
A2R
FR L1α1  L2 α2 L(α)  2 L(2α)

Y  (i) α 
S  R L1  L2 3L
11. (2) Consider the diagram

1 2 L
x L  L20  0 (1  T )2  1
2 2
L0 L
x 1  2T  1  0 2T
2 2
3.77
x  2  25  106  32
2
x  7.5 cm
Applying Pythagorean theorem in right angle
triangle in figure 13. (3)
2 2
From the right angle triangle PRO
 L  L   L  2
     x 2
 2 2
   l
 OR    PR    PO   l 2   
2 2 2
2
 
1
 x ( L2  L2  2LL)  L2 After expansion
2
2
1 2 1 
 (L2  2LL) (OR) 2  l 1   2t     l 1  1t  
2 2 
If L is very small, then, neglecting (L2 ) , we l2 l2

l2   l 2 1   22 t 2  2 2 t   1  12t 2  21t 
get 4 4
1 Neglecting  22 t 2 and 12 t 2
x   2 LL (i)
2
But L  L  t (ii) l2 2
0  l 2  2 2 t    21t   2 2  1  1  4 2
4 4
Substituting value of L in Eq. (i) from Eq. (ii)
14. (2) The initial positions of the rod and scale
1 1 are shown in the figure.
x 2 L  L  t  L 2  t
2 2
10
  2  1.2  105  20
2
 5  4 1.2 104  20 30 103 cm

Given that Cu  steel
L
12. (4) Let is the length of each part after
2 Let xth centimeter mark on the scale coincides
rising temperature. From the right angle triangle with the right end of the cu-rod.
2 2
 L 2  L0  After expansion x(1   s T )  80cm(1  Cu T )
   x  
2
  2
x  80.0096cm
L2 L2 15. (1) As we know that coefficient of linear
 x2  0
4 4 expansion is
1 L L 5  105
x 2  ( L2  L20 )       5C.
4 L0  L0 10  106  1
303
16. (4) According to given conditions T1  T2 , Hence Y1 A11T  Y2 A2 2 T
 40 A1 Y2 2
86412  2 (i) 
g A2 Y11
 20 18. (2) Let l is the displacement of the junction

86396  2 (ii) towards right. If the rods are free to expand
g
without any constrains then the two rods should
 expand by l1 1T & l2  2 T as shown in the
86400  2 (iii) figure.
g
From equations (i) and (iii)
2 
12   40    (iv)
g
From equations (ii) & (iii)
2 
and 4    20  (v)
g
On dividing (iv) and (v)
 l40  l 
3  
fig (1)- Initial position of the combination before
l  l20
rising temperature.
fig (2)- Expected natural extensions in the rods
 l (1  )  l 
  after rising temperature.
l  l   fig (3)- Final position of the combination after
rising temperature.
1  (40  )  1
 By comparing fig (2) & fig (3) we can conclude
1  1  (  20)
that both rods are in compression.
40   l1 1T  l
3 (by Binomial theorem) Strain 
  20 l1
   25 Force exerted by the first rod at the junction
From eq’s (i) and (iii) Y1 A(l1 1T  l )
F1 
86412 l1
 1  15    1.85  105 / C
86400 Strain in the second rod
Stress T /A l2  2 T  l
17. (4) Y  Strain 
Strain  l /l l2
Y  l Force exerted by the second is

T A  Y  A T
l Y2 A(l2  2 T  l )
F2 
In both the rods tension will be same so l2
As the system is in equilibrium F1  F2 V  V0  L   g  T

l1l2 T (Y11  Y2 2 )
 l  Given V0  1000 cc,  g  0.1  104 / C
Y1l2  Y2l1
  g  3 g  3  0.1  10 4 / C  0.3  10 4 / C
19. (3) V  V0 (1   )
 V  1000 1.82  104  0.3  104   100
L3  L0 (1  1 ) L20 (1   2  )2
 15.2cc
 L30 (1  1 )(1   2  )2
23. (1) Let A0 is the base area of the container
Since L30  V0 and L3  V
initially. After increasing the temperature its
Hence 1    (1  1 )(1   2  )2 base area is becomes A.
 (1  1 )(1  2 2  )
 (1  1  2 2  )
   1  2 2
20. (1) The linear velocity
v  r
dv dr The area of the base after increasing
 (  constant) temperature is
v r
dv rdT A  A0 (1  2T ) (i)
  dT
v r The volume of the liquid after increasing
dv temperature is
100  dT 100  3.6 102 %
v V  V0 (1  T )
21. (3) As we know, Bulk modulus
Ah0  A0 h0 (1  T )
P V P
K   A  A0 (1  T ) (ii)
 V  V K (i)
  From eq’s (i) & (ii)
 V 
V   2
 t (ii)
V 24. (3) Let VS is the volume submerged, VT is
P the total volume of the solid
  t
K VS b
The temperature that should be rised to make   0.8
VT  L
P P
same strain is t   b  density of block; L  density of liquid
 K 3 K
T  50C
22. (1) Due to volume expansion of both liquid
and vessel, the change in volume of liquid VS'  'b b 1   L T
  
relative to container is given by VT'  'L  L 1   b T
305
 1  18 105  50 
 0.8   5 
 1  6 10  50  F  32 80
1. (3) 
 1  9 103  9 5
 0.8   3 
 1  3 10  F  32  144 or F  176 F .
2. (1) Fractional loss of time per second
 0.8  (1  6  103 )
1
VT'  VS'   T
 1  0.8  0.8  6  103 2
'
VT
Therefore
= 0.1952
1
 (T0  15)  (24 hrs)  5 (1)
25. (1)  ag   r  3 g (  ag  apparent expansion 2
of glass) 1
and  (30  T0 )  (24 hrs)  10 (2)
 as   r  3 s (  as  apparent expansion of 2
steel) from eqn (1) and (2)
From the above two eq’s on solving T0  20C
 ag   as  3( s   g )   g  9  106 / C
26. (3) Let d is the density of the liquid at 0C 1. (1) Given triple point of water on scale
then A=200A
d d Triple point of water on scale B=350 B
d1  & d2 
1  t1 1  t2
We know that triple point of water on absolute
d1 1  t2 scale
 
d 2 1  t1 = 273. 16 K
27. (2) If   3 then for a same temperature 200 A  350 B  273.16 K
rise the liquid expands more than solid and the 273.16 273.16
liquid density decreases. So the cube should 1A  K and 1B  K
200 350
submerge more to balance its weight
If TA and TB are the triple point of water on two
scales A and B then
1. (4) The given temperature in 0C is 173.4 0C. 273.16 273.16

TA  TB
The value of this temperature in fahrenheit is 200 350
9 TA 200 4 4
 173.4  45  45  279.4F   or TA  TB .
5 TB 350 7 7
2. (3) Given l1  l2 or l1 a t  l2 s t 2. (1) Let V be the volume of the sphere and 
be the density of water. Buoyancy (F) on the
l1  s l1 s sphere due to water is
   
l2  a l1  l2  a   s
F  V g
s 1
   0.33
2 s   s 3 Since, 0 C  4 C , so, F0 C  F4 C .
3. (4) Since,  Al   steel , so in bimetallic strip on Initial temperature

heating, aluminium strip will expand more than T1  27  273  300 K
steel strip. Due to it, aluminium strip will bend
more on convex side and steel strip on concave Final temperature T2  227  273  500 K
side.
Coefficient of linear expansion
4. (1) Triple point of neon, T1  24.57 K    1.70 105 / C
Triple point of CO2 , T2  216.55K
Coefficient of areal expansion    2
On celsius of neon,
 3.40  105 / C

t C  24.57  273.15  248.58C
1
d12

Triple point of CO2 , t C  216.55  273.15 Initial area of hole at 27C  A1   r 2 
1 4
 56.60C  2
The temperatures in Fahrenheit scale are

 4.24   4.494 cm2
4
9
F1  32  (248.58)  415.44F
Area of hole at 227C  A2   A1 1  .t 
5
 4.494  1  3.40  105   227  27  
9
F2  32  (56.6)  69.88F
5  4.495  1.0068
If diameter of hole becomes d2 at 227C , then

d 22
1. (2) Absolute humidity A2 
4
10
  12.5 g / m3
800 103 d 22
4.525 
2. (1) With increase in temperature, the effective 4
length (l) of simple pendulum increases even
through its centre of mass still remains at the or d 2  4.2544cm
centre of the bob. Time period, T  2 l / g or Change in diameter  d   d2  d1

T  l . T increases as temperature increases  1.44 102 cm
3. (4) Let V be the volume sphere of radius R 5. (4) Given, coefficient of volume expension
4 3
at temperature T then V  R .     49 105 / K
3
Increase in volume of sphere with rise in Rise in temperature  T   30C
temperature T is
Let initial volume of glycerine be V0 .
4
V  V T  3  R3 T Volume of glycerine when temperature is
3
increased by 30C
V  4R 3  T
V  V0 1  T 
4. (1) Given, diameter of the hole
 d1   4.24 cm  V0 1  49  105  30
307
 V0 1  0.01470  1.0147V0
1. (2) As the temperature of rod rises,

V0 1 Extension in the rod, L  LT
 (i)
V 1.0147
F/A
 m / V V0 1 Young’s modulus of the rod, Y 
   L / L
0 m / V0 V 1.0147 (ii)
FL FL F
Fractional change in density Y  
AL ALT A  T
   0 
   1 2. (3) Due to thermal expansion, change in
0 0 0 length
Substituting value from Eq. (ii), we get F/A
Y
l / l
1
Fractional change in density  1
1.0147  l 
F  AY  T    T 
 l 
 1.45 102
3. (4)
Negative sign shows that the density of glycerine
decrease with rise in temperature
1. (1) As metal A contracts more than metal

B    A   B  , the bimetallic strip bends more
1. (1) Standard equation of line y=mx+c we get
towards the side of A.
100 5
slope of the line AB is m  
212  32 9 F/A YAl
2. (4) Y F 
l / l l
F  YA  T
1. (4) Slope of the line
1006  1000 L where  T is the thermal strain
A   L0  A
T T
 2 1011  40 104 1.2 105 10
6
 1000 mm  A (i)  9.6 104  1105 N
T
similarly for line B 3. (4) The initial time period is
2 l0
 1002 mm  B (ii) t  2  2
T g
From (i) & (ii)  A  3 B The initial time period after rise in temperature
2. (2) Density of water is maximum at 4C and is
is less on either side of this temperature.
l (1 T ) Vl  Vc  x  l T  V0  c T

t  2
g
x c 1
 
 2(1 T )1/2  2  T V0  l 4
t  t0  T  t  T (x- initial volume of the liquid)

( V0 - initial volume of the container)
5. (3) The two scales are shown in the figure.
From the figure we can write
c 1 50Y  (160Y ) 37 K  273K

4. (2) Given that   4 
l x  (160Y ) 340K  273K
If the vacant space remains constant then  x  86.3Y

change in volumes of liquid and total container
must be equal
Kinetic Theory of Gases 309
10. (1) PV  nRT
m
 PV  RT
M
1. (3) Where m = mass of the gass
2. (4) At low temperature the molecules K.E is PM m

 
low and they come closer at high pressure. So RT V
intermolecular forces start acting among the
particles and hence the gas deviates at low Where  = density
temperature and high pressure.
PM P  m0  N A Pm0 Pm0
3. (1)    
RT RT  R  kT
 T
V  NA 
4. (1) At constant pressure  constant
T
11. (3) The ideal gas equation is
If T increases V also increases.
N
As n remains same so number of molecules per PV  nRT  RT  NKT
NA
cm3 decreases.
Where K is the Boltzmann’s constant
5. (3) At low pressure and high temperature a
real gas behaves like an ideal gas and hence it Given that P  106 KT
obeys ideal gas equation PV = nRT
6. (2) PV  nRT N
  106
V
PV
n 12. (3) At constant temperature
RT
As P, V and T are same then n is also same in PV  constant
both balloons
P1 V2 70 V
7. (3)     2  V2  700 ml
P2 V1 120 1200
8. (1) When T is constant
13. (4) According to ideal gas equation
PV= constant
RT RT
P P M 
If P2  1  V2  4V1 M P
4
m 5 M A  A TA PB 1 M 3
PV  nRT  RT  RT  . .  (1.5)(1)    A 
9. (3) M B  B TB PA 2
  M 4
M 32 B
310 Kinetic Theory of Gases
14. (2) At constant pressure

P  RT
19. (4) v 
V2 T2 V   M
V T    T2   2  T1
V1 T1  V1  M - Average molecular weight of air
 3V  0
T - absolute temperature
 T2     273  819 K  546 C
V 
v1 T
15. (2)  1
v2 T2
1 V2 P1 100 100
P     V2  V1  0.953V1
V V1 P2 105 105 1 T
 1
2 T2
1 V  V2
% change in volume  V  100
1 T1  300 K  T2  1200 K
V  0.953V1 20. (2) v rms  v av  v mp ,

  100  4.76%
V1
3RT 2 RT
vrms  , vmp 
16. (2) Mean velocity of gas molecules is, M M
8RT
v where, T = Temperature of the gas 8RT
M and vav 
M
in kelvin M = Molar mass, R = Universal gas
constant 21. (2) At NTP, T  273K , p  1.01  105 Nm2
For the same gas, v  T Here, d  2.4  1010 m
v1 T 300 v1 300 kT
  1    
v2 T2 T2 2 v1 T2 2 d 2 p
1 300
Squaring both sides, we get, 4  T 
(1.38  1023 )  273
2
1.414  3.14  (2.4  1010 ) 2  1.01 105
T2  1200 K  1200  273  927C
 1.46 107 m
1/ 2
1 3  3RT 
17. (2) As, Mc 2  RT or c    f
2 2  M  22. (2) CP  CV  R  CP  R  CV  R  R
2
1
 c 3 5
M R R  R
2 2
1/ 2
c1  M 2 
So,   CP 2 2 4
c2  M1  23. (4)   1   1 
CV f 6 3
12  22  32  42
18. (3) c 4
4 CP  CV  R  CV  CV  R
3
1  4  9  16 15 1
  kms 1  CV  R  CV  3R
4 2 3
24. (1) As the given conditions are at NTP

N P 4 1010
   1011 per m3
m=M=4.0 g (As it occupies 22.4 liters) 23
V kT 1.38 10  300
vrms  999m / s R  8.3 J .k 1mol 1  105 / cm3 .
 RT Mv 2 31. (4) Here temperature remains constant so

vrms     1.6
M RT PV
1 1  PV
2 2  76  5  P2  35
CP 76  5
  CP   CV  1.6  5  8 J .k 1mol 1  P2   10.85 cmof Hg.
CV 35
25. (2) We define specific heat capacity at 1 2 1 
32. (4) PV  mNc 2  N  mc 2 
constant volume 3 3 2 
R 2
CV  PV  E
(  1) 3
R 3
For gas A, (CV ) A  (5 / 3)  1  2 R
1. (4)
R 5
For gas B, (CV ) B  (7 / 5)  1  2 R 2. (4) Hydrogen gas can be liquified at 240C.
So this gas is more preferable among the others
For gaseous mixture,
as it liquifies at very low temperature.
R 13
(CV )mix   R 3. (1) Intermolecular forces should not be
(19 / 13)  1 6
present among ideal gas particle. So it can’t be
We define specific heat capacity of mixture liquified.
n1 (CV )1  n2 (CV )2 4. (1)
(CV )mix 
n1  n2 5. (2)
3 5 6. (3)
1 R  n  R
13
R 2 2  (3  5n) 7. (1)
6 1 n 2(1  n)
8. (4)
or 13  13n  9  15n, i.e., n  2 mole.
9. (3)
26. (1)
27. (3)
m V
28. (2) PV  nRT  RT  RT
M M
29. (3) According to law of equipartition of RT
energy, kinetic energy per degree of freedom P
M
1
of a gas molecule is kT . When temperature is constant
2
N P1 P2 P
  2  2 1  21
30. (1) Pressure PV  N RT  NkT 1 2 P1
A
N P 11. (2)
 
V kT 12. (3)
13. (2) When P is constant T  c (P  constant)

V where c is a constant
 constant
T
Let T1  T0 & T2  1.04T0
If T2  4T1  V2  4V1
c c
14. (1) If pressure of water vapour increases at 
constant temperature then condensation begins  2  1 T T1
 100  2  100  3.84%
and pressure remains constant. 1 c
T1
15. `(3)
PV  RT (ideal gas equation) (i) 20. (2)
V 2 P  C (additional law) (ii) 12  22  32  42

21. (3) c
4
Dividing both the equations
1  4  9  16 15
  kms 1
C 1 4 2
V V 
RT T
22. (2) vrms  T
1
 T
V 3RT
23. (4) vrms 
When volume (V) is doubled, temperature is M
T
halved i.e., T '  3RT2 3RT1
2 
vrms  M M  100
m d % increase in 3RT1
16. (1) PV  nRT  RT  P  RT
M M M
20  17.32
m  100  15.5%
[Density d  ] 17.32
V
P v1 d2
P1 P 24. (1) Speed of sound v  d

v2

d1
P  2
  constant or
dT d1T1 d 2T2
[ P  constant]
17. (4) At constant pressure V  T
1 v1 M2
25. (3) vrms   
V1 T1 V T 300  280 M v2 M1
   V2  1 2   280ml.
V2 T2 T1 300 v M2
   M 2  16
2v 32
18. (1) 1 mole of H 2 mass is 2g. So 3 moles of
H 2 are present. So the given gas is CH 4
19. (2) The ideal gas equation is 26. (1) RMS velocity depends only on
temperature. So the ratio of rms velocities is
V  RT
PV  nRT 
M 1:1
RT So, average translational kinetic energy is also

27. (3)  doubled.
2 d 2 N A P
36. (1) The average kinetic energy of a gas
RT particle is
28. (2) Mean free path   2 d 2 N P
A 3
K .E  KT
2
1
 ; If  is doubled then K.E is independent of nature of gas
P
37. (2) At constant volume when pressure is
P becomes half. doubled then temperature is also doubled. As
29. (2) CP  CV  R  Universal gas constant K .E  T so K.E is also doubled
38. (4) Kinetic energy depends only on
CP 2
30. (1)   1 temperature.
CV f
39. (1) Internal energy of n1 moles of He is
31. (3) A diatomic molecule has three
translational and two rotational degrees of 3
freedom. U1  n1 R(10T )
2
Hence total degrees of freedom f  3  2  5 Internal energy of n2 moles of hydrogen is
32. (4) CP  CV  R and
5
U 2  n2 R (6T )
C R 2
 P  CV 
CV  1
Given that U1  U 2
33. (1) For monoatomic gas
n1CV1 T1  n2CV2 T2
5
CP  R 3 5
2 CV1  R, CV2  R
2 2
For diatomic gas
3 5
n1 R (10T )  n2 R(6T )
5 2 2
CV  R
2
n1
1
34. (1) When two gases are mixed then their total n2
internal energy remains constant.
40. (4)
n1CV1T1  n2CV2 T2  (n1  n2 )CV T
41. (2) Pressure will be less in front portion of
(CV1  CV2  CV , As both are diatomic)
the compartment because in accelerated frame
molecules will feel pseudo force in backward
m1 m direction. Also density of gas will be more in the
n1  , n2  2
M1 M2 back portion.
T
n1T1  n2T2 42. (2) PV  nRT  P  . If T and V both
T  305 k  32c V
n1  n2
doubled then pressure remains same, i.e.,
35. (2) If volume is doubled at constant pressure,
then absolute temperature of the gas is doubled. P2  P1  1atm  1  105 N /m 2
1 mn 2 dp  Mgdh
43. (1) Let P0  c 
3 V p RT0 (1  ah)
m p dp Mg h dh
n   
1  2  p0 p RT0 0 (1  ah)
P'  (2c)2  2P0
3 V
p Mg
 ln  ln(1  ah)  ln(1  ah) Mg / aRT0
p0 aRT0
PV PV PV Hence, p  p0 (1  ah)Mg / aRT0

1. (2)  R(constant)  1 1  2 2
T T1 T2
5. (1) Let 1 be the density after first stroke.
200  V P  1.02V As the mass remains constant
  2
(273  22) (273  42)
V   (V  V ) 1 or 1  V  / (V  V )
[V2  V  0.02V ]
where  is the initial density of the gas.
200  315
 P2   209 kPa Similarly, if 2 be the density after second stroke,
295  1.02
then
2. (3)
PV PV V 1  (V  V ) 2
1 500 0.5  V2
1 1
 2 2   V2  900 m3
T1 T2 300 270 2
 V   V 
or 2    1    
 V  V   V  V 
Normal stress p pV
B   So, after nth stroke,
3. (4) Volume strain  V  V
 
 V  n n
 V   V 
n     or  n    
 V   V  V   V  V 
p  B 
 V  As temperature is constant in this process
V m V
Where normal stress = p, volume strain   pV  RT  RT
V N M
4. (2) We know that the variation of pressure
with height of a fluid is given by 
p RT  p  
M
From gas law dp   gdh (i)
n n
m  pn  V  1  V 
pV  RT or p  RT    or  
M M p  V  V    V  V 
pM
or  (ii) By taking log on both sides
RT
Putting Eq. (ii) in Eq. (i)
log e 
n
 pMg  (V  V ) 
dp  dh log e  
RT  V
3 2 2 f 1
6. (4) Their average translational K.E is k B T. 13. (2)   1 ,  1   
2 f f 2  1
k B  Boltzmann’s constant 2
 f 
 1
1 1 3
m H2 c H2 2  m He c He
2
 k BT
2 2 2
n1CV1  n2CV2
c H 2 : c He  m He : m H 2  4 : 2  2: 2 CPmix n1  n2
 mix  
14. (1) CVmix n1CP1  n2CP2
7. (4) At NTP or STP n1  n2
T=273 K
R R
3kT 3  1.38  1023  273 Where CV  & CP 
As, c    1  1
m 5  1017
5 7
 15  103 ms 1  1.5 cms 1 1
1
3  5
n1 1 n2 2  5  7 
 1 1
vrms 
3RT  1  1  2  1  3   5  3
8. (2)  mix     1.5
M n1 n 1 1 2
 2 
1  1  2  1  5   7 
3  8.314  300   1    1
1930  3  5 
M
15. (2) Average time between collisions
M  2g
Mean free path

So the given molecule is H 2 vrms
9. (4) Argon is a monoatomic gas so it has only
1 CV  M 
translational energy. t 2
;t  where C  2

d N V T  d N 3R 
Translational : Rotational =100% : 0% 3RT
M
3kT 3kT
10. (2) vB  v  , vA   
2m m
V2
T 
2 t2
2
v2 Given that TV  1  k
TA T TA T V 2  1 V  1
11. (3)  4 B  2 B V  k  k
MA MB MA MB t2 t2
1  1
3RTA 3RTB c q
 2  cA  2cB  A  2 t V 2
 2
MA MB cB
' Q
12. (3) Here, the degrees of freedom F  6, 16. (2) Molar specific heat C 
nT
 F  6 Q
CP   1   R   1   R  4 R Specific heat C 
 2  2 mT
21. (3) Here, work done is zero.

C' m
  M (Molecular weight)
C n So, loss in kinetic energy= Change in internal
energy ofgas
C'
C
M 1 2 R
mv  nCV T  n T
CP'  CV1  R 2  1
R 1 2 m R
MCP  MCV  R  CP  CV  mv  T
M 2 M  1
For nitrogen M=28
Mv 2 (  1)
 T  K ( K  Kelvin)
R 2R
CP  CV 
28 22. (3) The number of moles of the system
17. (2) Translational kinetic energy of all remains same,
3 PV PV P(V1  V2 ) P(V1  V2 )TT
molecules of diatomic gas will also be PV , 1 1
 2 2 T  1 2
2 RT1 RT2 RT ( PV T
1 1 2  PV
2 2T1 )
because all the gases have same translational
(degree of freedom 3). 23. (3) Length of air column on both sides is 45
cm when one side is at 0C and the other is at
18. (3) We know that 273C .
1 2 1  The pressure must be same on both sides. Hence

PV  mNv 2  N  mv 2 
3 3 2  T
 constant
2 V
PV  NK
3
T
 constant (V  Al )
3 PV l
N
2 K
l1 l2 l l2 l
N 3 PV   1   l1  2
n  T1 T2 273 (273  273) 2
N A 2 KN A
5 5 Also l1  l2  90 cm  l1  30 cm and l2  60 cm
19. (2) K .E  nRT  PV
2 2
Applying gas equation to the side at 0C , we
5 m 5 1 get
K .E  P     8  10 4   5  10 4 J
2  2 4
Pl
11 Pl P  30 76  45
20. (4) (Q)V  nCV T  W T  00 1   P1  102.4 cm
T1 T0 273 273  81
[W  Total heat capacity]
Where P0 &T0 are the initial pressure and
(Q )V U temperature.
 W  [ (Q)V  U ]
T T
24. (3) On combining the two vessels the total
80
W  4 J /K number of moles remains constant, i.e.,
(120  100)
n  n1  n2
4 J
Molar heat capacity is Using gas equation, we can write,
5 K .mol
PV
1 PV
n1  ; n2  2
RT1 RT2
PV
1 1 PV PV
1. (4)  2 2  T2  2 2  T1
P (2V ) T1 T2 PV
1 1
and n 
RT
1 10
where V is the volume of each vessel.  T2    300  100K  173C
30 1
P(2V ) PV PV P 1 P P 
Thus  1  2 or   1  2  N A PAVA TB
RT RT1 RT2 T 2  T1 T2  2. (4) PV  NkT   
N B PBVB TA
25. (4) The number of gas particles remain
constant N A P  V  (2T ) 4
  
NB V 1
ni  n f 2P   T
4
P(2V ) PV' '

PV 2P P '  T1  T2  3. (3) According to Boyle’s law
     
RT1 RT1 RT2 RT1 R  T1T2  ( PV
1 1 ) At top of the lake  ( P2V 2 ) At the bottom of the lake
2 PT2 2  1  600 4  PV
1 1  ( P1  h)V2
P'    atm
(T1  T2 ) (300  600) 3
3
4  5r  4 3
26. (2) After reaching the equilibrium states the  10      10  h   r
3  4  3
temperature and pressure in three parts must
be same. 610
h  9.53 m
64
V RT 1
i.e. PV  nRT    = constant 4. (1) As temperature is constant
n P K
PV
1 1  PV
2 2
n
Let  K
V 0.8  5  P  (3  5)  P  0.5 m .
By conserving the number of moles 5. (3) Given that
n1  n2  n3  K (V1  V2  V3 ) T  T0  V 2
Let T0 is the initial temperature of the gas in For one mole of the gas, V  RT / p
three regions.
P(V / 3) 2P(V / 3) P(V / 3)  R2T 2 
   KV T  T0    2 
RT0 RT0 RT0  p 
Where T0 is the temperature in three parts. Solving it for p, we get
4P p   RT (T  T0 )1/2 (i)
K 
3RT0
Differentiating it with respect to T, we get
The volume of the first part is
dp 1
   R[(T  T0 )1/2  T (T  T0 )3/2 ]
n1 V dT 2
V1  
K 4 Conditions for minimum pressure is that
(dp / dT )  0. Hence,  f  5  7
11. (4) CP    1 R    1 R  R
 2   2  2
 1 
 R  (T  T0 ) 1/ 2  T (T  T0 ) 3/ 2   0
 2  12. (1) We treat water like a solid. According to
Dulong-Petit law the heat capacity of many
1 solids at room temperature is equal to 3R. Water
(T  T0 )1/2  T (T  T0 )3/ 2  0 (  R  0) molecule has three atoms, two hydrogen and
2
one oxygen.
Now, Solving we get,
 Heat capacity per mole of water is
T  2T0
C  9 R.
Substituting the value of T  2T0 in Eq. (i) we
R
get 13. (3) We know that CP  CV 
J
pmin   R(2T0 )[2T0  T0 ]1/2 R
 J
CP  CV
or pmin  2 R ( T0 )
cal
CP  CV  1.98
3kT g  mol  K
vrmsHe 2mHe mAr 40
   J
6. (2) vrmsAr 3kT mHe 4 R  8.32
2mAr g  mol  K
8.32
 10  3.16 J  4.20 J /cal
1.98
8 RT
7. (1) vrms  3RT / M ,Vavg  3 5
M n1CV1  n2CV2 1 R  3  R
14. (1) CV mix   2 2
8. (4) As temperature is constant n1  n2 1 3
1 v M SO2 4 64  18.7Jk 1mol 1

vav   Gas   
M vSO2 M Gas 1 M Gas
15. (1) For mixture of gases, let specific heat be
 M Gas  4 , i.e., gas is He. CV
9. (2) Change in pressure will not affect the rms n1 (CV )1  n2 (CV ) 2
velocity of molecules. Only temperature will CV 
n1  n2
have effect
v1 T 300 3 5R 3R
 1   where (CV )1  ,(CV )2 
2 2
v2 T2 400 2
5R 3R
400 2  8
v2  m/s  2 2  17 R  1.7 R
3 28 10
R R R
CV      1.5 16. (2) CP  CV 
10. (3) M
0.5   1
The given gas is monoatomic Where CP & CV are specific heat capacities.
M is the molecular weight, M=32 for oxygen

5 3
gas 2 RT  4 RT  5RT  6 RT  11RT
2 2
Q
17. (1)
'
Molar specific heat C  [As F1  5 (for oxygen) and F2  3 (for argon)].
nT
3 5
Q 20. (2) U  n1CV1 T  n2CV2 T  1 RT   2  RT
Specific heat C  2 2
mT
13RT
C' m 
  M (Molecular weight) 2
C n
F 3
C' 21. (1) Average kinetic energy E  kT  kT
C 2 2
M
3
 E   (1.38  1023 )(273  30)
CP'  CV'  R 2
MCP  MCV  R  6.27  1021 J
R  0.039 eV  1eV
CP  CV 
M 22. (3) Kinetic energy  Temperature. Hence
For hydrogen M=2 if temperature is doubles, kinetic energy will
also be doubled.
R
CP  CV  a n
2 1
23. (1) Number of moles of first gas  N
A
For nitrogen M=28
R 2 n
 CP  CV  b Number of of moles of second gas  N
28 A
a
 14
b 3 n
Number of of moles of third gas  N
A
18. (4) (Q)V  nCV T  W T
By conserving total internal energy
[W  Total heat capacity]
U f  Ui
(Q )V U
 W  [ (Q)V  U ]
T T n1 n n n n n
RT1  2 RT2  3 RT3  1 2 3 RTmix
NA NA NA NA
80
W  4 J /K
(120  100)
n1T1  n2T2  n3T3
Tmix 
n1  n2  n3
4 J
Molar heat capacity is
5 K .mol
3
24. (4) K .E  RT
19. (4) Total internal energy of system 2
F1 F K.E is independent of molecular mass for a

 U oxygen  U argon  n1 RT  n2 2 RT
2 2 diatomic gas.
8RT dV  xS (i)
25. (2) vavg  (T- same for both gases)
M Where S  5mm2  5 102 cm2
Pressure and density depend on number of In equilbrium, pressure on both sides is say P ' .
particles in the container Then
P '(V  dV )  nRT (ii)
26. (2) and P '(V  dV )  nR(T  dT ) (iii)

Dividing (iii) by (ii) we get
When the piston is allowed to move the gases V  dV T  dT


are kept separated but the pressure has to be V  dV T
equal. ( P1  P2 ) and final volume x and (6V--x),
Here V = l litre  103 cm3
the no of moles are same in initial and final
position at each parts. T = 293 K and dT = 0.1 K
 P1  P2 PV  n1 RT Solving, we get
dV  0.17cm3
n1 RT n2 RT

x 6V  x dV 0.17
Hence x    3.4cm
S 5  102
5 PV 10PV
where n1  , n2 
RT RT m
28. (1) PV  RT
M
n1 n2

x 6V  x As the gas is in equilibrium the pressure in both
flasks must be same.
5PV 10 PV 1 2
    V1 m1 T1
xRT (6V  x) RT x 6V  x  V  mT   
V2 m2 T2
 6V  x  2 x  x  2V and
6V  x  6V  2V  4V 2V m 100 m
    m2 
V m2 200 4
 (2V ,4V )
29. (1) Final pressure in the containers will be the
27. (3) The initial and final positions of the two
same
gases are shown in the figure below :
PVA PVB PV PV
  0 A 0 B
RTA RTB RT0 RT0
P P P P
   0 0 VA  VB 
2T0 T0 T0 T0
4P0
P
3
30. (2) The number of moles present on both sides
Let the mercury is displaced by x. Then are same. After reaching mechanical equilibrium
the pressures on both sides are same. 5 5

2. (1) K .E  nRT  PV
The temperatures on both sides are given as 2 2
T1  273K & T2  400 K 5 mP 5 1 8  104  m

    5  104 J V  
2 d 2 4  d
P1  P2
3. (1) Neglecting bond length, the volume of an
nRT1 nRT2 oxygen molecule has been taken as 2 times that

V1 V2 of one oxygen atom.
Let x is the displacement of the mercury column In 22.4 litres i.e., 22.4 103 m3 , there are
towards the cooler side.
V1  A(40cm  x) N A  6.23  1023 molecules
V2  A(40cm  x) Total volume of oxygen molecules

4
A - Area of cross section of the tube.  2   r3  NA
3
273 400
  22.4 103 m3 is occupied by N A molecules.
40  x 40  x
 x  7.55cm  Fraction of volume occupied
The length of air column on cooler side is 32.45 4 3

2     1.5  1010   6.2  1023
cm  3  8  104.
3
(22.4  10 )
31. (1) By conserving the total internal energy
(n1  n2  n3 )CV T  n1CV T1  n2 CV T2  n3CV T3
T
 n1  CV T1   n2  CV T2   n3  CV T3 1. (3) The pressure inside the tube after closing
 n1  CV  CV  n2    n3  CV the tube is 76 cm of mercury.
1. (2) Given: Ti  17  273  290K
T f  27  273  300 K
Atmospheric pressure, P0  1  105 Pa
Volume of room, V0  30m3 We assume that T is constant

PVN A PV
1 1  PV
2 2
N
RT After pulling the tube up the pressure of the air
Given that number of molecules N=n is 76 cm - x, where x is the rise in the level of
mercury
PV
0 0
 1 1
 n f  ni     NA (76)(8)  (54  x)(76  x)
R  T f Ti 
1105  30  1 1   x 2  130 x  3496  0

  6.023 1023   
8.314  300 290  x  38 cm
 2.5  1025 Length of air column= 54- 38 = 16 cm
2. (1) The speed of sound in a gas is given by Final p2  11  1  12atm  12 1.013 105 Pa
 RT
v
M V2  30 L  30  103 m3
vO2  O2 M He
   T2  273.15  17  290.15K
vHe M O2  He
Number of moles
1.4 4 p1V1 12 1.013 105  30 103
   0.3237  
32 1.67 RT 8.314  300.15
vO2 460 = 14.61
 vHe    1421m / s
0.3237 0.3237
Hence, mole removed = 19.48 - 14.61 =4.87
Mass removed  4.36  32g  0.1558 kg
1. (3) As piston is able to move up or down
without friction, therfore, when temperature is
increased, piston moves out, increasing the 1. (1) When pressure of an ideal gas is constant,
volume V, The pressure p on the gas remains Charles’ law is obeyed i.e.,
the same, because of fixed mass.
2. (2) The pressure of the gas inside the vessel, V 1
V  T or  constant 
as observed by us, on the ground remains the T p
same. This is because the rocket is moving with
constant velocity. Hence pseudo force does not From the slope of curve shown in figure p1  p2 .
act on the gas particle. 2. (4) Applying standard gas equation,
3. (2) As the vessel contains both hydrogen and
oxygen, therfore, as per Maxwell’s law of speed p1V1 p2V2

n1T1 n2T2
distribution, f1  v  and f 2  v  will obey the law
separately.(In general p2 V1 T2 n2
  
f  v   constant  v e
 2
2  mv / 2 KT  p1 V2 T1 n1
Where m = mass of one molecule) 2 T 3000

As T  300  10 and every molecule of
4. (3) Absolute pressure p1  (15  1)atm 1
H 2 splits into hydrogen atoms, doubling the

[ Absolute pressure = Gauge pressure +
number, therefore n2  2n1
atmospheric pressure]
p2
 16 1.013  105 Pa   2  10  20.
p1
V1  30 L  30  10 3 m3
3. (4) Here, p1  p,V1  V , T1  T
T1  273.15  27  300.15K
1 1 pV
2 2 pV
Using ideal gas equation from T  T
1 2
pV
1 1 16 1.013 105  30 103
n 
RT 8.314  300.15 V1 T2
p2  p1 .
= 19.48 V2 T1
V T p The graph between V & TC is a straight line

 p 1.1  1.1
1.05V T 1.05 having positive slope and it does not pass through
origin
p2  1.05 p.
2. (1) For an ideal gas at constant temperature
4. (2) Root mean square speed of argon atom PV = copnstant i.e . PV doesn’t vary with V.
3RTAr 3. (4) At constant temperature PV = constant
 vrms  Ar  (i)
M Ar 1
P .
V
Root mean square speed of helium atom
PM
3RTHe 4. (3) 
 vrms He  (ii) RT
M He
Density  remains constant when P / T or
Dividing Eq. (i) by Eq. (ii) volume remains constant.
 vrms  Ar 3RTAr M He In graph (I), pressure is increasing at constant

  temperature hence density is increasing. In graph
 vrms He M Ar 3RTHe
(II) density is decreasing.In graph (III)
Given  vrms  Ar   vrms  He P
 Constant so density remain constant.
T
 T  M  5. (2) As the slope is constant for any line

1   Ar  He 
 THe   M Ar  V1  V2 &V3  V4
M He THe From the given graph it can be observed that

  the slope of line 34 is more than the slope of line
M Ar TAr
12
 39.9  2
TAr  253.15    2.52 10 K . m34  m12
 4 
nR nR

V34 V12
V12  V34
PV = n RT
6. (1) According to ideal gas equation PV =
When P is constant nRT
 nR  m   1
V  T PV  RT , P  RT  
 P  M M P T
Where T is in kelvin

Here, represents the slope of the graph.
T  TC  273 P
TC  Temperature in degree centigrade  

As    
 P 1  P  2
 nR 
V   TC  273
 P  Hence T2  T1 .
7. (4) From ideal gas equation m m m

   M1  M 2  M 3
PV = nRT (i) M 3 M 2 M1
or PV  nRT (ii)
M 1 corresponds to CO2 , M 2 corresponds to He
Dividing equation (ii) by (i) we get and M 3 corresponds to H 2
V T V 1 10. (4) The mean free path is given by
    [Given]
V T V T T
1

1 2nd 2
  . So the graph between  and T
T n - number of particles per unit volume
will be rectangular hyperbola.
d - diameter of a particle
N PN A P
PV  nRT n  
V RT K BT
m
PV  RT 1 K BT
M  2
2 d P
The given curve is an isothermal curve.
Temperature is same for both curves. So the graph between  & T is a straight line
passing through origin. Graph between  & P is
PV
 Constant hyperbolic.
m
At a given pressure 1 m0 m0
  
2nd 2
2  m0 n  d 2
2d 2
Where m0  mass of a gas particle
m0 n  Total mass per unit volume
 m0 n  
1
 The graph between  &  is hyperbolic.

V2  V1 (from graph)
V 1. (2) According to ideal gas equation

 constant (P is constant)
m
V nR
 m2  m1 PV  nRT or 
T P
The correct graph is (3). At constant pressure
V
 constant
PV = nRT T
As the graph is between PV and T so nR Hence graph (2) is correct.
represents slope.
2. (2) As the straight line has negative slope the
From the graph n3  n2  n1 general expression for the equation of line is
P  C1V  C2 6. (3) For a given pressure, volume will be more

if
Where C1 & C2 are (+) ve constants
V
From ideal gas equation we can write the above temperature is more  Constant
T
equation as
nRT
 C1V  C2
V
C1 2 C2
T V  V
nR nR
The equation represents a curve.
Differentiating the equation w.r.to V
From the graph it is clear that V2  V1  T2  T1.
dT 2C C
 1V  2
dV nR nR 7. (3) (i) the dotted line in the diagram shows
d 2T 2C PV
 1 that there is no deviation in the value of for
dV 2
nR nT
Which represents a maxima. different temperatures T1 and T2 for increaing
The correct graph is (2). pressure , so this gas behaves ideally.
3. (2) As (ii) We know at high temperature , the deviation

of the gas is less and at low temperature the
T  T  deviation of gas is more. In the graph, deviation
 2  1  tan  2  tan 1     
 P  2  P 1 for T2 is more than for T1 . Thus, T1  T2 (iii) The
T two curves intersect at dotted line, so the
From PV  nRT ;  V  V2  V1.
P PV
value of at that point on the y - axis is same
4. (1) When Pressure is constant nT
PV= nRT for all gases.
8. (3) At high pressure intermoleculer attraction
m m
P  RT among molecules increase and hence the gas
 M
deviates from ideal behaviour. Path (1) is not
 possible as it represents ideal gas behaviour.
P RT  T  Constant Path(4) is also not possible as it deviates before
M
the pressure increases to a higher value. Path
PM (2) is more possible than path (3) since in path
5. (4) =    (P / T ) (2) the gas deviates slowly from ideal behaviour
RT
as pressure increases.
3
 A  0 and  B   0
2 3RT 2
9. (1) vrms   vrms  T.
3 M
( PB / TB )  ( PA / TA )
2
dN
3 10. (4) The maximum value of represents
( x / 2T0 )    ( P0 / T0 ) dv
2 most probable speed of gas particles. Area of
x  3P0 dN
the graph is A   dv  Total no of particles.
dv
It is always true that area of the graph on the 3

 n RT
right side of vmp is more than the area on the 2
left side of vmp . Total energy when n moles of diatiomic gas

converted into monoatomic
3 5
(U f )  2n RT  ( N  n) RT
2 2
n1CP1  n2CP2
 max  1 5
1. (2) n1CV1  n2CV2  nRT  NRT
2 2
   Change in total kinetic energy of the gas
CP   R
  1 1
U  nRT
2
R
CV  1 mN 2
 1 5. (3) Pressure, P  vrms
3 V
1
n2 2 P  (vrms ) 2  T
n1 
1 1  2 1
 mix 
n1 n So, force  (vrms )2  T
 2
1  1  2  1 i.e., Value of q=1
5 4 6. (4) Kinetic energy of each molecule,
n1  2,  1  , n2  3,  2 
3 3
3
We get K .E  K BT
2
 mix  1.42 Given that T  273K
3RT Height attained by the gas molecule, h=?
2. (4) vrms 
m 3 819 K B
K .E  K B (273) 
v  40 2 2
rms He
  10  3.16 K.E = P.E
v 
rms Ar
4
819 K B 819 K B
3. (1) Mean time   between two   Mgh  h 
2 2Mg
successive collisions
1 T T 1 P2 T1
   
TP P 2 P1 T2
1. (1) 1
6  108 500 mean 
 2  N
2 300 2 d 2  
V
and, PV = NK B T
 15  108  4  108 s
K BT
4. (1) Energy associated with N moles of  mean 
diatomic gas, 2d 2 P
5 1.38  1023  300

U i  N RT 
2 2  3.14  0.64  1018  1.1 105
Energy associated with n moles of monoatomic  132 1010 m  13.2 nm
gas
2. (4) The ratio of specific heats at constant Ideal gas equation is

pressure (CP ) and constant volume (CV ) N
PV  nRT  RT
CP  2 NA
   1  
CV  F  Given that
where F is degree of freedom J
R 8 , T  300K
CP  2  7 K .mole
 1   
CV  5  5 N A  6  1023 mole 1
3. (4) According to Dulong-Petit law many P  4.0  1015  105 N
solids at room temperature CV  3R. For solids m2
when heat is given their volume approximately V RT 8  300
 
remain constant. N PN A 4.0  10  105  6 1023
15
U  Q V
3   1011 m3
N
For on mole
Using equipartition of energy, we have   2.15 104  0.215mm
3RT  MCT
6. (3) In isothermal process temperature
remains constant.
3R 3  8.314
C  3RT 3PV 3PV
M 27 103 vrms    (i)
M Mn m
C  925J / kgK
Isothermal work done is
3RT V2
4. (1) c  W  nRT ln  PV ln 2
M V1
3  8.314  300 Given that W  575J  PV ln 2
(1930)2 
M
575
3  8.314  300  PV  (ii)
M  2g ln(2)
1930  1930
From eq’s (i) & (Ii) we get
The gas is H 2 .
3  575 3  575
5. (2) From the given information we calculate v rms   2
 499m / s
the average volume occupied by each molecule m(ln 2) 10  0.693
(not the volume of molecule). The average 7. (2) In isobaric process the heat given to a
volume occupied or share of each molecule is
gas is
equal to the ratio of total volume divides by total
number of molecules. This volume can be Q  nCP T
approximated as a cube of side length ( ). The
cube root of the average volume per molecule The graph between Q and T is a straight line
is equal to ( ). Here the value ( ) is also called passing through origin.
as mean free path and it is equal to average
distance between two molecules of a gas as slope  nCP
As n same for all the gases
We know that
C P( Poly )  C P( Di )  C P ( Mono)
a,b,c correspond to P, D and M

328 Calorimetry
Given that s1  s2  s3
m(40)  2m(50)  3m(60) 320

   53.6C
m  2m  3m 6
1. (1) The coversion factor between  F and 1 3 s1 4
C is 8. (2)  , 
 2 4 s2 3
9
 F  C    32 Thermal capacity per unit volume is
5
m s  s 
9   1  1  1  1:1
 F   C  m/ 2 s2 2
5
As  F  C the heat capacity is not same in 9. (4) Energy supplied by the heater to the
both scales. system in 10 min
1 Q Q1  P  t  90  10  60  54000 J
since s
m T Now if  is the final temperature of the system,
As Q and m are same energy absorbed by it to change its temperature
from 10C to C is
sT  constant
sc Tc  sF TF Q2   ms T  water   ms T coil  calorimeter
As TF  Tc  sF  sc  360  1   10   40    10   4200

2. (1) When water is cooled at 0C to form ice
then 80 calorie/gm (latent heat) energy is released.  400    10 4200
Because potential energy of the molecules
decreases. Mass will remain constant in the Q1  Q2    10C
process of freezing of water. 10. (4) With rise of altitude pressure decreases
3. (3) and boiling point decreases.
4. (4) 11. (1)
5. (3) 12. (3)
6. (1) Q  m  c  T ; if T  1K then 13. (2) Let L be the latent heat and using principle
of calorimetry.
Q  mc  Thermal capacity..
2L  2 100  54.3  40   54.3  25.0
m s t  m2 s2t2  m3 s3t3
7. (1) t 1 11
m1s1  m2 s2  m3 s3  L  540.3 cal / g
Calorimetry 329
TA  T  4, T  TB  4C
1. (4) cA
 1
cB
1 Q
2. (3) Heat capacity c 
m T 14. (4) Heat lost by hot water = Heat gained by
(water + copper calorimeter)
when T  0 c   ( Q can be (+)ve or
msw (50  40)  (1.5  sc  0.2sw )(40  25)
(-)ve)
3. (4) sw  4200 J / kgk , sc  390 J / kg.k
4. (3) Due to large specific heat of water, it solving m  508 g

releases or absorb large heat with very small
temperature change. 15. (1) Steam at 100 C contains extra
5. (3) 2260 J gm energy as compared to water at
6. (1) 100 C. So it’s more dangerous to burn with
steam than water.
 1 Q

7. (3) Specific heat capacity c  m  t  16. (3) As the block is very large the pressure at
 1 the base is also large and hence melting point of
ice at high pressure decreases. Ice starts melting
1  Q  at the bottom.
Molear heat capacity C  n  t 
 2
17. (1)
n  Q  18. (3)
c nM  T 1 19. (2) Work done in converting 1 g of ice at 10C
 (m  nM )
C  Q  to steam at 100C = Heat supplied to raise
 
 T 2 temperature of 1 g of ice from
8. (3) Latent heat is independent of its physical 10C to0C  m  cice  T  . + Heat supplied to
structure. melt 1 g of ice at 0C mLfus + Heat supplied to

raise temperature 1 g of water
9. (2)
0C to100C  m  cwater  T  + Heat supplied to
10. (2) For solids and liquids CP  CV . Whereas
for gases CP  CV  R convert 1 g water into steam at 100C  m  Lvapour 
11. (1) At low pressure melting point increases   m  cice  T    m  Lfus    m  cwater  T 
and at 0C it can’t exist like water so it freezes.
  m  Lvapour 
Q  mc  Thermal capacity..
 1 0.5 10  1 80  11100  1 540
Q J
12. (1) c 
m   kg  C  725cal  725  4.2  3045J .
13. (2) Heat lost by A = Heat gained by B 20. (3) Suppose m gm ice melted, then heat
required for its melting  mL  m  80 cal
 mA  cA  TA  T   mB  cB  T  TB 
Heat available with steam for being condensed
mA  mB and then brought to 0C
330 Calorimetry
 1 540  11 100  0   640cal 3. (2) Heat gained by the water = (Heat
supplied by the coil)  (Heat dissipapted to
 Heat lost  Heat taken environment)
 640  m  80  mc   PCoil t  PLoss t
 m  8 gm  2  4.2 103   77  27   1000t  160t

Short cut : You can remember that amount of
4.2 105
steam  m ' at 100C required to melt mgm ice  t  500 s  8min 20 s
840
at 0C is 4. (3) We are given that power of the machine,
m P  10kW  104 W
m'  .
8
Time for which the machine is used,
Here, m  8  m '  8 1  8gm t = 2.5 min = 150 s
Mass of the aluminium block,
1. (3) When A,B are mixed: m  8kg  8  103 g
ms1 (15  10)  m(25  15)s2 Specific heat of aluminium, c  0.91J / gC
Since 50% of the energy is used up in heating
5s1  5s2  s1  s2
Energy absorbed by the block, Q = 0.5 Pt
When B & C are mixed:
 0.5 104 150 J  7.5 105 J
ms2 (30  25)  ms3 (40  30)
Let T be the rise in temperature of the block.
5s2  10 s3
As Q  mcT
s2  2 s3  s1
Q 7.5 105  J 
When A & C are mixed: T    103C
mc 8  103  g   0.91 J / g C 
ms1 (T  10)  ms3 (40  T )
5. (1) Let the final temperature be T C.
2s3 (T  10)  s3 (40  T )
The net heat change of the system is zero.
2T  20  40  T  T  20C
q1  q2  q3  0
2. (3) When two gases are mixed together then
m1c1 (T  T1 )  m2 c2 (T  T2 )  m3c3 (T  T3 )  0
Heat lost by the Helium gas = Heat gained by
the Nitrogen gas m1c1T1  m2 c2T2  m3c3T3
 T m1c1  m2 c2  m3c3
7 
nB   CV  He   T0  T f   n A   CV  N   T f  T0 
 3  2
6. (1) Given that
3 7  5
 1  R   T0  T f   1  R  T f  T0  T  T2  T1  T 
T1  T2 
2 3  2 2
where given T is temperature of the mixture
3
By solving we get T f  T0 . Apply heat lost = heat gained
2
Calorimetry 331
m1c1 T1  T   m2 c2 T  T2  As Q1  Q2, so ice will not completely melt and

final temperature  0C.
m1 c2 T  T2  c2 As heat given by water in cooling up to 0C is
 
m2 c1 T1  T  c1 only just sufficient to increase the temperature
of the ice from 20C to0C, hence mixture in
7. (4)
equilibrium will consist of 10 g of ice and 10 g
mgh 2 L 2  3.4  105 of water, both at 0C.
 mL  h    68 km.
2 g 10
11. (1) In the entire process heater is heating
8. (1) Initially ice will absorb heat to raise it’s water and cooler is removing heat from water.
temperature to 0C then it’s melting takes place As Pheater  Pcooler the difference will make the
water to rise its temperature.
If mi  Initial mass of ice, mi'  Mass of ice that
melts and mw =Initial mass of water  Pheater  Pcooler   t  msT
By law of mixture, Heat gained by ice = Heat
  3 103  P   3  3600  120  4.2 103  20
lost by water
 p  2067 W
 mi  c   20  mi'  L  mwcw  20
12. (4) Assume that radius is decreased by R
 2  0.5 20  m  80  5 1 20  m  1kg
'
i
'
i mL  AT
So final mass of water = Initial mass of water + where m is the decrease in mass of the drop.
Mass of ice that melts  5  1  6 kg. A is decrease in surface area. T is the coefficient
of surface tension.
9. (2) Heat received by ice is
2
Q1  mL  mcT  10700 cal.

4R2 RL  4T  R2   R  R  
 
Heat lost by the container is  R 2 RL  T  R 2  R 2  2 RR  R 2 
500 500
 BT 2   R 2 RL  T 2 RR  Risvery small 
Q2   m  A  BT  dT  m  AT 
300  2 300
2T
= 21600 m R
L
By principle of calorimetry,
Q1  Q2  m  0.495kg
10. (1) Heat lost by water is Q1 1. (3) The heat input in time t is Pt
Pt
Q1  10  1  10  100 cal PT  ML  L 
M
Heat gained by ice is Q2 2. (3) Let final temperature of water be 
Heat taken = Heat given
Q2  10  0.50 0   20    10  80
110 1  10  10   10  220 1 70   
 100  800 cal  900 cal    48.8C  50C.
332 Calorimetry
3
mAcA  4 / 3 rA  A cA  rA   AcA
3
1 2
3. (3)     7. (1) W  JQ  I   J  Ms 
mB cB  4 / 3 rB3B cB  rB  B cB 2
3
1  2 1 1 12  1 R 2 2
         MR 2   2  J  Ms    
 2   1   3  12 2 5  5 Js
4. (1) Heat required to raise the temperature 8. (4) Ice  0C  converts into steam 100C 
of m grams of substance by dT is given as
in three steps.
dQ  mcdT  Q   mcdT
The heat required Q  Q1  Q2  Q3
Therefore, to raise the temperature of 2 g of
substance from 5C to15C  5  80  5 1 100  0  5  540  3600 cal
15 Q Q
Q   2   0.2  0.14T  0.023T 2  dT 9. (1) c  
mt 0
5
15
Since during phase transition T  0
 0.14T 2 0.023T 3 
 2  0.2T    81.83 cal 10. (1) Heat given out by water = heat absorbed
 2 3 5
by the melted ice
5. (4) Let the radii of the spheres be R,
mc
R  a, R  2aand R  3a where a is a constant and  mc    0   ML  M 
L
the specific heat capacities be s, sr,
11. (1) Heat gained by ice + heat for melting ice
sr 2 and sr 3 where r is another constant.
= heat lost by water
Given that
 1 cal   cal  40
 heat capacity of D   heat capacity of C 
 m   T C    m   80 
 2 g  C   g  100
 :   8 : 27
 heat capacity of B   heat capacity of A 
 cal 
Heat capacity = mass  specific heat   m  1  T C 
 g  C 
  R  3a 3 sr 3    R  2a 3 sr 2 
 :   8 : 27 (T / 2)  32  T  T  64C
3
  R  a  sr   R3 s 
dm
 Ra 12. (3) Let be the rate of evaporation and
dt
3 dm
m2  4 / 3   R  R   8 M  M0  t is the amount of liquid left in
 3
 dt
m1  4 / 3   R   1 the pitcher at any instant.
6. (4) As the system is thermally isolated, the The heat reqiured for evaporation is equal to
heat required for vaporisation is extracted from heat gained
water getting frozen. Let x be the maximum
fraction of water that gets soldified. The (1 - x)  dm 
Mcd    tL
is the fraction that gets vaporised.  dt 
 x  3.36 105  1  x   21105  dm  dm

 M0   t  cd   Lt
 x = 0.862.  dt  dt
Calorimetry 333
 5  5 Heat lost = Heat gain

 10000   t 1  5   540  t
 60  60
 0.1103  swater   80  T 
5
 50000  45 t  t  0.3 103  swater  T  60
12
 1  80  T   3  T  60  T  65C
50000  12
 t  18min 26 s
545 2. (4) According to principle of calorimetry,
13. (2) The total heat absorbed by water. Heat lost = Heat gain
H  mL 100  0.1T  75  100  0.1 45  170 1 45
10 cal
The radiation power is P   540  5400 10 T  75  450  7650
1 s
14. (2) The mixture has both ice and water. When T  885C
the metal piece is placed some part of ice melts.
Let m is the mass of ice that melts. The change
in the volume of ice (or the mixture) is
1. (3) Given that Qc  QW
m m
  0.15  m  0.15  0.92 mc cc  c  mW cW W
ice  H 2O 0.08
mc c. c 50 103  420 10
0.15  0.92  W    5C
 Hlost  H gained  15  s  100   80 mW cW 10  103  4200
0.08
S  0.092cal / gmC. 2. (1) U  Q  mcT
15. (4) Required work = Energy released = 100 103  4184  50  30  8.4kJ
Here, Q   mc dT
4
 T3  1. (1) Given, mass of the child (m) = 30 kg
  0.1 32   3 
dT  0.002kJ
20  400  Time taken (t) = 20 min
Therefore, required work = 0.002 kJ. Fall in temperature  101  98 F
5 5
T  3F  3  C  C
1. (2) Density of water  10 kg / m3 3 9 3
Specific heat of human body
Let the final temperature of the mixture be T
Assuming no heat transfer to or from container.  s   4.2 103 J / kg  C
Heat lost by water at 80C Latent heat of evaporation
 0.1103  swater   80  T   L   580cal / g   580 103  4.2 J / kg
Heat gained by water at 60C Heat given by body during fall in temperature
 0.3 103  swater   T  60 Q1  msT

334 Calorimetry
Let m ' be the mass of sweat evaporates from Q1  Q2  Q3  0
the human body.
Heat required for evaporation M1 s(T  T1 )  M 2 s(T  T2 )  M 3 s(T  T3 )  0
Q2  m ' L M1T1  M 2T2  M 3T3  ( M1  M 2  M 3 )T
But Q1  Q2  msT  m ' L M1T1  M 2T2  M 3T3

 T
M1  M 2  M 3
msT 30  4.2 103  5 / 3
m'  
L 580  4.2 103
10 1. (4) The horizontal parts of the curve, where

  0.0862 kg
116 the system absorbs heat at constant temperature
must represent changes of state. Latent heats
0.0862 are proportional to lengths of the horizontal parts.
Rate of evaporation of sweat  In the sloping parts, specific heat capacity is
20
inversely proportional to the slopes.
= 0.00431 kg/min
2. (1) Initially, on heating temperature rises from
= 4.31 g/min
10C to 0C. Then ice melts and temperature
does not rise After the whole ice has melted,
temperature begins to rise until it reaches 100C.
1. (2) Then it becomes constant, during vaporisation
Given, mass of copper block (m) = 2.5 kg
Change in temperature  T   500C 1. (2) The volume of matter in portion AB of

the curve is almost constant and pressure is
Specific heat (s)  390J / g  K
decreasing. These are the characteristics of liquid
Latent heat of fusion of water state.
( L)  335  103 J / kg 2. (4)
Heat energy absorbed by copper block
Q1  msT  2.5  390  500J
1. (4) According to principle of calorimetry,
Let m ' kg of ice be melted.
Qgiven  Qused
Heat energy required to melt ice, Q2  m ' L
0.2  s  150  40
Q1  Q2
 0.15 1 40  25  0.025  40  25
ms T  m ' L
msT 0.175(15) cal J
m'  s  0.1193  499.46
L 0.2 110 g C kg C
2.5  390  500 2. (3) Energy given by heater must be equal to
  1.5kg
335 103 the sum of energy gained by water and energy
2. (2) Let T is the equilibrium temperature of lost from the lid.
the system. The total change across the system
is zero Pt  msT  energylost
Calorimetry 335
1000t  2   4.2  103   50  160t  dQ   dQ 

   
 dt  water  dt liquid
840t  8.4  103  50
 t = 500 s = 8 min 20 s. (mwcw  W )T (ml cl  W )T

t t
(W= water equivalent of the vessel)
1. (2) As Pt  mcT or, mw cw  ml cl
P  10  60  mc  1100 (i) mW cW
 Specific heat of liquid, cl  ml
and P  55  60  mL (ii)
Dividing equation (i) by (ii) we get 50  1
  0.5 kcal / kg
100
10 100

55 L 3. (2) The heat released by steam when it
converts from 100C to 0C.
L  550cal. / g.
Q  10  540  10 1100
2. (4) As the surrounding is identical, vessel is
The mass of the ice that melts is
identical time taken to cool both water and liquid
(from 30C to 25C ) is same 2 minutes, Q  6400  m80  m  80 g
therefore
Water in calorimeter  500  80  10  590g
336 Thermodynamics
W Q  U 1 2
 1 
Q Q  7
 Q : U : W  7 : 5 : 2
1. (4) 14. (4) The root mean square velocity of a gas
2. (3) With rise in temperature, internal energy molecule is given by
also increases.
3RT
3. (2) vrms 
M
4. (2)
Gas is compressed isothermally, so T remains
5. (1) constant and hence, root mean square velocity
6. (1) The area under P-V curve for isothermal will remain same
curve is less than for adiabatic curve. V 
15. (4) W  nRT ln  2 
7. (3) As dW  0 so it is an isochoric process  V1 
16. (2)
P
 constant
T 17. (2) Due to compression the temperature of
the system increases to a very high value. This
Given dq  dU  0  T decreases and hence causes the flow of heat from system to the
P also decreases. surroundings, thus decreasing the temperature.
8. (3) In a cyclic process, a system starts from This decrease in temperature results in decrease
one point and ends at the same point. In that in pressure.
case, the change in the internal energy must be 18. (1) Differentiating PV = constant w.r.t V
zero and therefore, the thermal energy added to
the system must be equal to the work done during P V
 PV  V P  0  
the cycle, i.e., in a cyclic process, P V
 E 0 1
19. (2) PV  const  V  
P
9. (4) Q  U  W  2  103  4.2  U  500
1
 U  7900 J TV  1  const  V  1 
T
10. (2) 20. (1) Given that
11. (1)
PT 3 = constant (i)
12. (4) Work done  PV  P V2  V1  For an adiabatic process,
U nCV T 1 5 
13. (3)    (ii)
Q nCP T  7 PT 1  constant
Thermodynamics 337
Comparing (i) and (ii), we get 33. (1) From the given problem,

 3 or  3  3   Q  Q1  Q2  Q3  Q4
1 
 5960  5585  2980  3645
3
2  3 or    1.5.
2 Q  9605  8565  1040 J
21. (4) In adiabatic process PV   constant Efficiency of a cycle is defined as
 RT    1 Network W Q
   V  TV  cons tan t.   
 V  Input heat Q1  Q4 Q1  Q4
 
22. (3) PV
1 1  P2V2
Putting Q  1040 J
 
V   V 
P2  P1  1   P1    P1 (23 )5/3  32 P1 and Q2  Q1  5960  3645  9605 J
 V2  V / 8 
23. (3) 1040
   0.1082  10.82%
24. (4) For adiabatic process 9605
 1
TV
1 1
 T2V2 1  T2 
34. (1)   1    100
 1  1  T1 
T2  V1  V  1
  2 2  
T1  V2   V1  2  300 
25  1    100
 T 
1
V2  1   1 1  5
         T  400 K
V1  2  2 3
35. (2) In a reversible cycle
25. (2)
26. (2) Q Q'

T T'
 350   300 
27. (2)  A  1   , B   1   36. (4) For 1st case
 700   650 
B   A  T2 
efficiency    1    100
 T1 
28. (1)
29. (4) T2 

30. (4) 1   100  40
 500 
TL TH  TL T2  300K
31. (3)  1 
TH TH
for 2 nd case
TL  100 T  TL
'  1   H  300 
TH  100 (TH  100)   1   100  60
 T1 
  '
T1  750 K .
32. (3)
338 Thermodynamics
19. (2) Change in internal energy is U  nCV T

3
1. (3) For monoatomic gas CV  R
2
2. (1)
3 3
U  n RT  1  8.31 (100  0)
3. (3) 2 2
4. (1)  12.48 102 J .
5. (1) Average K.E of a gas particle depends 20. (4) U  nCV T  n(5 / 2) RT
on absolute temperature. Temperature decides
the direction of heat flow. Q  nCP T  n(7 / 2) RT
6. (4) Work is a path function so it depends on
path. n7 n5
W  Q  U  RT  RT  nRT
2 2
7. (4)
W 2
8. (3) 
U 7
9. (2) 21. (4)
10. (4) For process to be reversible it must be
22. (3) At constant temp dU  0
quasi-static. For quasi static process all changes
take place very slowly. Isothermal process occur  dQ  dW
very slowly so it is quasi-static and hence it is
reversible. 23. (2) In isothermal process temperature
remains constant. i.e., T  0. Hence according
11. (4)
Q
12. (2) to C   Ciso  
mT
13. (2) The K.E of the (gap+container) converts
24. (1)
into increase in internal energy.
25. (4)
14. (4)
15. (3) Change in internal energy does not depend 1 Q
26. (3) Heat capacity c 
upon path so U  Q  W remains constant. m T
If T  0 & Q  0
16. (4) Here Q  50 J , W  15J
c
 U  Q  W  50  (15)  65J
27. (2) In adiabatic expansion as volume
17. (2) increases pressure and temperature decreases.
W  P(dV )  2.02  105 (3342  1) 106  334.2J In adiabatic expansion work done by the gas is
at the expense of its own internal energy so
Q  U  W temperature decreases.
 U  Q  W 28. (4) For adiabatic process pV  = constant
 mL  167  540  4.2  2  334.2  4202 J 
M 
18. (4) When heat is supplied at constant pressure, p   = constant
a part of it goes in the expansion of gas and   
remaining part is used to increase the p.M  .  = constant
temperature of the gas which in turn increases
the internal energy. p.   = constant
Thermodynamics 339
29. (1)  V 
1.4 1
T2     300
30. (2) In adiabatic process V / 4 
Q  0  U  W  0 T2  (4)0.4  300  300  (4)0.4 K
(Q  U  W ) 35. (1) Internal energy is a function of (T). As

internal energy is a state function the working
31. (3) substance final temperature must be equal to
intial temperature.
32. (3) For adiabatic process PV   constant
36. (4)
P V  1 dV  V  dP  0 37. (2)
P 38. (1) For cyclic forces U  0 So, Q  W
dV  dP  0
V 39. (2)
 dP 40. (1)
Bulk modulus  P
dV / V 41. (1) Internal energy and entropy are state
function, they do not depend upon path but on
2 the state.
33. (1) R  CV
3
 0
42. (2)   1   1
We know, C p  CV  R  T2 
R T2
 1   R  CV (   1) 43. (3) C  1 
CV T1
44. (2)
5
Comparing   W  T 
3 = 1- L 
45. (4) ΔQgiven  TH 
P1T   constant
TL  400 K , TH  500 K
5
P  T (  /  1) , Given   3 Q  6  104 given
 5 W  1.2  104 J

(   1) 2 46. (3) The efficiency of carnot engine is
So, P  T 5/ 2 T1  T2 T
 1 2
T1 T1
34. (2) For adiabatic change the relation between
temperature and volume is TV  1 = constant The value of efficiency of Carnot engine can
be 1 or 100%, if the sink temperature must be
where  is ratio of specific heats of the gas. ok. As practically absolute zero temperature
cannot be achieved. Hence carnot engine cannot
V give 100% efficiency.
Given, T1  27  273  300 K ,V1  V ,V2 
4
47. (4) As,
 1
 V1  T2 (47  273) 320 1
TV
1 1
 1
 T2V2 1  T2   V   T1  1 1 1   20%
 2 T1 (127  273) 400 5
340 Thermodynamics
T2 (273  69) Now work done is  dW   PdV

48. (4)  1 1  0.5
T1 (273  411)
From (i) PdV  nRdT  dV
 Work done   Q  0.5 1000  500J
 dW   nRdT   dV
200 200
49. (4)
W
 COP  3  W 
3
Joule.
  nRdT 
 (nRdT )
P
3T0 0 3T

 nR  dT  nR  dT ( P  P0  constant)
T0 P0   T0
1. (4) Q  nCP T
 nR 2T0 P0
 nR 2T0   nR 2T0 
U  nCV T P0   ( P0  )
U CV 1 5 hence n  1
  
Q CP  7
R 2T0 P0
 W
2. (4) Q  U  W ( P0  )
7. (2) In isothermal compression, there is

U  Q  W  mL  P V2  V1 
always an increase of heat which must flow out
of the gas.
1.013 105 (1671  1)106
 1 540 
4.2 Q  U  W  Q  W [ U  0]
 540  40  500cal 1.5 104

 Q  1.5 104 J  cal  3.6 103 cal.
4.18
CP
3. (4) Q : U  nCP T : nCV T  C   :1 8. (1) Slope of adiabatic curve    (Slope of
V
isothermal curve)
4. (1) QP  nCP T
9. (1) We generally assume that a sound wave
QV  nCV T propagates under adiabatic condition.
T   P (1 )  const.
QP
    QV  50cal
QV Differentiating
5. (1) At constant pressure   T 1  dT  P (1 )  T   (1   ) P    dP  0
W  PV  nRT  1 8.31 100  831J (  1)T

 dT  dP
6. (2) P
Given PV  nRT  V (i)  1.5  1   273 105 N 

 dT    5 2
 0.001 
 1.5   10 ( N / m ) 104 m2 
 (P  )V  nRT
 8.97  108 K
nRT
V 10. (1) For adiabatic process, TV  1  constant
( P  )
 1
nRdT m
dV  T   constant
( P  ) 
Thermodynamics 341
T The equation of motion for the piston

 constant
  1
 p0 x0 
Mg     A  Frestoring
1  ( x0  x) 
  T 1/  1  3  4/3
 1
 x0 
2 2 p0 A  1   
 Frestoring
f   6  ( x0  x ) 
 1  4 
  1 
3   
 
11. (4) In adiabatic process Q  0 1
p0 A 1     Frestoring
  x 
Q  U  W  1   
  x0  
W  U   nCV (T2  T1 ) 
  x 
p0 A 1  1     Frestoring
W  nCV (T1  T2 )   x0  
 
3
6R  R (T  T2 ) 
2  p0 Ax   x x
F  Ma    1   1 
x0   x0  x0 
T2  (T  4) K  
12. (3) When piston is slightly displaced through

 p A
equilibrium position, it will oscillate under a net a    0  x   2 x  4 2 2 x
restoring force. The piston will perform SHM.  Mx0 
Pressure applied by piston
1  p0 A 1  p0 A2
  
Mg 2 x0 M 2 MV0
 P0 (At equilibrium)
A
 V0 
Mg  P0 A (i)  x0  A 
 
13. (3) For isothermal process
P1 V2
 2
P2 V1
For adiabatic process
5/3 5/3
P2  V3  P  16V  5
   2    2  32
Let piston be moved by a distance x P3  V2  P3  2V 
Here, the system is completely isolated so it is
adiabatic. P2 P
 P3   P2 
32 64
p0V0  pV 
1U 
p0 Ax0  pA (x0  x) 14. (1) As, P   
3 V 
p0 x0 U
p But  KT 4 (K is a constant)
( x0  x) V
342 Thermodynamics
1 From ideal gas equation

So, P  KT 4
3 PV=nRT
nRT 1
  KT 4 [As PV  nRT ]  TV 1 3  const
V 3
1
4 3 3  4 3
T V 3
R T  constant  V   R 
3  3  As V increases T also increases.
1 18. (2) In the free expansion W  0 , Q  0
Therefore, T 
R (adiabatic vessel) then U  0
T 2 19. (4) When the bigger piston displaces towards

15. (2) The given relation is p  left by 1m the smaller piston also displace by
V
1m towards left.
T 2 The length of the left side cylinder becomes 2m
 V
p and right side cylinder becomes 0m.
As pressure is kept constant, 2  R2

The initial volume V   R (1m)  (1m)
4
 2 T 
 dV    dT
 p  The final volume V   R 2 (2m)
As the number of moles are equal initially and
Now, W   pdV   0 p  2T  dT
2T
finally.
T 0
 p 
ni  n f
2T0
T 2 
 2    3T02 PV PV
 2 T0  5P
P 
T 2T 4
16. (2) PV  nRT
20. (3) An external agent displaces the system
 P0  (1  V )V  nRT2 to the equilibrium position & we can’t use
conservation of energy for the system.
 PV  (1  )V 3 
 T As the changes take place under adiabatic
0
nR conditions.
dT P0 3V 2 (1  ) PV 
1 0  PV1

& PV 
2 0  PV2

  0
dV nR nR
P1 P  
 2  PV
1 1  PV
 P0  3(  1)V 2 
V1 V2
2 2
P3 P11/  V1  P21/  V2 (i)

 V2 
3(  1)
We know that V1  V2  2V0 (ii)
P0 P 2P
 P  P0  (1  )  P0  0  0 2V0 P11/ 
3(  1) 3 3 V1 
P11/   P21/ 
17. (2) Given that
21. (2) In equilibrium position, P1 A  Fagent  P2 A
2
PV 3
 constant Fagent  ( P2  P1 ) A
Thermodynamics 343
24. (2)
PV
2 2 3R
 TT ( P  P ) 
Q    Cv (T2  T1 )   T2  1 2 1 2 
RT2 2  PT
1 2  PT
2 1 
3 PPV
1 2 (T2  T1 )
 Q
4 ( PT
1 2  PT
2 1)
Elementary workdone by the agent
25. (3) The input electrical power and output
Fagent dx  ( P2  P1 ) A  dx  ( P2  P1 ) dV (i)
mechanical powers must be same mg  v  i 2 R.
Applying PV = constant for two parts, we have
26. (4) Final volume of gas
P1 (V0  Ax)  PV
0 0
and P2 (V0  Ax)  PV
0 0
 h h  9 Ah
PV PV V '  A h    
P1  0 0
P2  0 0  2 16  16
(V0  Ax) and (V0  Ax)
As the gas has to balance the additional spring
PV force
0 0 (2 Ax) 2 PV V
 P2  P1   0 0 (V  Ax)
V02  A2 x 2 V02  V 2 Final pressure of the gas is
When the volume of the left end is  times the
K h 6
volume of right end, we have P '  pa  16  105  37 10  h  N / m2
 
A  16  27 
V0  V    V0  V 
P 'V ' PV
 1 Now, 
V  V0 (ii) T' T
 1
The workdone by the agent is given by  5 37  106  h   9  Ah 

10  16  27    16  5
   10  Ah
V V
2 PV   
0 0V 4  273
W   ( P2  P1 )dV   dV
2
0 0 V
V 2   273 
0 3 
  PV 2 2 V Solving, h  1.6m.
0 0 [ln(V0  V )]0
2 2 2
27. (3) As heat Q supplied to right part, the piston
  PV
0 0 [ln(V0  V )  ln V0 ] moves towards left thus compressing the left
  2   1  2   spring and elongating right spring. To keep the
2
  PV
0 0 ln V0   V0   ln V0  temperature of left part constant some energy
   1  
say Q ' is transferred from gas to external
P1 (v / 2) P (v / 2) arrangement (which is used to keep its
22. (3) (T  T1 )  2 (T2  T ) temperature constant).
T1 T2
As temperature of left part is constant so
T T (P  P ) change in internal energy of left part is zero.
 T 1 2 1 2
( PT
1 2  P2T1 )
Let T f is the final temperature of right part
P11 (v / 2) P1 (v / 2) PT ( P  P )
23. (4)   P11  1 2 1 2  U  nCV T  nCV (T f  T )
T T1 ( PT
1 2  P2T1 )
P21 (v / 2) P2 (v / 2) P T (P  P ) Where T f is the final equilibrium temperature

  P21  2 1 1 2
T T2 ( PT
1 2  PT
2 1)
of right part.
344 Thermodynamics
T2
30. (4)  1
T1
Coefficient of performance
T2 T2 / T1 (1  ) 1
Using ideal gas equation     1
T1  T2 1  (T2 / T1 )  
3l
p1  A 
4  p0 Al (Left part)
RT RT
5l 31. (1)
p2  A 
4  p0 Al (Right part)
RT f RT
So 90 J heat is absorbed at lower temperature.
From equilibrium of piston,
2 T
kl kl kl 32. (2) Coefficient of performance   T  T
p2  p1    1 2
4A 4A 2A
(273  23) 250 250
kl 4 P kl     5.
P2  P1   0 (273  27)  (273  23) 300  250 50
2A 3 2A
P2 4 kl T1  T2
  33. (3) Initially   T
P0 3 2 AP0 1
1 T1  280
5T  kl 2 4 nRT    T1  560 K
 Tf     where p0  2 T1
4  2nRT 3  Al
T1 ' T2 ' T ' (273  7)
5kl 2 5T Finally 1 '   0.7  1
Tf   T1 ' T1 '
8nR 3
 T1 '  933K
3R 3nR  5kl 2 2T 
U  n   (Tf  T )    Increase in temperature  933  560  373K
2 2  8nR 3 
1g 2kcal
28. (1) Input energy    2kcal / s.
s g TC
 1
TH
10
Output  10kW  10KJ / S  kcal / s.
4.2
TC - Temperature of sink
output energy 10
   1, TH - Temperature of source
input energy 4.2  2
29. (2) Given that
T2 3  300  T T
 1  1    T2  318.75K . 1  1  2  T  TT
600 4  800  T2 T
1 2
Thermodynamics 345
35. (2) No heat is added or removed from the 7. (1) Temperature remains constant
system.
PV
1 1  PV
2 2
dQ dU  PdV dT dV
36. (1) dS    nCV  nR 0.8  5  P2  (3  5)  P2  0.5m
T T T V
2V 8. (4) For cylinder A, Q  nC p T1

T2 dT dV
 S2  S1  nCV   nR 
T1 T V1 V For cylinder B, Q  nCV T2
T2 V Hence, nC p T1  nC V T2

 S  nCV ln  nR ln 2
T1 V1
(CV  R)30  CV T2
37. (4) The entropy change of the body in the 5
two cases is same as entropy is a state function. For diatomic gas, CV  R
2
So the option is (4).  T2  42K
V2
9. (3) W  nRT log e
V1
1. (3) As on heat is lost, therefore
Loss of kinetic energy = Gain of internel energy  22.4 
 1  8.31 (273  0) log e  
of gas  11.2 
1 2 1 m R  831  273  log e 2  1572.5 J

i.e., mv  nCV T  mv 2  . T
2 2 M  1
[ log e 2  0.693]
Mv    1
2
10. (3) For adiabatic process PV   constant
 T  K
2R
P V  1dV  V  dP  0
3R
2. (3) Q  U  nCV T  2   100  300R
2 P
dV  dP  0
V
Q Q CP 
3. (2)   
W Q  U CP  CV (   1)  dP
Bulk modulus  P
dV / V
4. (4) For process at constant pressure
11. (3) Sudden bursting of a tyre can be treated
5 as adiabatic expansion
Q  nCP T  nRT and
2 T1 P11  T2 P21
 1 3/2
2  T1   P1   300 
3/2 1
W  PV  nRT  Q  3.375 
5        
 T2   P2   T2   1 
5. (3) W  PV  1.01 105 (3.34  2  103 ) 300
T2   200 K  73C
(3.375)1/3
 337  103 J  340kJ
 1
12. (4) Use TV
1 1  T2V2 1
6. (1) 3
W  P  V  400 10  0.3 10 10 2
 1
TV
1 1 273
T2    207 K
 400 10  3  12kJ . V2  1
(2) 0.4
346 Thermodynamics
Change in internal energy Since area under adiabatic process (BCED) is

greater than that of isothermal process
R
u  (T1  T2 )  8.31(273  207)  1369.5J (ABDE). Therefore net work done
(1   ) 1.4  1
W  WAB  WBC
13. (3) T1V1 1  T2V2 1
5 2 WAB  ( )ve ; WBC  ()ve  W  ()ve
 1 1
T1  V2   L A 3 L  3
    2   2  .
T2  V1  5
 L1 A   L1  17. (4)  =  monoatomic gas.
3
14. (4) For adiabatic process,
From first law of thermodynamics
PV   constant (i)
Q  W  U
Ideal gas relation is
nRT W  Pi V  105  7  103  700J
V = (ii)
P
U  nCV T
From Eqs. (i) and (ii), we get
3 900
 RT 

T   Pf V f  PV 
i i  J.
P   constant   1  constant 2 8
 P  P
So, Q  W  U  588J
Given, P  T C
18. (2) The work done by the gas


 P  T  1  C 
 1 f
W   PdV
5 5 i
For a monoatomic gas    C 
3 2
f
1
15. (2) TV  1  constant W   V 2 dV  (V f3  Vi 3 )
i 3
5/31
 8V 
(300)V 5/31  T2  
 27  V f  2Vi  2(1.00m 3 )  2.00m 3
2 2/3
 8V 
(300)V 3  T2   1
 27  W  [5.00atm / m6 )(1.013 105 Pa / atm)]
3
4
(300)  T2   [(2.00m3 )3  (1.00m3 )3 ]  1.18MJ
9
 T2  675K RT
19. (1) W   PdV   dV
Rise in temperature = 675 - 300= 375 K V
Since V  KT 2/3
16. (4) From graph it is clear that P3  P1 .
2 1/3 dV 2 dT
Hence, dV  KT dT   
3 V 3 T
Hence
T2
2 RT
W dT  2 R(T  T )  2 R(30)  20R
T1 3 T
2 1
3 3
Thermodynamics 347
20. (3) The polytropic equation is PV x  constant 24. (1) At thermal equilibrium the final
temperatures on both sides are equal. Let T is
x=-3 the final temperature. The pressures on both
sides are also equal and P is the final pressure
PV 3  constant
on both sides.
R R 5 The internal energy of a monoatomic gas is
C  
 1 x 1 3 3
U  nCvT  n RT
2
R R 3R R 7
C     R By conserving the energy of the system initially
5
 1 3  1 2 4 4 and finally.
3
3 3 3 3
R nRT0  nR(2T0 )  nRT  nRT
21. (2) CP  CV  2 2 2 2
M (No. of moles are same on both sides)
Where CP & CV are specific heat capacities. 3T0
M is the molecular weight, M=32 for oxygen T 
2
gas
The heat that has flown from right side to the
22. (3) In free expansion Q  0, T  0, but it left side is
doesn’t represent isothermal and adiabatic 3 3 3 3PV
processes. Temperature remains constant and   nRT  nRT0  nRT0  0 0
2 2 4 4
it does not represent isobaric process.
25. (1) Let A denote the cross-sectional area of
23. (3) From adiabatic relation between the piston and y be the vertical displacement
temperature and volume between its initial and final equilibrium positions
(  1)
TV
i i  T f V f(  1)
Let V0 and T0 are the volume of the gas and

temperature initially in each part. After displacing
the adiabatic separator the new volumes are V1
and V2 and temperatures are T1 and T2
1 V 7
Given that V  3
2
14V0 6V
V1  V2  2V0  V1  & V2  0
10 10
 1
 1  14V0 
 T0V0  T1   (i) The decrease in potential energy of the weight
 10 
is equal to increase in internal energy of the air
 1 inside the cylinder. Conservation of energy
 6V 
T0V0 1  T2  0  (ii) between the initial and the final states gives
 10 
5
Wy  [ p1 A(h  y )  p2 A(h  y)]
T1 3 2
From (i) & (ii) we get 
T2 7 [2 p0 Ah] (i)
348 Thermodynamics
where p1 is the final pressure in the lower part The final temerature of compartment B is
of the cylinder and p2 that in the upper part.
PBVB
TB 
The internal energy of a gas made up of diatomic nR
5
molecules has been written in the form pV . 125P0
2 PB  PC  (Mechanical equilibrium)
27
As W is very large the decrease in potential
energy will be converted into internal energy. 41V0
VB  , n 1
25
The initial internal energy of the gas is negligible
in comparsion with final energy.
125P0 41V0

5  TB  27 25  205PV
0 0
Wy  [ p1 A(h  y )  p2 A(h  y)] (ii) R 27 R
2
When the load is finally at rest, 205PV
0 0
TA  TB 
27 R
 p1  p2  A  W (iii)
Since the wall between A & B is conducting in
The temperatures and the masses of the gases nature.
in the two halves are identical, and so their
internal energies must be equal : 28. (2) The efficiency of Carnot engine is given
by
5 5
p1 A(h  y )  p2 A(h  y ) (iv)
2 2 T2
 1
T1
5
Equation (i), (ii), (iii) & (iv) yield y  h for
7 Given, T1  600K , T2  300K
the displacement of the piston, i.e., the gas in
the lower part is compressed to 15 percent of 300
  1 
its original volume. 600
26. (4) Volume of the ideal gas is constant so 1 1
W  P V  0   1 
2 2
Q  U  U  i 2 Rt
W 800 1
 12 100  5  60  30 103  30kJ . Also,   Q  Q  2
27. (3) The final volume of the compartment C Thus, amount of heat supplied from source to
is
engine is Q  800  2  1600 J
 125P0  27 
PV
0 0  VC  VC  V0 29. (2) Here, Q1  Heat absorbed per minute
27 125
Q2  Heat rejected per minute
1 2
 27    27  3 9V0
VC    V0    V0  VC  25 W
 125   125  We know that %   100
Q1
The final volume of compartment B is
9V0 41V0 5.4 108 J 3
VB  2V0   %  9
100  100  15%
25 25 3.6 10 J 20
Thermodynamics 349
T1  T2 TL W 1
30. (1) In first case 1  35. (2) 1  1   
T1 TH Q1 6
2T1  2T2 T1  T2 or 5TH  6TL  0 (i)

In second case 2  2T1

T1

TL  62 1
2  1   21 
31. (1) Heat received (or produced by the burning TH 3
of petrol) in hour will be
 (2.4kg / hour)(35.5MJ / kg) or 2TH  3TL  186 (ii)
 85.2 106 J / hour  TH  372K  99C
 The rate at which heat is received 5 5

TL  TH   372K  310 K  37C
6 6
85.2  106 J
  2.37  104 J / s  23.7kW
(3600s) Q
36. (1) S 
T
The rate of heat rejection = rate at which
heat is produced- rate at which work is done In isothermal expansion T is constant and Q
is (+)ve since the gas absorbs heat so
= 23.7 kW-10kW =13.7kW
Q
S  0
 300  11 T
32. (1)   1  
 850  17
37. (1) S  S1  S2
Work done in each cycle is
To melt ice + to rise the temperature
W 1200
  Qin   17 T
Qin 11 Q 2 mcdT mL T2 T
    mc 
T T1 T T T1 T
1200  17
Qrej  Qin  W   1200  654.54
11
1 80 313
  11 2.303log
500 3 273 273
33. (1) In first case, (1 )  1  
800 8  0.28  0.1366  0.42calC 1
600
and in second case, (2 )  1 
x
TL
3
Since 1  2 , therefore  1 
600 1. (600)   1  T
H
8 x
300
600 3 5 600  8 1
or  1   or x   960K 900
x 8 8 5
W 2
34. (4) Coefficient of performance 
Qs 3 ( Qs - heat absorbed)
T2 (273  13) 260
 5  3
T1  T2 T1  (273  13) T1  260 Qs  W
2
 5T1  1300  260  5T1  1560
3
  1200
 T1  312 K  39C 2
350 Thermodynamics
Qs  1800 J V 
 1
5
T2  T1  1  monoatomic gas :  
Rejected  Qs  W  1800  1200  600 J  V2  3
2. (2) Given that 5
1
 V 3
W  30 J  T2   300 K   
 2V 
Heat given to the system, Q  40 J
=189 K (final temperature)
Change in internal energy is given by
U  Q  W  40  30  10 J
1. (1) Given that
Cp
3. (3)  mix  mix
CV V  T 3 (1)
mix
For adiabatic process

n1CV  n2 CV
CV  1 2
1
mix
n1  n2 (2)
V  T  1
3 5 From (1) & (2)
5 R  2 R
 2 2  25R
5 2 14 1 4
 3   
39 R  1 3
C p  R  CV 
mix mix
14 2. (1) In adiabatic process
Cp 39 Q  0,W  35 J
mix
  1.56
CV mix
25
From 1st law of thermodynamics,
4. (1) For an adiabatic process
Q  U  W ,
1
V T  Constant
 0  U  35  U  35 J
1 1
V1 T1  V2 T2 In the second case
1  1 
1
Q  12cal  12  4.2 J  50.4 J
V1 T1   V1  aT1
 32  W  Q  U  50.4  35  15.4 J
2

1
7 3. (4) For adiabatic process, U  100 J
 a   32    32  5  4    
 5 Which remains same for other processes also.
5. (2) In an adiabatic process
Let C be the heat capacity of 2nd process then
PV  constant

  C  5  U  W
and,PV = nRT, gives
 100  25  75
 1 1
V 
T  C  15 J K
 1
 V1  T  4. (1) The average time of collision between
   2 
 V2   T1  two gas particle is
Thermodynamics 351
1 4. (3) This is an adiabatic expansion. So


  3RT
N temperature falls.
2d 2  
V  M 5. (1, 3) In the given one complete cycle,
1  2  3  1, the system returns to its initial
V
 state.
T
dU  0 and dQ  dW , heat is completely
For adiabatic process TV 1  K
converted into mechanical energy, which is not
So,   V ( 1)/ 2 possible in such a process. Further, the two
adiabatic curves (2 3) and (3 1) cannot intersect
 1 each other. Choice (1) and (3) are correct.
Therefore, q 
2
1. (1) Amount of sweat evaporated/minute

1. (3) Let initial volume of the gas in the cylinder
be V. Sweat produced

Heat required for evaporation
 V1  V
and V2  V / 2 14.5 104 145

   0.25kg
580 103 580
As hydrogen is a diatomic gas
2. (1) When the compression is isothermal for
7
    1.4 gas in A,
5
For an adiabatic change, p1V1  p2V2 p2V2  p1V1
 p1V1 V
p V  p2   p1 1  2 p1
 2   1   (2)1.4  2.64 V2 V1 / 2
p1  V2 
For gas in B, when compression is adiabatic,
2. (2) Work done by the gas is equal to area of
rectangle ABCDA,  AB  BC  (2V0 ) p0  2 p0V0 p2 ' V2 '  p1V1
For anticlockwise direction work is (-)ve work 

V   V 
p2 '  p1  1   p1  1   2 p1
 Work done by the gas  2 p0V0 V '
 2   V1 / 2 
3. (4) Given, work done (W )  5.4  108 J / min p2 ' 2 p1

   2 1
Total heat energy taken from the boiler, p2 2 p1
Q  3.6  109 J / min 3. (1) Given, temperature of source

(T1 )  (36  273) K  309K
Efficiency of
W 5.4 108 Temperature of sink (T1 )  (9  273) K  282 K
( )   100  100  15%
Q 3.6 109
Coefficient of performance of a refrigerator
Heat wasted per minute  Q  W
T2 282
9 8 9    10.4
 3.6 10  5.4 10  3.110 J / min T1  T2 309  282
352 Thermodynamics
4. (2, 3) From the given initial state A to final  V  V1 

state B, Change in internal energy is same in all W12   2   P1  P2 
 2 
the four cases, as it is independent of the path
from A to B. 1
W13  V3  V1  P1   V3  V1  P3  P1 
As work done= area under p-V curve. 2
Therefore, work done is maximum in case 1.  V  V1 

Choice (2) and (3) are correct. W13   3   P1  P3 
 2 
5. (1, 3) Figure represents the working of
refrigerator, where 1
W12  W13   P1 V2  V3   V1  P3  P2       ve
2
Q1  Q2  W  W  Q1  Q2
 V3  V2 , P2  P3
Two possibilities are
W13  W12  Q13  U  Q12  U
If W  0, Q1  Q2  0.
 Q13  Q12
Both Q1 and Q2 are positive.
(2  105  1  105 ) 105
4. (2) a  N / m2
Q 2  Q1  0 2 2
If Q2 is negative, Q1 also negative (but less (200  100) 3

b cm b  50cm3
negative as W>0). 2
Wcycle  2.5  joules.
W = (-) ve As P is on x-axis and the gas follows

1. (3) Equation of line is
clockwise path.
PV0  PV
0  3PV
0 0
(i)
Wcycle  Area   ab , q  W  U cycle  0 
Also PV = nRT (ii)
5. (2) Path BC
From the above two eq’s
As volume decreases WBC  ()ve
3PV PV 2
T 0  0 Given that PB  2,VB  3 , PC  1,VC  2
nR nRV0
PBVB  PCVC TB TC
dT 3V 3P
for Tmax ,  0 V  0 ,P  0
dV 2 2 So the temperature decreases
U BC  nCV (T )  nCV (TC  TB )  ()ve
9 PV
 Tmax  0 0 QBC  U BC  WBC  ()ve
4nR
similarly W CD  (  ) ve as volume decreases
2. (1) , 3. (2) Let V1 , P1  , V2 , P2  and V3 , P3  are
the coordinates of the points 1,2 & 3. From the PDVD  PCVC TD TC
graph PV2 2  PV
3 3  nRT
(constant temperature U CD  nC (T )  nC (TD  TC )  ()ve
along the curve)
QCD  U CD  WCD  ()ve
1
W12  V2  V1  P1   V2  V1  P2  P1  QBCD  ()ve (heat flows out from the gas )
2
Thermodynamics 353
WABCDA  ()ve as it is clock wise cycle. dV V 1 dV 1

  
6. (4) The process AB is a rapid compression dT T V dT T
and hence it can be treated as adiabatic T
compression. The process BC is isochoric and 1
the process CA is isothermal expansion.
Area    dt    T  dT  ln(3)
T /3
7. (4) In cyclic process, change in total internal 10. (3) When temperature remains constant then
energy is zero. the process is isothermal. In isothermal process
 U cyclic  0 C  . so graph (3) is not possible.
11. (3) & 12. (1) Given that PV x  constant

5R
U BC  nCV T  1 T
2 we know that for an ideal gas PV  nRT
Where, CV  molar specific heat at constant From the above two equations we get
volume. For BC, T  200 K TV x 1  K (constant)
U BC  500R T ( x  1)V x  2 dV  V x 1dT  0
8. (1) Path AB
T ( x  1)
The path AB is isobaric dV  dT (i)
V
V From 1st law of Thermodynamics
P = constant   constant
T
dQ  dU  dW
V  increases  T  increases
nCdT  nCV dT  PdV
Path BC
The path BC represents isothermal process PdV PV
C  CV   CV 
T = constant ndT nT ( x  1)
P  Decreasing V  Increasing R R R
C  CV   
Path CD x 1  1 x 1
The path CD is isochoric
R R
C 
P  1 x 1
V = constant   constant
T
When the graph intersects x - axis
P  Decreases T  Decreases
R R 7
The graph between P and T is a straight line C 0  x
 1 x 1 5
passing through origin.
Path DA When the graph intersects y-axis x  0
The path DA isothermal R R 7

C R  R
T = Constant, P  increases V  Decreases  1  1 2
9. (4) Given that VT = K on differentiating the 13. (3) Given that P = Constant
equation
V
  K  constant   V  K  T
VdT  TdV  0  VdT  TdV T
354 Thermodynamics
V 1 K 1
  (K )  
V T V V T
1. (2) As the volume is continuously increasing
1 and the work of expansion is always positive,
The graph between   and T is hyperbolic.
T so the work done by the system continuously
T2 increases.
14. (1)  1
T1
4g
2. (2) Number of moles n  4 g  1
T2
When T1 increases T decreases and
1
1 3PV
0 0
 increases. As T1      1 WAB  PV
0 0  PV
0 0 
2 2
3
15. (4) Let slope of line AC is m1 & that of line U AB  ncv T   2P0 2V0  PV
0 0
2
T coordinate
DB is m2 , then slope(m)  9
P coordinate U AB  PV
0 0
2
T  coordinate
 P  coordinate  Q  W  U AB  6 PV
m 0 0
T1 T T T 3. (2)  Monoatomic   diatomic

PA  , PB  1 , PC  2 & PD  2
m1 m2 m1 m2 The graph 2 is more vertical than graph 1
1- corresponds to diatomic
2- corresponds to monoatomic.
4. (1) Wcycle  Area of the cycle  r 2  rV rP
(20  10)
Pressure radius  rP   kPa  5kPa
2
(40  10) 3
Volume radius  rV   m  15m3
2
Wcycle  75 KJ

Given that WAB  2WCD
W  ()ve (clock wise direction when P is on y-
P P axis)
nRT1 ln A  2nRT2 ln C
PB PD
W  ()ve (clock wise direction when P is on x
m2 m - axis)
nRT1 ln  2nRT2 ln 2
m1 m1
5. (1) The clockwise cycle area is more than
T1 the area of anticlockwise cycle.
2
T2 6. (1)
Thermodynamics 355
7. (3) & 8. (4) W 0

 
 PV   R  3 QCD  3PV
0 0
QAB   0 0    2T0  T0   PV
0 0
5
 RT0   1  2 Path DA
3 
PV
0 0  2 PV
0 0 3PV
0 0
 PV   5R  U  
2 2
QBC   0 0    4T0  2T0   5 PV
0 0
 RT0   2  3
The heat absorbed in the total cycle is W   PV

0 0
13 5 PV
 Qa  QAB  QBC  PV
0 0 QDA  0 0
2 2
In the paths AB and BC gas is absorbing the
W PV 2
  net  0 0  heat.
Qa 13 13
PV
0 0
2 3PV
0 0 13PV
0 0
Qabsorbed   5PV
0 0 
2 2
9. (2) Consider the cyclic process ABCDA
10. (3) Path AB
The path AB passes through the origin V  mT
( m - slope)  P = constant
V  increases, T - increases.
Path BC
Path AB
The path BC is isochoric and hence V = constant
PV
2 2  PV 2 PV  PV  P = mT (m - constant )
U  1 1
 0 0 0 0
 1 5 T - decreases  P - decreases
1
3
Path CA
3PV
0 0 The path CA is isothermal
U  ,W  0
2
PV = constant
3PV
0 0 V - decreases  P - increases
QAB  U  W 
2
1 dV
Path BC 11. (3), 12. (2) T 2
V dT
4 PV
0 0  2 PV
0 0 dV 2
U   3PV
0 0  dT
2 V T
3
2
ln V  ln T 2  V  T
W  2 P0 (2V0  V0 )  2 PV
0 0
V ( PV )2
QBC  5 PV
0 0
P 2V  constant
Path CD
 PV 1/2  constant
2 PV  4 PV
U  0 0 0 0
 3PV
0 0
2
 P V  constant
3
356 Thermodynamics
If a gas undergoes a relation PV x  constant value of f=3 so slope of the line is less for
isobaric process. Line -4 represents isobaric.
 1 1  Line- 2 represents adiabatic ( N 2 ) . Line -3
The heat capacity C  R   
  1 x 1 represents adiabatic (Ar)
1 7
hence x  ,  
2 5
   
 1 1  1. (3) PV
i i  Pf V f
C  R  9
7 1  R
 1 1 2
5 2  Pi 105
Pf  
3 31.4
13. (1) & 14. (2) we know that dQ  nCdt
Pf V f  PV
i i
Q   nCdt  n  Cdt W
1 
Q  n (Area of the graph)
105
1.4
 3  103  105  103
1 9T R
 n(  T0  R  T0 R)  0 3
2 2 1  1.4
W  90.5 J
The work done W  PV  nRT  3RT0
3 2. (3) Q  U  W
U  Q  W  RT0
2
U  Q  W
15. (2) & 16. (3) As the graph is hyperbolic the
relation between U and  is  U  ACB
  U ADB
U  constant (i)
 60  30  Q  10
5
U  nCV T  n RT (ii)  Q  40J
2
3. (2) Work done on gas
  m /V (iii)
 pf 
From (i) (ii) & (iii)  nRT ln    R (300) ln(2)  300 R ln 2
 pi 
5 m T
n RT  constant   constant  CP  n1CP1  n2 C P2
2 V V   
4. (3)
 CV  mip n1CV1  n2 CV2
 P  constant (isobaric Process)
W  PV  nRT n1  2, n2  n
2U U 0 5 5 5 7
W  (U  nRT  U  nRT ) CP1  R CP2  R
5 5 2 2 2 2
7U 0
Q  U  W  3 5
10 CV1  R CV2  R
2 2
nf
17. (3) For adiabatic process, WA  RT and
 CP 
2 3
  
for isobaric process WI  nRT . As minimum  CV  mix 2
Thermodynamics 357
3 10  7n 9. (1)

2 6  5n Given PV  nRT  V (i)
n2
 (P  )V  nRT
5. (3) From P-V graph,
nRT
V
1 ( P  )
P  , PV  constant  T= constant
V
nRdT
The correct graph is option (3) dV 
( P  )
6. (4) Volume of water does not change, no
work is done on or by the system (W = 0) Now work done is  dW   PdV
According to first law of thermodynamics
From (i) PdV  nRdT  dV (P=constant)
Q  U  W
 dW   nRdT  dV
For Isochoric process Q  U
U  mcT  2  4184  20  16.7kJ .   nRdT 

 (nRdT )
P
7. (2) Energy of microwaves lie in range of
2T0 2T
vibration energy of water molecules.  0
 nR  dT  nR  dT ( P  P  constant)
P0   T0 0
8. (2) Area of the piston is A  8  103 m2 T0
T1  300K ; V1  2.4  103 m3  nRT0 P0

 nRT0   nRT0 
P0   ( P0  )
V2  V1  A x
RT0 P0
 2.4 103  8 103  0.1  3.2  103 m3  W
( P0  ) ( n  1 )
K  8000N / m ; T2  ? ; P1  105 N / m 2
10. (2) Given Q1  1000 J ; Q2  600 J
Initially the piston is in equilibrium. Let P0 is the
atmospheric pressure. T1  127C  400 K
P0  P1 (pressure of the gas) Efficiency of carnot engine,
After expansion again the piston is in equilibrium W

  100%
Q1
Kx
P2  P0 
A Q2  Q1
  100%
Kx 8000  0.1 Q1
P2  P1   105  3
 2 105 N / m2
A 8 10
1000  600
P1V1 P2 V2   100%  40%
 1000
T1 T2
105  2.4  103 2  105  3.2  103 2 2 Q T
 Now, for carnot cycle Q  T
300 T2 1 1
3.2  300 600 T

T2   800K  2
1.2 1000 400
358 Thermodynamics
600  400 PV
2 2 - PV
PV 3PV
T2   1 1
0  0 0  0 0
 240K  240  273 2
 -1 5 2
1000 1
3
T2  33C QBC  U  W
4 PV
0 0  2 PV
0 0
  2 PV
0 0  5PV
5 0 0
1. (1) Work and heat are path functions. 1
3
In the paths CD & DA, Q is (-)ve
 16 28 44 
2. (4) PV      RT
 32 28 44  Wcycle PV
0 0 2
    0.15
Qabsorbed 3PV
0 0 13
 5 PV
0 0
5 RT 2
or, P
2 V 6. (2) H  mL  5  336  103  Qsink
 T2  Qsink T
3. (3) Carnot efficiency   1  T  ,  sink
Qsource Tsource
 1 
where T1 and T2 are temperature of source and Tsource

 Qsource   Qsink
sink and respectively. Tsink
T2 Energy consumed by freezer

 0.6  1 
127  273 T 
Wouput  Qsource  Qsink  Qsink  source  1
 Tsink 
T2
or,  0.4
400 Given: Tsource  27C  273  300 K ,
 T2  160K Tsink  0C  273  273K
 T2  160  273 C  300 

Wouput  5  336  103   1  1.67  105 J
 273 
or, T2  113C
7. (1) Efficiency of heat engine is given by
T2 W
4. (2) Efficiency  1  T  Q  W W PV nRT R R 2
1 2      
Q nCP T nCP T CP 5R 5
250 W 2
 1 
300 Q2  W
5
For monoatomic gas CP  R.
Q2 500  4.2 2
W  J  420 J
5 5 8. (2) In VT graph
5. (2) The net work done in the cycle is equal ab-process: Isobaric, volume and temperature
to the area enclosed in the graph  Wcycle =P0V0 increases.
Path AB bc-Process : Adiabatic, volume increases and

pressure decreases.
QAB  U  W
cd-process : Isobaric, volume decreases.
Thermodynamics 359
da-process: Adiabatic, volume decreases and

pressure increases.
The above processes correctly represented in
P-V digram (2).
9. (3) Work done W =830J
No.of moles of gas, n =2
So the correct option is (3)
For diatomic gas   1.4
12. (2) Volume of the gas
Work done during an adiabatic change
m
nR(T1  T2 ) V
W d
 1
PV   constant
2  8.3(T ) 2  8.3(T )
 830  
1.4  1 0.4 
P'  V   d '

   
P V '   d 
830  0.4
 T   20K
2  8.3 128  (32) 
10. (1) From 1st law of thermodynamics
7
   1.4
Q  U  W  Q  U  PV 5
150  U  100(1  2) 13. (2) In ideal gases the molecules are

considered as point particles and for point
 U  150  100  250 J particles, there is no internal excitation, no
vibration and no rotation. For an ideal gas the
Thus the internal energy of the gas increses by
internal energy can only be translational kinetic
250J
energy and for real gas both kinetic as well as
11. (3) P-V indicator diagram for isobaric potential energy. The internal energy of an ideal
gas is a function of temperature only so when
ideal gas undergoes free expansion W=0, Q  0
(insulated chambers)  U  0 so
temperature of the ideal gas remain constant.
Whereas internal energy of a real gas is a
function of both temperature and pressure.
When a real gas undergoes free expansion the
P-V indicator diagram for isochoric process gas gets cooled.
14. (3) Q  U  W
U  Q  W
U  mL  P(V2  V1 )
 J
 1g  2256   105  (1671  1)  106 m3
P-V indicator diagram for isothermal process  g
 2089
360 Heat Transfer
T
5. (1) The temperature gradient is
l
30  
 80  30    40    10C
1. (3) For cooking low specific heat is perferred 0.5
for utensil material as it should need less heat to 6. (1) Let T1 & T2 be the temperatures
raise it’s temperature and it should have high
conductivitiy, because, it should transfer heat across the two layers. As the two layers are in
quickly. series the rate of heat flow is same in both.
2. (3) Rate of melting ice  rate of heat K A AT1 K B AT2


transfer (dQ/dt)  
dQ T 2K B T1  K B T2
Further 
dt ( / KA)
 2T1  T2 (1)
dQ T
 A Given that T1  T2  60K (2)
dt l
From (1) & (2)
If T , A and  are all doubled, then (dQ/dt)
and hence rate of melting of ice, are doubled. T1  20K & T2  40K
3. (2) Q = mL in both cases 7. (4) As rate of heat flow is same in both rods
K1 A  T  t1 K 2 A  T  t2  1    A1     2  A2
  l1 l2
d d
 K1 K 2  t 2 t1 (100  )2 A (  70) A


l l
4. (3)
   90C
2
Q KA Q A d
    [d=Diameter of rod] 8. (2) Let  is the temperature of the junction.
t l t l l
As heat flow is same in both rods
2
(Q / t )1  d1  l2 K1 A1 (100  ) K2 A2 (  0)
    
(Q / t )2  d2  l1 18cm 6cm
2 Given that A1  A2 , K1  9 K 2
1 1 1
     
2 2 8    75C
Heat Transfer 361
KA(T1  T2 )t 2 K1 K 2 2.K .2 K 4
9. (1) Since Q . For same K'    K.
L K1  K 2 K  2 K 3
insulation the rate of heat flow must be same
and even A & (T1  T2 ) are also same. Hence, KA(T1  T2 )t
14. (1) Since Q . For same
L
K
remain constant. If K1 and K 2 be the insulation the rate of heat flow must be same
L
thermal conductivities of brick and cement K
and even A & (T1  T2 ) are also same. Hence,
L
respectively and L1 and L2 be the required
thickness then remain constant. If K1 and K 2 be the thermal
conductivities of brick and cement respectively
K1 K 2 1.7 2.9
 or  and L1 and L2 be the required thickness then
L1 L2 20 L2
K1 K 2 1.7 2.9
 or 
2.9 L1 L2 20 L2
 L2  20  34.12cm
1.7
2.9
10. (1) In parallel combination equivalent thermal  L2  20  34.12cm
1.7
resistance is
15. (3)
1 1 1 16. (3) When element and surrounding have same
 
R R1 R2 temperature heat will not flow from the filament
and it’s temperature remains constant.
K ( A1  A2 ) K1 A1 K2 A2
  17. (4) Open window behaves like a perfectly
l l l
black body.
18. (2) According to Wien’s law  mT  constant
K A  K 2 A2 K1  K 2
K 1 1  [As A1  A2 ].
A1  A2 2  m1 0 4
  m1 T1   m2 T2  T2  T1   T1  T1
 m2 3 0 / 4 3
dQ dQ1 dQ2
11. (1)  
dt dt dt Now
4 4
4 R 2 (1   2 )  R 2 (1   2 ) P2  T2  P  4 / 3T1  256
K  K1 P T4     2   
L L P1  T1  P1  T1  81
3 R 2 (1   2 )
 K2 m T ' 3000 3
L 19. (1) mT  m' T '     .
m' T 2000 2
12. (4)
20. (3)
13. (2) The equivalent thermal resistance is
T1 2 104
mT  constant     200.
R  R1  R2 T2 1 0.5  106
2l l l 21. (2) From Stefan’s law E   AT 4

 
k ' A K1 A K2 A
( E / A) 6.3 107
T4    1.105 1015
 5.7 108
 0.1105 1016
362 Heat Transfer
T  0.58 104 K  5.8 103 K . E T 4
Q E1 T14
22. (4) For a black body  P  A T 4 Here, E  T 4
t 2 2
4 4 4 4
P T  P  273  727   300   1  1
 2  2   2       
P1  T1  20  273  227   1200   4  256
Hence, E1 : E2  1: 256
P 4
 2   2   P2  320W .
20 27. (3) Rate of cooling  
23. (4) Rate of cooling  T 4  T04  28. (2) The heat from hot milk spread on the table
is transferred to the surroundings by conduction,
H T1  T0 
4 4
convection and radiation. According to Newton’s
  4 law of cooling the temperature of mlik falls
H ' T2  T04 
exponentially with time.
H'
T  T  H    400    200
2
4
0
4 4 4
 3
 H  H.
29. (2)
T  T    600    200 
1
4
0
4 4 4
 16 30. (4) According to Newton’s law of cooling
2 4 Rate of cooling  Mean temperature difference
Q  r  T 
2 4
24. (3) Qr T  2  2   2 
Q1  r1   T1  Fall in temperature  1  2 
   0 
Time  2 
2 4
  2    2   64.
        
25. (3) Radius of sphere r =3cm=0.03m  1 2    1 2    1 2 
 2 1  2 2  2 3
The energy absorbed
 T1  T2  T3
E  30kW  30  103 watt
31. (2) Initially at t  0
Energy absorbed by black body when the
temperature of surroundings is T, is given by Rate of cooling  R  Difference intemperature
E   AT 4 of body    0 
or 30  103  5.672  108  4 (0.03)2  T 4 R1 1   0 100  40 3

   
R2  2   0 80  40 2
30 103  0.064 108  T 4
30 103
T4   46.875 1012
0.064 108 1. (4) Heat energy always flow from higher
temperature to lower temperature. Hence,
 2.616  103  2616K
temperature difference w.r.to length
26. (2) Here, initial temperature, (temperature gradient) is required to flow heat
T1  27  C  300 K from one part of a solid to another part
2. (3) Mud is a bad conductor of heat. So it
Final temperature, T2  927  C  1200 K
prevents the flow of heat between surroundings
According to Stefan’s law, the radiant energy is and inside.
Heat Transfer 363
3. (2) Thermal conductivity is independent of 9. (4) Temperature of interface

temperature of the wall, it is a constant for the
1 2
material, so it will remain unchanged. 
R1 R2

4. (4) The densest layer of water will be at 1 1

bottom. The density of water is maximum at R1 R2
4C. So the temperature of bottom of lake will
be 4C.
1  0 C, 2  100 C
5. (2) The thermal resistance of a rod of length 1 2

R1  , R2 
l and area of cross section A is kA 3kA
R
l    60 C
kA 10. (4) Let T be the temperature of the interface.
K - Thermal conductivity of the rod. As the two sections are in series, the rate of
flow of heat in them will be equal.
A1 k2 3
  
A2 k1 5
K1 A1 1   2 
6. (4) i1  and
l
K1 A(T1  T ) K2 A(T  T2 )
K 2 A2 1   2    ,
i2  l1 l2
l
where A is the area of cross-section.
given i1  i2  K1 A1  K 2 A2
K1 A(T1  T )l2  K 2 A(T  T2 )l1
7. (1) Rate of heat flow
K1T1l2  K1Tl2  K 2Tl1  K 2T2l1
 Q  k r 1   2  r
2 2
 ( K 2 l1  K1l2 )T  K1T1l2  K 2T2 l1

 
t  L L
K1T1l2  K 2T2l1
T
2
Q1  r1   L2   1   2  1
2 K 2 l1  K1l2
           
Q2  r2   L1   2   1  2 11. (2) The equivalent thermal resistance is
R  R1  R2
 Q2  2Q1
2l l l 4k
8. (2) Let  is the temperature of the mid point    keq 
keq A kA 2kA 3
(1   ) A1 (   2 ) A2
 12. (3) Thermal resistance
l1 l2
l l 3l
(100   )3 A (  120)2 A R  
 kA 2kA 2kA
1 1
13. (2) Given, K B  K A 2
300  3  2  240
300  240  2  3  540  5    108C and K B  3K C
364 Heat Transfer
 KC  K A 6 22. (4) From Wien’s law
Rods are in series form, so 1T1  b
R eq  R1  R 2  R 3 n1  T2  b  T2  T1 / n
 P1  CT14
Where R 
kA P2  CT24  C (T1 / n) 4  1 / n4
L l l l 23. (1) The Wien’s radiation law is given by

 1  2  3  l  l  l  l 
K K A K B KC 1 2 3
mT  b
3l l 1 l For the given problem, b  2.88  103 mK .

  
K KA KA 2 KA 6
Hence, mT  2.88  103
3l 9l K
  K A 2.88  103
K KA 3 T 
m
14. (4)
2.88  103
15. (2)   6000 K
480  109
16. (3) When light incident on pin hole enters into
the box and suffers successive reflection at the 24. (3) Heat radiations are electromagnetic waves
inner wall. At each reflection some energy is of high wavelength.They can travel in vacuum
absorbed. Hence the ray once it enters the box 25. (1)
can never come out and pin hole acts like a
perfect black body. 26. (2) Convection and conduction mode of heat
transfer require physical contact between the
17. (2) A perfectly black body is a good absorbers bodies to transfer heat but radiation does not
of radiations falls on it. So it’s absorptive power require any medium. As they transfer heat as
is 1. electromagnetic waves/radiation. This is
18. (1) explained by stefan’s law
19. (1) By Stefan’s law dQ dQ
27. (1)   AT 4  A
dt dt
Q   AT 4
28. (3) Power radiation is
The rate of dissipation of heat by a body of
emissivity e is given by Q T 4
4 4
Q '  e AT '4 Q1  T1   T 
   
Q 2  T2   3T 
Here, T '  2T , e  0.25
 Q 2  81Q1  81Q
4
 Q '  0.25    2T   4 T 4  4Q
E1 T14 E
29. (2) E T4   4  24  E2 
20. (2) E2 T2 16
m2 T1 2000 Q1 T14 E 

21. (2)    m2   4  3.33 m. 30. (1) QT4   4  T24   2  T14
m1 T2 2400 Q2 T2  E1 
Heat Transfer 365
1  1000 
4
dT  A 4
 T24 
4
 1000     38. (3)  T  T04 . If the liquids put in
16  2  dt mc
exactly similar surrounding then the rate of
 T2  500 K . cooling will be same if A, m, c & T must be
A
Q T 
4
Q T  2   2 
4
same or
mc

T 4  T04  must be same for both
31. (2)
Q1  T1 
39. (3) From Newton’s law of cooling
4
Q1  T  16 81 (80  60)  80  60 
     Q2  Q1  K  30 
Q2  T  T / 2  81 16 10  2 
Q2  Q1 60  T  60  T 
% increase in energy  Q1
 100  400%.  K  30 
10  2 
32. (1) PR  AT 2 T  48C
4
40. (1) The newton’s law of cooling equation is
 5.67  108  1.5  310   785 J .
t
2 d
4   k  dt
Q1 r12T14 4 2  2000  1
  0 0
33. (2)      1.
Q2 r22T24 12  4000 
  
34. (1) From Stefan’s law, the emissive power of n  2 0   kt
a body at absolute temperature T is  1   0 
E  e T 4 A  2   0  1   0  e  kt
' T Given that

When R '  100 R and T  then energy
2 1  0
emitted is   2  0
2
4
T 
E '   4 (100 R ) 2   1 n  2 
2   e  kt  t 
2 k
 100 
'
2
41. (2) Black bulb absorbs more heat in
E    E
 4  comparison with painted bulb. So air in black
bulb expands more. Hence, the level of alcohol
E' in limb X falls while that in limb Y rises.
  625
E
35. (3)
1. (3) The heat required to melt ice is
36. (4) Newton’s law of cooling is used for the
determination of specific heat of liquids. KA  1  2 
Q t  K1t1  K 2 t2
l
37. (2) For small temperature difference,
Newton’s law of cooling is the special case of K1 t2 35 7
   
Stefan’s law. K 2 t1 20 4
366 Heat Transfer
2. (1) Q = it where i = heat flow rate dQ  dmL  (Ady)L

T 100
  L y t 1 L 2
R R ydy   dt  t  y

K 0 0 2 K
 100   100 
Q   30  R1     30 5. (4) Consider a circular disc of radius r present
 R1   Q  at a distance x from the front end of the cylinder
as shown in the figure
 100   100 
Also, Q   R   30  R2   Q   60
 2   
When the rods are connected in parallel
1 1 
Now, Q   R  R 100t
 1 2 
1 1 Q Q Q
    The rate of heat flow across the disc is
R1 R2 3000 6000 2000
K r 2 dT
Q  100t i
Q  t  20min dx
2000
dx K
3. (2) Here, x  4 mm  4  103 m, T  32C r  dT
2
i 
dQ 200 1000  4.2 From the figure
 200kcal / h  J/s
dT 60  60
r  r1 r2  r
 233.33J / s tan   
x L
A  5cm2  5  104 m2  r  r1  x tan 
dQ  T  dr  dx tan 
We know that,  KA  
dt  x 
r r
1 2 dr K  2
 Thermal conductivity of material,  dT
tan  r1 r 2 i r1
Q / T
K
A  T / x 
1  1 1  K
3    [T2  T1 ]
233.33  4 10 tan   r1 r2  i
K 4
 58.33W / m  C.
5 10  32
4. (1) When water gets converted into ice heat L  r2  r1  K 
  [T2  T1 ]
liberates and this heat flows through ice slab. r2  r1  r1r2  i
The heat flow through the ice slab of thickness

K r1r2 [T2  T1 ]
dQ
 KA
  () i
y is L
dt y
The small heat dQ comes when water converts 6. (3) Consider a concentric spherical shell of
into ice radius r and thickness dr as shown in the figure.
Heat Transfer 367
The radial rate of flow of heat through this shell Let T be the temperature of the junction
in steady state will be
dQ k1 A  T1    k 2 A    T2 
 
dQ dT dt d1 d2
H   KA
dt dr
 k1d 2 T1     k2 d1    T2 
dT
  K  4r  2
dr k1 d 2T1  k 2 d1T2

k1 d 2  k 2 d1
r2 dr 4K T2
 2
  dT
r1 r H T1 9. (2) Thermal resistance of a rod is
l
R
KA
Which on integration and simplification gives The equivalent resistance between B and C is
As rate of heat flow is same
dQ 4Kr1r2 T1  T2  dQ rr
H    12 TA  TB TB  TC TC  TD
dt r2  r1 dt  r2  r1  R

R

R
7. (2) Heat current:
60  TB  TB  TC (1)
dT
i  kA TC  240  TB  TC (2)
dx
idx  kAdT Solving (1) and (2)
Given that TB  120C

K  T
10. (1) TB  TA  Heat will flow from B to A via
l T2
i  dx   A  TdT two paths (i) B to A (ii) and along BCA as shown.

0 T1
il   A
T 2
2
 T12 
i 
A T12  T22 
2 2l
8. (3) In series both walls have same rate of
heat flow
Rate of flow of heat in path BCA will be same
Q Q
i.e.,    
 t  BC  t CA
368 Heat Transfer
13. (1) Every body at all time, at all temperature


K  
2T  TC A

K TC  T  A
emits radiation (except at T = 0). The radiation
a 2a emitted by the human body is in the infra-red
TC 3 region.
 
T 1 2 14. (3) According to Wien’s displacement law
11. (1) Let interface area be A, then the thermal b 0.0029

m   4
 58  109  58nm
1 L T 5  10
resistance of wood, R1  K A and that of brick
1 15. (1) Maximum amount of emitted radiation
wall
b
L 5 L1 corresponding to  
R2  2   R1 T
K 2 A 5 K1 A
Let thermal resistance of the each sandwitch layer 2.88 106 nmK
m   500nm
5760K
= R. Then the above wall can be visualised as a
circuit So U 2 corresponds to maximum emitted
radiation U 2  U1
16. (2) In steady state the radiation absorbed by
Thermal current through each wall is same. the sphere is equal to radiation looses by it.
As the parallel beam falls on the sphere the
25  20 20  T3 T3  T4 T4  20
Hence R  R  R  R radiation that falls is I (R 2 )
1 1
 25  20  T4  20  T4  150 C I R 2   T 4  T04  4R 2
20  T4
also, 20  T3  T3  T4  T3   2.50 C  I   T 4  T04  4
2
12. (3) Radius of cylinder = R  T 4  T04  40  108
Outer radius of shell = nR
 T 4  81108  40 108
l l

RR K1R K2  n  1 R2
2 2  T 4  121 108  T  330 K .
Req  1 2 
R1  R2 l l 17. (3) The three plates are shown in the figure

K1R K2  n  1 R 2
2 2
Where l is the length of the cylinders.

l
But Req 
K n 2 R 2
 K1  K 2  n 2  1 
On solving, K   
n2
4K1  5K2 K1  5K2 / 4 Let T ' is the temperature of the middle plate.
 
9 9/ 4 Here T ' should by lower than 3T and higher
n = 3/2 than 2T
Heat Transfer 369
In steady state energy absorbed by middle plate 21. (3) Let radius of sun =R
is equal to energy released by middle plate.
Distance of the earth from the sun = d
4 ' 4 ' 4 4
A 3T   AT   A T   A  2T  Power radiated from the sun
4 4 4 4   4 R 2   T 4  P
 3T   T '   T '    2T 
P
4
2 T '   16  81 T 4 Energy received per unit area is S 
4 d 2
1/ 4
 97  T4 4 R
2
T '    T.  4 R 2   T
 2  4 d 2 d2
18. (1) At steady state condition 2
1  2R 
 T 4  
P  AT 4 4  d 
1 Angle subtend by sun at the earth,
4 P  P 4
T   T  
A  A  2R

d
19. (1) P  A T 4  T04 
S  constant  T 4   2
 2rl T 4  T04 
S 2
4 4
 5.68 108  2103 1  900   300   dT 
  22. (1) P   4r 2  T 4   ms   
 dt 
 73.6W
Where m is the mass and s is the specific heat of
L the body
P  I 2 R  I 2
A
4 dT
4r 2 T 4   r 3s.
L 3 dt
 I2 0 1  
A
dT 3T 4

I 2  2 108 1  7.8 103  900  300  dt sr
73.6  2
103  T2 dT t
  4
  constant   dt
T1 T 0
or I = 36A.
20. (2) As heat is given both expands by the 1 1
or c 3  3   t
same values. C.M of Sphere B raises up and  T2 T1 
C.M of sphere A goes down slightly. The raise 23. (2) The rate of loss of heat of a hot body is
in P.E of B should come from the expense of
heat supplied. Whereas decrease in P.E of A dQ
 T 4  Ts4 
will be converted into heat. dt
 TA  TB T  Ts  T (T is small)
370 Heat Transfer
4
 T  P P
4 4T  1 1
T 4  (Ts  T )4  Ts4 1    Ts 1   Given, log 2 P  1  P0  2
 Ts   Ts  0
4
At 2767C; P2  eA  3040 
 4T 
T 4  Ts4  Ts4 1  4 3 3
  Ts  4T Ts  4Ts (T  Ts )
 Ts 
 P2   eA  3040 4  2 
Reading  log 2    log 2  4 
dQ  P0   eA  760  
  T  Ts 
dt
 log 2  44  2   9
24. (2) From Newton’s law of cooling,
1  2  1  2 
  0 
t  2 
1 K 5
where, 0  temperature of surroundings 1. (1) Given A1  A2 and K  4
2
50  49  50  49  l1 l l K 5
   30   R1  R2   2  1  1 
t1  2  K1 A K 2 A l2 K 2 4
(i)
2. (4) the time required to increase the thickness
40  39  40  39  of ice from x1 to x2
and   30 
t2  2 
L 2
(ii) t
2K 
 x2  x12   t   x22  x12 
Dividing Eq. (ii) by Eq. (i) , we get
Where  is the temperature of atmosphere
39
 t2   5  10 s
t  x2  x1  7 1  0 
2 2 2 2
19
 '  ' 2 '2   2 2  t '  21 hours.
t  x1  x2  t '  2  1 
d
25. (1)  k    0  , where k = constant.
dt 3. (2) Rate of flow of heat is constant
throughout the rod.
 d t
    k .dt Temperature difference is given by
i    0
0
T  iR

ln    0    kt
i l 1
where R   R
KA A
ln    0   ln  i  0   kt
Area across CD is less. Therefore, T across
  0 CD will be more.
 e  kt
i  0
4. (2) In steady state, the net outward heat flow
   0   i  0  e .  kt
is constant, and does not depend on radial
position.
26. (2) Power radiated P  eAT 4
2 dT
At 487C; P1  eA  760 
4 Let c1  i   K (4r )
dr
Heat Transfer 371
c
The total extension is
dT
  1 2
dr 4Kr 1
130
l  130  xdx   1.2  10 5  1
2
T c1 r 1 0
 dT    dr
2T0 4K a r2    0.78 mm.
c1 1 1  6. (1) In configuration I
T  2T0  r  a (i)
4K   total thermal resistance
When r  2a  T  T0   3 
R1   
KA 2KA 2 KA
c1  1 1 
T0  2T0   (ii)
4K  2a a  In configuration II
From eq’s (i) & (ii) we get  1

 1   1 
4T0 3a R2   2  
T at r   1  1  KA 3 KA
3 2  2
5. (2) From the given data;
 T  t
We have Q 
R
As Q is constant
l
Thermal resistance R  tR
KA
t2 R2 2
RPQ K 1   t2   9  2min.
 RS  t1 R1 9
RRS K PQ 2
7. (1) The rate of heat flow is same in all the
Let T is the temperature of the junction. As rate three rods.
of heat flow is same in both parts.
100  T1  2KA T1  T2  KA T2  0  KA
400  T  
so, T  10   T  140C l l 2l
2
The temperature of the first rod as a function of T2
2 100  T1   T1  T2  
distance x is 2
T  x   10  130 x 2T1
T1  85.7C and T2   57.1C
3
Each part of the rod has different change in
temperature w.r.to the initial value. 8. (4) Let T is the temperature of the interface.
The change in temperature of the rod as a K Copper  4 K & K brass  K
function of x w.r.to the initial temperature of the
rod is As the rate of heat flow is same in two slabs
 T  130 x 4K .A(100  T ) KA(T  0)


Extension in a small element of length dx is l l
dl  dx T  130 xdx T  80C

372 Heat Transfer
9. (2) For the two sheets, shown in figure, rate 13. (1) According to Wien’s displacement law a
of heat transfer is same, i,e., hot body emits different wavelengths at different
temperatures.
dQ1 dQ2 1   2
  
dt dt R1 R2 b 2.93  103
14. (3) T   107 K .
 m 2.93  1010
 1 R2  R2  R1  2 R1
15. (2) let P = power radiated by the sun, R =
1 R2  2 R1 radius of planet.

R1  R2
P
Energy received by planet   R2
10. (1) As it is clear from figure. 4d 2
dQ dQ1 dQ2 Energy radiated by planet   4R 2  T 2 .

 
dt dt dt
P
For thermal equilibrium,  R2  4R2 T 4 .
K  A1  A2  dT dT dQ 4d 2
 K1 A1  K 2 A2
dx dx dx
1
 T4  or T 4  d 1/ 2
K A  K 2 A2 d2
K 1 1
A1  A2
16. (3) Let r is radius of planet
11. (1) The equivalent thermal conductivity is
T1  Temperature of planet at which energy is
R  R1  R2 radiated
l1  l2 l l Power radiated by Sun, P  4R 2 T04
 1  2
K eq A K1 A K 2 A
Power received by planet
l l x  4x 5 P 2 R2T04 r 2 
K eq  1 2   K.   r 
l1 l2 x 4x 3 4d 2 d2
 
K1 K 2 K 2 K
Power radiated by planet  4r 2 T14
Hence rate of flow of heat through the given
combination is For thermal equilibrium,
5 R2T04 r 2 
K A T2  T1   4r 2 T14
Q K eq . A T2  T1  3 d 2
 
t  x  4x 5x
R 2T04  4d 2T14
1
K A T2  T 1
R
3 T1  T0
x 2d
On comparing it with given equation we get 17. (3) The heat radiated per unit time is
1
f  . dQ 2 4
3 R T
dt
12. (3) Highly polished mirror like surfaces are
good reflectors, but not good radiators.  R 2T 4  constant
Heat Transfer 373
4 2 c - heat capacity
 T1   R2 
   
 T2   R1  dT k
The rate of cooling 
dt Rc
1/ 2
T1  R2  where ( k - is a constant)
 
T2  R1  For the material D the value of RC is maximum
1/2 1/2
and rate of cooling is minimum.
R   R 
T2  T1  1   T1  1  20. (4) Rate of
 R2   R1 / 2 
 A T  T0 
4 4
cooling  
T2  2T1  1.414T1 t mc
18. (3) Let R = radius of the sun, d = distance of m
the earth from the sun.
 t [  , t ,  , T 4  T04  are constant]
A
Power radiated by the sun   4R 2  T 4  P. m density  Volume 3
 t   2  t 
A Area 
t1  1
 
t2  2
100 1cm
  t2  200sec
Energy received per unit area per second t2 2cm
normally on the earth.
21. (2) Let W is the water equivalent of the
P 4R2T 4 m1 s1  Ws1 m2 s2  Ws2
  
4d 2 4d 2 calorimeter. t1 t2
2 2
R 1  2R  where W is the water equivalent
  T 4     T 4  
d
  4  d 
40 1  W 1 100 1  W 1
Angle subtended by the sun at the earth is    80  2W
10 20
2R

d  100  W  W  20 g
2
   . 22. (1) From Newton’s law of cooling
19. (3) The rate of heat flow is 1  2  0 
t log  
dQ K  1  0 
 e (4 R2 )T
dt
Where K is a constant and 0 is the temperature
where R is the radius of the sphere and of the surroundings
T  T 4  T04 T & T0 are the temperature of the
body and surrounding 1  50  20 
5 log  
K  80  20 
dT 4 e R 2 4 e R 2 3eT
 T  (T )  1  30  20 
dt mc 4  Rc t log 
  R 3c 
3 K  60  20 
374 Heat Transfer
1
log e 2
5 K td  4.5hr
 
t 1 log 22
K i.e., the death happened around 7:30PM
 t  10minutes.
23. (4) Let T is the temperature of the heater. 4 4
T   900 
The rate of change of temperature is 1. (40.5) Q2  Q1  2   0.5    40.5J
T
 1  300 
dT 2. (16) Amount of energy radiated 
  K (T  T0 )
dt
(Temperature) 4
Here heat flows from heater to the room and
and then from room the atmosphere. So at 3. (4.8) Rate of heat flow is given by,
thermal equilibrium K1 and K 2 are KA(1  2 )
i Where, K= coefficient of
corresponding constants related to heater to room l
and room to atmosphere thermal conductivity
In case (1) l= length of rod and A = area of cross-section of
rod
  
K1 T  Tr1  K 2 Tr1  Tout1 
and In case (2)
  
K1 T  Tr2  K 2 Tr2  Tout 2 
Dividing these equations
T  Tr1 Tr1  Tout1


T  Tr2 Tr2  Tout1 Let the junction temperature is T,
iCopper  iBrass  iSteel
T  20  20   20

T  10 10   40
0.92  4(100  T ) 0.26  4  T 0.12  4  T
 T  60C  
46 13 12
 200  2T  2T  T  T  40C
T T  0.92  4  60
t   K ln  2 0   iCopper   4.8cal / s
 T1  T0  46
From the given information
 75  60  1. (320) From Stefan’s law the rate at which an

2hr   K ln   (i)
 80  60  object radiates, energy ( E) per unit surface
The time of death is area is
P  T 4;
 80  60 
td   K ln   (ii)
 98  60  Given, P1  20 cal m 2 s 1
Heat Transfer 375
T1  227  273  500 K Since area of plate is largest so it will cool fastest
and sphere has the least surface area so the
 20   (500)4 (i) cooling is slowest.
Also T2  7270 C  727  273  1000 K 2. (1) Given, initial temperature T1  80C
 P2   (1000)4 (ii) Final temperature T2  50C
20 (500) 4 Temperature of the surroundings T0  20C
 
P2 (1000) 4
t1  5 min
(1000) 4  20
P2   320 cal m2 s 1 According to Newton’s law of cooling.
(500) 4
2. (4) According to the question, dT T  T 

Rate of cooling,  k  1 2  T0 
dt  2 
When two identical rods are connected in
parallel, the rate of melting of ice is
 80  50   80  50 
So, rate of heat transfer is given by  k  20 
5  2 
Ldm dQ 2KA T1  T2  30

x    k  65  20
dt dt l 5
2 KA 100  0  200 KA 6 2
  (1)  k 
l l 45 15
Similarly, for series connection In second case,

Ldm KA Initial temperature T1'  60C
y  100  0 
dt 2l
Final temperature T2'  30C
KA
  50
l t'?
Dividing Eqs. (i) and (ii), we get
 60  30  2  60  30 
Now,    20 
x 200 t' 15  2 
 4
y 50
30 15
3. (0.18) Heat emitted in time t is  t'  9min.
2  25
Q  e T 4 At
4
300  e   5.67  108  1000   5.00  104   60 
Q kA  
 e  0.18 1. (1) As, t  l
 
 l / kA R
(R- Thermal resistance)
1. (3) Rate of cooling Gi ven that Q and  are same for two
combinations.
Q
 A
t tR
376 Heat Transfer
tp Rp R/2 1  T   T 
      
ts Rs 2R 4   l  steel  l Cu
ts 4
The slope of steel is more than copper. The
tp    1min. correct option is (3)
4 4
3. (1) Temperature does not change when a
2. (3) Given, base area of boiler  A  0.15m2 body is in thermal equilibrium. Hence energy
radiated remains same.
Thickness  d   1 cm  1102 m
1
Rate of water boils 4. (1) According to Wien’s law m  and
T
6.0 from the figure
 6.0kg / min  kg / s  0.1kg / s.
60
(m )1  (m )3  (m ) 2 therefore T1  T3  T2 .
Thermal conductivity of brass
 dT 
K  109J / s  m  K 5. (4) Rate of cooling     emissivity (e)
 dt 
Latent heat of vapourization of water
L  2256  103 J / Kg  dT   dT 
From graph,   dt     dt   ex  e y
 x  y
Let 1 be the temperature of the part of the
boiler in contact with the stove. As good absorbers are good emitters
Rate of heat energy supplied So emissivity (e)  Absorptive power

(a)  ax  a y
KA
 mL 6. (1) When a body cools by radiation, the rate
d
of cooling is given by
KA  1  2  d eA 4
 mL  (  04 )
d dt mc
As the water boils at 100C ve sign shows that temperature decrease, c is
the specific heat of the material and  0 is the
2  100C and 1  238C
surrounding temperature.
d 1
 
1. (3) According to Newton’s law of cooling, dt c
the temperature goes on decreasing with time  d 
non-linearly and finally it reaches temperature i.e., rate of cooling  R   is inversely
 dt 
0 . proportional to the specific heat of material. For
A, rate of cooling is large. Therefore, specific
2. (3) The rate of heat flow is
heat of A is smaller.
T 7. (3) As emissivity of sliver is zero (e=0) so
i  KA  constant
l the total radiation that falls on the sphere will be
T completely absorbed and hence its temperature
K  constant will remain constant since the sphere acts like a
l
good emitter (Good absorbers are good
K Cu  K steel emitters).
Heat Transfer 377
7. (2) The rate of heat flow

dT
1. (1) According to Newton’s law of cooling i  KA is same in steady state condition,
dx
d dT
  K (  0 ) A  constant  Hyperbolic graph
dt dx
d
   0  K  dt 8. (3) According to Stefan’s law E   T 4
log e (  0 )   Kt  C (C- is a constant)  ln E  ln   4ln T  ln E  4ln T  ln 

The garph between loge (  0 ) and t is straight The graph between ln E and ln T will be
line. straight line, having positive slope ( m = 4 ) and
2. (1) Let 0 is the temperature of the hot end intercept on ln E axis is equal to ln 
and  is the temperature at a distance x from d
hot end 9. (2) For   t plot of cooling   slope of
dt
dQ kA(0   ) the curve.

dt x d
where  (  0 )
x dQ x dQ dt
 0       0 
kA dt kA dt At P
dQ
Hence is constant throughout the rod. The d
dt  tan 2  k  2  0  , where k = constant
graph between  and x is a straight line. dt
3. (1) As the slabs are connected in series the At Q

rate of heat flow across them must be same d tan 2  2  0 
 tan 1  k  1  0   
K1 AT1 K 2 AT2 dt tan 1  1  0 

l1 l2
K1  10 K2  30 K 1
  1 
3 3 K2 3 1. (2) Assume R be the thermal resistance of
4. (1) According to Wien’s displacement law uniform rod of length L.

1
 m2  m1 [ T1  T2 ]
m 
T
There fore I   graph for T2 has lesser
wavelength (m ) and so curve for T2 will shift
towards left side.
5. (4) A  T 4 by stefan’s Boltzmann law not by
Wien’s displacement law.
1
As  m   T3  T2  T1
T
6. (3) AFrom Wien’s displacement law T 120 75
Heat current: I  R  8R / 5  R
T m  b  T  m eq
The graph between T & m is straight line passing  3R  75 3R

 T  PQ
 I    45C
through origin.  5  R 5
378 Heat Transfer
2. (3) Let T is the temperature of sphere at 50  42  50  42 

some arbitrary time. The rate of heat looses by  K   0  (ii)
10  2 
the sphere is
Dividing equation (i) by (ii) we get
dQ
 AT 4  T04 
dt  0  10C
2. (2) From Newton’s law of cooling,

M (T 3 )dT
 AT 4  T04 
dt 1    
t log e  2 0 
T 3 dT Adt k  1   0 
 T 4  T04    M 
From the given information
Let x  T 4  T04
1 (40  30)
5 log e (i)
dx  4T dT 3
k (80  30)
2T
1 01 A t 1 (32  30)
4 3T0 x
dx  
M  0
dt And, t  k loge (62  30) (ii)
Dividing equation (ii) by (i),

2T0 2T0 4At
ln( x)3T  ln T  T  
4
0
4

0 3T0 M 1 (32  30)
log e
4
 16T  T  4
4At t k (62  30)
ln  0 0 
4  4 5 1 log (40  30)
 81T  T 
0 0M e
k (80  30)
 15  4At On solving we get, time taken to cool down from

ln    
 80  M
62C to 32C , t  8.6 minutes.
M   16  M  16  3. (1)
t ln    2
ln  
4A  3  16R   3 
1. (2) By Newton’s law of cooling Let conductivity of steel K steel  k

1  2    
  K  1 2  0  Conductivity of copper K copper  9k
t  2 
 copper  100 C ;  steel  0C
where  0 is the temperature of surrounding.
L
Now, hot water cools from 60C to 50C in 10 lsteel  lcopper 
2
minutes,
Let  is the temperature of the junction. As
60  50  60  50  heat flow is same in both rods
 K   0  (i)
10  2 
9KA(100  ) KA(  0)
   90C
Again, it cools from 50C to 42C in next 10 L/2 L/2
minutes.
Oscillations 379
dy
  cos t   sin t
dt
d2 y
2
  2 sin t   2 cos t
1. (1) dt
2. (2)   2 (sin t  cos t )
3. (2) We know that during SHM, the restoring
force is proportional to the displacement from d2 y
  2 y
equilibrium position. Hence restoring force is dt 2
maximum when the displacement is maximum. The other two functions do not obey
4. (1) Angular frequency of SHM is equal to
d2 y
angular velocity of particle   v / R 2
  2 y
dt
5. (3)
8. (1)
6. (1) Time taken by particle to move from x =
9. (1)
0 (mean position) to x = 4 (extreme position)
T 1.2 2 2
   0.3s vmax  a  a   (50  103 )   0.15m / s
4 4 T 2
Let t be the time taken by the particle to move 10. (2) Maximum velocity,
from x =0 to x =2 cm
2
vmax  a  vmax  a 
2 1 2 T
y  a sin t  2  4sin t   sin t
T 2 1.2
2 a 2  3.14  7  103
2 T    0.01s
  t  t  0.1s. Hence time to move vmax 4.4
6 1.2
from x = 2 to x =4 will be equal to 0.3- 0.1=0.2s
11. (3) We know that v   A2  x 2
Hence total time to move from x=2 to x=4 and
back again  2  0.2  0.4sec 2
2  2A 
Initially v   A   
 3   3 
7. (3) The function y  5cos   3 t 
 4  After increasing the velocity
represents equation of SHM with phase constant
2
3  2A 
3v   A '2   
4  3 
y  sin t  cost Where A ' is the final amplitude.

380 Oscillations
2 20. (2) In S.H.M. if rest energy is zero

 2A 
A '  
1  3  max P.E  max K .E  K 0 (cos 2 t ) max  K 0  1  K0
 2
3  2A 
A '2   
 3  ka
21. (4) 2 f  ; 2 
2 2
m
 4A  4A 7A
9  A2  2
  A '  9  A' 
 9  3 2 1 2
Potential energy U  m x  2 kax
2
12. (2) Maximum acceleration  a 2  a  4 2 n2
2 2 1
 0.01 4  ( )2  (60)2  144 2 m / s 2 22. (4) Potential energy, V  m x
2
13. (3) vmax  A ,  max  A 2 1
2 2 2
and kinetic energy T  m ( A  x )
2
Given that vmax   max    1 & T  2 sec
T A2  x2
aT  2 xT 4 2 4 2  
14. (2)  = 2 T =  Constant V x2
x x T T
23. (3) The displacement of particle, executing
15. (2) A  10 102 m and   10rad / s SHM,
amax   2 A  10  102  102  10m / s 2 .  
y  5sin  4t  
 3
16. (1) a  16 2 x
 
Standard equation of SHM is a   2 x v  20cos  4t  
 3
Hence, comparing two equation, we get   4
T 
2
2 1 Velocity at t   
4  
and T    sec
 4 2
 T 
17. (4) The average acceleration of particle v  20cos  4   
 4 3
performing SHM over one complete oscillation
is zero. Because acceleration is a sinusoidal  
 v  20cos  T  
function and average value of sinusoidal function  3
for complete cycle is zero.
Comparing the given equation with standard
18. (3) We can find the velocities,
equation of SHM y  a sin(t   ),
dy1
v1   2 10cos(10t   ) we get   4.
dt
and v2  3 10sin10t  30cos(10t   / 2) 2 2 
As T T  T 
 4 2
 Phase difference
Now, putting value of T in Eq. (ii), we get
 (10t   )  (10t   / 2)     / 2
   
v  20cos     20sin
19. (4) y  a sin(2 nt   ). 2 3 3
Its phase at time

3
 20   10  3
t  2 nt   2
Oscillations 381
The kinetic energy of particle,

g
29. (4) As the bob executes SHM  
1 2 L
KE  mv
2 x  a cos t (since it is released from extreme
position)
 m  2 g  2  10 kg 3
30. (1)
1
  2  103  (10 3)2 1 l   
1/ 2
2 T '  2  2   T
g eff (    )g     
 103 100  3  3 101  K .E  0.3J .
 l 
a
2
 T  2 
1 2
Ky 2    g 
U 2 y 2 1 E
24. (3)   2   2   U  .
E 1 Ka 2 a a 4 4

2 31. (3) T1  2 (1)
g
25. (1) Total energy in SHM is always constant
1 
and it is equal to m 2 a 2 . So T2  2 (2)
2 ga
1
T .E  avg
 m 2 a 2 
2 T3  2 (3)
g a
26. (2) As we go down to the bottom of mine g
value decreases. So T increases and we say that By eliminating (2) & (3) we get
it looses time since it makes lesser number of
oscillations in a given standard time. g T32  T22

27. (2) Since, frequency  g , and g decreases 4 2  2T22T32
therefore frequency also decreases.
2T22T32
 T12 
l T32  T22
28. (1) T  2
g'
T2T3 2
 T1 
g ' is the effective acceleration due to gravity T22  T32
Inside the liquid the effective weight is 32. (3) The time period of a simple pendulum of
mg '  mg  Fb   sVg  lVg a length l is given by
l
s T  2 (1)
sVg '  sVg  Vg g
n
Where g is acceleration due to gravity..
 1
g '  g 1   Where length of pendulum increases by 21%
 n
We have l '  1.21l
l
 T  2
 1 1.21l
g 1   T '  2 (2)
 n g
382 Oscillations
Dividing Eq. (1) by Eq. (2), we get

4 2 m 4 2 m 4 2 m
  2  2
T 1 t02 t1 t2
 '
T' 1.21  T  1.1T
1 1 1
or  
Hence, increase in time period to2 t12 t22
1.1T  T  0.1T 37. (3) The initial shift in mean position of the
Percent increase in time period Mg
block of mass M is x1  (from natural
K
0.1T
  100  10% length)
T
The new mean poistion of the combined mass is
33. (4) Maximum velocity, vmax  a
Mg  mg
x2 
 2  K
or vmax  a  
 T 
mg
shift 
v T 15  3.14 K
a  max   1.5cm
2 2  3.14  5
 
34. (4) Time period of a spring block system does 38. (1) x  Asin t & y  Asin    2   Acos t
 
not depend on acceleration of the frame. So
correct option is (4) x2 y 2
  1  x 2  y 2  A2 (circular)
35. (4) Effective spring constant K eff  2 K . A2 A2
39. (2) A2  a 2  a 2  2a 2 cos 

1 2K
f 
2 m 2
 a 2  2a 2  2a 2 cos    
3
When one spring is removed, K eff becomes K
new frequency 40. (2) The coin will leave contact when it is at
the highest point and for that condition
1 K Maximum acceleration = Acceleration due to
f '
2 m gravity
f' 1 f g
  f '  2 A  g  Amplitude A  .
f 2 2 2
m
36. (1) As, t1  2 k
1
1. (3) vmax   A
4 2 m
or k1  2
t1 2. (3) a   2 x
4 2 m 3. (4)
Similarly, k2  t22 4. (4) The motion of planets around the sun is
periodic but not simple harmonic motion.
4 2 m
and (k1  k2 ) 
t02 5. (3)
Oscillations 383
6. (4) Equation of motion y  a cos t Motion is simple harmonic:
1
a 1  2
So,   and T  2 4  4
  a cos t  cos t   t  4
2 2 3
a
 12. (1) x  a sin 2 t  (1  cos 2t ).
T 2
2 t  4 2
  t  3   sec 2 
T 3 2 3 2 3 Frequency  
2 
7. (2) The amplitude is the maximum
displacement from the mean position. a a
13. (2) If x is position, x   cos 2t. Here
2 2
8. (3) In one complete oscillation particle comes
displacement x varies simple harmonically with
to its starting position, so displacement is
angular frequency 2.
zero.
14. (3) In S.H.M at mean position velocity is
9. (2) The amplitude is the radius of the circle
maximum
0.8
R m  0.4m. 15. (3)
2
T /2
The frequency of the shadow is the same as   sin tdt
vavg  0 2 T /2
that of angular velocity of the circular motion, T /2  cos t 0  2
 dt T
so 0
16. (2)
  30rev / min  0.5rev / s  rad / s
17. (2)
 2  1rev 
   0.5rev  18. (2) Velocity is same. So by using v  a
 
 A11  A22  A33 .

and f   / 2   0.5Hz 19. (3) A   2 y at mean position y=0
2
10. (4) Let the displacement equation of particle So acceleration is minmum (zero).
executing SHM is y  a sin t 20. (2) Amax   2 a.
As particle travels half of the amplitude from
21. (4) In SHM if v  vmax then a  0
the equilibrium position, so
a if v  0 then a  amax
y
2 22. (1) Maximum velocity v   a
a 1  v2
Therefore,  a sin  t or sin t   sin 2
2 2 6 Maximum acceleration f   a  f 
a
T 23. (1) The speed
t 
12 dy
v  0.15sin(0.50t )m / s
11. (4) Given 4v  25  x
2 2 dt
Differentiating with respect to t on both sides, dv

a  0.075cos(0.50t )m / s 2
dt
x
8va  2 xv  a  
4  y  0.30m, v  0 and a  0.075m / s 2
384 Oscillations
24. (4) Velocity v   A2  x2 and acceleration 1

30. (4) PE  m 2 x 2
2
 2x
1
Now given, KE  m 2 (a 2  x 2 )
2
 2 x   A2  x2   2 1   22  12 1
TE  PE  KE  m 2 a 2
2
2 2
   3 T  
 3 Since PE is maximum at x =a and KE is
maximum at x = 0, therefore TE remains
25. (4) Force and acceleration will be always constant throughout the motion.
towards the origin. Position vector defined from
the origin is always away from the mean position. 31. (1) Kinetic energy= Potential energy
v can be towards or away from the mean 1 1

    m 2 (a 2  y 2 )  m 2 y 2
position. So F .r & a.r are (-) ve. 2 2
26. (1) Given that a  bx a2  y2  y2
Comparing a   2 x   2  b    b a
2 y2  a2  y  
2
27. (3) y1  4sin(10t   ) ;
32. (3) The potential energy of a particle
y2  5cos10t T
executing SHM is periodic with time period .
2
dy1
v1   40cos(10t   )
dt 1
33. (3) KE  PE
n
dy2
v2   50sin10t  50cos(10t   / 2) 1 11
dt 
m 2  A2  x 2    m 2 x 2 
2 n2 
  n
Phase difference between v1 and v2      x A
 2 n 1
34. (4) KE =PE

28. (1) Phase for 1st SHM: 1  10 t 
3 1 1 A
m 2 (A 2  y 2 )  m 2 y 2  y 
 2 2 2
Phase for 2 nd SHM: 2  8 t 
4 35. (4) Let x  a cos t
 1
Phase difference:   1  2  2 t   KE  ma 2 2 sin 2 t
12 2
Put t = 0.5s 1
PE  ma 2 2 cos 2 t
2
 13
  2 (0.5)   when KE  PE , sin 2 t  cos 2 t
12 12
29. (4) Total energy of particle in S.H.M.  tan t  1
1   T
E KA2 (independent of x)  t   tmin   .
2 4 4 8
Oscillations 385
36. (4) In SHM restoring force is

1
2
45. (3) T  2 .
F  m x g
Given that F   kx 2 l l
T'  T '  2
Hence k  m 2 2 g 4g
the lift should be accelerated by 3g upwards
1 2 2 1 2 because the total acceleartion g+3g pseudo
Total energy is TE  m A  kA .
2 2 acceleration.
Hence total energy depends upon K and A.
L
37. (2) 46. (2) Time period T  2
g
1 1 1
38. (3) m 2 x 2   ma 2 2 47. (3) The effective acceleration of a bob in
2 4 2 water
a 4
 x  cm  2cm
2 2  
 g '  g  1   where  and  are the density
39. (4) Total energy of SHM,  
of water and the bob respectively. Since the
1 period of oscillation of the bob in air and water
E  m 2 a 2 , (where, a = amplitude)
2 are given as
Kinetic energy, l l
T  2 and T '  2
g g'
1 1
K  m 2  a 2  y 2   K  E  m 2 y 2
2 2
T g' g (1   /  )  1
a 2     1  1
1 2  a  E T' g g  
When y   K  E  m  E 4
2 2  4 
T 1 1 1
3E Putting  . We obtain,  1     2
K T' 2 2 
4
1 l
40. (2) Eav  U av  m 2 A2 T  2  T  l , hence if l made 9
4 48. (2)
g
41. (3) times T becomes 3 times. Also time period of
simple pendulum does not depend on the mass
l
42. (2) T  2 of the bob.
g'
49. (4) Time period in liquid is given as
2 2
g' g a
s
T' T
Where a is the acceleration of the car  s  
As g '  g  T decreases s
T' T
43. (1) The CM of the (bob + liquid) system goes s
s 
down and hence l increases. So T also increases. 8
l 8
44. (4) T  2 T  l T' T
g 7
386 Oscillations
50. (1) Time period of a pendulum does not 57. (2) In series
depend on mass. Time period of seconds
pendulum is 2s. 1 1 1 4 2 m
  K
K K1 K 2 T2
51. (4) As T.E of the pendulum is conserved
As
K .Emax  P.Emax  mg  1  cos 
1
K  T 2  T12  T22
1 2 1 1 T2
52. (2) Kx   KA2
2 4 2
58. (1) At extreme position velocity of M will be
2
A A zero. So amplitude remains same.
x2  x
4 2 59. (3) x  Asin t &
53. (4) Given mg  kx  
y  sin       A sin(t )
 2
m x
 
k g x
  1  x   y (straight line)
y
m x
Now, T  2  2
60. (4) A  A12  A22  2 A1 A2 cos
k g
54. (3) As, T  m  1.1  250  

A  a 2  a 2  2a 2 cos  
and  2.2  250  x 3
A  3a
1 250
   x  750g
2 250  x 61. (2) y  a sin t  a cos t
55. (4) The time period of the block would be
given by,  
AR  a 2  a 2  2a 2 cos    2a
2
M
p  2
K
4M d2y
p '  2  2P 1. (2) For SHM,  y
K dt 2
56. (4) The upper two springs are in parallel their
d2 y
equivalent spring constant is 2K. Similarly the For the equation Asin t  B cos2t is not
dt 2
equivalent spring constant for two lower springs proportional to (-y).
is 2K. The equivalent spring constant of the
2. (1)
system is
x  12sin t  16sin 3 t  4[3sin t  4sin 3 t ]
1 1 1
   k eq  k  4[sin3t ] (By using sin 3  3sin   4sin3  )
keq 2k 2k
 Maximum acceleration
m Amax  (3 )2  4  36 2
T  2
k
Oscillations 387
3. (1) The two particles are shown in figure dv 


Acceleration, a    x0 2 cos  t  
dt  4
   
 x0 2 cos     t   
  4 
 3 
 x0 2 cos  t   (i)
 4 
Acceleration, a  Acos(t   ) (ii)

T 3T Comparing the two equations, we get
Where t '  t 
8 8
3
4. (3) Average velocity A  x0 2 and   .
4
t t
dx
0 dt  dt 0 dx x(t )  x(0) 7. (3) y1  4sin(10t   ) ; y2  5cos10t
v  
t t t dy1
v1   40cos(10t   )
dt
A(cos  / 6  1) 3 A
  ( 3  2)
 / 6  dy2
v2   50sin10t  50cos(10t   / 2)
since particle does not change its direction in dt
the given interval, average speed
 
Phase difference between v1 and v2     
3 A 2  
v  (2  3)

2
8. (3) Phase difference     2  1
5. (2) x  5.0cos  2 t    3
dx 
Velocity   10 sin(2 t   ) Let 1  ,
dt 6
d2x  2   5
Acceleration   20 2 cot(2 t   )  2    
dt 2  3 6 6
 At t  1.5sec, x  5.0cos 4  5 1
 sin 1  sin  2 
2
dx
Velocity   10  zero   0
dt Displacement is measured form starting point
(t=0)
d2x
Acceleration   20 2 1  20 2
dt 2   A
So, x  A sin(t  1 )  A sin  0   
 6 2
6. (1) Here,
x  x0 cos(t   / 4) 9. (1) x1  A sin(t  1 ), x2  A sin(t  2 )
dx             
  x0 sin  t   x1  x2  A  2cos  t  1 2  sin  1 2  
 Velocity, v  dt 4   2   2 

388 Oscillations
When x1  x2 is maximum the variable function As the maximum separation between the
 
particle is A,  2 Asin  A   
 1  2  2 3
cos  t   should be equal to 1.
 2 
14. (4) Given that k  10cos 2 t
  
A  2 A sin  1 2 
 2   1  cos  2t  
k  10    5 1  cos 2t 
 2 
1  2  
  1  2  .
2 6 3
k max  10  cos  2t   1
5T T
10. (1) T 
4 4 k min  0  cos  2t   1
In S.H.M T .Emax  K .Emax  10
By the time, the bigger pendulum makes one 15. (4) U  2  20 x  5x2
full oscillation, the smaller pendulum will make
dU
 1 F   20  10 x
1   oscillation. The bigger pendulum will be dx
 4
in the mean position and the smaller one will be At equilibrium position: F  0
in the positive extreme position. Thus, phase 20  10 x  0  x  2
difference  90
Since particle is released at x  3, therefore
2 2 amplitude of particle is 5.
11. (3) Let T1  T and T2  KT ; t  t  2
T KT
 K 1  5T  K  1 
t   1;    1 which gives K=5
 KT  4  KT 
If length of first pendulum=l, T1  2 l / g It will oscillate about x =2 with an amplitude of

5.
T2  5  2 l / g  2 25l / g
 maximum value of x will be 7.
So, ratio of lengths =1:25 16. (1) Potential energy =U = Vm
12. (1) Let T1 , T2 be the time period of shorter
length and longer length pendulum respectively.  U   50 x 2  100 102
As per question, nT1   n  1 T2 ; dU

F   (100 x)102
dx
1 4
So n 2   n  1 2
g g  m 2 x  (100  102 ) x
n 1   n  1 4  n  2 10  103  2 x  100 102 x

13. (1) Let, x1  Asin t and x2  Asin(t   )   2  100   10
    10 5
x2  x1  2 A cos  t   sin f   
 2 2 2 2 
Oscillations 389
17. (1) The potential energy function for the mass We can write
m in a undimensional potential field is given as
d2x B4
A B m   x
U ( x)   (i) dt 2 8 A3
x2 x
Since this is the potential energy of mass m in a d2x  B4 
potential field, so the force acting on the mass is    3 
x (iii)
dt 2  8mA 
obtained by the relation:
dU 2 A B d2x
F     x
(ii) dt 2
dx x3 x 2
Now, the equilibrium position is given by So from the above relations, it is clear that
particle will perform SHM. But for any SHM,
2A B
F  0 or  0
x3 x 2 d2x
2
  2  x (iv)
(2 A  Bx)  0 dt
2A 2 B4
So, x So, from eqs. (iii) and (iv), we have  
B 8mA3
Since x  2 A / B gives stable situation
2 8 A3 m 2mA
T  2  4 A
(because energy is minimum here; d 2U / dx2 is  B4 B4
positive), so stable equilibrium position is given 18. (2) Tension in the string when bob passes
2A through lowest point
as x 
B
mv 2
Now if the mass is displaced slightly away from T  mg   mg  mv ( v  r )
r
the equilibrium position, the force acting on mass
m is given as 2 2
Putting v  2 gh and    
T 2
2A B
F 3
 2
 2 A / B   x   2 A / B   x we get T  m( g   2 gh )
2A B
 3 3
 2 2
 2 A   B x   2 A   B x  l
  1     1   19. (4) Time period, T  2
 B   2A   B   2A  g
3 2
2A  B x  B  B x  when additional mass M is added to its bob
 1  2 A   4 A2 / B 2 1  2 A 
8 A3 / B 3    
l  l
So, using binomial expansion for small  x , we TM  2
g
get
Mgl
l 
B3  3B x  B
3
 2 B x  AY
F 2 1  2 A   4 A2 1  2 A 
4A    
Mgl
3 4 l
B  3Bx 2Bx   B   TM  2 AY
 1 1     3  x
4 A2  2A 2 A   8A 
g
390 Oscillations
2
 TM  Mg l
 T   1  AY 22. (3) The effective length of the bob is
  2
2
1 A  TM   2g l
    1    T  2
Y Mg  T  l 2g

20. (2) The vehicle goes down with acceleration 23. (4) Both will oscillate about their centre of
mass. the oscillation of M is same as a simple
g sin  . so the pendulum is in noninertial frame
pendulum of mass M and length l1. The time
and Pseudo force acts on it.
period of both particles are same. Let it be T.
So the only resultant force on the bob is

mg cos  . The effective acceleration is
g '  g cos  l1
T  2
g
l
T  2
g cos  where l1 can be given as Ml1  ml2 and
l1  l2  l
21. (1) Take the lowest point of the pendulum as
the reference line to measure P.E
ml
l1 
P.Ei  P.E f (when the bob is at extreme M m
positions)
ml
Then, T  2
L L  (M  m) g
mg ( L  L cos  )  mg   cos  
2 2 
24. (1) If spring has the mass then:
1 cos
1  cos    1 k
2 2 f 
2 M spring
m
cos  1 3
 cos  
2 2
Spring equivalent =3k (As all are in parallel)
3
As   300 cos  1 3k 1 3k
2 f  f 
2 (3m) 2 2m
m
cos  3  1 3
25. (3) The velocity of the spring increases

  cos 1 (0.732)  420
linearly from left to right. The velocity of the
    2 spring at a distance x from the left end is
Oscillations 391
v equilibrium position is x and linear acceleration

v x is ‘a’.
l
consider a small element of length dx as shown
in the figure.
The kinetic energy of the element is
1m 1 m v2 2
dx    dxv 2  x dx
2 l  2 l l2  2Kx0  mg (i)
l Let T1 & T2 are tensions in the strings

mv2 2 mv2  l 3  1  m  2
k  x dx     v
2l 3 x2 2l 3  3  2  3  (T1  T2 )  mg  ma (ii)
26. (3) If stick is rotated through a small angle a
For rotation, T1  T2  R  I  
L R  
 , the spring is stretched by a distance .
2
Ia
 (T1  T1 )  (iii)
L  R2
Therefore restoring force is spring  K    .
2 
From (ii) and (iii)
 KL L
Each spring causes a torque  ( Fd )    Ia
 2 2 2T1  ma   mg
R2
in the same direction.
Therefore equation of rotational motion is Ia
2(kx  kx)  ma   mg
R2
  I
2K
ML2  a x
 KL L  I 
2    I with I  m  2 
 2 2 12  R 
 6K  I
    m
M   T  2 R2
2K
Standard equation of angular S.H.M. is
   2 28. (3) When a mass m is placed on mass M, the
new system is of mass =(M+m), attached to the
6K spring. New time period of oscillation,
2 
M
M m
T '  2
 1 6K k
frequency f  
2 2 M M
T  2
27. (2) Let x0 be the static deflection of the k
pulley. It is slightly pulled downward and Let v = velocity of mass M while passing through
released at t=0. At time=t, its displacement from the mean position.
392 Oscillations
v '  velocity of mass (M+m), while passing 1 '2 1

through the mean position.  kA  ( M  m)v'2
2 2
According to law of conservation of linear
1/ 2
momentum M m mv mv
A'     
 k  ( M  m) ( M  m) k
Mv  (M  m)v '
At mean position, v  A  and v '  A' ' 32. (2) In CM frame both the masses execute
SHM with
 MA  (M  m) A' '
k 2k
 M  M T'  
or A'    A  A  m
 M  m ' M m T
Initially particles are at extreme
 M  M m M
  AA
M m M M m Initial extension in the spring is ( L  L )
k 2k
29. (2) Here   . The maximum static Distance between them  L0  ( L  L0 )cos t
3m m
frictional force is 33. (1) Suppose static deflection of the body of
f max   mg cos  2 tan  mg cos  2mg sin  mass m be x, that of spring (i) be x1 and spring
(ii) be x2 .
 x1 K1  x3 K3  mg (i)
K 2 x2  2 K1 x1 (ii)
The given arrangement can be obtained by
superimposing the following arrangements.
Applying Newton’s second law on the upper

block at lower extreme position,
f  mg sin   m 2 A  f  Am 2  mg sin 
3mg sin 
As f  f max ,  2 A  g sin  or A 
k
30. (4) Time period of a spring block system does
not depend on acceleration of the frame. So
correct option is (4)
31. (3) If v ' are the velocities of the block of
mass M and (M+m) while passing from the mean
position when executing SHM.
Using law of conservation of linear momentum,
we have
K 2 x2
K3 x3  mg  mg k1 x1  mg
mv  (M  m)v ' 2
or v '  mv / (M  m) mg 4mg mg
Now, x  x3  2 x2  x1   
Also, maximum PE = maximum KE K3 K2 K1
Oscillations 393
The given arrangement can be replaced by an If the density of the liquid is  , then the total
equivalent spring-mass system shown in fig -IV mechanical energy of the liquid column is
in the which static deflection of the mass is x.
The difference in the two levels is 2x. So the
 mg  Ke x column having length 2x produces restoring force
on the entire liquid.
 mg 4mg mg 
 mg  Ke      A2xg   A2ha
 K3 K2 K1 
g g
a    x   
1  1 4 1  h h
    
Ke  K3 K2 K1 
36. (4) Let at any instant, the body is at angular
position  with respect to the vertical line drawn
m  1 4 1 
 T  2  2 m     from the centre Q. If  is the angular
Ke  K3 K 2 K1 
displacement of the ball about its centre, then
34. (1) When the particle of mass m at O is
pushed by y in the direction of A the spring A  r 
 R  r    r    
will be compressed by y while spring B and C Rr
will be stretched by y '  y cos 45. So that the
total restoring force on the mass m along OA.
Fnet  FA  FB cos 45  FC cos 45 Restoring torque acting on the ball
  mg sin  r
 ky  2ky 'cos45  ky  2k  y cos 45 cos45  2ky
For small , sin  
Also Fnet  k ' y  k ' y  2ky  k '  2k
  mg   r
m m
T  2  2  r   r2 
k' 2k or   mg   r  mg    
R r  Rr 
35. (2) Suppose that the liquid is displaced slightly
from equilibrium so that its level rises in one
mg  r 2 
arm of the tube, while it is depressed in the Angular acceleration      
second arm by the same amount, x. I  Rr 
Now comparing above equation with standard

equation of SHM,
mg  r 2 
  2, we get    
I  Rr
394 Oscillations
Here I is the moment of inertia of the rolling ball a  2 x, we have

about the point of contact which is I  (7 / 5) mr2
g
2 
mg  r  2
5 g  k2 
     R  r  1  2 
7 2  Rr  7 R  r  r 
mr
5
2r 2
For solid sphere k 2  .
2 7 R  r  5
and T  2
 5g
5 g
Alternative method: Hence 
7 R  r
We can imagine that when the CM of the body
is shifted slightly by an angle  relative to the 37. (1) U  Cx 2
centre of curvature of the surface, the magnitude
of acceleration of the CM of the body can be dU 2c
F   ma  2cx  a   x
given as dx m
g sin  2c 2 m
a   T   2
k2 m  2c
1 2
r 38. (3) When the block is pushed down then the
(because the body is accelerating down the additional buoyancy force acts as restoring force.
inclined plane of instantaneous angle  )
 Fb  ma
x
Substituting sin     , we have l abxg  (abc) s a
Rr
l g  g 
 gx a x  x
a cs  dc 
 k2 
 R  r  1  2 
 r  g
So, acceleration    
x
 cd 
dc
Hence, time period, T  2
g
39. (3) Let x be the instantaneous downward

displacement of joint A.
The initial and final position are
 
(negative sign is given because a and x are
oppositely directed)
Comparing the above equation with
Oscillations 395
 (2l  4m  2l  m)
x  (4m  2x  m  2x) g  Frequency with which piston executes SHM.
2
3g 3  10  2  1  P0 A 1  P0 A2
 
x x  x    x f  
5l 150  5  2 x0 M 2 MV0
5 2
4
41. (4) vector diagram of amplitudes is given
2 below.
 T sec  5sec
2 / 5
Mg
40. (3) Initially at equilibrium  P0
A
 
PV
0 0  PV
Mg  P0 A (i)
P0 A x0  PA ( x0  x)

By treating the SHMs as vector
P0 x0
P Aˆ 3A ˆ A 3A ˆ
( x0  x) A1  Aiˆ ; A2  i j ; A3   iˆ  j
2 2 2 2
The resultant amplitude is AR  Aiˆ  3Ajˆ
AR  2 A
 
42. (1) y3  15cos t  15sin   t 
2 
 3 
 15sin  t    15sin(t  270)
 2 
Let piston is displaced by distance x
 Vector diagram of amplitudes is:
 P x 
Mg   0 0   A  Frestoring
 ( x0  x) 
 x0 
P0 A 1   
 Frestoring
 ( x0  x) 
 
  
 1    x 
F  P A 1     P A 1  1 
   
 1  x     x  
   
x  
   x 
 P A 1  1   Ax  3  5cos37  7
  x 
Ay  15  5sin 37   12
 P0 Ax
F   A  (Ax )2  (Ay )2  72  122  193 units
x0
396 Oscillations
1  cos10t  2
43. (2) y  25sin 4t   1
 2  a  a0    [ from Eq. (i)]
3
25 25
 sin 4t  sin 4t cos10t a0
2 2 a
9
25 25
or y  sin 4t  sin(14t )  sin(6t ) bt 2 m
2 4 46. (1) A  A0e
25 25 25 3 A0 
b4
4
or y  sin 4t  sin14t  sin 6t   A0 e 22  A0 e  b  b  ln  
2 4 4 4 3
25 25 25 bt 2 m
or y  sin 4t  sin14t  sin(6t ) 47. (2) A  A0e
2 4 4
So it is superposition of 3 SHMs. Thus n = 3 2m A0 2  0.25 A
 t ln  ln 0  4.95sec
b A 0.07 A0 / 2
44. (4) Amplitude resonance takes place at a
frequency of external force which is less than 48. (4) For damped oscillations
the frequency of undamped maximum vibration.

 bt 2 m 
At amplitude resonance 1  2  2 2 x  a0 e cos(t  )
Where 1 during frequency 1   

 bt 2 m 
 a  a0 e
Velocity-resonance takes place (i.e., maximum
a
energy) when frequency of external periodic Given that a  when t  
force is equal to natural frequency of undamped e
vibrations. a0   b  2m
  a0 e 2 m    .
e b
At velocity resonance   2
2  driving frequency
45. (3) In damped oscillation amplitude goes on A
1. (3) At t  0, x 
decaying exponentially 2
a  a0 e  bt / 2 m A  5
 A sin  or  = or  =
where, b=damping coefficient 2 6 6
dx
a0 

100bT 

v  A cos(t   )
 a0 e  2m 
 a0 e100 P dt
3
At t  0, v  A cos
where P  bT / 2m, T= time of one oscillation
 3 5 3
1 Now, cos  andcos  
or  e 100 P (i) 6 2 6 2
3
 200 bT 
As v is negative at t  0, must be 5 / 6.
 
Finally a  a0 e  2m 
2. (1) Given equation of motion is y  sin 3 t

2
a  a0 e 100 P   (3sin t  sin 3t ) / 4
Oscillations 397
 sin 3  3sin   4sin 3    a

2
3a
2 1    1  1 
 A A
dy  d d 
  (3sin  t )  (sin 3 t )  / 4
dt  dt dt  a 1
  A  2a
A 2
dy
4  3 cos t  3 cos3t 1
dt cos  
2
dy2
 4 2
 32 sin t  92 sin3t T  6
dt
5. (2) x  Asin t
2 2 2
dy 3 sin t  9 sin 3t
 2  a   2 A sin t
dt 4
T /4
dy 2 2
 2 is not proportional to y.   A sin(t )dt
4 2 A T / 4
dt  a  0
 0 sin(t )dt
T T
Hence, motion is not SHM. 0
4
3. (2) One of the condition for SHM is that
restoring force (F ) and hence acceleration (a) 4 2 A t /4
 a  cos(t )0
T
should be proportional to displacement ( y).
t/4
Let, y  sin t  cost 4 A  2 T 
 a   cos  cos0 
T  T 4 0
dy
  cos t   sin t
dt 4 A 8 A
 
T T2
d2y
  2 cos t   2 cos t 6. (1) Let the equation of simple harmonic
dt 2 motion be,
or a   2 (sin t  cos t ) x  Asin t
a   2 y  a   y Acceleration a   2 x
This satisfies the condition of SHM, other | a |  2 A sin t

equations do not satisfy this condition.
T /4
4. (2) The displacement of the particle is   2 A sin tdt 2
0
x  A cos  t 
Then, the reqiured ratio is T 
2 A
Given that A (1- cos  ) = a 4
A (1 - cos 2  ) = 3a 7. (1) For an SHM, the acceleration a   2 x
 a a
cos    1   where  2 is a constant. Therefore, is a
 A x
constant. The time period T is also constant.
 3a  aT
cos 2   1   Therefore, is a constant.
 A x
398 Oscillations
dy1   11. (1) x1  a sin(t  1 ) ; x2  a sin(t  2 )

8. (4) v1   0.1  100 cos  100 t  
dt  3
       
 x1  x2  2a sin  t  1 2  cos  1 2 
dy    2   2 
v2  2  0.1 sin  t  0.1 cos   t  
dt  2
 1  2 
Phase difference of velocity of first particle with To maximise x1  x2 : sin  t  1
 2 
respect to the velocity of 2 nd particle at t =0 is
   Given that x1  x2 max  a 2

  1  2    .
3 2 6
   
 a 2  2a 1 cos  1 2 
 2 
9. (2)
1    
  cos  1 2 
2  2 
 1  2 
   1  2 
4 2 2
12. (2) x  Acos(t   )
At t  0, x  0
For first S.H.M. For first S.H.M.
 0  cos 
Ratio of amplitude Phase difference is
 
cos  cos   
a1 3    2 2
   .
a2 2 4 6 12 13. (1) Smaller pendulum will be that for which
time period is less, here 1.5 s. So when this
10. (1) At phase 3 / 2,
completes one vibration, the bigger pendulum
x1  A sin
3
 A will complete its 3 4 oscillation (because its time
2
period is 2s) i.e., it will be at its negative extreme
x2  0 position. Hence the smaller pendulum will be
ahead by a phase of  / 2 than bigger pendulum
So, at t  0, x1   A cos t and x2  A sin t at this time.
x1  x2 14. (1) Let T1 and T2 are the time period of the
two pendulums
 cos t  sin t
100 121
tan  t  1 T1  2 and T2  2
g g
3
t  ( T1  T2 because l1  l 2 ).
4
Let at t = 0, they start swinging together. Since
 2  3 their time periods are different, the swinging will
 t   t  3T / 8
 T  4 not be in unison always. Only when number of
Oscillations 399
completed oscillation differ by an integer, the

d2x
two pendulum will again begin to swing together. and 2
 2 x  0
dt
Let longer length pendulum complete n
oscillation and shorter length pendulum complete d2x
(n+1) oscillation, for the unison swinging, then  Acceleration, a    2 x
dt 2
(n  1)T1  nT2
d2x
 F  ma  m  m 2 x (ii)
100 121 dt 2
(n  1)  2  n  2  n  10
g g
3kx
From Eqs. (i) and (ii) , we get  
1 2 2 E1 m
15. (1) E1  Kx  x  ,
2 K
2 m m
T   2  2
1 2 2 E2  3kx 3k (a sin t )
E2  Ky  y 
2 K
1
T 
1 2 2E a
And E  K ( x  y)  x  y 
2 K
19. (4) U ( x)  ax 2  bx 4
2 E1 2 E2 2E
    E1  E2  E dU
K K K F    2ax  4bx3
dx
16. (4) U  a  bx 2 At mean position F  0
dU  2ax  4bx3
F   [0  2bx]
dx
2 2a a
or x    x   a / 2b
 2b  4b 2b
F  ma  2bx  a     x
m
a
1/ 2 ‘U’ is minimum at x  
 2b  2b
  2 f   
m
d2x
Hence frequency depends upon b and m. Now, F  m
dt 2
17. (2) Potential energy of body in SHM at an
instant, d2x
 m  2ax  4bx3
1 dt 2
U  ky 2
2
d2x 2ax  2b 2 
where the displacement, y  ( x  a)
 2
 1  x 
dt m  a 
3 dU For small ‘x’,

18. (2) U ( x)  k x  F  
dx
d2x  2a 
   x
F  3k x
2
(i) dt 2 m
Also, for SHM, x  a sin t    2a / m

400 Oscillations
20. (4) Here, mass of the bob=M 23. (1) The bob is subjected to two simultaneous,
accelerations perpendicular to each other viz
Length of the pendulum =L
acceleration due to gravity g and radial
v2
acceleration a R  towards the centre of the
R
circular path.
From figure, h  L  L cos  L(1  cos )
Potential energy at A  Mgh  MgL(1  cos )

Potential energy at B  0
2
2  v2 
1  Effective acceleration aeff  g   
Kinetic energy at A  mv  0 [ As v  0]  R
2
2
 Time period of the simple pendulum
Mechanical energy at A  0  MgL(1  cos )
l
Let K B is the kinetic energy at B T  2
aeff
Mechanical energy at B  K B  0
l l
 2  2
According to conservation of mechanical energy v 2 2
v4
g2    g2 
at A and B, we get R2
R
0  MgL(1  cos )  K B  0  K B  MgL(1  cos )
T1 T2
21. (3) Time period of pendulum doesn’t depends 24. (4) The total time for oscillation is 
2 2
upon mass but it depends upon length (distance
between point of suspension and centre of L L
T1  2 , T2  2
mass). In first three cases length are same so g 4g
T1  T2  T3 but in last case centre of mass lowers
which in turn increases the length. So L
Given that T  2
option (3) is correct g
k k T T
22. (2) T1  and T2  T1  , T2 
g  d 2 4
g 1  
 R
3T
TNet 
2
4
T1 d T  d
 1   1   1 25. (1) The motion of M is SHM, with length
T2 R  T2  R
CM  L2  d 2
  T 2 
d  1   1   R ( L2  d 2 )1/ 2
T  2
  T2   g
Oscillations 401
The torque about O is

   K 2 xR  K1 (2 x)2 R  I 
x 2x
tan     
R 2R
d 2
K2 R2  4K1 R2  I
26. (1) Torque acting on the bob dt 2
 l  (mg )l sin
d 2 R   K2  4K1 
2
2

or (mi l 2 )  ( mg g )l dt  mR 2 
  mR 2 
 2 
 mg g  2
or       
m
 i l 2(4 K1  K2 )
 .
3m
2 g m g 2 mi l
where,   m l T 

 2
mg g 29. (2) Restoring torque
i
27. (1) The two springs are in parallel. mL2

  KyL   (Since y  L from figure)
3
 Effective spring constant,
k  k1 + k 2 mL2 d 2 3k
 KL2    2  
3 dt m
Now, frequency of oscillation is given by
k1 +k 2  m m
1  T  2  2  2
f  (i)  3k 3k
2 m
When both k1 and k 2 are made four times their

original values, the new frequency is given by
1 4k1 +4k 2
f '
2 m
1 4(k1 +k 2 )  1 k1  k2  3k
  2
 2   2 f 
2 m  m  m
30. (4) Angular frequency of system,
28. (1) Let x be the displacement of CM from
mean position. K K
 
mm 2m
Maximum acceleration, amax   2 A

Frictional force between P and Q
= force exerted on lower block
 K  KA
 m 2 A  m  A
 2 m  2
402 Oscillations
31. (1) Let A be the maximum amplitude of

Fr  F '  F  m2 g  (m1  m2 ) g
oscillation for the required situation.
Maximum force on block B is m1 g
kA  m1 g  A 
F  Mamax  m 2 A k
since the lower spring exerts force on the block
For no slipping Fmax  f L
only in the downward motion.
The maximum frictional force between the 35. (3) The reduced mass
blocks is f L   s mg
(m)(m) m
 
i.e., m 2 A  s mg mm 2
0.6  g The given system is equivalent to a system of a

 A  0.6  Amax  0.6m particle of mass m/2 connected to a spring of
4 2
stiffness k ragidly.
4
The required period of oscillation.
32. (1) For the gievn amplitude d the block looses
contact with the platform when maximum  m/2 2m
acceleration is equal to g at the highest position. T  2  2 
k k k
2
kd  mg  m d  mg 36. (3) Let us assume that in equilibrium
condition spring is x0 elongate from its natural
g 1 g length
  n
d 2 d
33. (3) The net force exerted by the spring is zero
at the mean point. Therefore, linear momentum
must be conserved.  Mv1  ( M  m)v2
k
As v  A  A
m
k k
 MA1  ( M  m) A2
M mM
 k 
  V  A 
 M 
In equilibrium T0  mg
 A1 M  A2 M  m
and kx0  2T0
A mM
 1  kx0  2mg (i)
A2 M
If the mass m moves down a distance x from its
34. (2) In equilibrium F  (m1  m2 ) g equilibrium position then pulley will move down
By removing mass m1 x kx
by . So the extra force in spring will be .
2 2
F '  m2 g From figure
Oscillations 403
k x BA2
Fnet  mg  T  mg   x0   
2 2 mV
  x 2 mV
 k  x     T  2
2  BA2
T  
 2  39. (1)
 
 
At equilibrium Fb  mg
kx0 kx
Fnet  mg    Al0 g  dlg
2 4
Restoring force,
from eq. (i), we get
Fr  mg  Fb'
 kx
Fnet 
4 F  mg   A(l0  x) g
Now compare eq. (ii) with F   KSHM x dAla  dAlg   Al0 g   gAx
K g
then KSHM  a x
4 dl
Therefore, wooden cube performs S.H.M.
m 4m
 T  2  2
KSHM K g ld
   T  2
dl g
2 2
m 1 m 1
I
 40. (1) When the ball m falls from a height h, it
37. (3) T  2  2 12 3
mgx m(10)(0.5) reaches the surface of earth in time, t  2h / g .
Where x is the distance of CM from A. Its velocity is v  2 gh . It then moves into the
tunnel and reaches on the other side of earth
 and goes again upto a height h on the other side
T sec
3 of earth. The ball again returns back and thus
executes periodic motion. Outside the earth ball
38. (4) Let the ball is pressed down by a small crosses distance h four times. When the ball is
amount y, then the volume of air decreases by in the tunnel at distance x from the centre of the
Ay. earth, then gravitational force acting on ball is
Then, excess pressure is
GMmx
 F
 dV   yA  R3
dp   B    B  
 V  V  i.e., F  x
A restoring force F(AdP) acts in the upward As, this force, F is directed towards the centre
direction. of earth i.e., the mean position. So, the ball will
execute periodic motion about the centre of
  yA   BA2 earth.
 F  A  B     y
  V  V
Here, inertia factor = mass of ball = m
BA2 GMm gm
 a y comparing this with a   2 y Spring factor  
mV R3 R
404 Oscillations
 Time period of oscillation of ball in the tunnel

d2 y g l
is 2
    y T  2
dt l g
inertia factor
T '  2 42. (4) The time period of a compound pendulum
spring factor
is
m R
 2  2 I
gm / R g T  2
mgL
Time spent by ball outside the tunnel on both the
sides will be  4 2h / g ml 2 ml 2 2mgl 2
Here I   
3 3 3
Therefore, total time period of oscillation of ball
is where L is the distance of centre of gravity from
peg P.
R 2h
 2 4
g g
41. (1) Let S be the surface tension of the soap
film. For equilibrium of rod
L
From figure, cos45 
l/2
l
L
2 2

2(l  y ) tan =Length of the boundary of soap
2 2ml 2 4 2l
film in contact with the rod.  T  2  2
l 3g
3 mg
mg  ( Fsurface )1 2 2
  
mg   2l tan  S  2; mg  4 Sl tan I
 2 2 43. (3) T  2
mgR
If the rod is displaced from its mean position by
small displacement y, then restoring force on the For side by side oscillations
rod is
I1  2mR 2 (Passing the centre and plane)
    
F   4S (l  y)tan  mg    4S tan  y
 2   2 mR 2
I2  (Passing through edge)
2
  
4S tan 4S tan   y As I1  I 2  T1  T2  f1  f 2
a 2 y 2
m  
 4Sl tan 2  44. (1) x  x1  x2  x3
 
 g 
   2sin t  2 3 sin t  2 cos t  3sin t
Oscillations 405
3 3 cos t Along Y-axis,

3 3
 (5  2 3)sin t  (2  3 3)cos t y  r2 sin 37  ( B sin t )  B sin t
5 5
 x  Asin(t  ) 3
or y  B sin t (ii)
5

1 2  3 3

Where A  68  32 3 and   tan  5  2 3  SHMs given by (i) and (ii) are perpendicular
 
with phase difference zero. So options (1) is
x  11.1sin(t  31.54)0 correct
47. (2) During the phenomenon of resonance the
45. (4) x  5sin t and amplitude of oscillation becomes large.
  Because applied frequency is equal to natural
y  5cos t  5sin  t  
 2 frequency.
So these two are perpendicular SHMs with  bt 
 
same amplitude (and frequency) and their phase 48. (3) For damped motion A  A0 e  2m 
 A0 e  Pt

difference is . So the path will be a circle of b
2 where P 
radius 5 units. 2m
46. (1) A0
For first case,  A0 e2 P
3
1
or  e 2 P
3
A0
For second case,  A0 e6 P
n
3
1 3 1
or  e 6 P   e 2 P      n  33
n 3
Along X-axis x  r1  r2 cos37 49. (3) Now, A  A0e
bt
2m
or x  Asin t  B sin t cos37

A0 b
  A0 e100b / 2500  A0 e 10
4 2
 A sin t  B sin t
5
 b  10ln2  6.93kg / sec.
This is the sum of two SHMs happenings on X-
50. (4) For damped oscillations
axis with phase difference   0. So, for this

 bt
2m 
2
x  a0 e cos(t  )
4  4
AR  ( A) 2   B   2. A. B cos 0
5  5  bt 2m 
 a  a0 e
16 2 8 a
x  A2  B  AB sin t
25 5 Given that a  when t  
e
2
 4  a0   b  2m
x   A  B  sin  t (i)   a0 e 2 m    .
 5  e b
406 Oscillations
l 2l
 l2   l1 
3 3
1. (27) Let the displacement of cylinder from its
3C 3k 3C
equilibrium position is ' x ' . k1   & k2   3k
2l 2 l
 ma    gAx 3. (1.414) x  a cos(t   ) (i)
  gA  dx
 a   x And v   a sin(t   ) (ii)
 m  dt
 gA Given at t  0, x  1cm and v   ,   

 cylinder 
m
Putting these values in equation (i) and (ii), we
k get
  spring block 
m 1 1
sin   and cos 
a a
 K   gA = 27 N/m
2 2
 1 1
2  sin 2   cos 2          a  2cm
2. (1) Here g  A  a a
g g gT 2 10  4 10
 A  2
   m
2  2  4 2 4 2  2
  1. (4) Given, y  a sin t  b cost
 T 
Let a  Acos and b  Asin (i)
then y  Acos sin t  Asin cost

bt

1. (0.729) A  A e 2m
y  Asin(t   )
0
after 5 second Which is in the form of SHM

b (5) From Eq. (i)
2m
0.9 A0  A0e ...(i)
a 2  b2  A2 cos2   A2 sin 2 
After 10 more second
 A  a 2  b2
b (15)

2m
A  A0e ...(i) 1
3
2. (1) Given, y  sin t  [3sin t  sin 3t ]
From (i) & (ii) 4
A = 0.7 2 9 A0
As this motion cannot be represented by single
sin or cos function, hence it is not SHM. As this
C motion involves sine and cosine functions, hence
2. (3) k
l it is periodic motion.
C C 3. (4) Motion of an oscillating liquid column in a

k1  k2 
l1 & l2 l
U tube is SHM with period, T  2 , where
g
l1  l2  l l is the height of liquid column in one arm of U
tube in equilibrium position of liquid. Therefore,
Given that l1  2l2 T is independent of density of liquid.
Oscillations 407
1. (1) Given,
 
y  3cos   2t  (i)
4 
dy  
Velocity, v   3  2 sin   2t 
dt 4 
Acceleration,
x(t )
sin t 
dv   B
a  4 2  3cos   2t   4 2 y
dt 4 
2 2
x(t )  B sin t  B sin t  B sin t
As A   y hence particle will execute SHM T 30
comparing Eq. (i) with equation
y  Acos(   ' t)
l
2  1. (2) For a simple pendulum, T  2
we have,  '  2 or  2 or T '  g
T' 
2. (2) Given, angular frequency of the piston, 4 2 L
T2  2 (constant) L
  200rad / min g T 
Stroke length =1m The graph between T 2 and L is straight line

 Amplitude of SHM, passing through origin.
2. (3) F  ma  m 2 x
Stroke length 1
A   0.5m
2 2 F  ( Asin t )
Now, vmax   A  200  0.5  100 m / min 3. (2) At A
3. (4) x  a cos t and y  a sin t vA  0  The particle is at extreme position

and hence a is maximum
2 2 2 2 2 2
 x  y  a (cos  t  sin  t )  a At B
It is an equation of a circle. Thus trajectory of vB is maximum  The particle is at mean
motion of the particle will be a circle. position and hence a = 0
4. (2) Given, T=30s, OQ=B. The projection of At C
the radius vector on the diameter of the circle
vC  0  The particle is at extreme position
when a particle is moving with uniform angular
and hence a is maximum.
velocity ( ) on a circle of reference is SHM.
Let the particle go from P to Q in time t. 4. (2) At time t  0, x  A, hence potential energy
should be maximum. Therefore graph I is correct.
Then POQ  t  OQP. The projection of Further in graph III. Potential energy is minimum
radius OQ on x-axis will be OR=x(t) say. at x  0, hence this is also correct.
408 Oscillations
5. (1) T  l T2 l
7. (1) From the graph T=4s
and vmax  5m / s;  A  5m / s
F 8
6. (3) From graph, slope K   4
x 2  2  5T 5  4 10
 A5 A   m
 T  2 2 
m 0.01
T  2  T  2  0.3s.
K 4 2 
Also,    rad / s
7. (2) 4 2
The equation of velocity can be written as
 
v  5sin  t  m / s
1. (3) In simple harmonic motion 2 
y  a sin t and v   cos t from this we have 
The phase   t
2 2
2
y v
2
 2 2  1, which is equation of ellipse. Phase change is
a a
2. (2) Acceleration  displacement   1
  t   t  t f  ti  t  s
2 6 3
3. (3) Let the equation of SHM is x  Asin t
   1 
dx v  5sin      2.5 m / s
v  A cos t   A2  x 2  2  3 
dt
8. (2) Time taken to reach the extreme position
dv 2 T
a   A 2 sin t   x from equilibrium position is . Velocity is
dt 4
maximum at equilibrium position and zero at
a
But x 2 1
2 extreme position. v  A  cos  t , K .E  mv
2
(m is the mass of particle and v is the velocity
a2 2 2 2 a2 
v   A  4  v    A  4 
2
of particle)
   
1
The graph between v 2 and a 2 is a straight line K.E  mA 2 2 cos 2  t Hence graph of K.E. v/
2
with (-)ve slope. s time is (b)
2
4. (3) Using acceleration a   x
At  xmax a will be maximum and positive.
5. (1) The body undergo SHM
6. (3) Potential energy of particle performing

2 2 1
SHM is given by: PE  m x , i.e., it varies
2
parabolically such that at mean position it 1. (2) The given equation is
becomes zero and maximum at extreme
positions. 
y = 5 sin 3t  3 cos3t 
Oscillations 409
In second case
1 3 
 y  2  5  sin 3t  cos3t 
In parallel k eq  2 k
2 2 
1 k2 1
    f   Hz
 10  cos sin 3t  sin cos3t  2 8 2
 3 3 
4. (4) The amplitude of oscillation is
   bt
y  10sin  3t   A  A0 e m (i)
 3
Energy of vibration drops to half of its initial value
Comparing it with standard equation
EA2  A E
y  Asin  t   
E0
   3 and A = 10 cm A 2 1
 
A0 E0 2
2 2
 Time period T   s
A0
 3
A , m  0.1kg , b  102 kgs 1
2. (2) In first collision mu momentum will be 2
imparted to system. In second collision when t
A0
momentum of (M  m) is in opposite direction mu  A0 e 10
2
momentum of particle will make its momentum
zero. t
ln 2   t  3.5s
After 13th collision 10
bt
5. (4) As we know, E  E e m
0
b15
mu  (M  13m)v 15  45e

m
mu u (As no. of oscillations = 15 so t = 15sec}

v 
M  13m 15 1 
m
b15
e
The amplitude is related with maximum velocity 3
as Taking log on both sides
v A b 1
 n3
m 15
u K 1 75 1 1
   A A  6. (2) x  sin t  sin 2t
15 M  13m 15 1 3 2
v  cos t  cos 2t
1 k
3. (4) In first case f  1 At t  0 x & v both have to be zero
2 m
7. (4) y  a 2 [2cos 2 t  1]
2 k
4 
m  x2 
 a 2  2  2  1
k  42 m=1  a1 
410 Oscillations
2a 2 2 2
y x  a2
a12 T'  T
3
at x = 0, y is negative and this is an equation of 2. (1) Amplitude of a damped oscillation:
parabola. Hence answer is (4)
bt

8. (3) Assume that the body starts its motion A  A0e 2m
from the mean position. The displacement of

1
the body is Given A  A0 ,
2
x  Asin(t )
bt
 ln 2  0.693
 2  2m
x  sin  t   sin(4 t )
 0.5 
2m
dx t  0.693
v   4 cos(4 t ) b
dt
When its displacement is half 500
t  2  0.693
20
1  2 
x  sin  t
2  0.5  t  50  0.693  34.6sec .
 1 3. (1)
4 t  t  s
6 24
a  A sin t0
The average velocity is
1 b  A sin 2t0
t 24
 vdt 4  cos(4 t)dt

c  A sin 3t0
0 0
 v  
t 0 1
a  c  Asin t0  sin 3t0   2 Asin 2t0 cos t0
24
96    1   ac
 v  sin  4 r     sin(0)   2cos t0
4    24    b
 v  12cm / s 1 ac 1 ac
 cos 1   f  cos 1  
t0  2b  2 t0  2b 
4. (4) Let a and v are maximum acceleration

 and velocity respectively.
1. (1) T  2
g
a 2 A
Given that v  10  
When the lift is accelerating upwards, effective  A
value of g is  g  g /2 

  10s 1 ,  & x  5m
4
 T' g
 T '  2  
g T 3g  
g Let x  A sin  t  
2 2  4
Oscillations 411
  7. (2) x  a sin t & y  a sin2t

5  A sin  0    A  5
 4
y  2a sin t cost
2
a   A  500 2
x2
5. (4) Projection of uniform circular motion on y  2x 1 
a2
along a line represents SHM. The two particles
are shown in the circular motion
2
y x (a  x)(a  x)
a
When y  0
x  0&(a  x)(a  x)  0
 x  a
so the correct graph is (2)
8. (4) Let T is the maximum tension in the spring
The force on the upper block is T  mg  m2 A
1.6
 1 625   10N
100
The two particles are moving in opposite
T  20N
direction along the circle. The time of meeting
is The force on the floor is N  T  40  60N
 2 / 3
r   d2x
t t 9. (1) F  m  bx
dt
 2  2 
r  2  2    10. (4) Let x0  Submerged length in equilibrium
T  T
x = Additional submerged length at time t
T
t  At equilibrium
6
6. (3) h = Length of block immersed in water  Mg  Ax0 g  Kx0 (i)
mg  FB
d2x
M  A( x0  x)g  Mg  K ( x  x0 ) (ii)
 Alg  l Ahg dt 2
650  A  54  102 g  900  A  hg d2x

From (i) and (ii)  M  ( Ag  K ) x
dt 2
 h  0.39m  39cm
Ag  K
h  2 
For a floating cylinder, T  2 M
g
M
 T  2
so for pendulum T  2 l  l  h ( Ag  K )
g
412 Waves on String
6. (3) y  f  x 2  vt 2  doesn’t follow the
standard wave equation f  x  t  .
7. (1) y1 =10-6 sin 100t   x 50   0.5 m

1. (1) Frequency of a stretched string is
n T  
f   106 cos   100t  x 50  0.5 
2L  2 
For f to be high L is small, T is large and  is  10-6 cos 100t  x 50  0.5   2  m

also small.
2. (2) y2  10-6 cos 100t  x 50  m
3. (3) y  104 sin  60t  2 x     2  0.5  1.07rad

A  10 4 m,   60 rad / sec, k  2m 1 8. (2)
 9. (2)
Speed of wave v   30m / sec
k TAB  64 N , TCD  32 N
 60 30
f    Hz 64 32
2 2  vAB  3
m / s, vCD  m/s
10  10 8  103
2
  m
k vAB 64 8 8
    x  5
vCD 10 32 5
T
4. (4)  0.14  T  4  0.14s
4 10. (4) Rate of energy flow is given by
1 1
 f  s1  1.79Hz 1 2 2
T 4  0.14 P a  .v
2
6 6
y  x, t  0   y  x, t    P v
5. (1) 2 then
2
x  x  2t 
11. (3) I max  I1  I 2  2 I1 I 2
y 24
 
t  x  2t 3 at x  2, t  2 I max  I1  I 2  2 I1 I 2
24  Sum of maximum and minimum intensities

vy  3
 3m / s.
 2  2  I1  I 2  .
Waves on String 413
12. (1) Superposition of waves does not alter the 2

'm'  r ' l '  r 2 
frequency of resultant wave and resultant   '   
amplitude l l' 4 4
 a 2  a 2  a 2  2a 2 cos   2a 2 1  cos  T
9
T' 
cos   1/ 2  cos2 / 3  2 / 3.
' 4  1 f  200 Hz
f ' 
13. (3) 2 '
2  2  3
0
14. (2) Particles have kinetic energy maximum

at mean position. n T
22. (3) The frequency f 
15. (2) 2L 
16. (3) Phase difference between any two points Given that
present in two adjacent loops is always equal to
. n T n 1 T

15 2L  2L   450  375
17. (4) Speed v  5
3
18. (4) Frequency of fundamental mode 1 T
f1   450  375  75
2L 
1 T
f0 
2l  1 360 300
L  3
  2m
If temperature increases, T decreases, f 0 2  75 4  10 150
decreases. m    2  8  103 kg
If temperature decreases, T increases, f 0
increases. 1 T
23. (2) f 0  fundamental frequency 
2l 
19. (2) Since both ends are free, (the free ends
are antinodes. f 0 1 T 2  f 0 2 15
   f0  

f0 2 T (T / T )  2 
100
Hence, 240     480cm
2
 1500Hz
240      240cm
T v 1 T 1
3 Also, v       2%  1%
240     160cm  v 2 T 2
2
For fundamental frequency,
20. (3)

21. (1) The fundamental frequency is  l    2l  constant.
2
T 24. (3) For the given standing wave

 2 A  6 or A  3cm
600   f0
2 25. (4) At x = 0 the phase difference should be
.
T
T' &  '  2
9 Thus, the correct option is (4).
414 Waves on String
26. (1) wavelength is maximum for fundamental 4. (3) Points B and F are in same phase as they
 are  distance apart.
mode  l   max  4l
4
5. (2) y  Asin(t  kx)
 vmax  f  max  4 fl.
  2 f , K  2 / 
27. (3) Sonometer is used to produce resonance
of sound source with stretched vibrating string.   2, k  1
28. (3) Middle paper does not fall. 6. (1) Given, y  a sin2 (bt  cx)
So there must be a node at the middle. On comparing with general equation
The standing wave patteren is
 2 t 2 
y  A sin   x
 T  
2
We get    2 b
T
1 1
 and A  a    and T 
4 x  2     4 x. c b
2
Maximum velocity ,
29. (1) The string vibrates in two segments in
the first overtone. Therefore, the amplitude of vmax   A  2 ba
vibration is maximum at  L 4  and  3L 4  .
 b
30. (3) Tension in sonometer wire = 4g Wave velocity , v  
T c
  mass per unit length Given, particle velocity  2  wave velocity
1 T 1 4g b 1
  416  2 ba  2   c 
2l  2l  c a
When length is doubled, i.e., 7. (1) Comparing the given equation with
Let new load = M  2 

y  A sin  2 ft  x
  
As,  '  
2
 3.14
1 Mg 1 4g 1 Mg 1 4g 
  
2l '  2l  4l  2l 
   2m
 M  2  2  M  16kg 8. (2) We have the progressive wave given by
y  2cos 6.284  330t  x 
As 6.284  2
1. (3)
2. (4) In transverse wave particle in the medium   6.284  330
oscillate perpendicular to the propagation. 2 1
Time period T   s
3. (3)  330
Waves on String 415
9. (1) x  A sin  2t  0.1x 

T
13. (4) v
 2 t 2  
also x  A sin   x
 T   When wave travels up tension increases
2 Hence ' v ' increases, and '  ' increases.
  0.1    20
 Frequency remains constant.
10. (1) Comparing the given equation with 14. (3) Suppose F is the tension in the string due
to its rotation. Choose a small element of the
y  a sin t  kx 
string of length l . If  is the mass per unit length
of the string, then mass of the element,

We get   3000  n   1500Hz
2 m  l    R
2 1
and k   12    m
 6
1
So, v  n  v  1500   250 m s
6
11. (1) y  60cos 180t  6 x  (1)
  180, k  6
 180
v   30 m s Using Newton’s second law for the element,
k 6
we have
Differeniating (1) w.r.t t,
mv 2
2 F sin  / 2 
dy R
v  60  180sin 180t  6 x 
dt
 
vmax  60  180  m s For small , sin 
2 2
vmax 60  180  mv 2
  3.6  104  2F 
v 30 2 R
12. (4) Compare the given equation with standard
 F  v 2
form
The speed of the disturbance
 2 x 2 t 
y  A sin  
  T  
F

v 2
 v.
 
A 1
15. (1) Equation of wave,
2 2 2
 3,   and  15 y  0.2sin 1.5 x  60t 
 3 T
Comparing with standard equation, we get
2 
T v 5
15 T y  A sin  kx  t   k  1.5 and   60
416 Waves on String
 60 27. (4)
 Velocity of wave, v    40 m s
28. (4) Frequency of fundamental mode
k 1.5
T 1 T
 v f0 
 2l 
Where,  is linear density and T is tension in If temperature increases, T decreases, f 0

the string. decreases. If temperature decreases, T
2 increases, f 0 increases.
So, T  v 2    40   3  10 4  0.48 N
29. (1) The frequency of the 3rd overtone is
16. (3) Here,   5.0m, n  2
(2n  1)v 7v
f   (n  3)
 v  n  2  5.0  10.0 m s 4l 4l
2 2 30. (4)
I max  I1  I 2   9  1 
17. (1)     31. (2)
I min  I1  I 2   9  1 
32. (1) Velocity of sound is inversely proportional
2
4 to the square root of density of the medium.
  4
2
T T
18. (1) Amax  A1  A2 ; Amin  A1  A2 v 
 A
 Amax  Amin  A1  A2  A1  A2  2 A2 . i.e., v.  constant
19. (1) Path difference   v1 2 2
2
   2
2
v2 1 
 aR  a 2   3a   2  a  3a  cos   2a.
2
f1 v1
  2
20. (2) y1  5sin t  kx  f 2 v1
y2  5sin  t  kx  150   33. (2) y  x, t   2sin  0.1x  cos 100t 
y2  5sin  t  kx  120  compare with y  A sin  kx  cos t
 Resultant  52  52  2.52 cos120  5 2

k = 0.1      20cm

21. (4)
The distance between a node and next antinode
22. (4) In stationary wave all the particles in one
particular segment (i.e., between two nodes)  20
   5cm
vibrates in the same phase. 4 4
23. (1) 34. (1) Equation of the component waves are;
24. (2) y  A sin  t  kx  and y  A sin  t  kx 
25. (3) For both waves the speed of component waves
26. (3) The string must be in fundamental mode 
since  is maximum    2l is
k
Waves on String 417
1 T 2 x
35. (4) f0  y  2a sin .sin t , we get
2l 

For elastic wire T = Kx
2 x
x  elongation from natural length  bx,

Hence frequency depends on K and original
tension 2
 
36. (1) Frequency of first overtone are second b
harmonic  n2   320Hz. So, frequency of first The distance between constructive nodes
harmonic
 
n 320  
n1  2   160 Hz. 2 b
2 2
37. (2) First overtone of string A = Second 41. (2) f  T  effective weight of load
overtone of string B.
1 T T
 Second harmonic of A = Third harmonic of 42. (4) n 2
n
B. 2l r  lr
 1 T  n1 T l r
 f1     1 2 2
 2l r 2   n2 T2 l1 r1
 1 T   1 T  T 3l 3r n
 2   3      3 3  n2 
 2l A rA    2lB rB   3T l r 3 3
l A 2 rB l 1
  A 5 9g 3 Mg
lB 3 rA lB 3 43. (2) f0  
2l  2l 
1 T T  M  25kg
38. (3) f  f
2l  l
2 2 2 44. (2)
T2  f 2   l2  2 3 9
        2   
T1  f1   l1    4
4
n1 : n2 : n3  3: 2 :1
39. (1) When the string vibrates in n loops, its
nv 1
frequency is f n  n (T &  are same)
2L 
Where L is the length of the string and v is the
velocity of the wave. 1 1
1 :  2 :  3  : :1  2 : 3: 6
3 2
 When the string fixed at its both ends vibrates
in 1 loop, 2 loops, 3 loops and 4 loops, the  1   2   3  110
frequencies are in the ratio 1: 2 : 3 : 4
 2 x  3 x  6 x  110  x  10
40. (2) y  a sin bx sin t
 The two bridges should be set at 2x i.e., 20
On comparing with standard equation of cm from one end and 6x i.e, 60 cm from the
stationary wave other end.
418 Waves on String
2
6. (3) At t=0, y  Asin kx  A sin x

 x
1. (2) y  A cos 2  2 nt  2 
  2  25
From the figure, kx= x  4   x 
 6 12
A A  x
  cos  4 nt  4  7. (2) y  f  kx  t 
2 2  
A 2
 Amplitude  , and frequency =2n. For 1, k   4 and   8
2 
 
Wavelength  . The wavelength of wave -2 is and frequency
2 2
is double the initial frequency since v is same
b d for both
2. (1) Speed of pulse   .
a c
2
3. (2) For 2, k '  ( / 2)  4 and  '  2  8  16
4. (2) Given wave equation is y  x, t   phase of 2  (8 x  16t )


 ax 2 bt 2  2 abxt  8. (1) Tension in string at P
e
2
e  
 a x bt
It is a function of type y  f  x  vt 
Coefficient of t b
 Speed of wave  
Coefficient of x a
5. (1)   mass per unit length
  0.8  106  12.5  103  0.01kg / m.
  2 20  40 rad / s  Velocity of pulse at P
Wave speed is T
v  v2   T

v  T /   64 / 0.01m / s  80m / s
v 2 d   dT (v = constant)
v 80
    m  4m
f 20 v 2 d     dzg
 2  As the pulse goes down z increases

 y  1.0cos  40 t  x 
 4  so dz is (+)ve, where as d  has to be (-)ve .
   Since  should decrease to maintain the pulse

 (1.0cm) cos  40t  x  .
 2  velocity as constant.
Waves on String 419
d
 g z 11. (2) The tension in the pulley system is
     2   dz
0 
v  0  2  M  2 M  
T  g

 gz / v 2   M  2M 
   0e
9. (4) Consider a small element of length dx T

v0 
present at a distance x from the top end. 
(2M  M ) g g
a 
2M  M 3
Let t is the time at which the pulse reaches Q.
Here v t is the distance travelled by the pulse
1 2
v gx on the string. at is the distance travelled by
2
dv 1 g the string in time t.
 av  gx
dx 2 x 1
l  v0 t  at 2
 a  g / 2  constant 2
By solving the equation carefully for t and
10. (2)
keeping the values of T , v and a we get
3l  4 M 4M 
t  2 
g  m m 
1/4 1/4
A2  h1   4300 
12. (4)      4.55
A1  h2   10 
0.2 x
The tension at a distance x from the block is   4.55  x  11.37m
0.5
mg  mx 
T  Mg + x M  g
l  l  13. (3) The radius of the string at a distance x is
T  Ml  r r 
 v    x g r  r1   2 1  x
  m   l 
l dx 2
 t  r r 
0
 Ml  A    r1  2 1 x
  x g  l 
 m 
F F F 1
 v   
2   Ml  Ml   A   r2  r1 
   lg    r1  x
g   m  m   l 
l l0 dx   r1  r2 
2
mg
 M m  M   t
0 v
 
F  4 
 2l0 .
420 Waves on String
14. (1) Velocity of the pulse is 19. (4) In this case of sustained interference in
which position of maxima and minima remains
16 40 fixed all over the screen
v m/s  m/s
5  102 5
2
I min  a1  a2 
1 1 1 1  
 KE  mv 2  I cm   2  mv 2  mR 2 2 I max  a1  a2 
2 2 2 2
And both waves must be travelling in the same
1 1 direction with a constant phase difference
 mv 2  mv 2
2 2 (condition for coherence)
1 1600 1 T
K.E  mv2  2 R  v2  2   5 102  J 20. (4) The fundamental frequency, f 
8 5 2L 
 4 Joule. where  is the mass per unit length of the wire
15. (2) In reflected and incident wave all other
factors are common. So it is only amplitude 1 T 1 4T 1 T
f  2
 2

which can change the power. Given that 75% is 2L D 2 L D LD 

1 4
transmitted. hence th will be reflected back
4
1
f 
A LD
Ar  i
2
 As P  A2 
21. (3) y  A sin  kx  kct   A sin  kx  kct 
Ai v v v 1
 2or 1 2  2 or 1 
Ar v2  v1 v2 3  kx kct  kx  kct   kx kct kx kct 
 2Asin cos 
16. (1) Energy per unit time (i.e) power is  2   2 
constant at junction  Pi  Pr  Pt . The energy
 2 A sin  kx  cos  kct 
per unit area per unit time is not constant (i.e.
Intensity). Because in transmission area of
2 2
medium may change Thus  k ,  
 k
17. (3) a1  5, a2  10

2 2
The distance between adjacent nodes is
I max  a1  a2   5  10  9 k
    
I min  a1  a2  2  5  10  1
22. (1)
1 1   23. (2) In stationary wave y  2a cos kx sin t

18. (1) y sin t  sin  t  
a b  2
Amplitude of the stationary wave is

Here phase difference  A  2a cos  kx 
2
Resultant amplitude Intensity at certain point is maximum when
 1   1 
2
1 1 ab
2 amplitude is maximum. Intensity is maximum
      a  b  ab
 b  b when cos(kx) is maximum
Waves on String 421
2 29. (3) Velocity of the particle is

24. (3) Wave number k   0.6cm 1
 dy x
 4cos   cos  40 t  .40
dt  3 
 
  cm
2 0.6 1
here x  3, t 
8
In third harmonic the length of the string
dy
3      v p  160 cm / s
   3  cm  15.7cm dt
2  0.6 
30. (3) The distance between an adjacent node
25. (1) Displacement node occurs at

 3 and anti node is
10x  , e.t.c… 4
2 2
 6
  1.5cm  0.015m
1 3 4 4
x  , ...
20 20
31. (3) The positions of the points
x  0.05m and 0.15 m
 
Displacement anti node occurs at are x1  
3k 6
10x  0, , 2,3, e.t.c..
3 3
x2  
x  0,0.1m,0.2m,0.3m 2k 4
  2 (distance between two consecutive nodes 7
(or) anti nodes.) x  x2  x1 
12
  2  0.1  0.2m The points x1 and x2 are shown the figures.
 50
wave speed    50 / sec
k 10
 D is wrong
2
26. (2) Wave number k   0.6cm 1

1  
 
  cm
2 0.6
3   
l   3  cm  15.7cm
2  0.6 
2 7
2  x   
  6
27. (2) From the given equation k  ,   40.
3 If two points are present in the same loop of a
 standing wave then   0 and if they are
Wave velocity v   120cm / s
k present in adjacent loops then   
 6 1 6
28. (1) x   3cm 
2 2  2 7
422 Waves on String
32. (2) The clamped square plate is shown in the  x y

figure    n (n = 0, 1, 2,....)
 2 
 x y 2
   x y 
 2  
2
y  x

2
The displacement of the plate must be zero at All the points that lie on the line y  x  are

all the edges of the plate that is always at rest.
x = 0 (on y - axis) 34. (1) The wave is in fundamantal mode of
x = L (parallael to y - axis and present on the vibration as shown in the figure
edge) The angular frequency
y = L (parallel to x - axis and present on the
  2 v  10
edge)
y = 0 (on x axis)
For all the above four conditions u = 0
The equation that satisfies is
 2 
 l k  
x y 2  l
u  a sin( )sin( )
L L
The velocity of the particle at the mid point of
33. (4) The given equation represents the the string is v p  A  3.14m / s  A  0.1m
superpostion of two progressive waves travelling
along x and y axes given by From the above information
A 
r1   sin(t   x) & y  0.1sin( x )sin(10 t )
2 2
35. (4)
A
r2  sin(t   y ) 1 T 1 T f0l
2 f0  , f1  
2l  l   l 
r  r1  r2 2   l    l 
 2   2 
 ( x  y)   ( x  y) 
 A sin   cos t   2 f0  2  l 
 2   2   f1   2 f 0 1  
 2 l   l 
 1  
from the given equation the amplitude of the  l 
  x y 
wave is AR  A sin   2   1 T  2  l 
   f2    2 f 0 1  
Again, l    l 
The amplitude becomes zero 2   l 
2 
  x  y 
AR  A sin     0 8 f0  l
  2   No. of beats  f1  f 2 
l
Waves on String 423
5. (1)  / 2  10    20m
1. (4) The given equation is 1 1

f  Hz.  v  f    20m / s  5m / s
4 4
 x
y  y sin 2  nt  
   6. (1) a  amplitude  0.04m,   0.2m
2 2v 2 200

  2 n, k     rad / s  2 103 rad / s.
  0.2
 The correct option is (1)

The velocity of the wave is v   n
k 7. (2) y  0.1sin(200 t  20 x)
Given that v p  4 v
y y x
 vp     (tan  )  v
 y t x t
y (2 n)  4n   
2
200
Now, v  wave velocity   10m / s
0.8 0.8 20
2. (1) y  2
(4 x  5t )2  5  5   v p  tan135   10 m / s  10 m / s.
42  x  t   5
 4 
8. (3) y  2sin(10t  5x)
y  f (n  vt)
y
 20cos(10t  5x)
5 t
v m/ s
4 at t  0, x  0  v p  20 cm / s
5 9. (4) The maximum of the pulse is y = A, when

 x  vt   2  2.5m / s
4
x t 
2 x t 
    0
wave is moving -ve x-direction e  T 
 1   a T 
5 x  a (t  T )
 Distance traversed in 2 sec   2m  2.5m
4 10. (3) For the same medium velocity of the
0.8 curve is constant.
At t  0, x  0, ymax ( x, t )  m  0.16m
5 1
2 (1.25 f1 )  1 f1  2 
At t  0 and x  a (a = a+ve constant). y will 1.25
have same value. So the pulse is symmetric.  Decrease in wavelength
3. (1) y( x, t )  f (kx  t ) 1  2
  100%  20%
1
 10 11. (2) For resonance, the amplitude should be
comparing, k  3,   10  V   m / s.
k 3 infinitely large. The combination ( b  0 and a =
  c x  d t 
2 c) best satisfises this condition.
4. (4) ye  f ( kx  t )
12. (1) y( x, t )  0.005cos(x t ) (Given)
So it represents a progressive wave moving in -
Comparing it with the standard equation of wave
 d
x direction with speed v  
k c y( x, t )  a cos(kx  t ) we get
424 Waves on String
k   and    Now, TAB  T cos and T sin   20
2 2 160
But k  and    TAB  20cot   N  38.8 N
 T 17
2 2 TAB
   and   Wave velocity in AB   107.89m / s
 T 
Given that   0.08m and T  2.0s
1.5
 Time  sec  13.9ms.
2 2 107.89
   25 and   
0.08 2 x
x  x2 
13. (3) 15. (3) T  g  dx  g  0 . .dx  0 g. 
0 l  2l 
T gx  x
v      
 2  l
dv g
av 
dx 4
At x distance from end dx gx

v 
dt 2
T
v
 l
dx g
t
  dt
0 x 2 0
 xg
v
 2l 2l 2l
t  2
a g/4 g
v  xg  v  x
If the string accelerates up with a  g. Then
3
10 10 1 the tension in the string is
14. (1)  kg / m  kg / m
3 300
 x2 
T  20 g  
 2l 
T
v  gx

Acceleration of the wave pulse relative to the

g
string is a  .
From the given figure 2
D/3 4 2 8 2 1
cos     Distance covered s  v0 t  a ' t
L/4 3 3 9 2
2
17 1 g  l 
 sin   0    g  .  2   3l
9 2 2   g 
Waves on String 425
16. (3) Consider a small element of length dx at 2

12 1  f 2   1 
a distance x from the end of the string.     As  
 22   f1   2 
The centrifugal force on the element is
1 1 f2 k
 . 
2 f
 1 
18. (2) Energy  Intensity  Area
 E  I  S , Also, I A2 , S is the area
E ' (2 A)2 (S / 2)
 E  A2 S   2
E A2  S
dT   dx x  2
E '  2E
l 2 l
x 
T   2  x dx   2   19. (1) The dimension of option (1) g  matches
x  2 x
the dimension of speed.
l 2  x 2  20. (2)
T   2 
2 21. (1) If wave reflects form rarer medium or
free end than the change of is phase zero.
T is the tension in the string at a distance x from
the end. The velocity of the pulse at the distance 22. (1) From the given waves it can be observed
that there is no phase inversion for the reflected
dx T  2 2 wave. It means the incident wave is travelling
x is v   l x
dt  2
in denser medium. I reflected  I incident (some energy
l t always gets transmitted)  a ' can not be greater
dx 

0 l 2  x2

2
 dt
0
than a.
23. (3) The resultant amplitude is given by
l
 1 x   Ar  A2  A2  2 AAcos   2 A2 1  cos 
sin l   2 t
 0
  2 
 1  l    Ar  2 A cos 1  cos   2 cos 
1 2  2
sin  l   sin (0)   t
    2
1
24. (1) Path difference  x   50cm  m
   2
 2  0  2 t
 
 Phase difference
 2 2 1
t x2    x      
 2  1 2
 2
T Total phase difference    
17. (3) v f  2
T  f  2
3 3

 A  a2  a2  2a2 cos  2 / 3  a

 f 1   22 f 22
2 2
1 1
2 25. (1)
426 Waves on String
1 T As length of one loop or segment is  2 , so

26. (2) n  n1l1  n2l2  k
2l  length of 2 segment is
k k k

l1  l2  ltotal   
 2  1.21Å   =1.21Å
n1 n2 n 2
31. (1) Let the string vibrates in p loops,
1 1 1 wavelength of the p th mode of vibration is given
  
n n1 n2
by
27. (4) y  a cos(kx  t )  a cos(kx  t) 2l
p  (1)
y  2a sin kx sin t p
28. (4) y  Asin(t  kx)  Asin(t  kx)  4 x 

Given, y  2sin   cos  96 t 
y  2 Asin t cos kx  15 
The amplitude is AR  2 A cos kx 4 2

k 
15 
For standing wave nodes
cos kx  0 15
 (2)
2
2 
.x  (2n  1) From (1) & (2) we get
 2
 p  16.
(2n  1)
x  , n  0,1, 2,3,.....
4 32. (1) From the given wave equation the
amplitude
29. (1) Fundamental frequency is
A = 2 mm sin[6.28x]
1 T 1 T
f   2 2
2  2 A k = 6.28     1m
 6.28
1 T For the wave fixed at two ends

f  
2  d 2  L  L  50cm
2
f1 T   d 22 33. (1) From the given equation the amplitude is

 1 2
f2 T2 1  d12  20 
AR  A sin  x
 3 
f1 2 1 2 2
    A  20   20  1
f2 1 9 1 3 AR   A sin  x   sin  x 
2  3   3  2
30. (2) The given standing wave is shown in the
20  1
figure. x1   x1  m  2.5cm
3 6 40
20  
x2    x2  12.5cm
3 2 3
Waves on String 427
at x = 0 it is a node 1
2
Element is dk  dmv p
The distance between points having amplitudes 2
A y
is 10 cm vp   A sin kx cost , dm  dx
2
t
34. (4)
1
35. (1) The given standing wave equation is y dk  ( dx) A2 2 sin 2 kx cos 2 t
2
=A cos kx sin t
(  linear mass density)
The velocity of the progressive wave is
1

78.5 dk   A 2  2 sin 2 ( kx ) dx (At t  0)
v   50.cm / s 2
k 1.57
The amplitude of the wave is. 1 L
k  A2 2  sin 2 kxdx
2 0
AR  Acos kx
1  1 
At antinode   A2 2  L  sin 2 kL 
4  2k 
A R = A = A coskx
T.E = K.E + P.E
kx = 0,  , 2 ...
12 2  1 
 =  A  L  sin 2kL  ( P.E  K .E )
2  2k 
x  0, ,  ..
2
For the first wave
The closest distance of antinode from the origin
1 
  E1   A2 2 L (k  )
is x    2cm 2 L
2 K
36. (4) The resultant of the two waves at the 1 2
E2   A2 (2 )2 L (k  )
given point is 2 L
         E1 1
y  y1  y2  2a cos  1 2  sin t   1 2    E2  4E1
 2    2  E2 4
 1  2  38. (4) Let y  A sin kx sin  t represents the

The resultant amplitude is AR  2a cos   standing wave fixed at both ends.The velocity
 2 
y
1   2  of a particle is v p 
Given that AR  a   t
2 3
v p  A sin kx cos  t
2
1  2  The small K.E of the element is
3
37. (3) Consider a standing wave 1 1
dk  dmv 2p   dxv 2p
y = Asin kx sin t 2 2
Consider a small element of mass dm in the 1

dk   A2 2 sin 2 kx cos2 tdx
string. The kinetic energy of the 2
428 Waves on String
L
1 41. (1) Let the two parts are L1 and L2 , then
k  A2 2 cos 2  t  sin 2 kxdx
2 0
L1  49cm and L2  51cm
L
1
k  A2 2 cos 2  t  1  cos 2kx  dx 1
4 0
For same tension f 
L
1  1 L  f1 L1  f 2 L2
k  A2 2 cos 2 t  L  sin 2kx 0 
4  2k 
 49 f1  51 f 2 (i)
given that the string is in Fundamental mode
 2
Also f1  f 2  1 (ii)
 L    2L  k 
2 2L From (i) and (ii); f1  25.5Hz
1 f 2  24.5Hz
k   LA2 2 cos2 t
4
1
2 2 2
= mA  cos t
4
  L  m
1. (40) The frequency produced in a string of
1 1 length l, mass per unit length  , and tension T
 k  mA2 2  cos 2 t  mA2 2
4 8 is
(for one time period) 1 T

n
2l 
1
2  cos 2 (t )  
 sin (t ) 

2
l1  50cm , n1  800 Hz ,
39. (2) Let T1 and T2 are tensions in the left and n2  1000 Hz
right strings.
n1 l2 n l 800  50
According to question,    l2  1 1   40 cm
n2 l1 n2 1000
1 T1 1 T2 2. (2) The tension in the string at a distance x

2l  l 
Mgx
from the lower end is T  x  
T2  T1 / 4 L
For rotational equilibrium,
T1 x  T2  L  x   x  L / 5
2
40. (3) Phase difference   x is not valid

for a standing wave. The phase difference
between any two points present in successive The velocity of the pulse is
loops is  .
The K .EMax B is more than that of A since the T
 v  x   gx
amplitude of B is more than that of A . 
Waves on String 429
dx L
dx t
Tl T Y l
  gx    g  dt Also Y    
dt 0 x 0
Al A l
1 Y l
L f 
t2  2 2s 2l l 
g
l
l  1.5m,  0.01,   7.7 103 kg / m3
v 1 T1 l
3. (1.26) f  
1 1 
Y  2.2 1011 N / m2
v 1 T2 After solving
f  
1 2 
2 103
f   Hz
7 3
1 T1 1 T2

1  2  f  178.2Hz
T1   T   t x 
 1  2 3. (6.25) y  0.02(m)sin  2  0.04(s )  0.50(m)  
  2     
T1   2.2  .T2
2 Comparing this equation with the standard wave
equation
Mg  4.84[ M g  Fb ] y  a sin(t  kx)
The buoyancy force
2 1
 v   25Hz
Mg Mg 0.04 0.04
Fb  lVs g  l 
S ( S .G )
2
k    0.5m
  0.50
M
Mg  4.84  Mg  .g 
  S.G    velocity, v  n  25  0.5 m / s  12.5 m / s
4.84 T
1  4.84 
S .G Velocity on a string is given by v 

4.84 4.84 T  v 2    (12.5)2  0.04  6.25 N
 3.84  S .G 
S .G 3.84
 S .G  1.26
 
1. (2) In fundamental mode, l  2   
4 2
T 60.5
1. (110) v    v   0.035 / 7 
 110m / s
n T 1 T
2. (178)  
2l  2l A
430 Waves on String
   2l
m 3.5 102
 l   0.875m
Given l  100cm, f  2.53kHz  4  102
Using v  f  Now, from Eq. (i),

v  45  2  0.875  78.75 m / s [  2l ] (ii)
 v  2.53  103  2  100  102
T
v  5.06 km / s As v

2. (3) Given, frequency of A, f A  324 Hz
 T  v 2 ()  (78.75)2  4.0  102
Now, frequency of B, f B  f A  beat frequency
 248.06N
 324  6
or f B  330 or 318Hz
1 1
Now, if tension in the string A is slightly reduced 1. (1) Here, n  500Hz, T    2 103 s
n 500
its frequency will also reduce from 324 Hz.
Phase difference corresponding to
Now if f B  330 and f A reduces, then beat
frequency should increaes which is not the case 2  103 s  2 rad
but if f B  318 Hz and f A decreases the beat Phase difference corresponding to

frequency should decrease, which is the case
21103
and hence f B  318Hz. 1103 s    rad
2 103
2.5 2. (3)
3. (2) Here,   kg / m, T  200 N
20
3. (2) On reflection from a denser medium,
T 200 there is a phase reversal of 180.
v 
 2.5 / 20
2
New amplitude   0.6  0.4
3
200  200 200
v   40m / s
25 5  Equation of reflected wave is
l 20  x 
y  0.4sin 2 t   180
Time taken, t    0.5s  2 
v 40
4. (3) In fundamental mode of vibration,  0.4sin2(t  x / 2)
 4. (1, 4) The given equation is
 l or   2l
2
 2x 
v f (i) y ( x, t )  0.06sin   cos(120t )
 3 
f  45Hz
As terms involving x and t are independent of
m each other, the given equation represents a
As   4 102 kg / m
l stationary wave. Compare the given equation
with the standard from of eqaution stationary
m  3.5  102 kg wave
Waves on String 431
y( x, t )  2 Asin kx cos t
2 2 1. (2)
k 
 3
   3m and   120
 120
 f    60Hz
2 2
and v  f   60  3  180m / s
Hence the given stationary wave is the result of dotted line is next position of wave is small
superposition of two waves of wavelength 3m interveal of time
and frequency 60 HZ each, travelling with a
velocity of 180 m/s in a opposite directions.  option (2) only correct
5. (1, 2, 3)  
2. (1) Speed  f   4n  ab  As ab  
 4 
3
1. (4) Acceleration is maximum at extreme Path difference between b and e is
4
position. Velocity is maximum at mean position.
At mean position displacement may not be zero. 2
So the phase difference  (x)
2. (4) y  a cos kx sin t 
2 3 3
T  2 T  Phase difference  . 
At t  y  a cos kx sin     4 2
4  T 4
3. (2) After two seconds each wave travel a
y  a cos kx distance of 2.5  2  5cm i.e., the two pulses will
so the correct graph is (4) meet in mutually opposite phase and hence the
amplitude of resultant will be zero.
3. (3) After 2 sec the pulses will overlap
completely. The string becomes straight and 4. (1) When a pulse reaches a boundary then
therefore the string does, not have any potential its shape can be determined by assuming an
identical pulse travelling opposite to it from right
energy and its entire energy must be kinetic.
to left and its distance is present at 10x as shown
4. (3) If wave is reflected from fixed end the in the figure.
shape of reflected wave will be reversed.
5. (4) The reflected wave suffers a phase

change of 180.
so at the boundary the wave shape is the

resultant of left and right (imaginary) pulses. So
the correct graph is (1)
432 Waves on String
5. (2) When the triangular phase gets reflected, 1/4

it suffers a phase change of 180. Interfering  v2   g 2  a 2  a2
    2   1 
pulses are shown below.  v1   g  4g 2
60.5 1 a2
1
60 4 g2
a2 1 g g
2
 a 
g 30 30 5
3. (4) Frequency of B is either 420 Hz or 430
Hz. As tension in B is increased its frequency
will increase.
If frequency is 430Hz, beat frequency will
T TA increase.
1. (4) v 
 A If frequency is 420Hz beat frequency will
decrease, hence correct answer is 420Hz.
T Y L
Stress =  Y  Strain  4. (2) The frequencies in both wires must be
A L same
Y L A  m T, r and l are same f1  f 2

v    
L m/ L  L Frequency of vibration
Y LA n T
v f
m 2l  r 2 
As T, r, and l are same for both the wires

v2m
L 
YA f1  f 2
90  90  6  103 n1 n2 n1 1 1
    
16  1011  1  106 1 2 n2 2 2
L  0.03 mm
T1 Mg
2. (1) When the car is at rest v1  
 200
  1. (3) v   160 m/s
k 5 / 4
when the car is in acceleration
2. (1) Given taht
2 2
T2 M a g
v2   v  2( v p ) max
 
f   2 A
v1 g
 1/2
10
v2 g 2
 a2    4 A  4  40 cm

Waves on String 433
3. (2) l1 : l2 : l3  6 : 3 : 2 5. (4) Total length of the wire, L = 114 cm
6 f1 : f 2 : f 3  1: 3 : 4
 l1   110cm  60cm, l2  30cm, l3  20cm
11
Let L1 , L2 and L3 be the lengths of the three
T 400
parts
v  m / s  200m / s
 0.01 1
As n
L
Now common ratio of the three lengths
l1 , l2 and l3 is 10. 1 1 1
 L1 : L2 : L3  : :  12 : 4 : 3
1 3 4

  10cm    20cm
2  12 
 L1    114   72cm
 12  4  3 
v 200
 fmin   Hz  1000Hz
 20  102
 4 
L2    114   24cm
4. (2) If ‘a’ be the amplitude, then  19 
2a  1cm  a  0.5cm
 3 
Also if ‘f’ be the frequency, and L3    114   18cm
 19 
1 5
Then  0.4  f  Hz Hence the bridges should be placed at 72 cm
f 2
and 72  24  96 cm from one end.
 5 5
 vmax   a   2    0.5cm / s  cm / s.
 2 2
434 Sound Waves
d t1 v2 T2
t also v  T  t  v  T
v 2 1 1
2 303
   t2  1.9sec.
t2 283
1. (1)
2. (3) 13. (4)
3. (4) v
14. (3) f 
4. (4) 2L
As v changes f also changes since L is constant.

5. (4)   100m  t   4sec.
v v
15. (2) Fundamental frequency f is f 
2
4
6. (4)   x
 when the pipe gets filled up  decreases and
hence f increases. When the inflow and outflow
Where  is wavelength.
of water are equal  becomes constant and
   hence f also becomes constant.
   
2 3 6 v
16. (2) f closed 
7. (2) A wavefront is a line or surface on which 4L
the disturbance has the same phase at all points.
v
8. (3) f open 
2L
 f open  2  512  1024 Hz
P
9. (2) v
d 17. (3) Series of resonant frequency must be
integral multiple of fundamental frequency. This
 series is integral multiple of 75Hz . Hence
10. (2) v
M missing frequency is 75Hz
nv
 18. (1) In case of open pipe, f  where n =
Since is maximum for H 2 so sound velocity 2l
M
order of harmonics= order of mode of vibration
is maximum in H 2 .
f  2l
n
11. (4) v
12. (2) Suppose the distance between two fixed 480

  2  1  2.9 [Here v=320m/s]
points is d then 320
Sound Waves 435
19. (2) In closed organ pipe. First resonance l  l1  l2

occurs at  / 4. So, in fundamental mode of
 v v v
vibration of organ pipe  (l  0.3d )  
4 4n 4n1 2n2
Where 0.3d is end correction. 1 1 1
  
2n 2n1 n2
v v
Frequency of vibration n    4(l  0.3d ) n1n2
n
n2  2n1
As l is same, wider pipe A will resonate at a
lower frequency, i.e., nA  nB . 23. (4) In a closed organ pipe the fundamental
frequency is
20. (2) From v  f 
v
f 
 340  340      1m 4L
 320ms 1
For 1st resonance, l1  f   80 Hz
4 4  1m
In a closed organ pipe only odd harmonics are
l1  25cm, so length of water column is 95 cm.
present. So, it can resonate with 80 Hz, 240 Hz,
3 400 Hz, 560 Hz.
For 2nd resonance, l2   75cm,
4
24. (3) Using   2(l2  l1 )
Separation between consecutive nodes is,
l2  l1  50cm  v  2n(l2  l1 )
21. (2) Fundamental frequency of open tube is,  v  2  512(63.2  30.7)  33280cm / s
v Actual speed of sound
f 
2L v0  332m / s  33200cm / s
where v is the velocity of sound in air and L is Hence error  33280  33200  80cm / s
the length of the tube
25. (4)
330ms1 26. (2)
 f   660Hz
2  0.25m
27. (3) I R  I1  I 2  2 I1 I 2 cos 
The emitted frequencies are f, 2f, 3f, 4f ....
i.e., 660Hz, 1320Hz, 1980Hz, 2640 Hz,... 7I  I  9I  2(3I )cos
22. (2) Frequency of closed pipe, 3I  6I cos
v v 1
n1   l1  cos       120
4l1 4n1 2
28. (4) The wavelength of sound source
v v
Frequency of open pipe, n2  2l2
 l2 
2n2 330
  3 metre.
110
When both pipes are joined, then length of closed The phase difference between interfering waves
pipe at Pis
436 Sound Waves
2 2 2 v
 x   S2 P  S1P    5  4  35. (4) 1  2l , 2  2l  2l  n1 
 3 3 2l
 Resultant intensity at P is
v
2 and n2 
I R  I 0  4I 0  2 I 0 4I 0 cos  3I 0 2l  2l
3
29. (3) Persistence of human ear is 0.1s  v 1

 1 vl
 No. of beats  n1  n2  2  l  l  l   2l 2
 
 More than 10 beats can’t be heared
36. (1) Velocity of sound increases if the
30. (4) Beats are observed when intensity at a temperature increases. So with v  n , if v
point varies with time and beat frequency is increases n will increse at 27C , v1  n, at
equal to the number of times the intensity 31C , v2  (n  x)
reaches maximum per second.
 RT 
31. (1) 262  f  2(256  f ) Now using v  T  v  
 M 
f  250Hz
300  x (273  31) 304 300  4
32. (4) The time interval between two successive    
300 (273  27) 300 300
1 1/ 2
maximas is equal to s x  4 
6  1  1  
300  300 
1
For every th sec  1 4 
12  1     x  2.
 2 300 
The minimum will be observed.
 (1  x)n  1  nx 
33. (1) The number of beats will be the difference
of frequencies of the two strings. 37. (2)
Frequency of first string 38. (1)
1 T 1 20 39. (3) Reverberation time T  V .

f1    137.03Hz.
2l1  2  51.6  10 103
2
I
40. (4) L1  10log10 decibel
Similarly, frequency of second string I0
2I
1 20 L2  10log  L  3.01dB
f2   144.01 I
2  49.1  102 103
41. (4) Let I1 & I 2 are initial and final intensities
Number of beats  f 2  f1  144  137
 7beats  I1 
1  10log  
 I0 
v v
34. (4) f1  f 2  6 or 2l  2l  6
1 2 I 
2
Later,  2  10log  I 
 0 
v v
  6 or v  306 ms 1 .
2(0.5) 2(0.51) Given, 1   2  20
Sound Waves 437
I  v
 20  10log  1  For situation 2, f ap2  v  (v )  f  336 Hz
 I2  s
I1  100I 2 f  fap1  fap2  96Hz
P f2 9
42. (3) I 51. (4) 
4r 2 f1 8
 P  I  4r 2
 v  v0 
2 2
 (0.008W / m )(4  10 )  10 W
f2    f1  vs  v0 
 v  vs 
43. (3)
9 v  v0
44. (2) Wavelength of sound depends on wind  
and source velocity. 8 v  v0
45. (3)  v0  20 m s
 v 
46. (3) f ' f  
 v  vS 
1. (4)
47. (2)
2. (4)
 v  vS   330  220  3. (4) Speed of sound, doesn’t depend upon
48. (2) n'  n    2000  
 v  vS   330  220  pressure and density of the medium at constant
temperature.
550 4. (2)
 2000   10000Hz
110
5. (2)
49. (1) The apparent frequency
6. (4)
 v 7. (1)
 v 
f' f 5  6 f  1.2 f 8. (2)

 v  5
  9. (4)
10. (4)
v = Velocity of the sound
As the source is stationary wave length remains 2
11. (1) Phase difference  (Path difference)
unchanged for observer. 
 v0   2
50. (3) For situation 1, fap   v  v   f   (Path difference)
1
 s  3 
320

  378  432Hz  Path difference =
320  40 6
12. (1)
13. (3) Speed of sound, doesn’t depend upon
pressure and density of the medium at constant
temperature.
438 Sound Waves
14. (3) v
28. (3) Let is the fundamental frequency
1 v1 2 4 2L
15. (1) vsound     2 as 510 Hz & 680 Hz are successive frequencies
 v2 1 1
v1  v 
 v2  n   510 (i)
2  2L 
v 
vHe  He M H2  n  1    680 (ii)
16. (3)    L
2
vH 2  H2 M He
From (i) and (ii) we get n  3
5 5 2 25 v
      170
3 7 4 42 2L
P v d2
Given that v  340 m s
17. (2) Speed of sound v   1 
d v2 d1  L  1m
[ P  constant] 5v
29. (4) Second overtone of the closed pipe is
4L
v 330
18. (3) v f     1.29m
f 256 2v
First overtone of the open pipe is
19. (4) 2L
20. (2) Given that

21. (3) 5v 2v
  100
22. (1) 4L 2L
23. (2) At pressure antinode displacement node v 5 

forms   2   100
2L  2 
24. (4) Pressure change will be minimum at both v
 200
ends. 2L
25. (1) In open pipe at both ends, the natural 30. (3) For one end closed organ pipe,
frequencies of oscillation from a harmonic series fundamental frequency is
that includes all integrals multiples of the v 340
fundamental frequency, i.e. all even odd f    100 Hz . In one end closed
4l 4  0.85
harmonics are present.
pipe only odd harmonics of fundamental
Therefore, if fundamental frequency is n, then frequency can be generated hence
other frequencies are n, 2n, 3n, 4n,..., frequencies below 1250Hz are only six, which
v are 100Hz, 300Hz, 500Hz, 700Hz, 900Hz,
26. (2) Open pipe f1  1100Hz.
2L
v f v v
1
Closed pipe f 2  4 L  f  2 31. (4) fA  , fB 
2
2l 4l
27. (2) l2  3l1 fA 2


l2  3  20 = 60 cm fB 1
Sound Waves 439
 35. (4)
32. (2) 1  x   22.7 (1)
4 36. (4)
3 37. (4) Amax  a  b and Amin  a  b

2  x   70.2 (2)
4 Amax  Amin  2b
5 38. (4) Here, A1  A, A2  A,   120
3  x  (3)
4
The amplitude of the resultant wave is
From equations (1) and (2)
AR  A12  A22  2 A1 A2 cos 
  3 1 70.2  68.1
x 2   1.05cm
2 2  A2  A2  2 AA cos120
3  x  AR  A.
From equations (2) and (3)   x  5
1
39. (1) Since the source are incoherent there will
be no sustained interference.
 3  5 1  4 x  5  22.7  4  1.05  117.7cm
Hence at the given point
33. (1) In fundamental mode,
Average intensity  I1  I 2  I 3  .....  I10  10 I
 40. (1) For minimum,
 l    2l
2

x  (2n  1)
v v 2
f   (1)
 2l The maximum possible path difference =
distance between the sources= 3m.
For no minimum

3  6
2
v 330
f    55.
 6
In half length dipped in water mode,
If  f  55Hz. No minimum will occur..
l 
    2l 41. (4)
2 4
42. (4)
vv
 f '   f 43. (4)
 2l
44. (4)
34. (2) Frequency of the 3rd harmonic is
45. (2) Beats are produced on account of
3v interference of sound waves of slightly different
 300 Hz
4l frequencies.
Frequency of the 2 nd overtone is 46. (3)

47. (1) Decrease in tension decreases the
v v 5v frequency of wire. The frequency of the wire
 2n  1   2  2  1   500Hz
4l 4l 4l initially is 504 Hz..
440 Sound Waves
48. (4) Number of beats per second  n1  n2 guitar may have 435 Hz or 445 Hz frequency
for 2nd tuning fork of 437 Hz is used and 8 beats
1  2000  2n1  n1  1000 are heard this condition is satisfied if guitar’s
frequency is 445 Hz.
and 2  2008  2n2  n2  1004
Number of beats heard per sec  1004  1000  4 v
54. (3) We know that f  (Fundamental
2l
49. (4) 1  2l , 2  2l  2l frequency of an open pipe)
v v v v
 f1  and f 2  f1  f 2   3
2l 2l  2l 2l1 2l2
 No. of beats v  303m / s
v 1 1  vl 55. (3) As number of beats/sec = diff. in

 f1  f 2     frequencies has to be less than 10, therefore
2  l l  l  2l 2
0   n1  n2   10
50. (1)
56. (3) The sounds of different source are said
51. (4) Velocity of wave v  n
to differ in quality. The number of overtones and
v their relative intensities determines the quality
Where n = frequency of wave  n 
 of any musical sound.
v1 396 57. (2) Intensity  (Amplitude)2

n1    400 Hz
1 99  102
P 1
58. (1) I 2
 I 2
v2 396 4 r r
n2    396 Hz
2 100  102 I1 r22 I
 2  4  I2  1
I 2 r1 4
No. of beats  n1  n2  4
52. (1) The number of beats will be the difference  I1 
of frequencies of the two strings. 1  60  10 log  
 I 
Frequency of first string
 I1 
1 T 1 20  2  10log    54 dB
f1    137.03Hz.  4 I 
2l1  2  51.6  102 103
59. (3) Intensity,
Similarly, frequency of second string
P 200
I  2
 0.5
1 20 A 4  10 
f2   144.01
2  49.1  10 103
2
No. of decibels is given by

Number of beats  f 2  f1  144  137
I 0.5
10log10  10log10 12
 7beats I0 10
53. (2) When 5 Beats are heard with 440 HZ
then  10 log10  5  1011   84dB
Sound Waves 441
60. (4) Intensity level is decibel is given by 63. (3)

I1  v  f
L  10log10 f ' f 
I0 64. (3) 
vv 2
65. (4) Because of wind flow from B to A the

I
L  1  10log10 2 velocity of sound wave w.r.to ground is more
I0 from the waves that reach A.
I2 I  v 
 1  10log10  10log10 1
I0 I0 66. (3) (1) f '  f  v  v 
  
1 I  v 
  log10 2 (2) f '  f  v  v 
10 I1   
I2  v  v 
  1.26  I 2  1.26 I1 (3) f '  f  
I1  v 
 v  v 
I (4) f '  f  
 100  26%  v 
I
fv
I f ' f  for cases (C) and (D).
61. (4)   10log v
I0
67. (4)
I 68. (4) As the observer moves away from source
1  4  10log
I0 frequence decreases and hence pitch decreases.
69. (4)
 2I 
 2  10log    4  10log  2   7
 I0  70. (4) From Doppler’s shift, we know for this
case
I 
62. (4) dB  10log10   ,  v  vs 
 I0   330  220 
f '  f0    1000  
 v  vs   330  220 
where I 0  1012 wm 2
 550 
 1000    5000 Hz.
1
I 
1 4 I  110 
Since, 40  10log10  I   I  10
 0 0
71. (4) Let v be the velocity of the sound and be
2
I  the velocity of the observer.
Also, 20  10log10  I 
 0   v
 v 
 v  v0  5  n 6
I2 n'  n    n  5
  102  v   v   
I0
 
I2 r2 n ' 6 n ' n 6  5
So,  10 2  12  r2  10m  ;   100  20%
I1 r2 n 5 n 5
442 Sound Waves
72. (2) When source is approaching the

observer, the frequency heard v1 m2

v2 m1
 v   340 
na    n    1000  1063Hz 3. (4) Distance travelled by both the waves is
 v  vs   340  20 
same.
When source is receding, the frequency heard Let the time taken by the S and P waves to
reach the seismograph be t1 and t2 then
 v  340
nr   n   1000  944
 v  vs  340  20 t1  t2  4min
 na : nr  9 : 8  t1  t2  240s (i)
73. (1) Since, the person heard no beats Let distance of epicentre be s. Then
therefore their apparent frequency are the same.
s  v1t1  v2 t2
330 330
So, n   504   4.5  t1  8t2 (ii)
330  15 330  30
From eq’s (i) and (ii)
504  315
n  529.2 Hz 240  8
300  t1   548.58s
3.5
 v 
74. (3) f1  f    s  v1t1  4.5  548.5  2500km
 v  v 
4. (1) Velocity of sound wave is
fv
n1  f  f1   RT 5 RT
v  v v= 
M0 3M 0
 v  v 
f2  f  
 v 
v
n2  f 2  f  f  n2  n1
v
1. (1) We can assume that velocity of sound

wave is same since both waves travel in same The temperature as a function of position is
medium.
(TL  T0 ) x
T  T0 
1 f1  1 f 2 L
1 f1 5R   TL  T0  
f2   133Hz dx  v. dt 
2 3M 0 T0   L  x  dt
   
2. (2) Speed of sound in gases is given by
L dx t
 RT 1 
0
5R  (TL  T0 ) x 
  dt
0
v  v  (m  M )
M m 3M 0 T0  L 
 
Sound Waves 443
1
3Mo L   T T   2 3
 T0   L 0  x dx  t l2  x 
 
5R 0   L   4
L l2  3l1
 2L   x
3M o  (TL  T0 )   2
   T0   x  t
5R  TL  T0    L  
0
3M o  2L  8. (2)

5R
 
 TL  T0 
 TL  T0  t
v
2L 3M0 Fundamental frequency of wire  f wire  
t 2l
 TL + T0  5R
v 3v 5v
5. (3) Frequency of first overtone of closed (1) f  , , cannot match with
4l 4l 4l
pipe= Frequency of first over tone of open pipe
f wire
3v1 2v2
  v 2v 3v
4 L1 2 L2 (2) f  , ,
2(2l ) 2(2l ) 2(2l )
3 P 1 P  P 
   v   2v
4 L1 1 L2 2   
Second harmonic 2(2l )
4 L1 1 4 L 1
 L2  
3 2 3 2 matches with f wire .
6. (2) Maximum pressure at closed end will be Similarly other pipes frequencies will not match
atmospheric pressure adding with acoustic wave with fundamental frequency
pressure
9. (2) In winter fundamental frequency
So Pmax  PA  P0 and Pmin  PA  P0
v
f1 
P P P 4l1
max A 0
Thus P  P  P
min A 0 3v
First overtone f 3  4l
7. (3) Let x be the correction l1  x   / 4 3
v 3v
As f1  f 3 or 
4l1 4l3
l3  3l1  54cm
3v '
In summer f3' 
4x
As f 3  f 3'
3v 3v '
f3'  
4(54cm) 4 x
444 Sound Waves
v' 15. (4) The wavelength in closed pipe is

x (54cm)
v 2L
 , n  1, 2,3,....
n
In summer v '  v  x  54cm
10. (2) Pressure antinode forms at the 160cm 1.6
Given L = 80 cm    
displacement node. n n
End Correction in organ Pipes  v n n

v      v   330
11. (1) Let e be the end correction then T  1.6 1.6
according to question.
330
v 3v
As n =1    206HZ   1.6m
  e  0.025m. 1.6
4(l1  e) 4(l2  e)
 = 1294 k = 3.93
 5
16. (2) 6  1.25    m
4 6
12. (4) v v 1 2  1011 1

f   .    6000 Hz
 d  8  103 5
6
v v
f1  , f2  17. (1) & 18. (2) & 19. (1)
2  l  0.6d  4  l  0.3d 
The given standing wave equation is
f2 l  0.6d 2
 S  0.1mm cos  y  1cm cos(400t )
f1 2  l  0.3d  0.8
13. (1) Path difference r  2r  n 2

At y  1cm cos (0)  1
0.8
nv nv
r (  2)  thus f 
f r (  2) Displacement antinode corresponds to open end.
14. (1) The figure shows variation of 2

At y  99cm cos (1)  0
displacement of particles in a closed organ 0.8
pipe for 3rd overtone. For third overtone
so displacement node corresponds to closed
7 4l  l
l or   or  end.
4 7 4 7
2 2
k     0.8m
 0.8
4l 4l
 0.8  (2n  1) 
(2n  1) 0.8
7
l (2n  1)  5
4
This corresponds to second overtone.
l
Hence the amplitude at P at a distance from As pressure wave leads the displacement wave
7
closed end is a because there is an antinode at 
that point. by
2
Sound Waves 445
 2  or (2  1 )t  0, , 2,...
pex  p0 cos  ( y  1cm)   cos(400t )
 0.8 2
1 1
t  0, , ,....
 2  2   2  1
p0  BKS0  B  5105 N / m2 , k  , S0  0.1mm
 0.8 
1
2
T  t2  t1   103 s.
p  125 N / m 2  1
2  I 
Pex  125 N / m2 sin ( y  1cm)cos(400t ) 24. (3) 1  10log  
0.8  I 
20. (4) Beat period  2I 
1 1 1 After adding 50 more students 2  10log  I 
T   sec. Hence minimum   
n1  n2 384  380 4
time interval between maxima and minima 2I I
2  1  10log  10log
I0 I0
T 1
t  sec
2 8   10log2  3dB
21. (4)
 I 
25. (4) Given that B  10log  I 
  
 4I    I  
B  10log    10 log    log 4
 I    I  
B  B  10log(4)  B  20log 2   B  6 dB
Using n L ast  n First  9  4 I
26. (4)   10log10
I0
Given that 2 n first  nlast
If P is power of sound source then
 n first  36 nlast  72
( P / 4r 2 )
22. (1) Apparent frequency is equal to the   10log10 where I 0  1012 w / m 2
I0
frequency since relative velocity is zero. So
number of beats will be zero.
P
 10log10  10[log P  log(4I 0  r 2 )]
23. (3) y1  A sin 21t and y2  A sin 22 t 4r 2  I 0
y  y1  y2  A sin 21t  A sin 22t  10[log P  log 4I 0  2log r )]  a  b log r
(2  1 ) (  1 ) I
 2 A sin 2 t cos 2 2 t 27. (3) Loudness   10log10 I
2 2 0
The amplitude of the wave is

I2 P
 2  1  10log10  I
A '  2 A cos  (2  1 )t I1 and 4r 2
I  A '2  4 A2 cos 2 (2  1 )t
I 2 r12
 
For I to be maximum (2  1 )t  1 I1 r22
446 Sound Waves
r12 r 31. (2) vS  u  at  0  10 10   100

 (  20)    10log10  20log10
r22 r2
Apparent frequency
r1  v   340 
  10  r2  0.1r (r1  r )   f  1700  1313.6Hz.
r2  v  vs   340  100 
32. (3) Motorist will hear two frequencies:
9r
 shift  r  0.1r  0.9r  (i) directly from the band and (ii) reflected from
10
the wall. The frequency heard directly by the
28. (3) Frequency heard by observer (stationary) motorist from band
will be maximum , if the source is moving
 v  vm 
towards the observer, which is the case when f1  f   (i)
the source is at B. Frequency heard by observer  v  vb 
(stationary) will be mininum, if the source is The frequency of sound reaching the wall,
moving away from the observer, which is the
case when the source is at A.  v 
f"  f   (ii)
 v  vb 
The frequency of reflected sound from the wall,
heard by motorist,
 v  vm 
f2  f "  
For point C, frequncy heard is equal to the  v 
original frequency  v  vm 
f2  f   [from (ii)]
29. (1) From droppler effect.  v  vb 
The frequency received by the detector is Hence the frequency heard by the motorist
 v  vD  f  f 2  f1
f ' f  
 v   v  vm   v  vm 
 f  f   f  
Given that f  5 ( 5 - waves per second )  v  vb   v  vb 
 vv 
18200 91  f  f  2 m2   2vb .
f ' 
3600 18  v  vb 
33. (4) The frequency of reflected sound heard
5
vD  m/s by the driver
3
By keeping these values in the above equation The frequency received by the wall is
 v 
we get v  150 m / s f1  f  
 v  vs 
30. (2) The frequency received by the observer The frequency received by the car from wall is
is
 v  vs   v  vs 
 v   330  f 2  f1    f2  f  v  v 
f ' f    100    v 
v  v cos60  330  19.4cos60   s 
 s 
 330   330  (72  5 / 18) 

f '  100    103Hz
 124    140Hz
 330  9.7   330  (72  5 / 18) 
Sound Waves 447
34. (1) 2. (1)
 RT
3. (4) v
M
v1 M
 1 2
v2  2M1
2 d1  2 d 2  v  t1  v  t2
M 1  28 gm, M 2  4 gm
 2  d1  d 2   v  t1  t2 
7 5
v  t1  t2  340  1.5  3.5   1  , 2 
d1  d 2    850m 5 3
2 2
35. (4) Large vertical plane acts as listener. v1 7 3 4 3
   
The frequency received by the listener is v2 5 5 28 25
(c  v)n 4. (4) At given temperature and pressure

 n' 
c
This is the number of waves striking the surface 1 v1 2 4
per second. v     2 :1.
 v2 1 1
vu  v u 
36. (1) f1    f 0 and f 2    f0 5. (4) Density of mixture
 v   v 
Vo2  o2  VH 2  H 2
Number of beats per second   mix 
Vo2  VH 2
vu   vu 
f1  f 2    f0    f0
 v   v  V ( O2   H 2 ) O2   H2
 
2V 2
vu vu 
  f0 where VO2  VH2  V
 v 
 f  2u  H 2  16  H 2
 2u  0     8.5  H 2
2
v  
The velocity is related with density as
1
v
E 
1. (3) I
A
vmix  H2  H2 2
I '  2I    
vH 2  mix 8.5  H 2 17
A '2  2 A
n1 M 1  n2 M 2
A  2 A  A  A  
2 1 6. (4) M mix 
n1  n2
A 1 4  2  32 68
A
%  
2  1  100  41% 
1 2

3
448 Sound Waves
n1CV 1  n2CV 2 (2n  1)340 (2n  1) (2n  1)  100

CV (mixtue)  l  m cm
n1  n2 4  340 4 4
 3 5 For n =1,2,3,.. we get l=25cm, 75 cm, 125 cm...
1  2   R 13
2 2
  R As the tube is only 120 cm long, length of air
1 2 6
column after water is poured in it may be 25cm,
19 CP 19 75 cm, only The corresponding length of water
CP (mix)  CV  R  R or  R
6 CV 13 column in the tube will be (120-25)cm =95 cm
or (120-75) cm =45 cm.
19 8.31 300 Thus minimum length of water column is 45 cm.
v   400.9ms 1.
13 68  103
3 v 3v 1 1
11. (1)  ,  
4 1 2 2 2 6
7. (3)  v 
v1  v2 

 T1   T2 
2 2
v
12. (1) For closed pipe f 0   210 Hz
4L
2l
 After filling with water the frequency of the first
 Time taken   T1  T2  overtone is
3v 3v
v v f    1260 Hz
8. (2) f0 (closed)    L  2L
 4l 4 
2
3v 13. (3) Length of pipe = 85 cm = 0.85 m
third harmonic of closed pipe  3 f0(closed) 
4l Frequency of oscillations of air column in closed
pipe is given by,
3v v v
  100 or  100 and
4l 2l 4l (2n  1)v
f 
4L
v v
f 0(open)    200 Hz. (2n  1)v (2n  1)  340
2l 2l f   1250   1250
4L 0.85  4
9. (3) Using   2(l2  l1 )
 2n  1  12.5  n  6
 v  2n(l2  l1 ) 14. (1) The fundamental frequency of open tube
 v  2  512(63.2  30.7)  33280cm / s v
f  (i)
Actual speed of sound 2l
v0  332m / s  33200cm / s where l is the length of the tube
Hence error  33280  33200  80cm / s v = speed of sound
10. (3) Because the tuning fork is in resonance For closed pipe the fundamental frequency is
with air column in the pipe closed at one end, v v
f '   f
(2n  1)v  l  2l
the frequency is f  where n = 1,2,3... 4 
4l 2
corresponds to different mode of vibration
putting f=340 Hz, v=340 m/s, the length of air nv n(330)
15. (2) f   n3
column in the pipe can be 4l 4  0.17
Sound Waves 449
16. (2) For end correction x  3 constructive interference will take place when
  , 2....., the minimum distance x for a
v 3v maximum corresponds to

4(l1  x) 4( I 2  x)
 (i)
l2  3l1 70.2  3  22.7 v 360m / s
x   1.05cm
2 2 the wavelength is   f  180 s 1  2m
17. (3) Let x be the end correction
 2 x2 
l1  x   / 4 thus, by (i) 2  2    x  2
 4
3
l2  x  1/2
4  x2  x
4  4   1  2
l2  3l1  
x
2
x2 x2
4 1  x  x  3
4 4
thus the detector should be placed at a distance
of 3m from the source.
20. (2) &21. (1) & 22. (3)
As the rod is clamped at its mid point .the
midpoint C always behaves like a node.when
18. (2) Since
both sides of C have two antinodes the wave

  A  a12  a22  (4)2  (3)2  5. pattern is
2
19. (2) The situation is shown in fig. Suppose
the detector is placed at a distance of x meter
from the source. The direct wave received from
the sources travels a distance of x meter. The     2l
   l 
wave reaching the detector after reflection from 4 2 2 4 3
the wall has travelled a distance of
1/2 v Y3
2 f  = 3000HZ
2  2  x 2 / 4 meter. The path difference
     2l 
between the two waves is
When the rod is in Fundamental mode
  2 x 2  
  2  2     x  meter
  4 
 
  l    2l
4 4
v Y 1
f     1000 HZ
  2l
450 Sound Waves
vmax  A  2 Af  1.25  102 m / s second time. Or we can say all A,B and C are
again in phase after 1 s.
If the rod is clamped at an end and when it has
four antinodes then the mode of vibration is  Beat time period Tb  1s
1
Or Beat frequency fb  T  1Hz
b
7 10 26. (4) According to the question frequencies of

l  m
4 7 first and last tuning forks are 2n and n
respectively.
v
f   3500 HZ Hence frequency in given arrangement are as

follows
23. (2) There are two possible frequencies n+x
and n-x. But only n+x satisfied the given n Last  n first  ( N  1) x
conditions.
 2n  24  3  n  n  72 Hz
24. (3)
So, frequency of 21st tuning fork
f 1 T f 1 1 1
    n21  (2  72  20  3)  84 Hz.
f 2 T f 2 100 200
27. (2) The three possible frequencies are shown
f in the figure
f  1
200
 no.of beats =1
So 30 beats are heard in 30 s.
25. (1) Let us make the following table
28. (1) At night, amount of carbon dioxide in

atmosphere increases which raises the density
of atmosphere. Since, intensity is directly
proportional to density, intensity of sound is more
at night.
29. (1) Let intensity due to a single person = I
Beat time period for A and B is 1 s. It implies then 10log I / I 0  60  1
that if A and B are in phase at time t = 0, they
are again in phase after 1 s. Same is the case  8I   I 
2  10log    10  log  log8 
with B and C. But beat time period for A and C  I0   I0 
is 0.5 s.
 60  10log8  60  30log2  69
Therefore, beat time period for all together A,B
and C will be 1 s. Because if, at t = 0, A,B and 30. (3) Let a be the amplitude due to S1 and S2
C all are in phase then after 1 s. (A and B) and individually.
(B and C) will again be in phase for the first
time while (A and C) will be in phase for the Loudness due to S1 =I1  Ka 2
Sound Waves 451
Loudness due to S1 +S2 =I  K (2a)2 =4I1  v   11v 

v    
10 
 I  n '  n    10 
1  10log    1   9v 
 I   v    10 
 10 
 4I  11 n ' n 2
2  10log    n'  n 
 I  9 n 9
n   2  1  10log10 (4)  6 n ' n 2 200

 100%  100%  %  22.22
n 9 9
I
31. (3) L1  10log 34. (4) The situation is shown in the fig. Both
I
the source (engine) and the observer (Person in
the middle of the train) have the same speed,
2I
L2  10log but their direction of motion is right angles to
I
each other. The component of velocity of
observer towards source is v cos 45 and that
2I I
L2  L1  10log  log of source along the line joining the observer and
I I source is also v cos 45. There is no relative
motion between them, so there is no change in
L2  L1  3 dB
frequency heard. So frequency heard is 200Hz.
32. (2)
 v  v0  vwind 
 f '   sound f
 vsound  vsource  vwind 
 f ' f .
(v  vsource  0)
33. (1)
Speed of sound  v (say)
v
Speed of source = speed of observer 
10
 v  vt cos 45 
Let natural frequency of sound = n f ' f  
 v  v cos 45 
Apparent frequency of sound,
f ' f
 v  vo  35. (3) The maximum velocity of the observer
n'  n 
 v  vs 
k
Since source is moving in the direction of sound (block) is v  A  A
m
and observer moving towards source, then
v v 100
vo  and vs  v  2   10m / s
10 10 4
452 Sound Waves
The maximum and minimum frequency is

received by the receiver when it is at the mean
position
Y
1. (5) Velocity of wave 
 v  v0  
f max   f0  1020 Hz
 v 
9.27 1010
 v  v0   v  5.85  103 m / sec.
f min   f 0  960 Hz 2.7  103
 v 
Since rod is clamped at middle fundamental
Frequency band width received by the receiver wave shape is
ranges between 960 Hz – 1020Hz
36. (4) The frequency of reflected sound heard
by the driver.
The frequency received by the wall is
 v  
 L    2L
f1  f   2
 v  vs 
 L  60cm  0.6m
The frequency received by the car from wall is
 L  60cm  0.6m
 v  vs   f  f  v  vs 
f 2  f1   2     1.2m
 v   v  vs 
v 5.85 103
 330  (72  5 / 18)  f    5KHz
 124    140 vibration/sec  1.2
 330  (72  5 / 18) 
2. (6) Length of pipe = 85 cm = 0.85 m
37. (3) From Doppler’s effect
Frequency of oscillations of air column in closed
 340  pipe is given by,
f1  f   (Direct)
 340  5  f 
(2n  1)v
4L
 340 
f2  f   (By wall) (2n  1)v (2n  1)  340
 340  5  f   1250   1250
4L 0.85  4
Beats   f1  f 2   2n  1  12.5  n  6
3. (98) vm2  u 2  2as
 340 340 
5 f     f  170 Hz.
 340  5 340  5   vm2  2  2  s (u  0)
38. (2) On reflection frequency does not change vm  2 s

so not beats are heard.
According to Doppler’s effect
Relative separation
39. (4) t  v  vm 
Relative velocity f ' f  
 v 
a  330  2 s 
t 0.94 f  f    s  98.01m
vs  v2  330 
Sound Waves 453
9
4. (2. 04) Frequency of beats, f   3Hz
1. (12) When the train approaches him 3
1 1 1 1 
 v  320 f  v     3  300   
f1  f   f  Hz  1 2   2 2 
 v  vs  300
When the recedes 1 1 1 1 1 1

     
100 2 2 2 2 100
 v  320
f2  f   f  Hz
 v  vs  340 100
 2   2.04m
49
 f2 
  1  100  12%
 f1 
2. (1) For v and v-1 no. of beats are 1, i,e., the v

1. (3) In the first medium, frequency, f 
intensity becomes maximum once in the time 
interval of one second. Frequency remains the same in second medium,
i.e., f  f '
For v and v+1 no. of beats are 1, i,e., the intensity
becomes maximum once in the time interval of v ' 2v v
  
one second. ' ' 
For v-1and v+1 no. of beats are 2, i,e., the   '  2
intensity becomes maximum two times in the
time interval of one second. 2. (2) Height of tower (h)  300m
So all the three pairs intensities become Initial velocity u  0

maximum once in one second. So the no. of Acceleration due to gravity g  9.8m / s 2
beats =1
Speed of sound in air  340m / s
 I1 
3. (100) We have, 1  10 log  I  ; Time taken by stone to reach the pond  t1
 0
 I2  Using equation of motion,
 2  10 log  
 I0  1 1
h  ut  gt12  300  0   9.8t12
I  I  2 2
 1  2  10 log  1   10 log  2 
 I0   I0  300  2
 t1   7.82s
9.8
1
I  1
I 
or,   10 log  I  or, 2  log  I  Time taken by the sound to reach the top of the
 2   2  tower
I1 h 300
or,  102 or, I 2  I1 . t2    0.88s
I2 100 v 340
 Intensity decreases by a factor 100.  Total time t  t1  t2  7.82  0.88  8.7 s

454 Sound Waves
3. (4) Length at which first resonance occurs
l1  25.5cm  25.5  102 cm 1. (4) When tension decreased frequency

Length at which second resonance decreases, beat frequency decreased it means
that
occurs l2  793m
fB  f A
 Wave length   2(l2  l1 )
 f B  318
 2(793  25.5)  2  53.8
2. (4) There is no transfer of heat from
 107.6cm  1.076m compression to rarefaction as air is a bad
Using v  f , f  340Hz conductor of heat. Since time of compression
rarefaction is too small. Density of the medium
 Speed of sound in air  365.84m / s is maximum and minimum at compression and
rarefaction points. Bulk modulus remain
4. (1, 2, 4 ) The given equation is
constant.
  3. (4) Whistling train is the source of sound, vs .
y( x, t )  3.0sin 36t  0.018x  
 4
Before crossing a stationary observer on station,
As positive direction is from left to right and x vn
frequency heard is n '  (v  v )  constant and
is positive, therefore, wave is travelling from s
right to left. n '  n.
Compare the given equation with the standard Here, v is velocity of sound in air and n is actual
frequency of whistle.
from
After crossing the stationary observer, frequency
 2t 2x  vn
y  A sin     heard is n ''   constant and n ''  n.
 T   (v  vs )
2 2 Therefore, the expected curve is (4).

 36,  0.018
T  4. (1, 2) According to Newton’s formula for
Speed of wave, velocity of sound in a fliud,
 36 B
v   2000cm / s  20m / s v
T 0.018 
The wavelength of the wave is
1
2 v  B and v  
  3.48m
0.018
2 
Again, T  
36 18
1. (2) Let m be the total mass of the rope of
1 18 length l. Tension in the rope at a height h from
Frequency v   Hz  5.7 Hz
T  lower end = weight of rope of length h
Sound Waves 455
mg  v 
ie.s., T   h f '  f0  
l  v  vs cos  
T  300 
As v f '  f0 
 m / l   gh 
 300  30cos  
v 2  gh When    / 2, i.e., at N,
Which is a parabola. Therefore, h versus v graph f '  f 0  1000 Hz ,

is a parabola option (2) is correct.
When   0, i.e., cos   1,

2. (2) The path difference x  v 1000Hz  300m / s 
2 f max  f 0 
v  vs  300m / s  30m / s 
2 
  k x   
 2 10
 1000Hz  1111Hz
Destructive interference will occur at point P. 9
2 v 1000  300
Aresul tan t   2a   a 2  4a 2  a. f min  f 0   909 Hz
v  vs 330
It will remain constant with time.
5. (1) For observer approaching a stationary
 v  v  v v  v0
3. (2) f   f1  f1  f1  source n '  .n and given
 v  v v
 an 
 f g v0  at  n '    t  n this is the equation of
So, f  f1   1  t (v  gt )  v 
 v 
straight line with positive intercept n and positive
f1 g  2 10   f1  f1 10 n

3
Slope of graph    slope   . So the correct graph is (1)

v 30 300 v
 f1  103 Hz
4. (1) This frequency-time curve corresponds 1

2 2
to a source moving at an angle to a stationary 1. (2) Intensity of a sound wave is I   a
2
observer.
where  is the density of the gas and a is the
amplitude of the wave.
Intensity  a 2 2
Here
2 2
aA 2  1 I 2 1 1
 and A   A       
aB 1 B 2 IB  1   2  1
In the region SN, the source is moving towards 2. (3) Wavelength remains same during
the observer, i.e., the apparent frequency approach and recede.
456 Sound Waves
3. (2) As the apparent frequency is changing

 v 
non linearly with time so it must be a two v 
dimensional Doppler effects as shown in the f ' f  5 
figure. Where v is velocity of sound. v 4v 
 
 5 
 v 
For AP, f  f 0  v  v cos   Clearly, f ' is constant but f '  f . This is shown
 S 
is curve (3).
 v  5. (4)   Asin(kx  t )
For PB f  f 0  v  v cos  
 S 
d
Pex   B   BAk cos(kx  wt )
dx
amplitude of Pex  BAk
 2 
   5  105 104   2
  5  10 Pa
 0.2 
 v   v   2  1
f min  1800  f 0    2000   6. (2) P  B.K .S  B   S  P 
 v  vs   v  vs     
Thus, pressure amplitude is highest for minimum

 v   v 
f max  2250  f 0    2000   wavelength, other parameters B and S being
 v  vs   v  vs  same for all. From given graphs.
On solving = 33.3 m/s.
3  2  1 .  P3  P2  P1
4. (3) Let speed of observer be v along y-aixs
and speed of source be vS  2v along x-axis.
1. (3) Due to jagged end there is an end
correction  x 
v
 512 (1)
4 11  x   102
v
 256 (2)
4  27  x   10 2
OP  vt , OS  2vt From eqn. (1) and (2)
1 2 2 (11–x) = (27–x)
cos   & cos  
5 5
x = –5 cm
Now, apparent frequency f ' is given by From eqn. (1)
 v  v cos   v
f ' f    512  v  328ms 1
 v  vs cos  
2
4  16  10
Sound Waves 457
2. (2) Organ pipe will have frequency either 255

340
or 257 Hz Frequency received the wall 
340  5
Using 255Hz
Frequency received by the person from the wall
3v 3  340  340 
255  l m
2l 2  255 is f (from wall)  f    f2
 340  5 
l  200 cm
Beats   f1  f2 
3. (4) We know that the apparent frequency
 340 340 
5 f   
 v  v   340  5 340 5
f '   f from Doppler’s effect
 v  vS 
 f  170Hz
Speed of sound v  330m / s
2. (3) According to Doppler’s effect,
Frequency of whistle  f   540Hz.
0
vv 
Apparent, frequency f   v  v  f 0
330  30  s 
 f '  540  648Hz.
330  30
  f0  vf 0
4. (1) Reflected frequency of sound reaching bat Now, f   v0 
 v  vs  v  vs
 v  ( v0 )   v  v0  v  10
  f  f  f  f0
 v  v s   v  v s  v  10 So, slope 
v  vs
 320  10 
   8000  8516 Hz 3. (2) Given f A  1800 Hz & f B  2150 Hz
 320  10 
5. (4) Let f1 & f 2 be the frequencies received  u 

fB  f A  
by the pearson from two factories. u v
here, u  343m / s
v  55.8372 m / s
 v  v   320  2  Now, for the reflected wave,

f1  f    800  
 v   320 
 u  vt 
 f A'    fA
 v  v   320  2   u  vt 
f2  f    800  320 
 v   
 343  55.83 
   1800
f 2  f1  10 Hz  343  55.83 
 2499.44  2500Hz
 100
1. (4) From Doppler’s effect 4. (2) The velocity of the wave is v  
k 0.5
 340  92
f (direct)  f    f1 v  200m / s or v   200m / s
 340  5  0.46
458 Sound Waves
5. (3) f  500Hz
6. (4)
Case 1: When source is moving towards

stationary listener Let after 5 sec engine reaches point C
 v  AB BC
t 
apparent frequency f '  f  v  v  330 330
 s 
0.9 1000 BC
 340  5 
 500    506 Hz 330 330
 336 
 BC  750m
Case 2: When source is moving away from
the stationary listener Distance travelled by engine in 5 sec
 900m  750m  150m
 v   340 
f ''  f    500    494 Hz
 v  vs   344  150m
Therefore velocity of engine   30m / s
5sec
In case 1 number of beats heard is 6 and in
case 2 number of beats heard is 18 therefore
frequency of the source at B = 512Hz

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Units & Dimensions 11

v v.e v.e (ve)t   4.58105

  h    ML2T 1  . Total mass = (2.3 + 0.02015 + 0.02017) kg

4 12. (4) Angular momentum = mvr

 8  102  8% Which represent work.

22. (1) Given, L= 2.331 cm  m 

Round off up to 3 significant figure = 11.7cm3 0.1

26. (1)  0.2 %

X A B In other words, this has units of charges divide

T    M a b c   L a 3b  2c  T 2a 2c   ML2T 2 

a bc 0 11. (2) According to the method of dimensional

1 Hence, dimension of time are [G1 2 h1 2 c 5 2 ]

Gh   M 1 L3T 2   ML2T 1    M 0 L5T 3  t  1.23G 1/ 2  1/ 2 r n

The dimensions of t are [T], the dimensions of

As rounding off the number, we get density g , 2 and 10 3

15. (4) T  102  0.01 102  0.01 

aE 11. (1)  Mass   M   [ L T ]

 Weber   Volt  Sec   ( Joul / coul )  Sec 

2 Breadth b  (10.1  0.1)cm

Mean value of time, Impulse is equal to change in momentum.

 6.023 1023  5.234 1031

5. (2) Area of object  1.7cm2  1.75  104 m2

1.55  4.234 1.005  0.0201

Linear magnification  Area magnification Rounding-off up to three significant figures, we

1N  105 dyne,1m2  104 cm 2 = 500 new unit of length

Y  1.9  1012 dyne / cm2 Observed width

3. (1) Let mass, related as M  T x C y h z   M 2 L4T 6 A4    R 2 

M 1 L0T 0  M z Ly  2 z  T x  y  z and t  d 1/ 2 . S 1/2

5. (4) Let 0 related with e, m, c and h as 3 3

a  2 (i) log e  R   log e  V   log e  I 

a  2, b  0, c  1, d  1.   .04  .01  100%

Solving equations (1) and (2) we get

Now the difference vector is nˆd  nˆ1  nˆ2 or

5. (3) F  F12  F22  2F1F2 cos900  F12  F22  R 2  2 P 2 (1  cos )

4. (4) 10. (2) Given, P  iˆ  ˆj  kˆ , then we have

The Y component of force F is Given that A2  B2  2 AB cos   A  B

Dot product of these vectors should be equal to 

13. (1) Let the given vectors be  

 A  12  12  12  3, B  12  12  2 17. (4) For the resultant,

Thus the ratio of magnitudes A / B  1 B sin 

 (Triangle) since the two vectors are having same

1 Angle between B and C,   135

6. (4) 10. (4) A  3N , B  2 N

From triangle law, three vectors having R  9  4  12 cos  (i)

11. (1) R 2  F12  F22  2 F1 F2 cos   

 (12  2)iˆ  (4  6) ˆj  (3  12)kˆ

Where  is the angle made by the vector 1   5 17

B sin B sin   102  122  142

As the metal sphere is in equilibrium under the 15. (1)

From the figure T cos   W (i)  tan   3    60

and T 2  P2  W 2 AB sin   3 AB cos  tan   3

Thus, option (2) is false.

2. (2) (4) If C 2  A2  B2 . then angle between A and B

i.e., r is along positive x-axis. Thus, option (4) is false.

s 9. (2) vA  tan 30  and vB  tan 45 

v 15 17. (4) According to given relation acceleration

Given that acceleration, s2  at   10t  5t

a 1. (3) When the drunkard walks 8steps forward

2 2 140 sec, athlete will complete 3.5 revolution, i.e.,