I PUC PHYSICS ANNUAL EXAMINATION FEB-24

SOLUTIONS/SCHEME OF EVALUATION
MAX MARKS: 70 DATE: (24-02-24)

Subject code:33
PHYSICS
[Time: 3 Hours 15 Minutes] [Total No. of questions: 48] [Max.Markls:70]

Instructions: 1) All parts are compulsory

2) For Part-A questions. first written answer will be considered for
awarding
marks.
3) Answers without relevant diagram/figure/circuit, wherever necessary
will not carry any marks.
4) Direct answers to the numerical problems without detailed solution will not
carry any marks.

PART-A

I. Pick out the correct option among the four given options for ALL the following questions:
(15x1=15)
1) The dimensional formula of pressure is
a) [𝐌𝐋𝐓 −𝟐 ] b) [𝐌𝐋𝟐 𝐓 −𝟐 ]
c) [𝐌𝐋−𝟏 𝐓 −𝟐 ] d) [𝐌𝐋𝐓 −𝟑 ]
(c) [𝑴𝑳−𝟏 𝑻−𝟐 ]
2) A ball is thrown vertically upwards and allowed to move freely under gravity. The a-t graph of the
motion of a ball is

(grace marks may be awarded)

3) In a projectile motion, the horizontal range is maximum for angle of projection

a) 𝟎𝐨 b) 𝟒𝟓𝐨
c) 𝟔𝟎𝐨 d) 𝟗𝟎𝐨
(b)𝟒𝟓°

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 1

4) The product of force and time is
a) Time b) Torque
c) Impulse d) Acceleration
(c)Impulse
5) The recoil of a gun is an example for conservation of
a) Mass b) Charge
c) Energy d) Momentum
(d) Momentum
6) The scalar product two vectors is zero (𝑨. ⃗⃗⃗ 𝑩
⃗⃗ = 𝟎), the angle between two vectors is
𝐨 𝐨
a) 𝟎 b) 𝟒𝟓
c) 𝟗𝟎𝐨 d) 𝟏𝟖𝟎𝐨
(c)𝟗𝟎𝟎
7) Consider a system of two identical particles, one of the particle is at rest and other has an
acceleration ‘a’. The center of mass has an acceleration.
𝒂
a) Zero b) 𝟐
c) a d) 2a
𝒂
(𝒃)
𝟐
8) The value of gravitational constant is
a) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟎 𝑵𝒎−𝟐 𝒌𝒈−𝟐 b) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵𝒎−𝟐 𝒌𝒈−𝟐
c) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟐 𝑵𝒎−𝟐 𝒌𝒈−𝟐 d) 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟎 𝑵𝒎−𝟏𝟑 𝒌𝒈−𝟐
(𝒃)𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵𝒎𝟐 𝒌𝒈−𝟐
9) Which of the following materials is/are close to ideal plastic?
i) Putty ii) Mud iii) Steel
a) (i) and (ii) b) (i) and (iii)
c) (ii) and (iii) d) (i), (ii) and (iii)
(a) (i) and (ii)
10) Dynamic lift due to spinning of a ball is
a) Magnus effect b) Doppler’s effect
c) Pascal’s effect d) Torricelli effect
(a)Magnus effect
11) When a piece of iron is heated in a hot flame, it first becomes dull red, then reddish yellow and
finally white hot. This phenomenon can be explained by
a) Stefan’s-Boltzmann’s law b) Green house effect
c) Wien’s displacement law d) Newton’s law of cooling
(c)Wein’s displacement law
12) The efficiency of a Carnot’s engine working between the temperature 𝟏𝟐𝟕𝐨 C and 𝟐𝟕𝐨 C is
a) 0.24 b) 0.5
c) 0.75 d) 1.0
(a) 0.25
13) The total internal energy of a mono atomic gas is
𝟏 𝟏
a) 𝟐 𝑲𝑩 𝑻 b) 𝟑 𝑲𝑩 𝑻
𝟑 𝟓
c) 𝟐 𝑲𝑩 𝑻 d) 𝟐 𝑲𝑩 𝑻
𝟑
(c)𝟐 𝒌𝑩 𝑻
14) The motion which repeats itself at regular intervals of time is called
a) Projectile motion b) Curvilinear motion
c) Periodic motion d) Non-periodic motion
(c)Periodic motion

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 2

15) The longitudinal waves in a medium propagates due to
a) Shear modulus b) Bulk modulus
c) Young’s modulus d) Both shear and Bulk modulus
(b)Bulk modulus
II. Fill in the blanks by choosing appropriate answers given in the brackets for ALL the following
questions: (5x1=05)
𝒐 𝒐
(Surface tension, 𝟏𝟖𝟎 , Vector, Elliptical, 𝟗𝟎 , Absolute temperature)
16) A physical quantity having both magnitude and direction is called Vector
17) All planets move in Elliptical orbits with sun situated at one of the foci.
18) The Spherical shape of a liquid drop is due to Surface tension
19) At constant pressure, the volume of a gas is directly proportional to its Absolute temperature
20) At rigid boundary, there is a phase difference of 𝟏𝟖𝟎𝟎 between incident and reflected wave.

PART-B
III. Answer ANY FIVE of the following questions: (5x2=10)
21) Write any two rules of writing significant figures.
▪ All the non-zero digits are significant.
▪ All the zero’s between two non-zero digits are also significant.
22) A stone tied at one end of a string 80 cm long and is whirled in a horizontal circle with constant
speed. If the frequency of revolution of stone is 2Hz., then calculate magnitude of tangential
velocity.
Given,: 𝑟 = 80 𝑐𝑚 = 0.8𝑚, 𝑓 = 2𝐻𝑧, 𝑉 =?
𝝎 = 𝟐𝝅𝒇
𝜔 = 2 × 3.142 × 2 = 𝟏𝟐. 𝟓𝟔𝒓𝒂𝒅𝒔−𝟏
𝑽 = 𝒓 × 𝝎=0.8×12.56 = 𝟏𝟎. 𝟎𝟖𝒎𝒔−𝟏
23) Write any two advantages of friction.
▪ It enables to walk on the ground.
▪ Nails and screws are fix to wall due to friction.
24) What are conservative and non-conservative forces?
Conservative force:
If the work done by a force does not depend on the path taken and depends only on the initial and final
states
of a body, the force is called conservative force.
Non-conservative force:
If the work done by a force depends upon the path taken, the force is called non-conservative force.
25) Mention the expression for kinetic energy of a rotating body and explain the terms.
1
Kinetic energy of rotating body is 𝐾𝐸 = 2 𝐼𝜔2
Where, I-momentum of inertia and ω-angular rotation of a particle
26) State and explain Newton’s law of gravitation.
Statement: “The gravitational force of attraction between two masses is directly proportional to the
product of masses and inversely proportional to the square of the distance between them.”
Explanation: If ‘F’ is the gravitational force between two masses 𝑚1 and 𝑚2 separated by a distance ‘r’
in space , according to Newton’s law of gravitation
𝐹𝛼 𝑚1 𝑚2 ……….(1)
1
𝐹𝛼 ……………(2)
𝑟2

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 3

 From (1) and (2)
𝑚 𝑚
𝐹 = 𝐺 12 2 Where, G is universal gravitational constant.
𝑟
27) Mention any two factors on which thermal capacity of a body depends.
▪ Mass of a body
▪ Nature of material of body
28) State and explain first law of thermodynamics.
Statement: “The heat energy given to the system is equal to the sum of the increase in internal energy
and the external work done by the system.”
Explanation: If ‘dQ’ is the amount of heat added to the system, dU is the increase in internal energy of
the system and ‘dW’ is the external work done by the system then 𝒅𝑸 = 𝒅𝑼 + 𝒅𝑾
29) Draw a graph of kinetic energy and potential energy of an oscillating particle with displacement.

PART-C

IV. Answer any five of the following questions: (5x3=15)

30) Derive an expression for time taken to reach maximum height by a particle.
𝑣 = 𝑢 + 𝑎𝑡
𝐴𝑡 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ℎ𝑒𝑖𝑔ℎ𝑡 𝑣 = 0
𝑇𝑎𝑘𝑒 𝑎 = −𝑔, 𝑢 → 𝑢 𝑠𝑖𝑛𝜃, 𝑡 = 𝑡𝑎 = 𝑇𝑖𝑚𝑒 𝑜𝑓 𝑎𝑠𝑐𝑒𝑛𝑡
0 = 𝑢 𝑠𝑖𝑛𝜃 − 𝑔𝑡𝑎
𝑢 𝑠𝑖𝑛𝜃
𝑡𝑎 =
𝑔
31) Prove law of conservation of linear momentum using Newton’s laws of motion.
Proof: Consider two bodies A and B of masses m1 and m2 moving with initial velocities u1 and u2 in
the same direction. Let v1 and v2 be the velocities after collision. Let t is the time of collision.

Before collision:
Momentum of A = m1 u1
Momentum of B = m2 u2
Total momentum = m1 u1 + m2 u2
After collision:
Momentum of A = m1 v1
Momentum of B = m2 v2
Total momentum = m1 v1 + m2 v2
During collision,
Force on B by A = - Force on A by B

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 4

 FBA = - FAB
m2 a2 = - m1 a1
m2 (v2 – u2 )/t = - m1 (v1 – u1 )/t
m2 (v2 – u2 ) = - m1 (v1 – u1 )
m2 v2 –m2 u2 = - m1 v1 + m1 u1
m1 v1 + m2 v2 = m1 u1 + m2 u2
m1 u1 + m2 u2 = m1 v1 + m2 v2
“ie Total momentum before collision = Total momentum after collision”
32) Drive an expression for potential energy of a spring by graphical method.

33) To maintain a rotor at uniform angular speed of 120 𝐫𝐚𝐝 𝒔−𝟏 . Engine need to transmit to torque
of 180 Nm. What is the power required by the engine?
𝜔 = 120 𝑟𝑎𝑑/𝑠, 𝜏 = 180𝑁 − 𝑚
𝑃𝑜𝑤𝑒𝑟 𝑃 = 𝜏𝜔

𝑃 = 180 × 120 = 216 × 102 𝑁

34) Define:
i) Longitudinal strain
ii) Shear strain
iii) Volume strain.
i) Longitudinal strain: It is the ratio of change in length to the original length.
ii) Volume strain: It is the ratio of change in volume to the original volume.
iii) Shearing strain: It is the angular displacement of the plane perpendicular to the fixed plane.

35) Distinguish between streamline flow and turbulent flow.

Stream line flow Turbulent flow
Flow in which velocity of every particle at a Flow in which velocity of every particle at a
given point is same. given point is not same
Regular flow Irregular
Water flowing slowly in a pipe Water flowing rapidly in a river

36) On what factors does the rate of heat through a conductor depends on?
▪ Area of cross section
▪ Temperature difference between at the two ends
▪ Length of the conductor
▪ Nature of the conductor

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 5

 37) State and explain Boyle’s law.
Boyle’s law: At constant temperature(T), the volume(V) of the given mass of gas is inversely
proportional to its pressure(P).
1
𝑖𝑒 𝑉𝛼 𝑜𝑟 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃
38) Write newton’s formula for speed of sound in gas and give Laplace correction to Newton’s
formula.
𝐵
Newton’s formula 𝑣 = √𝜌 ℎ𝑒𝑟𝑒 𝐵 = 𝑃

𝑃
𝑣 = √ − − − − − − − (1)
𝜌

𝐴𝑡 𝑆𝑇𝑃 𝑃 = 101.3 × 103 𝑃𝑎, 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎𝑖𝑟 𝜌 = 1.293 𝑘𝑔𝑚−3

101.3 × 103
𝑣 = √ = 280𝑚𝑠 −1
1.293

“Thus the velocity of the sound obtained by Newton is not satisfying the experimental result.

Laplace correction
According to Laplace, when sound waves travel through gaseous medium, pressure and volume
changes under adiabatic conditions.
Therefore 𝑃𝑉 𝛾 = constant

𝛾𝑃
𝑣 = √ 𝜌 -----(2) here 𝛾 = 1.4

1.4×101.3×103
𝑣 =√ = 331m𝑠 −1
1.293

V. Answer ANY THREE of the following questions: (3x5=15)

39) Derive the kinetic equation of uniformly accelerated motion, 𝐯 𝟐 = 𝐯𝐨𝟐 + 𝟐𝐚𝐱, using v-t graph,
where terms have their usual meaning.

A C

O t D t

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 6

 Consider a particle moving with initial velocity 𝑣𝑜 . Let its velocity changes to 𝑣 in time t and a be
the acceleration. Let 𝑥 be the displacement.
𝑥 = area under v- t graph
𝑥 = area of rectangle OACD + area of triangle ABC
𝑥 = OA x OD + ½ ( BC x AC)
𝑥 = 𝑣𝑜 x t + ½ (𝑣- 𝑣𝑜 ) t
2𝑥 = 2 𝑣𝑜 t + (𝑣- 𝑣𝑜 ) t
2𝑥 = (2 𝑣𝑜 + 𝑣 – 𝑣𝑜 )t
2𝑥 = (𝑣 + 𝑣𝑜 )t
2𝑥 = (𝑣 + 𝑣𝑜 ) (𝑣- 𝑣𝑜 ) /a But a = 𝑣- 𝑣𝑜 /t ⇒ t = 𝑣- 𝑣𝑜 /a
2a𝑥 = 𝑣2 – 𝑣𝑜 2
⇒ 𝑣2 = 𝑣𝑜 2 + 2a 𝑥

40) Derive an expression for magnitude and direction of resultant of two vectors acting at a point.
Q S

R
B


 θ
A
N
O P
Consider two vectors 𝑨 𝒂 𝒅 𝑩 acting at a point O. Let θ be the angle between
the two vectors 𝑨 𝒂 𝒅 𝑩. Let be the resultant.

To find magnitude the of resultant ( )

From the triangle OSN,
𝟐
= 𝑵𝟐 + 𝑵𝟐
𝟐
=( + 𝑵)𝟐 + 𝑵𝟐 -----------(1)
From triangle SPN , sinθ = SN /PS ⇒ SN = PS sinθ = B sinθ
also cosθ = PN /PS ⇒ PN = PS cosθ = B cosθ
Eqn. (1) becomes
𝟐
= (𝑨 + B osθ )𝟐 +(Bs θ 𝟐

𝟐
= 𝑨𝟐 +(B osθ )𝟐 + 2AB osθ + (Bs θ 𝟐

𝟐
= 𝑨𝟐 + B 𝟐 os𝟐 θ + 2AB osθ + B 𝟐 s 𝟐
θ
𝟐
= 𝑨𝟐 + B 𝟐 ( os𝟐 θ + s 𝟐
θ) + 2AB osθ
𝟐
= 𝑨𝟐 + B 𝟐 + 2AB osθ

= 𝑨𝟐 + B 𝟐 B osθ

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 7

To find direction of resultant (𝜶 )

From triangle SON , tan 𝜶 = SN/ON

tan 𝜶 = SN/(OP + PN)

tan 𝜶 = B s θ /(A + B osθ )

−𝟏 𝑩 𝒔𝒊 𝜽
𝜶=𝒕𝒂 [ ]
𝑨+𝑩𝒄𝒐𝒔𝜽

41) Prove law of conservation of mechanical energy in case of freely falling body.
Proof: Consider a body of mass m dropped from a point A at a height h
above the ground. A
At point A: S
Potential Energy of the body 𝑃. 𝐸. = 𝑚𝑔ℎ
B h
Kinetic energy 𝐾. 𝐸. = 0
(h-S)
∴ Total Energy 𝑇. 𝐸. = 𝑃. 𝐸. +𝐾. 𝐸.

𝑇. 𝐸. = 𝑚𝑔ℎ + 0 C
𝑻. 𝑬. = 𝒎𝒈𝒉 … … … … … … (𝟏)

At point B:

Potential Energy of the body 𝑃. 𝐸. = 𝑚𝑔(ℎ − 𝑆) = 𝑚𝑔ℎ − 𝑚𝑔𝑆

1 1
Kinetic energy 𝐾. 𝐸. = 2 𝑚𝑣 2 = 2 𝑚 (2𝑔𝑆) = 𝑚𝑔𝑠 [𝑈𝑠𝑖𝑛𝑔 𝑣 2 = 𝑢2 + 2𝑎𝑆 = 0 + 2𝑔𝑆 = 2𝑔𝑆]

∴ Total Energy 𝑇. 𝐸. = 𝑃. 𝐸. +𝐾. 𝐸.

𝑇. 𝐸. = 𝑚𝑔ℎ − 𝑚𝑔𝑆 + 𝑚𝑔𝑠

𝑻. 𝑬. = 𝒎𝒈𝒉 … … … … … (𝟐)

At point C: Potential Energy of the body 𝑃. 𝐸. = 𝑚𝑔(0) = 0

1 1
Kinetic energy 𝐾. 𝐸. = 2 𝑚𝑣 2 = 2 𝑚 (2𝑔ℎ) = 𝑚𝑔ℎ [𝑈𝑠𝑖𝑛𝑔 𝑣 2 = 𝑢2 + 2𝑎𝑆 = 0 + 2𝑔ℎ = 2𝑔ℎ]

∴ Total Energy 𝑇. 𝐸. = 𝑃. 𝐸. +𝐾. 𝐸.

𝑇. 𝐸. = 0 + 𝑚𝑔ℎ

𝑻. 𝑬. = 𝒎𝒈𝒉 … … … … . . (𝟑)

“From Eq s. (1 , ( a d (3 t s fou d that total e ergy of the body rema s o sta t”.

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 8

42) a) Define torque
⃗
𝒅𝑳
b) Obtain the relation 𝝉
⃗ = .
𝒅𝒕

a) It is a turning force which tends to cause rotational motion.

Or It is the cross product of force and perpendicular distance.
⃗
𝑑𝐿
b) Derive =𝜏
𝑑𝑡
Angular momentum is given by
L = r  p
Differentiate, it w.r.t. time, we get,

dL d
= r  p
dt dt

dL dr dp
=  p + r 
dt dt dt
dL
= v mv + r  F
dt

dL
= m (v  v ) + r  F
dt

dL
= r  F =
dt
⃗
𝒅𝑳
=𝝉 ⃗
𝒅𝒕
43) a) What is isothermal process?
b) Obtain an expression for work done in isothermal process.
a) It is a process which takes place at constant temperature.

b)
Piston

Cylinder
Ideal gas

Consider ‘𝑛’ moles of an ideal gas enclosed in a cylinder having non conducting walls and
conducting bottom. Let the cylinder be fitted with a non-conducting frictionless piston. Let 𝐴 be
the area of cross section of the piston. Let P,V and T be the pressure, volume and temperature of
the gas.

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 9

 Let 𝑑𝑊 work done by the gas when piston moves through a small distance 𝑑𝑥 and is given by

𝑑𝑊 = 𝐹 𝑑𝑥

𝑑𝑊 = 𝑃 𝐴 𝑑𝑥

𝑑𝑊 = 𝑃 𝐴 𝑑𝑉 where 𝐴 𝑑𝑥 = 𝑑𝑉

The total work done to increase the volume of the gas from V1 to V2 at constant temperature is
𝑣
given by 𝑊 = ∫𝑣 2 𝑃 𝑑𝑣
1

𝑛𝑅𝑇
We know that, 𝑃𝑉 = 𝑛 𝑅 𝑇 ⇒ 𝑃 = 𝑉
𝑉2
𝑛𝑅𝑇
𝑊 = ∫ 𝑑𝑉
𝑉
𝑉1
𝑉2
1
𝑊 = 𝑛𝑅𝑇 ∫ 𝑑𝑉
𝑉
𝑉1
𝑉
𝑊 = 𝑛𝑅𝑇 [𝑙𝑜𝑔𝑒 𝑉]𝑉21
𝑊 = 𝑛𝑅𝑇 ( 𝑙𝑜𝑔𝑒 𝑉2 − 𝑙𝑜𝑔𝑒 𝑉1 )
𝑉2
𝑊 = 𝑛𝑅𝑇 𝑙𝑜𝑔𝑒 ( )
𝑉1
𝑽𝟐
𝑾 = 𝟐. 𝟑𝟎𝟑 𝑻𝒍𝒐𝒈𝟏𝟎 ( )
𝑽𝟏
44) Show that a stretched string vibrates with all harmonics.
Consider a stretched string of length l generating transverse wave. The velocity 𝑉 of transverse wave is
𝑇
given by 𝑉 = √𝑚 where T – tension in the string, m – mass per unit length of the string

The different modes of vibration ca be discussed as follows

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 10

Fundamental mode:
If 𝑓1 is the frequency of vibration in fundamental mode and 𝜆1 is the wavelength then,
𝑉 𝜆1
𝑓1 = 𝜆 but l = , 𝜆1 = 2l
1 2

1 𝑇
∴ 𝑓1 = 2𝑙 √𝑚 --- (1)

First overtone mode:

If 𝑓2 is the frequency of vibration in first overtone mode and 𝜆2 is the wavelength then,
𝑉 2𝜆2 2𝑙
𝑓2 = 𝜆 but l = , 𝜆2 =
2 2 2

1 𝑇
∴ 𝑓2 = 2 [2𝑙 √𝑚]

𝑓2 = 2𝑓1 --- (2)

Second overtone mode:

If 𝑓3 is the frequency of vibration in second overtone mode and 𝜆3 is the wavelength then,
𝑉 3𝜆3 2𝑙
𝑓3 = 𝜆 but l = , 𝜆3 =
3 2 3

1 𝑇
∴ 𝑓3 = 3 [ √ ]
2𝑙 𝑚

𝑓3 = 3𝑓1 --- (3)

It is clear that 𝑓1 : 𝑓2 : 𝑓3 = 1: 2:3.

“Thus in a stretched string both odd and even (All)harmonics are present”.

VI. Answer ANY TWO of the following questions: (2x5=10)

45) A ship of mass 𝟑 × 𝟏𝟎𝟕 𝒌𝒈 initially at rest is pulled by a force 𝟓 × 𝟏𝟎𝟒 𝑵 through a distance of 3m.
Assuming that resistance of water is negligible, find the speed of the ship after travelling 3m
distance.
Given: m=3× 107 kg, u=0 m/s, F=5× 104 N, S=3m, V=?
𝐹 = 𝑚𝑎
𝐹 5×104
𝑎 = 𝑚= 3×107 =1.66× 𝟏𝟎−𝟑m𝒔−𝟐
𝑉 2 = 𝑢2 + 2𝑎𝑠
: 𝑉 2 = 0 + 2 × 1.66 × 10−3 × 3
V=√9.96 × 10−3
𝑽 = 0.1 m𝒔−𝟏

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 11

46) Calculate the orbital velocity and period of revolution of an artificial satellite of the earth moving
at an altitude of 200 km.
Given, Radius of earth=6400 km, Mass of earth= 𝟔 × 𝟏𝟎𝟐𝟒 𝒌𝒈, 𝑮 = 𝟔. 𝟕 × 𝟏𝟎−𝟏𝟏 𝑵𝒎𝟐 𝒌𝒈−𝟐 .
Given: ℎ =200 km, R=6400 km, M=6× 1024 kg, 𝐺 =6.7× 10−11 𝑁𝑚2 𝑘𝑔−2, 𝑉0 =? , 𝑇 =?

𝐺𝑀
Orbital velocity:𝑉0 = √𝑅+ℎ

6.7××10−11 ×6×1024
=√ =7800 m𝒔−𝟏
(6400+200)×103

=√0.6096 × 108
𝑽𝟎 =7.8 km𝒔−𝟏
2𝜋(𝑅+ℎ)
Time period: T= 𝑉0

2×3.142×6600×103
T= = 𝟓𝟑𝟏𝟕 𝒔𝒆𝒄
7.8×103

T= 1.47 hours
47) A body cools from 𝟖𝟎𝐎 𝑪 𝐭𝐨 𝟓𝟎𝒐 𝑪 in 5 minute. Calculate the time it takes to cool from 𝟔𝟎𝟎 𝑪 𝐭𝐨 𝟑𝟎𝒐 𝑪.
The temperature of surrounding is 𝟐𝟎𝐎 𝑪.
i) 𝑇1 = 800 , 𝑇2 = 500 𝐶,t =5 min, 𝑇0 = 20℃
𝑇1 −𝑇2 𝑇1 +𝑇2
= -k ( − 𝑇0 )
𝑡 2
80−50 80+50
= −𝑘 ( − 20)
5 2

6=-k (65-20)
k = - 0.133
ii): 𝑇1 = 600 , 𝑇2 = 300 𝐶
60−30 60+30
= −(−0.133) ( − 20)
5 2
30
= 0.133(25)
𝑡

𝒕 =9 min
𝝅
48) A body oscillates with SHM according to the equation (in SI units) = 𝟓 𝐜𝐨𝐬( 𝟐𝝅𝒕 + 𝟒 ) at t=1.5s,
calculate the (a) displacement (b) speed and (c) acceleration of the body.
Given: 𝑎) 𝑥 =? , 𝑏) 𝑣 =? , 𝑐)𝑎 =?

𝑥 = 𝐴 cos( 𝜔𝑡 + 𝜑)……………..(1)
𝜋
𝑥 = 5 cos( 2𝜋𝑡 + 4 ) ……………..(2)

Comparing Eqn.(1) and (2) we get,

A= 0.4m, 𝜔 = 2𝜋
𝝅
a) 𝑥 = 𝟓 𝐜𝐨𝐬( 𝟐𝝅𝒕 + 𝟒 )

at 𝑡 =1.5s

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 12

 𝜋
𝑥 = 5 cos( 2𝜋 × 1.5 + 4 )
3𝜋 𝜋
𝑥 = 5 cos( + 4)
4
13𝜋
𝑥 = 5 cos( )
4

𝑥 = −𝟑. 𝟓𝟑𝟓𝒎
b) 𝒗 = 𝝎√𝑨𝟐 − 𝑥 𝟐
𝑣 = 2𝜋√(0.4)2 − (−3.535)2 = 𝟐𝟐𝒎𝒔−𝟏
c) 𝒂 = 𝝎𝟐 𝑥 = (2𝜋 )2 × 2 = 𝟏𝟒𝟎𝒎𝒔−𝟐

********* ********* *********

I PU PHYSICS ANNUAL EXAMINATION FEB-24 SOLUTIONS 13