CBSE Sample Papers for Class 11 Physics Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 2 with Solutions

Time Allowed : 3 hours
Maximum Marks : 70

General Instructions:

1. 1. All questions are compulsory. There are 40 questions.
   2. This Question paper has four sections : Section A, Section B, Section C, Section D
   3. Section A contains twenty five questions of one mark each, Section B contains five questions of two marks each, Section C contains seven questions of three marks each, Section D contains three question of five marks each.
   4. There is no overall choice. However, internal choices have been provided in seven questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks weightage. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:
      c = 3 × 10⁸ m/s,
      h = 6.63 × 10^-34 Js
      e = 1.6 × 10^-19 C,
      Radius of Earth, R_e = 6.4 × 10⁶ m
      Universal Gravitational constant. G = 6.67 × 10^-11 Nm2kg-2
      mass of electron, m_e = 9.1 × 10^-31 kg,
      mass of neutron, m_n = 1.675 × 10^-27 kg
      mass of proton, m_p = 1.673 × 10^-27 kg
      Avogadro’s number = 6.023 × 10²³ atom per gram
      Boltzmann constant = 1.38 × 10^-23 JK^-1

Section – A (25 Marks)

Question numbers 1 to 25 carry 1 mark each.

Question 1.
Which formula represents force [1]
(A) [ML²T^-2]
(B) [MLT^-2]
(C) [ML^-1T^-2]
(D) [ML²T^-2]
Answer:
Option (B) is correct.

Explanation:
Formula of force on any body,
f = ma
∵ Dimensional formula of force = [MLT^-2]

Question 2.
The ratio of displacement to the distance is : [1]
(A) always = 1
(B) always < 1
(C) always > 1
(D) always b ≤ 1
Answer:
Option (D) is correct.

Explanation:
∵ Displacement of a body is always less than or equal to distance covered by it

Question 3.
Newton’s second law of motion gives the measure of : [1]
(A) acceleration
(B) force
(C) momentum
(D) angular momentum
OR
’Newton Sec’ stands for the unit of
(A) force
(B) energy
(C) power
(D) momentum
Answer:
Option (B) is correct.

Explanation:
Since Newton’s second law of motion states that the rate of change of momentum of body is directly proportional to the applied force and takes place in the direction in which the force acts. [means greater the change in momentum, greater the force].
Thus, we can conclude,
Newton’s second law of motion gives measure of force.

OR

Option (D) is correct.

Explanation:
Since S.I. unit of momentum is kg.m/s and S.I. unit of force is Newton (N) or kg.m/s².
∴ Unit of momentum can also written \(\frac{\mathrm{kg} \cdot \mathrm{m} \cdot \mathrm{s}}{\mathrm{s}^2}\) or Newton’s

Question 4.
Phenomenon of beats is due to which of the following properties [1]
(A) interference
(B) diffraction
(C) polarisation
(D) refraction
Answer:
Option (A) is correct.

Explanation:
When two sound waves of nearby frequencies overlap and create a new resultant wave, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud producing beats.

Question 5.
Which of the following equations represents a wave : [1]
(A) y = A sin ωt
(B) y = A cos kx
(C) y = A sin (at – bx + c)
(D) y = A(ωt – kx)
OR
A tuning fork is struck against a rubber pad because it is
(A) light
(B) flexible
(C) cheap
(D) durable
Answer:
Option (C) is correct.

Explanation:
Since, y = A sin(at – bx + c) represents the equation of simple harmonic progressive wave because it describes the displacement of any particle (x) at any time (t) and satisfied the condition of wave equation
\(\frac{\partial^2 y}{\partial t^2}=v^2 \frac{\partial^2 y}{\partial x^2}\)

OR

Option (B) is correct.

Question 6.
Which of the following is correct: [1]
(A) Hydrogen is an ideal gas
(B) Oxygen is an ideal gas
(C) Air is an ideal gas
(D) None of the above gases is ideal
Answer:
Option (D) is correct.

Explanation:
Generally, Hydrogen, Oxygen, Air do not follow all the properties of ideal gas such as zero intermolecular force, zero molecular volume etc. So, these are not ideal gases.

Question 7.
A gas expands under constant pressure P from volume V₁ to V₂. The work done by the gas is : [1]
(A)P(V₂ – V₁)
(B) P(V₁ – V₂)
(C) P(V₁^r – V₂^r)
(D) P\(\left(\frac{1}{V_2}-\frac{1}{V_2}\right)\)
Answer:
Option (A) is correct.

Explanation:
Since work done by the gas
= P_constant × ∆V(In case of constant pressure)
= P(V₂ – V₁)

Question 8.
If the earth stops rotating the value of g’ at the equator will [1]
(A) increase
(B) remain same
(C) decrease
(D) none of these
Answer:
Option (B) is correct.

Explanation:
If the Earth stops rotating then value of g at equator and pole will be equal and the orbiting path of the Earth will be exactly to the
circle. (g ∝ \(\frac{1}{\mathrm{R}_e^2}\))
In general, the value of g at pole is greater than the equator.
So, when Earth stops the rotation, then value of g at equator will increase as compared to previous case.

Question 9.
Which of the following expression represents areal velocity. [1]
(A) 2m \(\vec{L}\)
(B) m \(\vec{L}\)
(C) \(\frac{\overrightarrow{\mathrm{L}}}{2 m}\)
(D) \(\frac{\overrightarrow{\mathrm{L}}}{m}\)
Answer:
Option (C) is correct.

Explanation: Areal velocity of any particle is defined as rate at which area is swept out by the particle as it moves along a curve.

Question 10.
Number of args contained in one joule is : [1]
(A) 10³
(B) 10⁵
(C) 10⁷
(D) 10⁹
OR
When a body moves with a constant speed along a circle
(A) its velocity remains constant
(B) no force acts on it
(C) no work is done on it
(D) no acceleration is produced in it
Answer:
Option (C) is correct.

Explanation:
∵ 1 Joule = 10⁷ erg
where joule and erg are units of energy in S.I. and C.G.S. System respectively.

OR

A body moving at constant speed in circular path experiences an acceleration directed towards the centre of the circular path. This acceleration is centripetal acceleration produces centripetal force. Due to this centripetal force, the work done on body is zero.

Question 11.
Why banking of road at turnings are required ? [1]
Answer:
Banking of road at turnings are require to provide the necessary centripetal force to avoid the chance of skid.

Question 12.
Which of the two linear velocity or the linear acceleration gives the direction of motion of a body ? [1]
OR
Why a one rupee coin placed on a revolving table flies off tangentially ?
Answer:
Linear velocity.
OR
This is due to the inertia of direction.

Question 13.
The top of a lake is frozen. Air in contact is at -15°C. What do you expect the maximum temperature of water ? [1]

1. in contact with lower surface of ice.
2. at the bottom of the lake.

Answer:

1. 0°C
2. 4°C

Question 14.
Does the work done in raising, a load on to a platform depend upon how fast is it raised ? [1]
Answer:
No, because the work done is independent of time.

Question 15.
What is equation of continuity ? [1]
Answer:
For incompressible fluids a₁v₁ = a₂v₂ = a₃v₃ = …………. where a₁, a₂, a₃, …. are the area of cross-sections and v₁, v₂, v₃, … are the magnitudes of velocities.
For compressible fluids :
a₁υ₁ρ₁ = a₂υ₂ρ₂ = a₃υ₃ρ₃ = …………
where ρ₁, ρ₂, ρ₃, are the densities of the fluids.

Commonly Made Error
Students could not comprehend the terms ‘incompressible fluids’ and ‘compressible fluids’ properly in the equation of continuity.

Answering Tip
Students should understand that the effect of pressure on fluid density is negligible in an incompressible fluid.

Question 16.
Where will the true weight of the body be zero ? [1]
OR
A comet move faster at aphelion or perihelion ?
Answer:
The true weight of a body will be zero where the gravitational effects are nil, e.g., at the centre of Earth.
OR
At the perihelion where it is close to sun, the comet moves in faster speed.

Commonly Made Error
Students could not comprehend the meaning of the terms ‘aphelion’ and ‘perihelion’ and thus could not answer the question.

Answering Tip
Students must possess brief understanding of important key terms.

Question 17.
Why do power transmission wires become tight in winter and slack in summer ? [1]
Answer:
∵ ∆l ∝ ∆θ
In winter, the wires contract and hence become tight. They expand during summer and become slack.

Commonly Made Error
Students could not reason out for the tight of power transmission wires in winters and slack in summer.

Answering Tip
Student should be familiarized with the concept of linear expansion.

Question 18.
What is the change in the internal energy of a gas, which is compressed isothermally ? [1]
Answer:
Zero.

Commonly Made Error
Students could not state the change in internal energy of gas compressed isothermally.

Answering Tip
Students should understand that for a perfect gas, internal energy (U) is a function of temperature (T) given by U = U(T).

Question 19.
Obtain the dimensional formula for R used in the ideal gas equation PV = RT. [1]
OR
At which temperature does all molecular motion case ?
Answer:
R = \(\frac{\mathrm{PV}}{\mathrm{T}}\)
= \(\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^3\right]}{[\mathrm{K}]}\)
= [ML²T^-2K^-1]

OR

At zero kelvin.

Question 20.
A man with a wrist watch on his hands fall from the top of a tower. Does the watch give correct time ? [1]
Read the following text and answer any 4 of the following questions on the basis of the same:

The ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins. A Ballistic Pendulum is a device for measuring a bullet’s momentum and speed by employing perfectly inelastic collision.

A large wooden block suspended by two cords serves as the pendulum bob. When a bullet is fired into the bob, it gets embedded in the bob and its momentum is transferred to the bob.

The bullet’s momentum and velocity can be determined from the amplitude of the pendulum swing. The velocity of the bullet, in turn, can be derived from its calculated momentum.
After collision, if the pendulum reaches a height h, then from principle of conservation of mechanical energy
\(\frac{1}{2}\)(m + M)υ_p² = (m + M)gh
where, m = mass of bullet, M = mass of the bob
υ_p = velocity of the bob-bullet combination
∴ υ_p = \(\sqrt{2 g h}\)
Now, Momentum before collision = Momentum after collision
mυ_B = (m + M) υ_p
where, υ_B = velocity of bullet
υ_B = \(\frac{m+M}{m} \sqrt{2 g h}\)
The ballistic pendulum used to be a common tool for the determination of the muzzle velocity of bullets as a measure of the performance of firearms and ammunition (Nowadays, the ballistic pendulum has been replaced by the ballistic chronograph, an electronic device).
Answer:
Yes, because the wrist watch is a spring wound watch and is independent of the ‘g but depends upon the P.E. stored in the spring.

Question 21.
In ballistic pendulum the collision is: [1]
(A) Elastic
(B) Perfectly inelastic
(C) Inelastic
(D) Partly elastic, partly inelastic
Answer:
Option (B) is correct.

Explanation:
A large wooden block suspended by two cords serves as the pendulum bob. When a bullet is fired into the bob, it gets embedded in the bob and its momentum is transferred to the bob. Hence the collision is perfectly inelastic.

Question 22.
Which two principles of Physics are applied to find the velocity of the bullet? [1]
(A) conservation of mechanical energy and conservation of momentum
(B) conservation of mechanical energy and conservation of mass
(C) conservation of mass and conservation of momentum .
(D) conservation of mechanical energy, conservation of momentum and conservation of mass
Answer:
Option (A) is correct.

Explanation:
Principle of conservation of mechanical energy, an expression for the bob- bullet combination after collision is derived. Then the principle of conservation of momentum is applied to find the velocity of the bullet before collision.

Question 23.
The ballistic pendulum was invented by a : [1]
(A) Chemist
(B) Physicist
(C) Mathematician
(D) Warrior
Answer:
Option (C) is correct.

Explanation:
The ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins.

Question 24.
Ballistic pendulum has been replaced by : [1]
(A) Seismograph
(B) Chronograph
(C) Gyrograph
(D) Tachograph
Answer:
Option (B) is correct.

Explanation:
The ballistic pendulum. has now been replaced by the ballistic chronograph, an electronic device.

Question 25.
A ballistic pendulum of 1 kg is fired with a bullet of mass 1 g. If the pendulum rises 2 cm, find the velocity of the bullet. [1]
(A) 12.65 m/s
(B) 6330 m/s
(C) 0.633 m/s
(D) 633 m/s
Answer:
Option (D) is correct.

Explanation: υ_B = \(\frac{m+\mathrm{M}}{m} \sqrt{2 g h}\)
Putting,
m = 1 g = 0.001 kg
M = 1 kg
g = 10 m/s²
h = 2 cm = 0.02 m
υ_B = \(\frac{0.001+1}{0.001} \sqrt{2 \times 10 \times 0.02}\) = 633 m/s

Section – B (10 Marks)

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Is \(|\vec{a}-\vec{b}|\) greater than or less than \(|\vec{a}|\) + \(|\vec{b}|\) ? [2]
Answer:

Question 27.
A body is moving with uniform velocity. Can it be said to be in equilibrium ? Why ? [2]
OR
Give three equations for the three components of vector force.
Answer:
Yes, it can be said to be in equilibrium when it moves with uniform velocity as no acceleration, i.e., no net force acts on the body.

OR

(i) F_x = \(\frac{d p_x}{d t}\) = ma_x
(ii) F_y = \(\frac{d p_y}{d t}\) = ma_y and
(iii) F_z = \(\frac{d p_z}{d t}\) = ma_z

Question 28.
Calculate the power developed by a person while eating 100 g of ice per minute. (Latent heat of ice = 80 cal g^-1.) [2]
OR
Why does the temperature remain constant, when solid melts or liquid boils ?
Answer:
Mass of ice eaten by the man per second,
m = \(\frac{100}{60}\) = \(\frac{5}{3}\) g/s
Latent heat of ice, L = 80 cal/g.
Therefore, energy acquired per second by the man in eating the ice, i.e., power developed by the man,
= mL = \(\frac{5}{3}\) × 80 cal/s
= \(\frac{5 \times 80}{3}\) × 4.2 J/s
= 560 W.

OR

Normally the heat supplied increases the kinetic energy of the molecules of the substance due to which temperature of the substance rises.

During melting or cooling, there is a large change in volume at large distances from each other against attraction between them.
Therefore only potential energy increases.

Question 29.
The periodic time of a body executing S.H.M. is 2 sec. After how much interval from t = 0, will its displacement be half of its amplitude ? [2]
Answer:
Here, T = 2s; t = ?, y = a/2
Now, y = asin ωt = asin \(\frac{2 \pi}{T}\) t
∴ \(\frac{\mathrm{a}}{\mathrm{2}}\) = asin \(\frac{2 \pi}{T}\) t = asin πt (∵ T = 2 s)
or sin πt = \(\frac{1}{2}\)
∵ sin 30° = sin \(\frac{\pi}{6}\)
or πt = \(\frac{\pi}{6}\) or t = \(\frac{1}{6}\) s.

Question 30.
What is periodic wave function ? [2]
Answer:
A wave function (x, t) which satisfies the periodicity conditions of position and time is called a periodic wave function, i.e.,

1. y(x + mλt) = y(x, t)
2. y(x, t + nT) = y(x, t)

where λ = wavelength of wave, T = period of the wave, n and m are integers.

Section – C (21 Marks)

Question numbers 31 to 37 carry 3 marks each.

Question 31.
The velocity-time graph for a particle is shown in figure. Draw acceleration-time graph from it. [3]

Answer:
Acceleration-Time Graph

Commonly Made Error
Students may draw a linear line for the time between 2 s to 6 s, where actually the acceleration is zero.

Answering Tip
Students should first divide the graph into three portion, i.e., 0 to 2s, 2s to 6s and 6s to 8s. Then think about the change in the velocity as given in these intervals and finally, plot the acceleration-time graph for the same.

Question 32.
Calculate the force required to move a train of 2000 quintals up an incline of 1 in 50, with an acceleration of 2 ms^-2, force of friction being 0.5 newton per quintal. [3]
Answer:
Given : m = 2000 quintals
= 2000 × 100 kg (∵ 1 quintal = 100 kg)
sin θ = \(\frac{1}{50}\), acceleration, a = 2 ms^-2
F = force of friction
= 0.5 N per quintal
= 0.5 × 200
= 1000 N.

While moving up an inclined plane, force required against gravity = mgsin θ
= 2000 × 100 × 9.8 × \(\frac{1}{50}\)
= 39200 N.
Also, if f = force required to produce acceleration
= 2 ms^-2,
Then f = ma = 200000 × 2
= 400000 N.
∴ Total force required
= F + mgsin θ + f
= 1000 + 39200 + 400000
= 440200 N.

Question 33.
Three blocks are connected as shown below and are on a horizontal frictionless table. They are pulled to right with a force F = 50 N. If m₁ = 5 kg, m₂ = 10 kg and m₃ = 15 kg, calculate tensions T₁ and T₂. [3]
Answer:
Given : F = 50 N, m₁ = 5 kg, m₂ = 10 kg, m₃ = 15 kg,
Since, the three blocks move with an acceleration ‘a’
So, a = \(\frac{\mathrm{F}}{m_1+m_2+m_3}\)
or a = \(\frac{50}{5+10+15}\) = \(\frac{50}{30}\)
= \(\frac{5}{3}\) ms^-2
To determine T₂ : Imagine the free body

Here, \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{T}_2}\) act towards right and left respectively.
Since the motion is towards the right side, so according to Newton’s Second law of motion:

F – T₂ = m₃a
or 50 – T₂ = 15 × \(\frac{5}{3}\) = 25
or T₂ = 50 – 25 = 25 N.
To determine T₁ : Consider the free body

Here m₁a = T₁
or T₁ = m₁a = 5 × \(\frac{5}{3}\) = \(\frac{25}{3}\)
= 8.33 N

Question 34.
Which of the following symptoms likely to afflict an astronaut in space : [3]
(a) swollen feet, (b) swollen face, (c) headache, (d) orientation problem.
OR
Escape velocity of projectile on a planet’s surface is 11.2 km s^-1. If a body is projected at double the speed, find its speed at an infinite distance from the planet.
Answer:
The astronaut in space will suffer from (a) swollen face, (b) headache and (c) orientation problem.
(a) We know that legs carry the weight of the body in the normal position due to gravity pull. The astronaut in space is in weightlessness state. Hence, swollen feet may not effect his working. Due to weightlessness the astronaut may develop swollen face. As eyes, ears, nose are all embedded in face.
(b) Headache is due to mental strain. It will persist whether a person is an astronaut in space or earth.
(c) Space also has orientation, we also have the frames of reference.

OR

Given, υ_e = 11.2 kms^-1
From principle of conservation of energy
\(\frac{1}{2}\) mυ² = \(\frac{1}{2}\) mυ_i² – \(\frac{1}{2}\) mυ_e²
or υ = \(\sqrt{v_i^2-v_e^2}\)
= \(\sqrt{v_i^2-v_e^2}\)
= \(\sqrt{3}\) υ_e
= 1.732 × 11.2 km/s [∵ \(\sqrt{3}\) = 1.732]
= 31.68 km s^-1

Question 35.
Find the coefficient of thermal conductivity. [3]
Answer:
Consider a cube of side x and area of each face A. The opposite faces of the cube are maintained at temperature, θ₁ and θ₂ where θ₁ > θ₂. Heat gets conducted in the direction of the fall of temperature. The flow of heat depends upon following factors :
(i) Q ∝ A
(ii) Q ∝ \(\frac{d \theta}{d x}\)
(iii) Q ∝ t
Containing equations (i), (ii) & (iii)
Q ∝ A \(\frac{d \theta}{d x}\) × t
Q = KA \(\frac{d \theta}{d x}\) t
or, Q = K \(\frac{\mathrm{A}\left(\theta_1-\theta_2\right)}{x}\) t

Here K is a constant called the coefficient of thermal conductivity of the material of the cube and t stands for time interval.
If A = 1 m², x = 1 m, t = 1 s, ∆θ = 1°C then Q = K
Thus coefficient of thermal conductivity is defined as the amount of heat required to flow through a solid having unit area of cross section, unit length to rise 1°C in 1 s.
We can also write as
H = KA\(\left[\frac{\Delta \theta}{\Delta x}\right]\)
where, H = Heat flow per second
\(\frac{\Delta \theta}{\Delta x}\) = Temperature gradient
T = θ
when, A = 1 m²
(θ₁ – θ₂) = 1°C
t = 1 s
x = 1 cm
then H = K.

Question 36.
Estimate the average thermal energy of a Helium atom at (i) room temperature (27°C), (ii) the temperature on the surface of the sun (6000 K), (iii) temperature of 10 million kelvin (the typical core temperature in the case of star). [3]
OR
Calculate the number of degrees of freedom of molecules of Hydrogen in 1 cc of Hydrogen gas at NTP.
Answer:
(i) K.E._avg = \(\frac{3}{2}\) kT
= \(\frac{3}{2}\) (1.38 × 10^-23) (27 + 273)
= 6.21 × 10^-21 J

(ii) K.E.’_avg = \(\frac{3}{2}\) kT’
= \(\frac{3}{2}\) (1.38 × 10^-23) × 6000
= 1.242 × 10^-19 J

(iii) K.E.”_avg = \(\frac{3}{2}\) kT”
= \(\frac{3}{2}\) (1.38 × 10^-23) × 10⁷
= 2.07 × 10^-16 J

Commonly Made Error
Students could not evaluate the average thermal energy.

Answering Tip
Students should be familiar with the relation, K. E_avg = (3/2) kT where, Temperature (T) is in kelvin.

OR

Volume occupied by 1 gram mole of gas at NTP = 22400cc
∴ Number of molecules in 1cc of Hydrogen
= \(\frac{6.023 \times 10^{23}}{22400}\) = 2.688 × 10¹⁹
As each diatomic molecule has 5 degrees of freedom, Hydrogen being diatomic also has 5 degrees of freedom
∴ Total no. of degrees of freedom
= 5 × 2.688 × 10¹⁹
= 1.344 × 10²⁰

Commonly Made Error
Students could not evaluate the degrees of freedom for the molecules of Hydrogen.

Answering Tip
Students should be familiar with the concept of degrees of freedom for monatomic, diatomic and triatomic molecules.

Question 37.
A particle in linear simple harmonic motion has a velocity of 4 ms^-1 at 3 m and 3 ms^-1 at 4 m from mean position. What is the time taken to travel half the amplitude from its positive extreme position ? [3]
Answer:

or 16a² – 256 = 9a² – 81
or a² = 25
∴ a = 5 m.
Step 2. Putting a = 5 in eqn (i),
4 = 4ω
or ω = 1 rad s^-1.

Step 3. Time taken to travel half amplitude from positive extreme position is given by displacement,
x = acos ωt
i.e., = \(\frac{5}{2}\) cos (1 × t) or, cos t = \(\frac{1}{2}\)
or t = cos^-1 \(\frac{1}{2}\) = 60° = \(\frac{\pi}{3}\)
or t = \(\frac{3.142}{3}\) = 1.047 s.

Section – D (15 Marks)

Question numbers 38 to 40 carry 5 marks each.

Question 38.
When a body slides down from rest along a smooth inclined plane making an angle of 45° with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance it is seen to take time pT, where p is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane. [5]
Answer:

For smooth inclined plane
Let s = length of inclined plane
or mg sinθ = ma
or a = g sin θ
Applying 2^nd kinematic equation of motion
s = ut + \(\frac{1}{2}\) at²
or s = 0 × T +\(\frac{1}{2}\) (g sin θ)T² (∵ t = T)
or s = \(\frac{1}{2}\) g sinθT²
or s = \(\frac{1}{2 \sqrt{2}}\) gT² ……………… (i) (∵ θ = 45°)
For rough inclined plane.
f = μN = μmg cos θ
mg cosθ – f = ma₁
or a₁ = (sin θ – μ cos θ)g = \(\frac{1}{\sqrt{2}}\) (1 – μ)g

Using, s = ut + \(\frac{1}{2}\) at²
or s = 0 × (pT) + \(\frac{1}{2}\) {\(\frac{1}{\sqrt{2}}\) (1 – μ)g} × p²T²
or s = \(\frac{1}{2 \sqrt{2}}\) (1 – μ) gp² T² …………….. (ii)
∴ by (i) & (ii), we get
\(\frac{1}{2 \sqrt{2}}\) gT² = \(\frac{1}{2 \sqrt{2}}\) (1 – μ) gp² T²
or 1 = (1 – μ) p²
or μ = (1 – \(\frac{1}{p^2}\))

Question 39.
A disc of radius R is rotating with an angular speed tag about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is μ_k. [5]
(A) What was the velocity of its centre of mass before being brought in contact with the table ?
(B) What happens to the linear velocity of a point on its rim when placed in contact with the table ?
(C) What happens to the linear speed of the centre of mass when disc is placed in contact with the table ?
(D) Which force is responsible for the effects in (B) and (C) ?
(E) What condition should be satisfied for rolling to begin ?
(F) Calculate the time taken for the rolling to begin.

OR
Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities ω (anti-clockwise) and ω(clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by (3R+ 8 ). They are now brought in contact (δ → 0). [5]
(A) Show the frictional forces just after contact.
(B) Identify forces and torques external to the system just after contact.
(C) What would be the ratio of final angular velocities when friction ceases?
Answer:
(a) Before being brought in contact with the table the disc was in pure rotational motion (only) about its axis passing through centre :
∴ υ_cm = 0 as the point on axis are considered at rest.
(b) When the rotating disc is placed in contact with table’s surface, due to friction, linear velocity of a point on the rim decreases.
(c) When disc is placed in contact with the surface of the table, due to frictional force, centre of mass acquires some linear velocity. The linear velocity increases.
(d) Frictional force
(e) When rolling of disc starts on table, then υ_cm = ωR due to reaction force.

(f) Acceleration produced in centre of mass due to reaction force F because of frictional force in disc
F = mA or a_cm = \(\frac{F}{m}=\frac{\mu_k m g}{m}\) = µ_kg.
Angular retardation produced by the torque due to friction.

Here, frictional force helps in rolling motion.

OR

(a)

Let us consider the following figure, which shows frictional forces.
The direction of υ₁ and υ₂ at point of contact are tangentially upward.
Then
⇒ υ₁ = ωR, υ₂ = ω2R,

(b) External forces acting on the system are equal and opposite, so net force is zero.
F’ = F = F” where F’ and F” are external forces through support
F_net = 0
External torque = F × 3R, anticlockwise.
So, as velocity of drum 2 is double, i.e. υ₂ = 2υ₂

(c) Let ω₁ and ω₂ be final angular velocities (anticlockwise and clockwise respectively) of smaller drum 1 and drum 2 respectively. Finally, when their velocities become equal there will be no friction., due to no slipping at this stage.
Hence, Rω₁ = 2 Rω₂ or \(\frac{\omega_1}{\omega_2}=\frac{2}{1}\)

Question 40.
A thin rod having length L0 at 0°C and coefficient of linear expansion a has its two ends maintained at temperatures θ₁ and θ₂ respectively. Find its new length. [5]
OR
We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If α_iron = 1.2 × 10^-5/K and α_brass = 1.8 × 10^-5/K, what should we take as length of each strip ?
Answer:
The rod’s temperature varies from θ₁ to θ₂ [from one end to another].
So, mean temperature of rod is
θ = \(\frac{\theta_1+\theta_2}{2}\) (at C)
So, rate of flow heat A → C and C → B are equal. θ₁ > 0 > θ₂
therefore, \(\frac{\mathrm{KA}\left(\theta_1-\theta\right)}{\mathrm{L}_0 / 2}=\frac{\mathrm{KA}\left(\theta-\theta_2\right)}{\mathrm{L}_0 / 2}\)
Here, K = Coefficient of thermal conductivity.

OR

From question.
l_iron – l_brass = 10 cm = constant at all temperatures.
Let l° be length at temperature 0° C.
l°_iron – l°_brass = 10cm at all temperature
∴ At a changed temperature,
l°_iron (1 + α_iron∆t) – l°_brass (1 + α_brass∆t) = 10 CM
l°_ironα_iron = l°_brassα_brass
∴ \(\frac{l_{\text {iron }}^o}{l_{\text {brass }}^o}=\frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}=\frac{1.8}{1.2}=\frac{3}{2}\)
Then, l°_iron – l°_brass = 10
3x – 2x = 10
x = 10
Length of iron rod = 3 × 10 = 30 cm
Length of brass rod = 2 × 10 = 20 cm
The difference between is 10 cm.