CBSE Sample Papers for Class 11 Physics Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Physics with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Physics Set 9 with Solutions

Time Allowed : 3 hours
Maximum Marks : 70

General Instructions:

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

Question numbers 1 to 25 carry 1 mark each.

Question 1.
The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is : [1]
(A) 663.821
(B) 664
(C) 663.8
(D) 663.82
Answer:
Option (C) is correct

Question 2.
Figure shows the orientation of two vectors \(\vec{u}\) and \(\vec{v}\) in the XY plane.
If \(\vec{u}\) = a\(\hat{i}\) + b\(\hat{j}\) and \(\vec{v}\) = p\(\hat{i}\) + q\(\hat{j}\)
Which of the following is correct ? [1]
(A) a and p are positive while b and q are negative.
(B) a, p and b are positive while q is negative.
(C) a, q and b are positive while p is negative.
(D) a, b, p and q are all positive.

OR
The component of a vector r along X-axis will have maximum value if
(A) r is along positive Y-axis
(B) r is along positive X-axis
(C) r makes an angle of 45° with the X-axis
(D) r is along negative Y-axis
Answer:
Option (B) is correct
Explanation: According to figure, in \(\vec{u}\) = a\(\hat{i}\) + b\(\hat{j}\), both a & b are positive while in \(\vec{v}\) = p\(\hat{i}\) + q\(\hat{j}\), p is positive and q is negative. Thus a, b and p are positive and q is negative.

OR

Option (B) is correct

Explanation: If \(\vec{r}\) is at angle θ with x-axis, then component of \(\vec{r}\) along x-axis = rcos θ. It will be maximum if cos θ = maximum = 1 or θ = 0° i.e., is along positive x-axis.

Question 3.
A metre scale is moving with uniform velocity. This implies
(A) The force acting on the scale is zero, but a torque about the centre of mass can act on the scale.
(B) The force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero.
(C) The total force acting on it need not be zero but the torque on it is zero.
(D) Neither the force nor the torque need to be zero.  [1]
Answer:
Option (B) is correct.

Question 4.
A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments : one in which the charged particle is also a proton and in another, a positron. In the same time f, the work done on the two moving charged particles is [1]
(A) same as the same force law is involved in the two experiments.
(B) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.
(C) more for the case of a positron, as the positron moves away a larger distance.
(D) same as the work done by charged particle on the stationary proton.
Answer:
Option (C) is correct

Explanation: Force between two protons = force between a proton and a positron.
Because of having much lighter weight than proton, positron moves away a larger distance as compared to proton.

As work done = force × displacement, therefore in the same time t, work done in case of positron is more than that of proton.

Question 5.
A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to +ve y-axis and intersecting z-axis at z = a show in given figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is:  [1]

(A) mυa\(\hat{e}_x\)
(B) 2mυa\(\hat{e}_x\)
(C) ymυ\(\hat{e}_x\)
(D) 2ymυ\(\hat{e}_x\)
OR
When a disc rotates with uniform angular velocity, which of the following is not true?
(A) The sense of rotation remains same.
(B) The orientation of the axis of rotation remains same.
(C) The speed of rotation is non-zero and remains same.
(D) The angular acceleration is non-zero and remains same.
Answer:
Option (B) is correct

Explanation: Initial velocity, \(v_i\) = \(\hat{v e}_y\)
After reflection, final velocity \(v_f\) = \(-\hat{v e}_y\)
The trajectory is at constant distance a on z-axis and as particle moves along y-axis, its y component changes.
Now, position vector, \(\vec{r}\) = y\(\hat{e}_y\) + a\(\hat{e}_z\)
Hence, change in angular momentum
\(\vec{r}\) × m(υ_f – υ_i) = 2mυa\(\hat{e}_x\)

OR
Option (D) is correct

Question 6.
As observed from earth, the sun appears to move in an approximate circular orbit. For the motion of another planet like mercury as observed from earth this would [1]
(A) be similarly true.
(B) not be true because the force between earth and mercury is not inverse square law.
(C) not be true because the major gravitational force on mercury is due to sun.
(D) not be true because mercury is influenced by forces other than gravitational forces.
Answer:
Option (C) is correct.

Question 7.
The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will  [1]
(A) be double.
(B) be half.
(C) be four times.
(D) remain same.
OR
Along a streamline
(A) the velocity of a fluid particle remains constant.
(B) the velocity of all fluid particles crossing a given position is constant.
(C) the velocity of all fluid particles at a given instant is constant.
(D) the speed of a fluid particle remains constant.
Answer:
Option (D) is correct

OR

Option (B) is correct

Question 8.
If an average person jogs, he produces 14.5 × 10³ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 × 10³ cal for evaporation) is [1]
(A) 0.25 kg
(B) 2.25 kg
(C) 0.05 kg
(D) 0.20 kg
Answer:
Option (A) is correct
Explanation: Amount of sweat evaporated per min calories produced per minute

Question 9.
Boyle’s law is applicable for an
(A) adiabatic process.
(B) isothermal process
(C) isobaric process.
(D) isochoric process  [1]
Answer:
Option (B) is correct.
Explanation: Since V ∝\(\frac{1}{P}\) at constant temperature

Question 10.
The displacement of a particle is represented by the equation y = sin³ ωt. The motion is [1]
(A) non-periodic.
(B) periodic but not simple harmonic.
(C) simple harmonic with period 2π/ω.
(D) simple harmonic with period π/ω.
Answer:
Option (B) is correct.
Explanation: As \(\frac{d^2 y}{d t^2}\) is not ∝(-y)
∴ It is not SHM but it is a periodic motion with period \(\frac{2 \pi}{\omega}\)

Question 11.
Define radian. [1]
Answer:
One radian is the angle subtended at the centre of the circle by an arc equal in length to the radius of the circle.

Question 12.
Under what condition the magnitude of average velocity of a particle is equal to the average speed ? [1]
OR
Find the angle between the two vectors :
\(\vec{A}\) = \(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\) and B = -4\(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\).
Answer:
The magnitude of the average velocity of a particle is equal to the average speed if the particle moves along a straight line in a fixed direction.

OR

\(\vec{A} \cdot \vec{B}\) = \(|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}| \cos \theta\)
∴ cos θ = \(\frac{A \cdot B}{|A||B|}\)
\(\vec{A} \cdot \vec{B}\) = (\(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\)).(-4\(\hat{i}\) + \(\hat{j}\) – 2\(\hat{k}\))
= (-4 + 2 + 2)
= -4 + 4 = 0
cos θ = 0
θ = cos^-10
Hence, θ = 90°

Question 13.
Why does a cricketer move his hands backwards when holding a catch ?  [1]
Answer:
In order to save himself from getting hurt, a cricketer moves his hands backwards while holding a catch. The cricketer increases the time interval by doing so. Increased time duration decreases the impulse, so the cricketer does not get hurt more.

Question 14.
Write the expression for elastic potential energy.  [1]
Answer:
Elastic potential Energy, U = \(\frac{1}{2}\)kx² where k is called spring constant.

Question 15.
What is areal velocity ?  [1]
OR
What is an isolated system ?
Answer:
Area swept out per unit time by particle is known as areal velocity, i.e., \(\frac{\Delta \mathrm{A}}{\Delta t}\)

OR

An isolated system is that on which no external force is acting.

Question 16.
What is the relation between orbital and escape velocity ? [1]
OR
Two satellites are at different heights. Which would have greater orbital velocity ? Why ?
Answer:
υ_e = \(\sqrt{2} v_{0^{\circ}}\)

OR

The satellite at the smaller height would have greater orbital velocity. This is because
υ₀ ∝ \(\frac{1}{\sqrt{r}}\)

Question 17.
Define compressibility. [1]
Answer:
Compressibility is the reciprocal of the bulk modulus, i.e., compressibility = \(\frac{1}{\mathrm{~K}}\)

Question 18.
What is column pressure ? [1]
Answer:
Pressure exerted by a liquid due to its height is called column pressure.

Question 19.
What is the ratio between the distance travelled by the oscillator in one time period and amplitude ? [1]
Answer:
In SHM,
The distance travelled by particle in one complete oscillation = AO + OB + BO + OA = 4AO

Amplitude = AO
∴ Ratio = \(\frac{4 \mathrm{AO}}{\mathrm{AO}}\) = 4

Question 20.
What is the phase relationship between displacement, velocity and acceleration in S.H.M. ? [1]

Read the following text and answer any 4 of the following questions on the basis of the same:
Railway track expansion joint:
Expansion and contraction of steel and concrete structure due to seasonal heating and cooling is a common problem found in civil engineering. To combat this problem, engineers put expansion joints to absorb these changes.

This problem is compounded on railway tracks. This could lead to rail buckling, known in the industry as “sun kink”, as shown below, and cause the derailment of train.
When exposed to temperature variations, the rail tends to vary its length. If this tendency is freely allowed, for a temperature variance At, the rail length L will vary by ΔL. This length variance can be computed as:
ΔL = αLΔt
where α = expansion coefficient of steel = 11.5 × 10^-6/°C.
The coefficient of thermal expansion is defined as the fractional increase in length per unit rise in temperature.

Traditional railway tracks are of standard lengths. When the tracks are laid, the lengths are joined end to end using “fishplates”—short lengths of steel plate overlapping the joint, and bolted to the ends of the rails.
At each joint there must be a short gap (≈ 1/8”) between the rail ends, to allow for longitudinal thermal expansion of the rails on hot days.
Answer:
In S.H.M. the velocity leads the displacement by π/2 radians and acceleration leads the velocity by π/2 radians.

Commonly Made Error
Some students are not able to give answer the problems related to S.H.M. .

Answering Tip
Students should learn about S.H.M. and its characteristics

Question 21.
“Sun kink” is the
(A) Buckling of railway tracks due to seasonal heating and cooling.
(B) Fracture of railway tracks due to seasonal heating and cooling.
(C) Buckling of railway track arising from the high pressure of loaded wagons.
(D) None of the above.  [1]
Answer:
Option (A) is correct
Explanation: Expansion and contraction of steel and concrete structure due to seasonal heating and cooling is a common problem found in civil engineering.
This problem is predominant on railway tracks. This could lead to rail buckling, known in the industry as “sun kink”, and cause the derailment of train.

Question 22.
What is the value of thermal expansion coefficient of steel? [1]
(A) 1.15 × 10^-6 /°C.
(B) 0.115 × 10^-6 /°C
(C) 11.5 × 10^-6 /°C
(D) 11.5 × 10⁶ /°C
Answer:
Option (C) is correct.
Explanation: Expansion coefficient of steel
= 11.5 × 10^-6 /°C.

Question 23.
What will be the expansion of a 20 m long railway steel track for 30°C variation of temperature? [1]
(A) 6.9 m
(B) 6.9 cm
(C) 6.9 mm
(D) 0.69 m
Answer:
Option (C) is correct.
Explanation:
ΔL = αLΔt
α = 11.5 × 10^-6 /°C.
L = 20 m
Δt = 30°C
Putting the values in the expression:
ΔL = 11.5 × 10^-6 × 20 × 30
= 6.9 mm

Question 24.
At each railway track joint a short gap of ………. (approximately) is left.  [1]
(A) 1/8 mm
(B) 1/8 cm
(C) 1/8 m
(D) 1/8 inch
Answer:
Option (D) is correct
Explanation: At each joint of railway track there must be a short gap (≈ 1/8″) between the rail ends, to allow for longitudinal thermal expansion of the rails on hot days.

Question 25.
Which of the following statement is true?
(A) Expansion joint maintain 1/8″ gap whatever be the temperature change.
(B) Expansion joint increases when temperature decreases and decreases when temperature increases.
(C) Expansion joint decreases when temperature decreases and increases when temperature increases.
(D) Expansion joints expands with the rise of temperature.  [1]
Answer:
Option (B) is correct.
Explanation: Normally 1/8″ gap is left when two railway track ends are joined with fishplate. When temperature rises, the rail tracks expand and this gap accommodates the expansion of the steel track. So, the gap contracts. When temperature decreases, the rail tracks contract and this gap increases.

Section – B

Question numbers 26 to 30 carry 2 marks each.

Question 26.
Find the number of times the heart of a human being beats in 10 years. Assume that the heart beats once in 0.8 s. [2]
Answer:
In 0.8 s, the human heart makes one beat.
∴ In 1 s the human heart makes
= \(\frac{1}{0.8}\) = \(\frac{10}{8}\) beats
∴ In 10 years the human heart makes
= \(\frac{10}{8}\) × 10 × 365 × 24 × 60 × 60 beats
= 3.942 × 10⁸ beats.

Question 27.
A person deep inside water cannot hear sound produced in air. Why ? [2]
Answer:
As speed of sound in water is roughly four times the speed of sound in air, therefore
µ = \(\frac{\sin i}{\sin r}\) = \(\frac{v_a}{v_w}\)
= \(\frac{1}{4}\) = 0.25
For refraction,
r_max = 90°
∴ (sin i)_max = 0.25
∴ i_max ≈ 14°. That is why most of sound produced in air and falling at ∠i > 14° gets reflected in air and person deep inside water cannot hear the sound.

Question 28.
What is rolling friction ? [2]
Answer:
The opposition offered to the circular motion of objects like ring, disc, sphere, cylinder etc. on another surface is called rolling friction. The coefficient of rolling friction (µ_r) is smaller than the coefficient of kinetic friction (µ_k). The sliding friction can be decreased a lot by converting it into rolling friction.

Question 29.
What are the essential points of difference between sound and light waves ?
OR
Explain why waves on strings are always transverse ? [2]
Answer:

1. Sound waves are mechanical waves which are longitudinal in nature. Light waves are electromagnetic waves which are transverse in character.
2. Sound waves need a material medium for propagation, whereas light waves do not need any medium.
3. Velocity of sound waves in air at 0°C ≈ 332 m/s, whereas velocity of light waves in air/vacuum = 3 × 10⁸ m/s. (any two)

OR

A string is non-stretchable, i.e., compressions and rarefactions cannot be produced in strings. Therefore, longitudinal waves in strings are not possible. Strings do have elasticity of shape. Therefore, waves on strings are transverse.

Question 30.
A simple harmonic motion of amplitude A, has a time period T. What will be the acceleration of the oscillator, when its displacement is half of the amplitude ?
OR
Will a pendulum gain or lose time when taken to the top of a hill ? [2]
Answer:
Acceleration,
a = -ω²y = –\(\frac{4 \pi^2}{T^2} \times \frac{A}{2}\) = –\(\frac{2 \pi^2 \mathrm{~A}}{\mathrm{~T}^2}\)
OR

Value of acceleration due to gravity decreases at the top of the hill. Time period of a simple pendulum is given by : T = \(2 \pi \sqrt{ }(l / g)\)
Decrease in g means, increase in T, i.e., the pendulum takes more time to complete vibration, it implies that it will lose time.

Section – C

Question numbers 31 to 37 carry 3 marks each.

Question 31.
Find resultant vector of the summation of two vectors A and B having angle 0 between them. [3]
Answer:
Let there be two vectors \(\vec{A}\) and \(\vec{B}\) with angle θ between them.

Using parallelogram law of vector addition resultant,
\(\overrightarrow{\mathrm{R}}\) = \(\overrightarrow{\mathrm{A}}\) + \(\overrightarrow{\mathrm{B}}\)
Draw DF ⊥ OC extend as OF
Considering right angled Δ OFD
OD² = OF² + DF²
= (OC + CF)² + DF²
= (A + Bcos θ)² + (Bsin θ)²
R² = A² + B² + 2ABcos θ
(∵ cos² θ + sin² θ = 1)

Eqn. (i) is called law of cosines.

Question 32.
State and prove impulse-momentum theorem.  [3]
Answer:
Impulse-momentum theorem states that the impulse of force on a body is equal to the change in momentum of the body.
i.e., \(\vec{J}\) = \(\overrightarrow{\mathrm{F}} t\) = \(\overrightarrow{p_2}-\overrightarrow{p_1}\)
Proof. According to Newton’s Second law of motion, we know that
\(\overrightarrow{\mathrm{F}}\) = \(\frac{d \vec{p}}{d t}\)
or \(\overrightarrow{\mathrm{F}}\) dt = d\(\vec{p}\) … (i)

When \(\overrightarrow{\mathrm{F}}\) = constant force acting on the body.
Suppose \(\overrightarrow{p_1}\) and \(\) be the linear moments of the body at time t = 0 and t respectively.
∴ Integrating equation (i) within these limits, we get
\(\int_0^t \vec{F} d t\) = \(\int_{\vec{p}_1}^{\vec{p}_2} d \vec{p}\)

Question 33.
Give conditions for no work.  [3]
Answer:
In physics no work is said to be done, if
(A) The applied force (F) is zero. A body moving with uniform velocity on a smooth surface has some displacement but no external force so in this case work done is zero.
(B) The displacement (s) is zero. A labourer standing with a load on his head does no work.
(C) The angle between force and displacement (θ) is π/2 rad or 90°. Then, cos θ = cos 90° = 0. Thus work done is also zero. In circular motion, instantaneous work done is always zero because of this reason.
(D) The change in kinetic energy (ΔKE) is zero.

Question 34.
A system consisting of two objects has a total momentum of (18 kgm/s)\(\hat{\imath}\) and its centre,of mass has the velocity of (3 m/s)\(\hat{\imath}\). One of the objects has the mass 4 kg and velocity (1.5 m/s)\(\hat{\imath}\). Find the mass and velocity of the other object.  [3]
OR
A dog of mass 10 kg is standing on a flat 10 m long boat so that it is 20 meters from the shore. It walks 8 rn on the boat towards the shore and then stops. The mass of the boat is 40 kg and friction between the boat and the water surface is negligible. How far is the dog from the shore now?

Answer:
Given,
Total momentum = (18 kgm/s)\(\hat{\imath}\),
Velocity of centre of mass = (3 m/s)\(\hat{\imath}\),
Mass of one object = 4 kg,
Velocity of this object = (1.5 m/s)\(\hat{\imath}\)
Let m be the mass of other object and v be the velocity.
Now we know,
Total momentum = Total mass × velocity of center of mass
(18kgm/s)\(\hat{\imath}\) = (m + 4)(3 m/s)/\(\hat{\imath}\)
or m = 2 kg
Now, υ_cm = \(\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\)
or, 3\(\hat{i}\) = (4 × 1.5\(\hat{i}\) + 2υ)/6
So, 18\(\hat{i}\) = 6\(\hat{i}\) + 2υ
Or, υ = 6\(\hat{i}\) m/s

OR

Take boat and dog as a system. Initially, centre of mass of the system is at rest. Since no external force is acting on the system, hence centre of mass of the system will remain stationary. Let initially distance of the centre of mass of the boat from the shore be x m.
Then, x_1c.m. = \(\left(\frac{40 \times x+10 \times 20}{40+10}\right) \mathrm{m}\)
Here, x_1cm = distance of the C.M. of the system from the shore.

Since dog moves towards the shore, for the centre of mass of the system to be at rest, the boat has to move away from the shore. Let distance moved by the boat be ‘x’. Then,

Hence, distance of dog from the shore is (20 – 8 + 1.6)m = 13.6 m

Question 35.
What is the relative humidity on a day in July in Mumbai when the partial pressure of water vapour is 0.013 × 10⁵ Fa and the temperature is 15°C ? The vapour pressure of water at this temperature is 0.0169 × 10⁵ Pa. [3]
Answer:
Relative humidity

Question 36.
You have a light spring, a metre scale and a known mass. How will you find time period of vibration of mass without the use of clock ? [3]
Answer:
Attach the given mass (m) to the spring and suspend the spring. Measure the increase in length (l) of the spring with the help of the scale.
Restoring force of the spring is given by,
F = -kl = – mg
or \(\frac{m}{k}\) = \(\frac{l}{g}\)
we know that time period of a vibrating system is given by,
T = \(2 \pi \sqrt{\frac{m}{k}}\)
or T = \(2 \pi \sqrt{\frac{l}{g}}\)

Question 37.
Define root mean square velocity of gas molecules. Give various relations for it.
OR
State Boyle’s law arid Charles’s law of gases from the relation P = \(\frac{1}{3} \rho v_{r m s}^2\) [3]
Answer:
Root mean square velocity is defined as the square root of the average of the squares of the individual velocities of the gas molecules i.e.,

where, υ₁, υ₂, υ₃, ……, vυ_n are individual velocities
υ_rms = \(\sqrt{\frac{3 P}{\rho}}\) = \(\sqrt{\frac{3 R T}{M}}\) = \(\sqrt{\frac{3 k_{\mathrm{B}} \mathrm{T}}{m}}\)
i.e., υ_rms ∝ \(\sqrt{\mathrm{T}}\)

OR

Boyle’s law :
Pressure exerted by one mole of gas
P = \(\frac{1}{3} \rho v_{r m s}^2\),
P = \(\frac{1}{3} \frac{\mathrm{M}}{\mathrm{V}} v_{r m s}^2\) [ρ = \(\frac{\mathrm{M}}{\mathrm{V}}\)]
PV = \(\frac{1}{3} \mathrm{M} v_{r m s}^2\)
Here M is fixed and v² ∝ T
If temperature T is fixed then PV = constant
or P ∝ \(\frac{1}{V}\)
If temperature of a given mass of a gas is kept constant its pressure is inversely proportional to its volume.

Charles’s law
Pressure exerted by one mole of gas
P = \(\frac{1}{3} \rho v_{r m s}^2\)
= \(\frac{1}{3} \frac{\mathrm{M}}{\mathrm{V}} v_{r m s}^2\),
or V = \(\frac{1}{3} \frac{\mathrm{M}}{\mathrm{P}} v_{r m s}^2\)
υ² ∝ T and M is fixed
V ∝ T
If pressure of a given mass of a gas is kept constant its volume is directly proportional to the temperature of the gas.

Section – D

Question numbers 38 to 40 carry 5 marks each.

Question 38.
A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s. How much will 5 J measure in this new system ? [5]
OR
(a) A particle has displacement equation
(i) x_A = 2t + 7
(ii) x_B = 3t² + 2t + 6
(iii) x_C = 5t³ + 4t
Which of them has uniform acceleration ?
Answer:
Dimension of energy = [ML²T^-2],
n₂u₂ = n₁u₁
⇒ n₂ = n₁\(\frac{u_1}{u_2}\)
= n₁\(\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^1\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^2\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^{-2}\)

Given:
n₁ = 5J
M₂ = α kg,
M₁ = 1 kg,
L₂ = β m, L₁ = 1 m,
T₂ = γ s, T₁ = 1 s,

In new system, 5J = \(\frac{5 \gamma^2}{\alpha \beta^2}\)

OR

(a)
(i) υ = \(\frac{d x_A}{d t}\) = 2
and a = \(\frac{d^2 x_A}{d t^2}\) = 0
As per this eqn. the particle has no acceleration at all.

(ii) υ = \(\frac{d x_B}{d t}\)
= 3 × 2t + 2 + 0
= 6t + 2
a = \(\frac{d^2 x_B}{d t^2}\) = 6
Here, acceleration is uniform.

(iii) υ = \(\frac{d x_C}{d t}\)
= 5 × 3t² + 4
= 15t² + 44
a = \(\frac{d v}{d t}\) = \(\frac{d^2 x_C}{d t^2}\)
= 15 × 2t = 30t
Here, acceleration depend upon time so it is not uniform.

(b) Two bodies are thrown with the same initial velocity at angle θ and (90° – θ) to the horizontal. Determine the ratio of the maximum heights reached by the bodies.
Answer:
Let the two bodies be projected with the same velocity u. Further, suppose that H₁ and H₂ are the heights reached by the two objects thrown at angles θ and 90° – θ respectively.

Question 39.
Derive an expression for K.E. of rotation. [5]
OR
Discuss the variation of g with height and depth.
Answer:
Kinetic energy of rotation :
K.E. of rotation of a body is the energy possessed by the body on account of its rotation about a given axis.
In figure, we have shown a rigid body rotating in xy plane about z-axis with a uniform angular velocity ω.

Let the body consists of particles of masses m₁, m₂, m₃, ……, m_n at distance r₁, r₂, r₃, …, r_n respectively.
Let the linear velocity of different particles are v₁, v₂, v₃, ……., v_n.
υ₁ = r₁ω,
υ₂ = r₂ω,
υ₃ = r₃ω, ………
K.E. of particles of mass m₁ is
\(\frac{1}{2} m_1 v_1^2\) = \(\frac{1}{2} m\left(r_1 \omega\right)^2\) = \(\frac{1}{2} m_1 r_1^2 \omega^2\)
Similarly, K.E. of other particles of the body are :
\(\frac{1}{2} m_2 r_2^2 \omega^2\), \(\frac{1}{2} m_3 r_3{ }^2 \omega^2\) ……….
∴ K.E. of rotation = \(\frac{1}{2} \mathrm{I} \omega^2\)
where, I = \(\sum_{i=1}^{i=n} m_i r_i^2\)

OR

Variation of acceleration due to gravity.

(a) Effect of altitude: g’ = \(\frac{g \mathrm{R}^2}{(\mathrm{R}+h)^2}\)
h = height above Earth surface.

when h is comparable with R and h < < R.
g’ = g\(\left(1-\frac{2 h}{\mathrm{R}}\right)\)
From these relations, we conclude that acceleration due to gravity decreases with increase in height from the surface of Earth.
(i) Fractional decrease in the value of g with height = \(\frac{g-g^{\prime}}{g}\) = \(\frac{2 h}{\mathrm{R}}\)
(ii) % decrease in the value of g , \(\left(\frac{g-g^{\prime}}{g}\right)\) × 100 = \(\frac{2 h}{R}\) × 100%

(b) Effect of depth :
g’ = g\(\left(1-\frac{d}{R}\right)\), d = depth below Earth surface
(i) The acceleration due to gravity decreases with increase in depth d and becomes zero at the centre of the Earth.

(ii) Decrease in the value of g with depth,
Δg = g – g’
= \(\frac{g d}{\mathrm{R}}\)

∴ Fractional decrease in the value of g with depth
= \(\frac{g-g^{\prime}}{g}\) = \(\frac{d}{R}\)

(iii) % decrease in the value 0f g with depth,
\(\frac{g-g^{\prime}}{g}\) × 100 = \(\frac{d}{R}\) × 100%

Question 40.
Derive an expression for molar heat capacity. [5]
Answer:
Molar specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree. By definition, one mole of any substance is a quantity of the substance whose mass in gram is numerically equal to the molecular mass M.
Thus C = Mc …(i)
n → No. of moles
m → Mass of substance in grams.

Commonly Made Error
Normally students are not able to distinguish between specific heat and molar specific heat.

Answering Tip
hey should learn formula –

C = Mc and C = \(\frac{\Delta \theta}{m(\Delta t)}\)