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**Question 1. Consider the P-V diagram given below for a cyclic process. Find the net heat supplied to the system during the process.**

a. 0.625п J

b. 0.25п J

c. 0.1п J

d. 0.2п J

**Answer: (b)**

It is a cyclic process so the net change in internal energy of the system will zero. i.e.,

ΔU = 0

From first law of thermodynamics,

ΔQ = ΔU + ΔW

Therefore, ΔQ = ΔW

ΔW is the area of the shaded region

Q_{cycle} = W_{cycle }= π(25)(10) Kpa-cc

= π(25)(10) x 10^{3} x 10^{-6}

= 0.250 πJ

**Question 2. A spring of force constant k = 100 N/m is compressed to x = 0.5 m by a block of mass 100 g released. Find the distance d where it falls.**

a. 5 m

b. 10 m

c. 15 m

d. 20m

**Answer: (d)**

Here, we can find the horizontal velocity with which the block will leave the surface. So, using the principle of conservation of energy we have,

0 + (½)kx^{2} = (½)mv^{2 }+0

Therefore,

Given k = 100 N/m

x = 0.5 m

m = 100 g

v = 5√10 m/s

Now, the horizontal distance moved by the block is given by d = vt, where t is the time taken by the block to fall the distance 2 m

Therefore, distance

Substituting (1) in (2)

Therefore,

= 20 m

**Question 3. In the given circuit find the current through 6Ω resistance**

a. 10 A

b. 7 A

c. 25 A

d. 30 A

**Answer: (a)**

Applying Kirchhoff’s law,

3(V-140)+10V+(V-90) =0

3V-420+10V+12V-1080 = 0

25V= 1500

V = 60

Current through 6Ω resistance, I = V/R = 60/6 = 10 A

**Question 4. Four molecules of a diatomic gas are heated from 0 ^{0}C to 50^{0}C. Find the heat supplied to the gas if work done by it is zero.**

a. 780 R

b. 500 R

c. 100 R

d. 650 R

**Answer: (b)**

For a diatomic molecule

C_{v} = 5R/2

C_{p} = 7R/2

Number of moles, n = 4

ΔT = 50

Sone Work done = 0, the process is isochoric

In an isochoric process, Q = ΔU = nC_{v}ΔT

Therefore, Q = (4)(5R/2)(50)

= 500 R

**Question 5. For the spherical interface shown in the figure, the two different media with refractive indices n _{1} = 1.4 and n_{2} = 1.25 are present as shown. The image will be formed at**

a. -125/3

b. -50/6 cm

c. -25/2 cm

d. -20 cm

**Answer: (a) **

Given:

Position of the object (u)= -40 cm

Refractive index of medium 1(n_{1}) = 1.4

Refractive index of medium 2 (n_{2}) = 1.25

Radius of the interface (R) = -30 cm

We know that for the spherical interface,

V = -125/3

**Question 6. When a disc slides on a smooth inclined surface from rest, the time taken to move from A to B is t _{1}. When the disc performs pure rolling from rest then the time taken to move from A to B is t_{2}. If **

**\(\begin{array}{l}\frac{t_{1}}{t_{2}}=\sqrt{\frac{3}{x}}\end{array} \)**

**find x.**

a. 2

b. 1

c. 5

d. 7

**Answer: (a) **

When disc slides a_{1} = gsinθ

So,

When disc do pure rolling

So,

From (1) and (2)

So, x = 2

**Question 7. An AC circuit consists of a series combination of an inductance L 1 mH, a resistance R = 111 and a capacitance C. It is observed that the current leads the voltage by 45°. Find the value of capacitance ‘C’ if the angular frequency of applied AC is 300 rad/s.**

a. 5.6 mF

b. 3.92 mF

c. 2.56 mF

d. 5.2 mF

**Answer: (c) **

**Question 8. An electron is projected into a magnetic field of B = 5 × 10 ^{−3 }T and rotates in a circle of radius of R = 3mm. Find the work done by the force due to the magnetic field.**

a. 0 J

b. 15 mJ

c. 14 mJ

d. 20 mJ

**Answer: (a) **

The work done by the force due to the magnetic field is 0.

**Question 9. A charge Q is divided into q and (Q – q). If Q/q = x, such that the repulsion between them is maximum, find x.**

a. 1

b. 2

c. 3

d. 4

**Answer: (b) 2**

As we know, F = (k(Q−q)q) /d^{2}

For F to be maximum,

dE/dq = 0

⇒ Q − 2q = 0

⇒ Q/q = 2

⇒ x = 2

**Question 10. Bird is flying in north-east direction with or v = 4√2 m/s with respect to the wind and the wind blowing from north to south with speed 1 m/s. Find the magnitude of the displacement of the bird in 3 sec.**

a. 5 m

b. 15 m

c. 10 m

d. 20 m

**Answer: (b)**

**Question 11. Deuteron and alpha particles having the same KE in a magnetic field. If the ratio of the radius of Deuteron and alpha particle is x√2. Then x =?**

a. 5

b. 8

c. 3

d. 1

**Answer: (d)**

KE = B^{2}q^{2}r^{2}/2m

As B and KE are the same

r^{2} ∝ m/q^{2} ⇒ r =√m/q

**Question 12. In the circuit shown, find the current through the Zener diode.**

a. 5 mA

b. 1 mA

c. 15 mA

d. 25 mA

**Answer: (d)**

Since 2000 Ω is parallel to the Zener diode

So, the current passing through it as shown in the circuit,

i_{3} =50/2000 = 25 mA

Potential difference across 1000 Ω,

𝑉_{1} = 100 − 50 = 50 𝑉

So, the electric current passing through it,

i_{1} = 50/1000 = 50 mA

So, current through Zener diode,

i_{2} = 50 − 25 = 25 mA

**Question 13. If **

**\(\begin{array}{l}\vec{A}.\vec{B}=\vec{A}\times \vec{B}\end{array} \)**

**find\(\begin{array}{l}\left | \vec{A}-\vec{B} \right |\end{array} \)**

a. A – B

b.

c. A + B

d.

**Answer: (d)**

Since,

Angle between the vectors, θ = 45^{0}

Hence,

**Question 14. An object moves from the earth’s surface to the surface of the moon. The acceleration due to gravity on the earth’s surface is 10 m/s ^{2}. Considering the acceleration due to gravity on the moon to be 1/6th times that of earth. If R be the earth’s radius and its weight be W and the distance between the earth and the moon is D. The correct variation of the weight W’ versus distance 𝑑 for a body when it moves from the earth to the moon is**

**Answer: (c)**

At the earth’s surface, the weight of the body is, W = mg

At the moon, the distance of the body from the earth’s is d = D

At the moon’s surface, the value of acceleration due to gravity is g’ = g/6

⇒ W’ = W/6

From the relation of acceleration due to gravity at height d above earth’s surface,

For d =0, ⇒g’ = g or W’ = W

At d = R,

⇒ g’ = gR^{2}/4R^{2} = g/4

Or, W’ = mg/4 = W/4

At distance d = D, when the body reaches the surface of the moon

W’ = mg/6 = W/6

Since Eq.(i) suggests that variation of g with distance(d) is non-linear, hence the graph of

𝑊′ vs 𝑑 will be non-linear as well. Also, d ↑, W’ ↓

Thus, the correct variation is represented by the following graph

**Question 15. For an element decaying through simultaneous reaction, the half-life for the respective decaying path is 1400 𝑠 and 700 𝑠. Find the time taken when the number of atoms becomes N _{0}/3 in the element sample. (N_{0} is the initial number of atoms in sample)**

a. (1400/5) In 3

b. (1400/3) In 3

c. (1400/3)ln 2

d. (700/3)ln 2

**Answer: (b)**

N = N_{0}e^{t/τ}

N_{0}/3 = N_{0}e^{t/τ}

Taking natural log on both sides,

Therefore, t = (1400/3) ln3

**Question 16: Consider a body of 800 kg moving with a maximum speed v on a road banked at θ= 30 ^{0} given cos 30^{0} = 0.87. Find the normal reaction on the body. Coefficient of friction μ_{s} = 0.2. [Take radius, r = 10 m]**

a. 10. 4 kN

b. 12.6 kN

c. 11.6 kN

d. 8.3 kN

**Answer: (a)**

Since the circular motion is such that v = v_{max} the tendency of the body is to move up the

inclined plane

[f_{s}]

_{max }= μsN = 0.2 N

Resolving along X and Y axis, we have N cos 30^{0} = mg + [f_{s}] max sin 30^{0}

⇒ N[0.87] = 8000 + 0.2𝑁 [1/2]

N[0.87 − 0.1] = 8000

N =8000/0.77 ≈10,400 newton

**Question 17: A spring with natural length l _{0} has a tension T_{1} when its length is l_{1} and the tension is T_{2} when its length is l_{2}. The natural length of spring will be:**

a.

b.

c.

d.

**Answer: (b)**

Let the natural length be L_{0}

Using hook’s law, Y= TL/AdL, where dL =L – L_{0}

Case 1: When tension is T_{1} length of wire =L_{1}

L_{1} – L_{0 }= T_{1}L_{0}/AY ——-(1)

Case 2: Tension is T_{2} and length of wire = L_{2}

L_{2} – L_{0} = T_{2}L_{0}/AY ——-(1)

Dividing both equations:

**Question 18: A conducting rod of length l is moving perpendicular to the magnetic field. The rod moves from 0 to 2b while the field exists only from 0 to b. Find the graph for emf and power dissipated w.r.t x.**

**Answer: **

From the given figure it is clear that the field exists from 0 to b only, therefore, the given conductor will experience a field only from 0 to b. Here the given conductor is moving in a

uniform magnetic field as long as field exists a constant emf (ε) will be induced in the conductor.

Induced emf in the conductor (ε) = Blv

Due to this emf current developed in the conductor as i = e.m.f/R = Blv/R

Power dissipation exists as long as emf exists in the conductor, p = i^{2}R= (Blv)^{2}/R

Hence, the graph of ε vs x and p vs x is as follows,

**Question 19: A travelling wave is found to have the displacement by y = 1/(1+x) ^{2 }at t = 0, after 3 sec the wave pulse is represented by equation y = 1/1+(1+x)^{2}. The velocity of the wave is:**

a. 1 m/s

b. ⅓ m/s

c. ⅔ m/s

d. ¼ m/s

**Answer: (b)**

Displacement of wave, Δx = 1 m

⇒ v x t = 1

t = 1/v

3 =1/v

v = ⅓ m/s

**Question 20: A ball of charge to mass ratio of 8 μC/g is placed at a distance of 10 cm from the ball. An electric field of 100 N/m is switched on in the direction of the wall. Find the time period of its oscillations. Assume all collisions are elastic.**

a. 1 sec

b. 2 sec

c. 3 sec

d. 4 sec

**Answer: (a)**

As the electric field is switched ON, the ball first strikes the wall and returns back.

One oscillation

Thus,

S = ut +(½)at_{1}^{2}

0.1 = (½) × 0.8t_{1}^{2}

t_{1} = (½) sec

Thus, time period,

T = 2 ×(½)

= 1 sec

**Question 21: The wavenumber of the spectral line in the emission spectrum of hydrogen will be equal to 8/9 times of the Rydberg’s constant. Then the electron jumps from**

a. 5 → 2

b. 5 → 3

c. 3 → 1

d. 4 → 2

**Answer: (c)**

If n_{L} = 1, n_{H} = 3

**Question 22: A vehicle moving with velocity v and releasing the sound of frequency 400 Hz. Listening to the reflected sound from a wall of frequency 500 Hz. Find the velocity of vehicle v**

a. 36.67 m/s

b. 30.12 m/s

c. 22.37 m/s

d. 20.25 m/s

**Answer: (a)**

Frequency received by wall

The reflected frequency received by man is

⇒

⇒ v = 330/9 = 36.67 m/s

**Question 23: A particle of mass moving with a speed v collide elastically with the end of a uniform rod of mass M and length L perpendicularly as shown in the figure. If the particle comes to rest after collision, find the value of m/M.**

a. 1/3

b. 1/2

c. 1/4

d. 1/5

**Answer: (c)**

Applying conservation of angular momentum about the Centre of mass of rod

Applying linear momentum conservation

mv = Mv_{1}……. (ii)

1 = [v_{1}+ ω (L/ 2)] / v ……. (iii)

Putting v_{1} from (ii) and ωL from (i) in (iii)

⇒ 1 = 4m/M

⇒ m/M = 1/4

**Question 24: Four planks are arranged in a lift going upwards with an acceleration of 0.2 m/s ^{2 }as shown in the figure. Find the normal reaction applied by the lift on 10 kg block: (g = 9.8 𝑚/𝑠^{2})**

a. 500

b. 700

c. 672

d. 800

**Answer: (b)**

N – 70g = 70 x 0.2

N = 70 (g + 0.2)

N = 700

**Question 25: A body of mass m emits a photon of frequency 𝑣, then loss in its internal energy?**

a.

b.

c.

d. Zero

**Answer: (c)**

Loss of energy =

**Question 26: In a magnesium rod of area 3m ^{2}, current I = 5𝐴 is flowing angle of 60^{0} from the axis of the rod. The resistivity of the material is 44 × 10^{−2} ohm x m. Find an electric field inside the rod**

a. 0.567

b. 0.367

c. 0.667

d. 0.767

**Answer: (b)**

E = 0.367

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