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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.
Solution:
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.
Solution:
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
Solution:
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.
Solution:
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
Solution:
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
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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.
Solution:
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 =?
Solution:
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.
Solution:
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
Solution:
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
Solution:
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)
Solution:
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]
Solution:
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:
Solution:
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.
Solution:
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:
Solution:
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.
Solution:
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
Solution:
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
Solution:
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.
Solution:
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})
Solution:
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?
Solution:
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
Solution:
Answer: (b)
E = 0.367