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July 20 Shift 1 – Physics Question Paper and Solutions

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

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

Qcycle = Wcycle = π(25)(10) Kpa-cc

= π(25)(10) x 103 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

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)kx2 = (½)mv2 +0

Therefore,

$$\begin{array}{l}v=\sqrt{\frac{k}{m}}x\end{array}$$
—-(1)

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

$$\begin{array}{l}d=v\sqrt{\frac{2H}{g}}\end{array}$$
—-(2)

Substituting (1) in (2)

$$\begin{array}{l}d=5\sqrt{10}\sqrt{\frac{2H}{g}}\end{array}$$

Therefore,

$$\begin{array}{l}d=5\sqrt{10} \sqrt{\frac{2(2)}{10}}\end{array}$$

= 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

Applying Kirchhoff’s law,

$$\begin{array}{l}\frac{V-140}{20}+\frac{V-0}{6}+\frac{V-90}{5}=0\end{array}$$

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 00C to 500C. 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

For a diatomic molecule

Cv = 5R/2

Cp = 7R/2

Number of moles, n = 4

ΔT = 50

Sone Work done = 0, the process is isochoric

In an isochoric process, Q = ΔU = nCvΔT

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 n1 = 1.4 and n2 = 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

Given:

Position of the object (u)= -40 cm

Refractive index of medium 1(n1) = 1.4

Refractive index of medium 2 (n2) = 1.25

Radius of the interface (R) = -30 cm

We know that for the spherical interface,

$$\begin{array}{l}\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R}\end{array}$$
$$\begin{array}{l}\frac{1.4}{v}-\frac{1.25}{-40}=\frac{1.4-1.25}{-30}\end{array}$$

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 t1. When the disc performs pure rolling from rest then the time taken to move from A to B is t2. 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

When disc slides a1 = gsinθ

So,

$$\begin{array}{l}S = ut_{1}+\frac{1}{2}a_{1}t_{1}^{2}=\frac{1}{2}gsin\theta t_{1}^{2}\end{array}$$
…………(1)

When disc do pure rolling

$$\begin{array}{l}a_{2}=\frac{gsin\theta }{1+\frac{k^{2}}{R^{2}}}=\frac{gsin\theta }{1+1/2}=\frac{2}{3}gsin\theta\end{array}$$

So,

$$\begin{array}{l}S= ut_{2}+\frac{1}{2}a_{2}t_{2}^{2}=\frac{1}{2}\times \frac{2}{3}gsin\theta t_{2}^{2}\end{array}$$
——-(2)

From (1) and (2)

$$\begin{array}{l}\frac{t_{2}}{t_{1}}=\sqrt{\frac{3}{2}}\end{array}$$

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

$$\begin{array}{l}tan \Phi =\frac{X_{c}-X_{L}}{R}=\frac{\frac{1}{\omega C}-\omega L}{R}\end{array}$$
$$\begin{array}{l}1(1)=\frac{1}{300C}-300(1 x 10^{-3})\end{array}$$
$$\begin{array}{l}\frac{1}{300C}= 1+0.3 = 1.3\end{array}$$
$$\begin{array}{l}C = \frac{1}{300\times 1.3}=0.00256 F = 2.56 mF\end{array}$$

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

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

As we know, F = (k(Q−q)q) /d2

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  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

KE = B2q2r2/2m

As B and KE are the same

r2 ∝ m/q2 ⇒ r =√m/q

$$\begin{array}{l}\frac{r_{D}}{r_{al}}=\sqrt{\frac{m_{D}}{q_{al}}}\times \frac{q_{al}}{q_{D}}\end{array}$$
$$\begin{array}{l}\frac{r_{D}}{r_{al}}=\sqrt{\frac{2}{4}}\times \frac{2}{1}=\sqrt{2}\end{array}$$
$$\begin{array}{l}\Rightarrow \sqrt{2x}=\sqrt{2}\Rightarrow x =1\end{array}$$

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

Since 2000 Ω is parallel to the Zener diode

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

i3 =50/2000 = 25 mA

Potential difference across 1000 Ω,

𝑉1 = 100 − 50 = 50 𝑉

So, the electric current passing through it,

i1 = 50/1000 = 50 mA

So, current through Zener diode,

i2 = 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.

$$\begin{array}{l}\sqrt{A^{2} + B^{2}-\sqrt{2AB}}\end{array}$$

c. A + B
d.
$$\begin{array}{l}\sqrt{A^{2} + B^{2}-\sqrt{2}AB}\end{array}$$

Since,

$$\begin{array}{l}\vec{A}.\vec{B}=\vec{A}\times \vec{B}\end{array}$$
$$\begin{array}{l}\Rightarrow \left | \vec{A} \right |\left | \vec{B} \right |cos\theta \Rightarrow \left | \vec{A} \right |\left | \vec{B} \right |sin\theta\end{array}$$

Angle between the vectors, θ = 450

Hence,

$$\begin{array}{l}\left | \vec{A}-\vec{B} \right |=\sqrt{A^{2}+B^{2}-2ABcos\theta } = \sqrt{A^{2} + B^{2}-\sqrt{2}AB}\end{array}$$

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/s2. 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 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,

$$\begin{array}{l}g’=\frac{gR^{2}}{(R+d)^{2}}—–(i)\end{array}$$

For d =0, ⇒g’ = g or W’ = W

At d = R,

⇒ g’ = gR2/4R2 = 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 N0/3 in the element sample. (N0 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

N = N0et/τ

N0/3 = N0et/τ

Taking natural log on both sides,

$$\begin{array}{l}ln(\frac{1}{3})=(\frac{-t}{\tau })lne\end{array}$$
$$\begin{array}{l}-ln3=(\frac{-t}{\frac{1400}{3} })\times 1\end{array}$$

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= 300 given cos 300 = 0.87. Find the normal reaction on the body. Coefficient of friction μs = 0.2. [Take radius, r = 10 m] a. 10. 4 kN
b. 12.6 kN
c. 11.6 kN
d. 8.3 kN

Since the circular motion is such that v = vmax the tendency of the body is to move up the

inclined plane

[fs]max = μsN = 0.2 N

Resolving along X and Y axis, we have N cos 300 = mg + [fs] max sin 300

⇒ 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 l0 has a tension T1 when its length is l1 and the tension is T2 when its length is l2. The natural length of spring will be:

a.

$$\begin{array}{l}\frac{T_{1}l_{2}-T_{2}l_{1}}{l_{1}-l_{2}}\end{array}$$

b.
$$\begin{array}{l}\frac{T_{2}l_{1}-T_{1}l_{2}}{T_{2}-T_{1}}\end{array}$$

c.
$$\begin{array}{l}\frac{T_{2}l_{2}-T_{1}l_{1}}{T_{1}-T_{2}}\end{array}$$

d.
$$\begin{array}{l}\frac{T_{2}l_{1}-T_{1}l_{2}}{T_{2}+T_{1}}\end{array}$$ Let the natural length be L0

Using hook’s law, Y= TL/AdL, where dL =L – L0

Case 1: When tension is T1 length of wire =L1

L1 – L0 = T1L0/AY ——-(1)

Case 2: Tension is T2 and length of wire = L2

L2 – L0 = T2L0/AY ——-(1)

Dividing both equations:

$$\begin{array}{l}\frac{T_{2}l_{1}-T_{1}l_{2}}{T_{2}-T_{1}}\end{array}$$

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. 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 = i2R= (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

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

$$\begin{array}{l}a =\frac{qE}{m}=\frac{8\times 10^{-6}}{10^{-3}}\times 100= 0.8m/s^{2}\end{array}$$

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

One oscillation

Thus,

S = ut +(½)at12

0.1 = (½) × 0.8t12

t1 = (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

$$\begin{array}{l}\bar{v}=Rz^{2}\left ( \frac{1}{n_{L}^{2}}-\frac{1}{n_{H}^{2}} \right )\end{array}$$

If nL = 1, nH = 3

$$\begin{array}{l}\bar{v}=R1^{2}\left ( \frac{1}{1^{2}}-\frac{1}{3^{2}} \right )=\frac{8}{9}R\end{array}$$

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

$$\begin{array}{l}f’ =\left (\frac{v_{s}}{v_{s}-v} \right )f_{0}\end{array}$$

The reflected frequency received by man is

$$\begin{array}{l}f” =\left (\frac{v_{s}+v}{v_{s}} \right )\left ( \frac{v_{s}}{v_{s}-v} \right )f_{0}\end{array}$$
$$\begin{array}{l}f” =\left (\frac{v_{s}+v}{v_{s}-v} \right )f_{0}\end{array}$$

$$\begin{array}{l}500 =\left (\frac{330+v}{330-v} \right )400\end{array}$$

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

$$\begin{array}{l}mv\left (\frac{L}{2} \right)=m\left ( \frac{L^{2}}{12} \right )\omega\end{array}$$
——–(i)

Applying linear momentum conservation

mv = Mv1……. (ii)

1 = [v1+ ω (L/ 2)] / v ……. (iii)

Putting v1 from (ii) and ωL from (i) in (iii)

$$\begin{array}{l}v =\frac{m}{M}v+\frac{6mv}{2M}\end{array}$$

⇒ 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/s2 as shown in the figure. Find the normal reaction applied by the lift on 10 kg block: (g = 9.8 𝑚/𝑠2) a. 500
b. 700
c. 672
d. 800 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.

$$\begin{array}{l}h\nu\end{array}$$

b.
$$\begin{array}{l}h\nu \left (1- \frac{h\nu }{2mc^{2}}\right )\end{array}$$

c.
$$\begin{array}{l}h\nu \left (1+ \frac{h\nu }{2mc^{2}}\right )\end{array}$$

d. Zero $$\begin{array}{l}h\nu = \frac{c}{\lambda }\end{array}$$
$$\begin{array}{l}mv = \frac{h}{\lambda }=\frac{h\nu }{c}\end{array}$$

Loss of energy =

$$\begin{array}{l}\frac{1}{2}mv^{2}+h\nu\end{array}$$
$$\begin{array}{l}= \frac{1}{2m}\left ( \frac{h\nu }{c} \right )^{2}+h\nu\end{array}$$
$$\begin{array}{l}h\nu \left (1+ \frac{h\nu }{2mc^{2}}\right )\end{array}$$

Question 26: In a magnesium rod of area 3m2, current I = 5𝐴 is flowing angle of 600 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

$$\begin{array}{l}\frac{I}{A_{effective}}=\frac{E}{\rho }\end{array}$$
$$\begin{array}{l}E = \frac{\rho I}{A}cos 60^{0}=\frac{44\times 10^{-2}\times 5}{3\times 2}\end{array}$$