JEE Main 2020 Physics Paper With Solutions Shift 2 Sept 5

Question 1: A ring is hung on a nail. It can oscillate, without slipping or sliding

(i) in its plane with a time period T₁ and,

(ii) back and forth in a direction perpendicular to its plane, with a period T₂.

The ratio T₁ / T₂ will be:

1) 3 / √2
2) √2 / 3
3) 2 / √3
4) 2 / 3

Answer: (3)

Question 2: The correct match between the entries in column I and column II are:

1) (a) – (ii), (b)-(i), (c)-(iv), (d)-(iii)
2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)

Answer: (3)

By theory

Question 3: In an experiment to verify Stokes law, a small spherical ball of radius r and density falls under gravity through a distance h in the air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface, then the value of h is proportional to (ignore viscosity of air)

1) r⁴
2) r
3) r³
4) r²

Answer: (1)

V_T = √2gh

r² ∝ √h ⇒ r⁴ ∝ h

h ∝ r⁴

Question 4: Ten charges are placed on the circumference of a circle of radius R with constant angular separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E at the centre of the circle are respectively: (Take V= 0 at infinity)

1) V = 0; E = 0
2) V = 10q / [4πε₀R]; E = 10q / [4πε₀R²]
3) V = 0; E = 10q / [4πε₀R²]
4) V = 10q / [4πε₀R]; E = 0

Answer: (1)

v_net = 5 (kq / R) + (5k [-q]) / R

v_net = 0 [Q_net = 0]

E_net = 0 by symmetry

Question 5: A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv² (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is:

1) –bv³(t)
2) –bv³ / M(t)
3) –2bv³ / M(t)
4) –bv³ / 2M(t)

Answer: (2)

in free space

no external force

so there in only thrust force on rocket

f_in = [dM / dt] (V_rel)

Ma = [– bv² / (t)] v

a = –bv³ / M(t)

Question 6: Two different wires having lengths L₁ and L₂, and respective temperature coefficient of linear expansion ɑ₁ and ɑ₂, are joined end-to-end. Then the effective temperature coefficient of linear expansion is:

1) [ɑ₁L₁ + ɑ₂L₂] / [L₁ + L₂]
2) 2 √ ɑ₁ɑ₂
3) 4 [(ɑ₁ɑ₂) / [ɑ₁ + ɑ₂])] [(L₂L₁ / (L₂ + L₁)²]
4) [ɑ₁ + ɑ₂] / 2

Answer: (1)

Question 7: In the circuit, given in the figure currents in different branches and the value of one resistor are shown. Then potential at point B with respect to the point A is:

1) +2 V
2) –2 V
3) +1 V
4) –1 V

Answer: (3)

Let V_A = 0

Applying KVL from A to B via ACDB,

V_A + 1 + 2 – 2 = V_B

V_B = 1V

Question 8: The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:

1) 37 / 3 m
2) 49 / 4 m
3) 12 m
4) 11 m

Answer: (1)

Distance = area under graph

= (1 / 2) (4) [(13 / 3) + 1] + [(1 / 2) (6 – [13 / 3]) × 2]

= 2 × (16 / 3) + [5 / 3]

= (32 / 3) + (5 / 3) = [37 / 3] m

Question 9: An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown in the figure. The connector has length l and resistance R. It slides to the right with a velocity v. The resistance of the conductor and the self-inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is:

1) [μ₀ / 2π] [Ivl /Rr]
2) [μ₀ / π] [Ivl /Rr]
3) [2μ₀ / π] [Ivl /Rr]
4) [μ₀ / 4π] [Ivl /Rr]

Answer: (1)

B = [μ₀I / 2πr]

Induced emf e = Bvl

= [μ₀I / 2πr] * V . l

= [μ₀Ivl / 2πr]

Induced current i = e / R = [μ₀ / 2π] [Ivl /Rr]

Question 10: Two Zener diodes (A and B) having breakdown voltages of 6 V and 4 V respectively, are connected as shown in the circuit below. The output voltage VO variation with input voltage linearly increasing with time is given by (V_input = 0 V at t = 0) (figures are qualitative)

Answer: (4)

t = 0

V_i = 0

V_i ∝ t

∵ Zenerdiode maintains constant breakdown voltage.

Question 11: In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n is:

1) 32
2) 1 / 32
3) 326
4) 128

Answer: (4)

PV^γ = constant

p (⍴^-γ) = constant

p₁(ρ₁^-γ) = p₂ (ρ₂^-γ) γ = 7 / 5 for diatomic

p₀ρ₀^-7/5 = (np₀) (32ρ₀)^-7/5

n = (2⁵)^7/5 = 2⁷ = 128

Question 12: A galvanometer is used in the laboratory for detecting the null point in electrical experiments. If on passing a current of 6 mA it produces a deflection of 2°, its figure of merit is close to:

1) 6 × 10^–3 A/div.
2) 3 × 10^–3 A/div.
3) 666° A/div.
4) 333° A/div.

Answer: (2)

Figure of merit = I / θ ⇒ A/div.

= [6 × 10^-3] / 2

= 3 × 10^–3 A/div.

Question 13: In the circuit shown, charge on the 5μF capacitor is:

1) 5.45 μc
2) 18.00 μc
3) 10.90 μc
4) 16.36 μc

Answer: (4)

(V – 6) × 2 + (V – 0) × 5 + (V – 6) 4 = 0

2V – 12 + 5V + 4V – 24 = 0

11V = 36

V = 36 / 11

q = CV = 5 × (36 / 11) = 16.36 μC

Question 14: A parallel plate capacitor has a plate of length ‘l’, width ‘w’ and separation of plates is ‘d’. It is connected to a battery of emf V. A dielectric slab of the same thickness ‘d’ and of dielectric constant k = 4 is being inserted between the plates of the capacitor. At what length of the slab inside plates, will the energy stored in the capacitor be two times the initial energy stored?

1) 2l / 3
2) l / 2
3) l / 4
4) l / 3

Answer: (4)

area of plate = lw

C = ε₀A / d = ε₀lw / d

U₁ = (1 / 2) cv² = [(1 / 2) ε₀lw / d] v²

C_eq = C₁ + C₂

When x length of the slab is inside the capacitor

C_eq = [ε₀wxk / d] + [ε₀w (l – x) / d]

C_eq = [ε₀w / d] [kx + l – x]

U_f = (1 / 2) C_eq V²

U_f = 2U_i ⇒ (1 / 2) (ε₀w / d) [kx + l – x] v² = 2 * (1 / 2) (ε₀w / d) v²

kx + l – x = 2l

4x – x = l

3x = l

x = l / 3

Question 15: A radioactive nucleus decays by two different processes. The half-life for the first process is 10 s and that for the second is 100 s. The effective half-life of the nucleus is close to:

1) 55 sec.
2) 6 sec.
3) 12 sec.
4) 9 sec.

Answer: (4)

T₁ = 10 sec; λ₁ = [ln 2 / T₁]

T₂ = 100 sec; λ₂ = [ln 2 / T₂], λ_eq = ln 2 / T_eq

λ_eq = λ₁ + λ₂

(ln 2 / T_eq) = (ln 2 / T₁) + (ln 2 / T₂)

(1 / T_eq) = (1 / 10) + (1 / 100) = (10 + 1) / 100 = 11 / 100

T_eq = 100 / 11 = 9 s

Question 16: A driver in a car, approaching a vertical wall, notices that the frequency of his car horn has changed from 440 Hz to 480 Hz when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is:

1) 24 km/hr
2) 36 km/hr
3) 54 km/hr
4) 18 km/hr

Answer: (3)

The car as the source, wall as an observer.

f₁ = {[v – 0] / [v – v_c]} × f₀ —- (i)

For reflected waves, the wall is a source, the driver is an observer.

480 = {[v + v_c] / [v – 0]} × f₁ ⇒ {[v + v_c] / [v] * [v] / [v – v_c]} × f₀

480 = [345 + V_C] * [440 / (345 – V_C)]

12 = {[345 + V_C] / [345 – V_C]} * 11

12 × 345 – 12 × V_C = 345 × 11 + 11 V_C

23V_C = 4200 – 3850 = 345

V_C = (345 / 23) m

V_C = (345 / 23) * (18 / 5) km / h

= [70 * 18] / 23

= 54.78

= 54 km/hr

Question 17: An iron rod of volume 10^–3m³ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be:

1) 0.5 × 10² Am²
2) 50 × 10² Am²
3) 5 × 10² Am²
4) 500 × 10² Am²

Answer: (3)

M = μ_rNiA

Here μ_r

= Relative permeability

N = No. of turns

i = Current

A = Area of cross section

M = μ_rNiA = μ_rnℓiA

M = μ_rniV = 1000(1000) 0.5 (10^-3)

= 500

= 5 × 10² Am²

Question 18: Two coherent sources of sound, S₁ and S₂, produce sound waves of the same wavelength, λ = 1 m, in phase. S₁ and S₂ are placed 1.5 m apart (see fig). A listener, located at L, directly in front of S₂ finds that the intensity is at a minimum when he is 2 m away from S₂. The listener moves away from S₁, keeping his distance from S₂ fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S₁. Then, d is :

1) 12 m
2) 2 m
3) 3 m
4) 5 m

Answer: (3)

For min at (L)

S₁ L – S₂ L = Δx = [λ / 2] (2n + 1);(n = 0, 12)

2.5 – 2 = [1 / 2] (2n + 1)

0.5 × 2 = (2n + 1)

2n = 0

n = 0 (first minima)

The adjacent maxima is the first maxima.

So at ‘p’ → first maxima

⇒ S₁P – S₂P = λ [n = 1] for first maxima

S₁P – 2 = 1

S₁P = 1 + 2

d = 3 m

Question 19: The quantities x = 1 / √μ₀ε₀, y = E / B and z = L / CR are defined where C – capacitance, R – Resistance, L – length, E – Electric field, B – magnetic field and ε₀, μ₀ – free space permittivity and permeability respectively. Then:

1) Only y and z have the same dimension
2) x, y and z have the same dimension
3) Only x and y have the same dimension
4) Only x and z have the same dimension

Answer: (2)

x = 1 / √μ₀ε₀ = (speed)

Z = l / CR = m / sec = m / s [y] = LT^-1

So, x, y, z has the same dimension.

Question 20: The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is : (h < < R, where R is the radius of the earth)

1) R²ω² / g
2) R²ω² / 8g
3) R²ω² / 4g
4) R²ω² / 2g

Answer: (4)

∵ The weight same at poles and at h (so g₁ = g₂)

g₁ = g – Rω²

g₂ = g [1 – (2h / R)]

∵ g₁ = g₂

g – Rω² = g [1 – (2h / R)]

Rω² = 2gh / R

h = R²ω² / 2g

Question 21: Nitrogen gas is at 300° C temperature. The temperature (in K) at which the rms speed of a H₂ molecule would be equal to the rms speed of a nitrogen molecule, is _________. (Molar mass of N₂ gas 28 g).

Answer: 41

V_rms = √(3RT / M)

√(3RT / 2) = √([3R * 573] / 28)

T / 2 = 573 / 28

T = 41 K

Question 22: The surface of a metal is illuminated alternately with photons of energies E₁ = 4 eV and E₂ = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _________.

Answer: 2

(V₁ / V₂)² = [4 – ɸ₀] / [2.5 – ɸ₀]

(2)² = [4 – ɸ₀] / [2.5 – ɸ₀]

10 – 4ɸ₀ = 4 – ɸ₀

3ɸ₀ = 10 – 4 = 6

ɸ₀ = 2eV

Question 23: A prism of angle A = 1° has a refractive index μ = 1.5. A good estimate for the minimum angle of deviation (in degrees) is close to N / 10. Value of N is

Answer: 5

A = 1°

δ = (μ – 1) A

= (1.5 – 1) A

= 0.5 × 1

= 5 / 10 = N / 10 so N = 5

Question 24: A body of mass 2 kg is driven by an engine delivering a constant power of 1 J/s. The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in m) _____________.

Answer: 18

P = Fv = mav

a = p / mv

dv / dt = p / mv

∫₀^u v dv = (p / m) ∫₀^t dt

u² / 2 = (p / m) t

u = √(2p / m) * (√t)

dx / dt = √(2p / m) * (√t)

∫₀^x dx = √(2p / m) ∫₀⁹ √t dt

x = (2 / 3) [(9)^½]³

= (2 / 3) * 27

x = 18

II-method

Pt = w = [1 / 2] mv² – 0

1 × t = [1 / 2] × 2 × u²

u = √t

(dx / dt) = √t = ∫₀¹ dx = ∫₀⁹ √t dt

x = [t^3/2]₀⁹ / (3 / 2) = 18m

Question 25: A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity 80 m/s hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be __________.

Answer: 20

Applying COAM about the Pivot,

L_i = L_f

0.1 × 80 × 1 = {[0.9 × 1²] / 3} × ω + (0.1) 1² ω

8 = (0.3 + 0.1) ω

8 = (0.4) ω

ω = 80 / 4 = 20