Final JEE-Main Exam June, 2022/28-06-2022/Evening Session
FINAL JEE-MAIN EXAMINATION – JUNE, 2022
(Held On Tuesday 28th June, 2022) TIME : 3 : 00 PM to 6 : 00 PM

PHYSICS
SECTION-A
TEST PAPER WITH SOLUTION

1. Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as v? = (n/m²) v? and a? = a?/(mn) respectively. Here m and n are constants. The relations for distance and time in two systems respectively are:
(A) (n³/m³) L? = L? and (n²/m) T? = T?
(B) L? = (n?/m²) L? and T? = (n²/m) T?
(C) L? = (n²/m) L? and T? = (n?/m²) T?
(D) (n²/m) L? = L? and (n?/m²) T? = T?
Official Ans. by NTA (A)
Sol. L?/T? = (n/m²) L?/T?
L?/T?² = L?/(T?² × mn)
(n/m²) × T?/T? = T?/(T? × mn)
n²/m = T?/T?
L?/L? = (n?/m²) × 1/(mn)
L?/L? = n³/m³

2. A ball is spun with angular acceleration α = 6t² - 2t where t is in second and α is in rads?². At t = 0, the ball has angular velocity of 10 rads?¹ and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:
(A) (3/2)t? - t² + 10t
(B) t?/2 - t³/3 + 10t + 4
(C) (2t?/3) - t³/6 + 10t + 12
(D) 2t? - t³/2 + 5t + 4
Official Ans. by NTA (B)
Sol. dω/dt = 6t² - 2t
∫??^ω dω = 2t³ - t²
ω = 10 + 2t³ - t²
dθ/dt = 10 + 2t³ - t²
∫?^θ dθ = 10 + 2t³ - t²
∫?^θ dθ = 10t + t?/2 - t³/3
θ = 4 + 10t + t?/2 - t³/3

3. A block of mass 2 kg moving on a horizontal surface with speed of 4 ms?¹ enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12 Nm?¹. The speed of the block as it just crosses the rough surface will be:
(A) Zero
(B) 1.5 ms?¹
(C) 2.0 ms?¹
(D) 2.5 ms?¹
Official Ans. by NTA (C)
Sol. a = -kx/2 = -12x/2 = -6x
vdv/dx = -6x
∫vdv = -∫6xdx
(v² - 4²)/2 = -(6[(3/2)² - (1/2)²])/2
v² - 16 = -6(9/4 - 1/4)
v² = 16 - 6 × 2 = 4
V = 2 m/s

4. A √34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If F_f and F_w are the reaction forces of the floor and the wall, then ratio of F_w/F_f will be: (Use g = 10 m/s²)
(A) 6/√110
(B) 3/√113
(C) 3/√109
(D) 2/√109
Official Ans. by NTA (C)
Sol. f = N?
N? = mg
N? × ? sinθ = mg ?/2 cotθ
N? = mg/2 cotθ
F_w/F_f = (mg/2 cotθ)/√((mg)² + (mg/2 cotθ)²)
= 1/√(1 + 4/cot²θ)
= 3/√109

5. Water fall from a 40 m high dam at the rate of 9 × 10? kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is: (Take g = 10 ms?²)
(A) 25
(B) 50
(C) 100
(D) 18
Official Ans. by NTA (B)
Sol. (9 × 10? × g × 40)/3600 × 0.5 = n × 100
(10? × 0.5)/100 = n
100 × 0.5 = n
n = 50

6. Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be:
(A) 2/9 F
(B) 16/9 F
(C) 8/9 F
(D) F
Official Ans. by NTA (C)
Sol. F = Gm²/r²
F' = G(4m/3) × (2m/3)/r²
F' = 8/9 F

7. A water drop of radius 1 μm falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is 1.8 × 10?? Nsm?² and its density is negligible as compared to that of water 10? gm?³. Terminal velocity of the water drop is: (Take acceleration due to gravity = 10 ms?²)
(A) 145.4 × 10?? ms?¹
(B) 118.0 × 10?? ms?¹
(C) 132.6 × 10?? ms?¹
(D) 123.4 × 10?? ms?¹
Official Ans. by NTA (D)
Sol. F_v = 6πmrv_t
mg = 4/3 πr³pg
6πmrv_t = 4/3 πr³pg
v_t = (4/3) × (πr³pg)/(6πmr)
v_t = (4/3) × (πr³pg)/(6πmr) = (2 × 10?¹² × 10³ × 10)/(9 × 1.8 × 10??)
= 123.4 × 10?? m/s

8. A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work 50 J is done on the gas during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:
(A) 20 J
(B) 30 J
(C) -30 J
(D) -60 J
Official Ans. by NTA (B)
Sol. ΔQ_cycle = 40 - 60 = ΔW
⇒ ΔW = -20 J = W_BC + W_CA
⇒ W_CA = -20 J - W_BC
= -20 - (-50)
= 30 J

9. What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
(A) The velocity of atomic oxygen remains same
(B) The velocity of atomic oxygen doubles
(C) The velocity of atomic oxygen becomes half
(D) The velocity of atomic oxygen becomes four times
Official Ans. by NTA (B)
Sol. V_rms = √(3RT/M)
T → 2T
M → M/2
V_rms ∝ √(T/M)
⇒ (V_rms)_atomic = (V_rms)_molecular × √(2/(1/2)) = 2(V_rms)_molecular

10. Two point charges A and B of magnitude +8 × 10?? C and -8 × 10?? C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 × 10? NC?¹. The distance 'd' between the point charges A and B is:
(A) 2.0 m
(B) 3.0 m
(C) 1.0 m
(D) 4.0 m
Official Ans. by NTA (B)
Sol. E? = 2 × Kq/(d/2)²
⇒ E? = 8 Kq/d²
⇒ d² = (8 × 9 × 10? × 8 × 10??)/(6.4 × 10?)
d = 3 m

11. Resistance of the wire is measured as 2Ω and 3Ω at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is:
(A) 0.033°C?¹
(B) -0.033°C?¹
(C) 0.011°C?¹
(D) 0.055°C?¹
Official Ans. by NTA (A)
Sol. R = R?(1 + αΔT)
3 = R?(1 + α(30 - 0))
2 = R?(1 + α(10 - 0))
3/2 = (1 + 30α)/(1 + 10α)
α = 1/30 = 0.033

12. The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 × 10??. What is fractional increase in the magnetic field inside solenoid with respect to air as medium inside the solenoid?
(A) 1.2 × 10??
(B) 1.2 × 10?³
(C) 1.8 × 10?³
(D) 2.4 × 10??
Official Ans. by NTA (A)
Sol. χ = 1.2 × 10??
μ_r = 1 + χ = 1 + 1.2 × 10??
Fractional Change
= ΔB/B = (μ?μ_r ni - μ?ni)/μ?ni = (μ_r - 1)
= 1.2 × 10??

13. Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 × 10?? N, then the value of x is approximately:
(A) 1
(B) 2.4
(C) 1.4
(D) 2
Official Ans. by NTA (C)
Sol. Force per unit length = μ?i?i?/(2πd)
= (μ? · x²)/(2π × 0.2)
F = 2 × 10?? = (4π × 10?? × x²)/(2π × 0.2)
⇒ 10?? = 10?? x²/0.2
⇒ x² = 10 × 0.2
⇒ 2

14. A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short circuited.)
(A) Halved
(B) Quadrupled
(C) The same
(D) Doubled
Official Ans. by NTA (D)
Sol. P = ε²/R = (NA dB/dt)² / (ρ?/A_C)
P' = (NA dB/dt)² / (ρ?/2)

15. An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm?¹. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:
(A) E_y = 60 sin[π/4 × 10³(x - 3 × 10?t)] j Vm?¹
B_z = 2 sin[π/4 × 10³(x - 3 × 10?t)] k T
(B) E_y = 60 sin[π/4 × 10³(x - 3 × 10?t)] j Vm?¹
B_z = 2 × 10?? sin[π/4 × 10³(x - 3 × 10?t)] k T
(C) E_y = 2 × 10?? sin[π/4 × 10³(x - 3 × 10?t)] j Vm?¹
B_z = 60 sin[π/4 × 10³(x - 3 × 10?t)] k T
(D) E_y = 2 × 10?? sin[π/4 × 10?(x - 4 × 10?t)] j Vm?¹
B_z = 60 sin[π/4 × 10?(x - 4 × 10?t)] k T
Official Ans. by NTA (B)
Sol. B? = E?/c = 60/(3 × 10?) = 2 × 10?? T
E × B must be direction of propagation.
So, B → z-axis
k = 2π/λ = π/4 × 10³ m?¹
E_y = 60 sin[π/4 × 10³(x - 3 × 10?t)] j Vm?¹
B_z = 2 × 10?? sin[π/4 × 10³(x - 3 × 10?t)] k T

16. In young's double slit experiment performed using a monochromatic light of wavelength λ, when a glass plate (μ = 1.5) of thickness xλ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:
(A) 3
(B) 2
(C) 1.5
(D) 0.5
Official Ans. by NTA (B)
Sol. Path difference at O = (μ - 1)t
If the intensity at O remains (maximum) unchanged, path difference must be nλ
⇒ (μ - 1)t = nλ
(1.5 - 1)xλ = nλ
⇒ x = 2n
For n = 1, x = 2

17. Let K? and K? be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ? and λ?, respectively are incident on a metallic surface. If λ? = 3λ? then:
(A) K? > K?/3
(B) K? < K?/3
(C) K? = K?/3
(D) K? = K?/3
Official Ans. by NTA (B)
Sol. hc/λ? - φ = K?
hc/λ? - φ = K?
λ? = 3λ?
3K? = 3hc/λ? - 3φ
3K? = hc/λ? - 3φ
3K? = K? - 2φ
3K? < K?
K? < K?/3

18. Following statements related to radioactivity are given below:
(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.
(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.
(C) Slope of the graph of log_e(no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time (τ).
(D) Product of decay constant (λ) and half-life time (T?/?) is not constant.
Choose the most appropriate answer from the options given below:
(A) (A) and (B) only
(B) (B) and (D) only
(C) (B) and (C) only
(D) (C) and (D) only
Official Ans. by NTA (C)

19. In the given circuit the input voltage V_in is shown in figure. The cut-in voltage of p-n junction diode D? or D? is 0.6 V. Which of the following output voltage (V?) waveform across the diode is correct?
Official Ans. by NTA (D)
Sol. In +ve half cycle D? → FB; D? → RB. 0 - 0.6 V V_out same as V_in In -ve half cycle D? → FB; D? → RB.

20. Amplitude modulated wave is represented by V_AM = 10[1 + 0.4 cos(2π × 10?t)] cos(2π × 10?t). The total bandwidth of the amplitude modulated wave is:
(A) 10 kHz
(B) 20 MHz
(C) 20 kHz
(D) 10 MHz
Official Ans. by NTA (C)
Sol. Bandwidth = 2f_m
= 2 × 10? Hz = 20 × 10³ Hz
= 20 kHz

SECTION-B

1. A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is x/121 %. The value of x is
Official Ans. by NTA (150)
Sol. X = 1.22 mm + 1.23 mm + 1.19 mm + 1.20 mm
X = 1.21 mm
Δx = (0.01 + 0.02 + 0.02 + 0.01)/4 = 0.06/4 = 0.015
Percentage error = 0.015/1.21 × 100
X = 150

2. A Zener of breakdown voltage V_Z = 8 V and maximum zener current, I_ZM = 10 mA is subjected to an input voltage V_i = 10 V with series resistance R = 100 Ω. In the given circuit R_L represents the variable load resistance. The ratio of maximum and minimum value of R_L is
Official Ans. by NTA (2)
Sol. V_i = 10 V = 2/100 mA
I = 2/100 = 20 mA
V_L = I_L R_L
8 = 10 × 10?³ × R_Lmax
(4/5) × 10³ = R_Lmax
800 = R_Lmax
R_Lmin = 400
800/400 = 2

3. In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1/α. The value of α is
Official Ans. by NTA (4)
Sol. β = (0.35 × 5)/7 = 0.25
1/α = 25/100
α = 4

4. In the given circuit, the magnitude of V_L and V_C are twice that of V_R. Given that f = 50 Hz, the inductance of the coil is 1/(Kπ) mH. The value of K is
Official Ans. by NTA (0)
Sol. V_L = V_C = 2V_R
X_L = X_C = 2R
X_L = 10 Ω
ωL = 10
2πfL = 10
L = 10/(2πf) = 1/(10π) H = 1000/(10π) mH
L = 1/(1/100 π); K = 1/100 = 0.01 ≈ 0

5. All resistances in figure are 1 Ω each. The value of current 'I' is a/5 A. The value of a is
Official Ans. by NTA (8)
Sol. R_eq = 15R/8 = 15/8 Ω
I = 3/(15/8) = 8/5 A
∴ a = 8

6. A capacitor C? of capacitance 5 μF is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C? of capacitance 10 μF as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C? is μC.
Official Ans. by NTA (100)
Sol. Before closing the switch
Q = C?V? = 5 × 30 = 150 μC
After closing the switch
V = Q/(C? + C?) = 150/(10 + 5) = 10 V
Q? = C?V = 10 × 10 = 100 μC

7. A tuning fork of frequency 340 Hz resonates in the fundamental mode with an air column of length 125 cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is cm. (Velocity of sound in air is 340 ms?¹)
Official Ans. by NTA (50)
Sol. Assumption : Ignore word "fundamental mode" in question.
λ = V/f = 340/340 = 1 m
First resonating length = λ/4 = 25 cm
Second resonating length = 3λ/4 = 75 cm
Third resonating length = 5λ/4 = 125 cm
Height of water required = 125 - 75 = 50 cm

8. A liquid of density 750 kgm?³ flows smoothly through a horizontal pipe that tapers in cross-sectional area from A? = 1.2 × 10?² m² to A? = A?/2. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is × 10?³ m³s?¹.
Official Ans. by NTA (24)
Sol. A? = A?/2
P? - P? = 4500 Pa
P? + 1/2 ρV?² + ρgh = P? + 1/2 ρV?² + ρgh
P? - P? = 1/2 ρ(V?² - V?²) ... (1)
And A?V? = A?V?
⇒ V? = 2V? ... (2)
4500 = 1/2 × 750 × 3V?²
V? = 2 m/s
Volume flow rate = A?V? = 24 × 10?³ m³s?¹

9. A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is N. (Take g = 10 ms?²)
Official Ans. by NTA (10)
Sol. 2g - T = 2a ...(1)
TR = MR²/2 α ...(2)
α = a/R ...(3)
T = 2a
2g - T = 2a
T = g = 10 N

10. A car covers AB distance with first one-third at velocity v? ms?¹, second one-third at v? ms?¹ and last one-third at v? ms?¹. If v? = 3v?, v? = 2v? and v? = 11 ms?¹ then the average velocity of the car is ms?¹.
Official Ans. by NTA (18)
Sol. <v?> = Displacement/time
(Let displacement be l)
= l / (l/V? + l/V? + l/V?) (1/3)
= 3 / (1/V? + 1/V? + 1/V?) = 3 / (1/11 + 1/22 + 1/33)
= 18 m/s

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