JEE Main 2022 June 26 – Shift 2 Physics Question Paper with Solutions

The JEE Main 2022 June 26 – Shift 2 Physics Question Paper with Solutions is available here for students’ reference. The physics curriculum for Classes XI and XII is the basis for the JEE question paper. Candidates can access the question paper in PDF format on this page. Students can also use the PDF file for future reference. These solutions were written by the JEE specialists of BYJU’S. Practising the JEE Main 2022 Question Papers will help candidates raise their JEE scores. Additionally, the applicant can efficiently prepare for the upcoming exam by using this JEE Main 2022 answer key as a guide.

JEE Main 2022 26th June Evening Shift Physics Question Paper and Solutions

SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. The dimension of mutual inductance is :

(A) [ML²T^–2A^–1]

(B) [ML²T^–3A^–1]

(C) [ML²T^–2A^–2]

(D) [ML²T^–3A^–2]

Answer (C)

Sol.

= [ML²T^–2A^–2]

2. In the arrangement shown in figure a₁, a₂, a₃ and a₄ are the accelerations of masses m₁, m₂, m₃ and m₄ respectively. Which of the following relation is true for this arrangement?

(A) 4a₁ + 2a₂ + a₃ + a₄ = 0

(B) a₁ + 4a₂ + 3a₃ + a₄ = 0

(C) a₁ + 4a₂ + 3a₃ + 2a₄ = 0

(D) 2a₁ + 2a₂ + 3a₃ + a₄ = 0

Answer (A)

Sol. From virtual work done method,

4T × a₁ + 2T × a₂ + T × a₃ + T × a₄ = 0

⇒ 4a₁ + 2a₂ + a₃ + a₄ = 0

3. Arrange the four graphs in descending order of total work done; where W₁, W₂, W₃ and W₄ are the work done corresponding to figure a, b, c and d respectively.

(A) W₃ > W₂ > W₁ > W₄

(B) W₃ > W₂ > W₄ > W₁

(C) W₂ > W₃ > W₄ > W₁

(D) W₂ > W₃ > W₁ > W₄

Answer (A)

Sol. W_a = 0, W_b = +ve, W_c = +ve > W_b, W_d = –ve

⇒ W_c > W_b > W_a > W_d

⇒ W₃ > W₂ > W₁ > W₄

4. A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is –

(A) ⅖

(B) 2/7

(C) ⅕

(D) 7/10

Answer (B)

Sol.

5. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude.

Reason R: At equator, the direction of acceleration due to the gravity is towards the center of earth.

In the light of above statements, choose the correct answer from the options given below

(A) Both A and R are true and R is the correct explanation of A.

(B) Both A and R are true but R is NOT the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

Answer (D)

Sol. g′ = g₀ – ω²Rcos²θ

θ = latitude.

6. If ρ is the density and η is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number R_e is:

Answer (C)

Sol.

Direct formula based.

7. A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively will be

(A) 3 : 2

(B) 9 : 4

(C) 2 : 3

(D) 1 : 1

Answer (D)

Sol.

As temperature is same for both the gases.

⇒ Both gases will have same average kinetic energy.

8. The charge on capacitor of capacitance 15 μF in the figure given below is

(A) 60 μC

(B) 130 μC

(C) 260 μC

(D) 585 μC

Answer (A)

Sol.

⇒ Charge on 15 μF capacitor = 60 μC

As all the capacitors are in series.

9. A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 μF. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be :

(A) 2 μF

(B) 32 μF

(C) 6 μF

(D) 8 μF

Answer (C)

Sol. Equivalent circuit is

Now

10. Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 μC are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be:

(A) 1 : 4

(B) 4 : 1

(C) 1 : 8

(D) 8 : 1

Answer (B)

Sol. q′ = 64q … (i)

A′ = 16 A … (ii)

Dividing (i) & (ii),

σ′ = 4σ

11. The equivalent resistance between points A and B in the given network is:

(A) 65 Ω

(B) 20 Ω

(C) 5 Ω

(D) 2 Ω

Answer (C)

Sol. Initially 5 Ω and 5 Ω are in series and then in parallel with 10 Ω this pattern continues thus

R_net = 5 Ω

12. A bar magnetic having a magnetic moment of 2.0 × 10⁵ JT^–1, is placed along the direction of uniform magnetic field of magnitude B = 14 × 10^–5 T. The work done in rotating the magnet slowly through 60° from the direction of field is:

(A) 14 J

(B) 8.4 J

(C) 4 J

(D) 1.4 J

Answer (A)

Sol.

So U_f – U_i = –MB(1 – cosθ)

= – 14J

So W = – ΔU = 14J

13. Two coils of self inductance L₁ and L₂ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be:

(B) L₁ + L₂ + M

(C) L₁ + L₂ + 2M

(D) L₁ + L₂ – 2M

Answer (D)

Sol. Self inductances are in series but their mutual inductances are linked oppositely so equivalent self inductance

L = L₁ + L₂ – M – M = L₁ + L₂ – 2M

14. A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s^–1. If the horizontal component of earth’s magnetic field is 0.2 × 10^–4 T, then emf induced between the two ends of the conductor is:

(A) 5 μV

(B) 50 μV

(C) 5 mV

(D) 50 mV

Answer (B)

Sol. Emf = _(½)Bωl²

= _(½)× 0.2 × 10^–4 × 5 × 1² V

= 0.5 × 10^–4 V

= 50 μV

15. Which is the correct ascending order of wavelengths?

(A) λ_visible < λ_X-ray < λ_gamma-ray <λ_microwave

(B) λ_gamma-ray < λ_X-ray < λ_visible < λ_microwave

(C) λ_X-ray < λ_gamma-ray < λ_visible < λ_microwave

(D) λ_microwave < λ_visible < λ_gamma-ray < λ_X-ray

Answer (B)

Sol. Wave length of microwave is maximum then visible light then X-rays and then gamma rays so the correct order will be

λ_gamma-ray < λ_X-ray < λ_visible < λ_microwave

16. For a specific wavelength 670 nm of light from a galaxy moving with velocity ν, the observed wavelength is 670.7 nm. The value of ν is:

(A) 3 × 10⁸ ms^–1

(B) 3 × 10¹⁰ ms^–1

(C) 3.13 × 10⁵ ms^–1

(D) 4.48 × 10⁵ ms^–1

Answer (C)

Sol.

For ν << C,

17. A metal surface is illuminated by a radiation of wavelength 4500 Å. The rejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately:

(A) 1.36 eV

(B) 1.69 eV

(C) 2.78 eV

(D) 2.23 eV

Answer (A)

Sol.

Putting the values,

18. A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:

(A) 3.75 hours

(B) 0.56 hours

(C) 0.26 hours

(D) 1.80 hours

Answer (D)

Sol.

⇒ λ_eff = λ₁ + λ₂

19. The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

(A) External input is necessary to sustain ac signal in output

(B) A portion of the output power is returned back to the input

(C) Feedback can be achieved by LR network

(D) The base-collector junction must be forward biased

Answer (B)

Sol. Feedback means a portion of the output power is fed to the inputs.

20. A sinusoidal wave y(t) = 40sin(10 × 10⁶ πt) is amplitude modulated by another sinusoidal wave x(t) = 20sin(1000πt). The amplitude of minimum frequency component of modulated signal is:

(A) 0.5

(B) 0.25

(C) 20

(D) 10

Answer (D)

Sol.

Modulate signal s(t) ≡ [1 + 20sin(1000πt)]sin(10⁷πt)

≡ sin(10⁷πt) + 10cos(10⁷πt – 10³πt) + 10cos(10⁷πt + 10³πt)

⇒ Required amplitude = 10

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. A ball is projected vertically upward with an initial velocity of 50 ms^–1 at t = 0 s. At t = 2 s, another ball is projected vertically upward with same velocity. At t = _____s, second ball will meet the first ball.

(g = 10 ms^–2)

Answer (6)

Sol At t = 2 s, v₁ = 50 – 2 × 10 = 30 m/s

v₂ = v₂

∴ a_rel = g – g = 0

∴ v_rel = 50 – 30 = 20 m/s

∴ Required time t = 2 + 4 = 6 s

2. A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms^–1. The impulse imparted to the ball is _________ Ns.

Answer (12)

Sol. I = mΔv

= 0.4 × 2 × 15 = 12 Ns

3. A system to 10 balls each of mass 2 kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the 7^th and 8^th ball is ______ N when 6^th ball just leaves the table.

Answer (36)

Sol. At given instant

∴ T₇₈ = (3m) × a_sys

4. A geyser heats water flowing at a rate of 2.0 kg per minute from 30°C to 70°C. If geyser operates on a gas burner, the rate of combustion of fuel will be __________ g min^–1.

Answer (42)

Sol. Q = msΔT

= 42 g/minute

5. A heat engine operates with the cold reservoir at temperature 324 K. The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is _____ K.

Answer (540)

Sol.

6. A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _______Hz.

Answer (152)

Sol. Given ν₂₀ = 2ν₁

Alsoν₂₀ = 4 × 19 + ν₁

Soν₂₀ = 152 Hz

7. Two 10 cm long, straight wires, each carrying a current of 5 A are kept parallel to each other. If each wire experienced a force of 10^–5 N, then separation between the wires is _____ cm.

Answer (5)

Sol.

So

= 50 mm

= 5 cm

8. A small bulb is placed at the bottom of a tank containing water to a depth of √7 m. The refractive index of water is 4/3. The area of the surface of water through which light from the bulb can emerge out is xπ m². The value of x is ________.

Answer (9)

Sol.

So

So A = πr²

9. A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is _______× 10^–6 m.

Answer (5)

Sol. 40 M = 1 cm

⇒ M = 0.025 cm ………..(1)

Also, 50 V = 49 M

⇒ Least count = M – V

⇒ LC = 5 × 10^–6 m

10. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0.42 V. If the threshold frequency is x × 10¹³/s, where x is ____ (nearest integer).

(Given, speed of light = 3 × 10⁸ m/s, Planck’s constant = 6.63 × 10^–34 Js)

Answer (35)

Sol.

⇒ f_th ≃35.11 × 10¹³ H

Download PDF of JEE Main 2022 June 26 Shift 2 Physics Paper & Solutions

JEE Main 2022 June 26th Shift 2 Question Paper – Solutions

JEE Main 2022 June 26 Shift 2 Question Paper – Physics Solutions

JEE Main 2022 June 26 Shift 2 Question Paper – Chemistry Solutions

JEE Main 2022 June 26 Shift 2 Question Paper – Maths Solutions