JEE Main 2022 June 26 – Shift 1 Physics Question Paper with Solutions

Students can find the JEE Main 2022 June 26 – Shift 1 Physics Question Paper with Solutions here. The questions are based on the Physics curriculum for Classes XI and XII. Candidates may access the PDF format of the question papers on this page. The PDF download can also be used by the students for future reference. The JEE experts at BYJU’S have authored these solutions. Candidates can improve their JEE marks by studying the JEE Main 2022 Question Papers. Additionally, by using this JEE Main 2022 answer key as a guide, applicants can effectively get ready for the upcoming exam.

JEE Main 2022 Physics Question Paper with Solutions June 26 – Shift 1

Download PDF

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. An expression for a dimensionless quantity P is given by

\(\begin{array}{l}P=\frac{\alpha}{\beta}log_e\left ( \frac{kt}{\beta x} \right )\end{array} \)

; where α and β are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of α will be

(A) [M⁰L^–1T⁰]

(B) [ML⁰T^–2]

(D) [ML²T^–2]

Answer (C)

Sol.

\(\begin{array}{l}\left [ \alpha \right ]=\left [ \beta \right ]=\left [ \frac{kt}{x} \right ]\end{array} \)

\(\begin{array}{l}=\left [ \frac{ML^2T^{-2}}{L} \right ]\end{array} \)

= [MLT^–2]

2. A person is standing in an elevator. In which situation, he experiences weight loss?

(A) When the elevator moves upward with constant acceleration

(B) When the elevator moves downward with constant acceleration

(D) When the elevator moves downward with uniform velocity

Answer (B)

Sol. Apparent weight = m(g – a)

⇒ Weight loss in downward accelerated elevator

3. An object is thrown vertically upwards. At its maximum height, which of the followingquantity becomes zero?

(A) Momentum

(B) Potential Energy

(D) Force

Answer (A)

Sol. At topmost position,

v = 0

⇒ momentum = 0

4. A ball is released from rest from point P of a smooth semi-spherical vessel as shown in figure.The ratio of the centripetal force and normal reaction on the ball at point Q is A whileangular position of point Q is α with respect to point P. Which of the following graphsrepresent the correct relation between A and α when ball goes from Q to R?

JEE Main 2022 June 26 Shift 1 Physics Q4

JEE Main 2022 June 26 Shift 1 Physics Q4 options

Answer (C)

Sol.

\(\begin{array}{l}N=mg\sin \alpha+\frac{mv^2}{R}\end{array} \)

and, v² = 2g × Rsinα

∴N = mgsinα + m × (2gsinα)

= 3mgsinα

∴

\(\begin{array}{l}\textup{ratio},~A=\frac{\frac{mv^2}{R}}{N}\end{array} \)

\(\begin{array}{l}=\frac{2mg~\sin\alpha}{3mg~\sin\alpha}\end{array} \)

\(\begin{array}{l}=\frac{2}{3}\end{array} \)

JEE Main 2022 June 26 Shift 1 Physics A4

5. A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rad s^–1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rad s^–1).

(A)

\(\begin{array}{l}\frac{M}{\left ( M+m \right )}\end{array} \)

(B)

\(\begin{array}{l}\frac{\left ( M+2m \right )}{2M}\end{array} \)

(C)

\(\begin{array}{l}\frac{2M}{\left ( M+2m \right )}\end{array} \)

(D)

\(\begin{array}{l}\frac{2\left ( M+2m \right )}{M}\end{array} \)

Answer (C)

Sol.

l₁ω₁ = l₂ω₂

⇒

\(\begin{array}{l}MR^2\times2=\left ( MR^2+2mR^2 \right )\omega_2\end{array} \)

⇒

\(\begin{array}{l}\omega_2=\frac{2M}{M+2m}\end{array} \)

6. The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by

(Given R = radius of earth)

JEE Main 2022 June 26 Shift 1 Physics Q6 options

Answer (A)

Sol. For r < R

\(\begin{array}{l}g=\frac{Gmr}{R^3}=Cr~\left ( C=\textup{Constant} \right )\end{array} \)

For r > R

\(\begin{array}{l}g=\frac{Gm}{r^2}=\frac{C^{‘}}{r^2}\left ( C^{‘}=\textup{Constant} \right )\end{array} \)

For the above equations the best suited graph is as given in option (A)

7. The efficiency of a Carnot’s engine, working between steam point and ice point, will be

(A) 26.81%

(B) 37.81%

(D) 57.81%

Answer (A)

Sol.

\(\begin{array}{l}\eta=1-\frac{T_C}{T_H} =\frac{T_H-T_C}{T_H}\end{array} \)

\(\begin{array}{l}=\frac{100}{373}\times100 %\end{array} \)

= 26.81%

⇒ option (A)

8. Time period of a simple pendulum in a stationary lift is ‘T’. If the lift accelerates with

\(\begin{array}{l}\frac{g}{6}\end{array} \)

vertically upwards then the time period will be

(Where g = acceleration due to gravity)

(A)

\(\begin{array}{l}\sqrt{\frac{6}{5}}T \end{array} \)

(B)

\(\begin{array}{l}\sqrt{\frac{5}{6}}T \end{array} \)

(C)

\(\begin{array}{l}\sqrt{\frac{6}{7}}T \end{array} \)

(D)

\(\begin{array}{l}\sqrt{\frac{7}{6}}T \end{array} \)

Answer (C)

Sol.

\(\begin{array}{l}T^{‘}=2\pi\sqrt{\frac{I}{g_{eff}}} \end{array} \)

\(\begin{array}{l}T^{‘}=2\pi\sqrt{\frac{I}{g+\frac{g}{6}}}=2\pi\sqrt{\frac{6I}{7g}}\end{array} \)

⇒

\(\begin{array}{l}T^{‘}=\sqrt{\frac{6}{7}}T\end{array} \)

⇒ Option ©

9. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed ν and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by

(R = universal gas constant)

(A)

\(\begin{array}{l}\frac{Mv^2}{7R}\end{array} \)

(B)

\(\begin{array}{l}\frac{Mv^2}{5R}\end{array} \)

(C)

\(\begin{array}{l}2\frac{Mv^2}{7R}\end{array} \)

(D)

\(\begin{array}{l}7\frac{Mv^2}{5R}\end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\frac{1}{2}mv^2=n\frac{5}{2}R\Delta T\end{array} \)

⇒

\(\begin{array}{l}\Delta T=\frac{mv^2}{5nR}\end{array} \)

\(\begin{array}{l}=\frac{MV^2}{5R}\end{array} \)

Option (B)

10. Two capacitors having capacitance C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is now connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is

JEE Main 2022 June 26 Shift 1 Physics Q10

(A)

\(\begin{array}{l}\frac{C_1C_2}{\left ( C_1+C_2 \right )}V\end{array} \)

(B)

\(\begin{array}{l}\frac{\left ( C_1+C_2 \right ) }{C_1C_2}V\end{array} \)

(D) (C₁ – C₂)V

Answer (A)

Sol.

\(\begin{array}{l}V_{\textup{common}}=\frac{C_1V}{C_1 + C_2}\end{array} \)

⇒ Charge on capacitor C₂

= C₂V_common

\(\begin{array}{l}=\frac{C_1C_2V}{C_1+C_2}\end{array} \)

⇒ Option (A)

11. Given below two statements: One is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) :When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it’s individual atom or molecule coincides with the centre of the negative charge distribution.

In the light of above statements, choose the most appropriate answer from the options given below.

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

(B) Both (A) and (R) are correct and (R) is not the correct explanation of (A).

(D) (A) is not correct but (R) is correct.

Answer (C)

Sol. Non polar material does not have any permanent dipole moment and when placed in an electric field the positive and negative charges displace in opposite directions and result into an induced dipole moment as long as the field is applied.

12. The magnetic flux through a coil perpendicular to its plane is varying according to the relation
φ = (5t³ + 4t² + 2t – 5) Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,

(A) 15.6 A

(B) 16.6 A

(D) 18.6 A

Answer (A)

Sol.

\(\begin{array}{l}\text{Emf}=-\frac{d\phi}{dt}=-\left ( 15t^2+8t+2 \right ) \end{array} \)

So,

\(\begin{array}{l}i=\frac{\left| \textup{Emf}\right|}{R}=\frac{\left ( 15t^2+8t+2 \right )}{5}\end{array} \)

at t = 2

i = 15.6 A

13. An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :

(A) 0.4%

(B) 0.2%

(D) 0.6%

Answer (C)

Sol. When the wire is stretched, volume remains constant. If length is increased by 0.4% area will decrease by 0.4% so

From

\(\begin{array}{l}R=\rho\frac{I}{A}\end{array} \)

\(\begin{array}{l}\frac{dR}{R}\times100=\frac{dI}{I}\times100+\frac{dA}{A}\times 100\end{array} \)

%R = 0.4 + 0.4 = 0.8%

14. A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is :

(A) 1:4

(B) 4:1

(D) 1:2

Answer (C)

Sol.

\(\begin{array}{l}R=\frac{mv}{qB}\end{array} \)

\(\begin{array}{l}\frac{R_\alpha}{R_p}=\frac{m_\alpha/q_\alpha}{m_p/q_q}=2\end{array} \)

15. If electric field intensity of a uniform plane electro magnetic wave is given as

\(\begin{array}{l}E=-301.6\sin\left ( kz-\omega t \right )\hat{a}_x+452.4\sin \left ( kz-\omega t \right )\hat{a}_y\frac{V}{m} \end{array} \)

Then, magnetic intensity ‘H’ of this wave in Am^–1 will be :

[Given : Speed of light in vacuum c = 3 × 10⁸ms^–1, Permeability of vacuum μ₀ = 4π × 10^–7 NA^–2]

(A)

\(\begin{array}{l}+0.8 \sin\left ( kz-\omega t \right )\hat{a}_y+0.8\sin\left ( kz-\omega t \right )\hat{a}_x\end{array} \)

(B)

\(\begin{array}{l}+1.0\times10^{-6}\sin\left ( kz-\omega t \right )\hat{a}_y+1.5\times10^{-6} \left ( kz-\omega t \right )\hat{a}_x\end{array} \)

(C)

\(\begin{array}{l}-0.8\sin\left ( kz-\omega t \right )\hat{a}_y-1.2\sin\left ( kz-\omega t \right )\hat{a}_x\end{array} \)

(D)

\(\begin{array}{l}-1.0\times10^{-6}\sin\left ( kz-\omega t \right )\hat{a}_y-1.5\times10^{-6} \sin\left ( kz-\omega t \right )\hat{a}_x\end{array} \)

Answer (C)

Sol. We know

\(\begin{array}{l}\vec{B}\times\vec{C}=\vec{E}\end{array} \)

Taking cross product of vector C both the sides

\(\begin{array}{l}\vec{C}\times\left ( \vec{B}\times\vec{C} \right )=\vec{C}\times\vec{E}\end{array} \)

\(\begin{array}{l}\vec{B}=\frac{\vec{C}\times\vec{E}}{C^2}\end{array} \)

\(\begin{array}{l}\vec{C}=C\hat{k}\end{array} \)

\(\begin{array}{l}\vec{E}=-301.6\sin\left ( kz-\omega t \right )\hat{a}_x+452.4\sin\left ( kz-\omega t \right )\hat{a}_y\end{array} \)

_and

\(\begin{array}{l}\vec{H}=\frac{\vec{B}}{\mu_0}\end{array} \)

On solving

\(\begin{array}{l}\vec{H}=-0.8\sin\left ( kz-\omega t \right )\vec{a}_y-1.2\sin\left ( kz-\omega t \right )\vec{a}_x\end{array} \)

16. In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size

\(\begin{array}{l}\frac{\lambda}{100}\end{array} \)

, whereλ is the wavelength of the wave in free space. The phenomenon, which happens there will be:

(A) Reflection

(B) Refraction

(D) Scattering

Answer (D)

Sol. Since size is of the order of

\(\begin{array}{l}\frac{\lambda}{100}\end{array} \)

, hence scattering will take place.

17. An electron with speed υ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?

(A)

\(\begin{array}{l}\frac{E_e}{E_{ph}}=\frac{2c}{\nu }\end{array} \)

(B)

\(\begin{array}{l}\frac{E_e}{E_{ph}}=\frac{\nu }{2c}\end{array} \)

(C)

\(\begin{array}{l}\frac{p_e}{p_{ph}}=\frac{2c}{\nu}\end{array} \)

(D)

\(\begin{array}{l}\frac{p_e}{p_{ph}}=\frac{\nu}{2c}\end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\lambda_e=\lambda_{ph}\Rightarrow\frac{h}{p_e}=\frac{hc}{E_{ph}}\end{array} \)

\(\begin{array}{l}\Rightarrow E_{ph}=p_e\times c =2E_e\frac{c}{\nu}\end{array} \)

⇒

\(\begin{array}{l}\frac{E_e}{E_{ph}}=\frac{\nu}{2c}\end{array} \)

18. How many alpha and beta particles are emitted when Uranium ₉₂U²³⁸ decays to lead₈₂Pb²⁰⁶?

(A) 3 alpha particles and 5 beta particles

(B) 6 alpha particles and 4 beta particles

(D) 8 alpha particles and 6 beta particles

Answer (D)

Sol.

\(\begin{array}{l}_{92}^{238}\cup\xrightarrow{~~~~~}_{82}^{206}Pb+x_2^4He+y\left ( _{-1}^0e \right )\end{array} \)

⇒ 206 + 4x = 238 …(1)

and 82 + 2x – y = 92 …(2)

⇒ x = 8 and y = 6

19. The I–V characteristics of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is :

JEE Main 2022 June 26 Shift 1 Physics Q19

(A) 1 : 2

(B) 5 : 1

(D) 20 : 1

Answer (B)

Sol. Dynamic resistance

\(\begin{array}{l}=\frac{dV}{dl}\end{array} \)

⇒

\(\begin{array}{l}r_1=\frac{2.1-2}{10-5}k\Omega\end{array} \)

\(\begin{array}{l}r_2=\frac{4.2-4}{250-200}k\Omega\end{array} \)

⇒ r₁ :r₂ = 5 : 1.

20. Choose the correct statement for amplitude modulation :

(A) Amplitude of modulating signal is varied in accordance with the information signal.

(B) Amplitude of modulated signal is varied in accordance with the information signal.

(D) Amplitude of modulated signal is varied in accordance with the modulating signal.

Answer (C)

Sol. In amplitude modulation, amplitude of carrier signal is varied according to the message signal.

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 fighter jet is flying horizontally at a certain altitude with a speed of 200 ms^–1. When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle θ with the horizontal, to hit the jet. If the bullet speed is 400 m/s, the value of θ will be ____°.

Answer (60)

Sol.

To hit the jet

400 cosθ = 200

⇒

\(\begin{array}{l}\cos\theta=\frac{1}{2}\end{array} \)

⇒ θ = 60°

2. A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ___ m.

[Use g = 10 m/s²]

Answer (5)

Sol. gt = g

⇒ t = 1 sec

\(\begin{array}{l}\Delta h=\frac{1}{2}gt^2=\frac{1}{2}\times5\times1^2=5~\text{m}\end{array} \)

∴ h = H – Δh

= 10 – 5

= 5 m

3. The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5 × 10^–4 is ____ kJ/m³. Assume that material is elastic upto the linear strain of 5 × 10^–4.

JEE Main 2022 June 26 Shift 1 Physics NQ3

Answer (25)

Sol.

\(\begin{array}{l}u_d=\frac{1}{2}\times Y\times\left ( \textup{strain} \right )^2\end{array} \)

\(\begin{array}{l}=\frac{1}{2}\times\left ( \frac{20}{10^{-10}} \right )\times\left ( 5\times10^{-4} \right )^2\end{array} \)

= 10¹¹ × 25 × 10^–8

= 25 × 10³ J/m³

= 25 kJ/m³

4. The elongation of a wire on the surface of the earth is 10^–4m. The same wire of same dimensions is elongated by 6 × 10^–5m on another planet. The acceleration due to gravity on the planet will be ___ ms^–2. (Take acceleration due to gravity on the surface of earth = 10 ms^–2)

Answer (6)

Sol.

\(\begin{array}{l}\Delta I=\frac{M^{‘}gI}{2A~y}\end{array} \)

⇒ Δl ∝ g

⇒

\(\begin{array}{l}\frac{g_p}{g_e}=\frac{\Delta I_p}{\Delta I_e}=\frac{6\times 10^{-5}}{10\times10^{-5}} \end{array} \)

⇒ g_p = 6 m/s² as g_e = 10 m/s²

5. A 10 Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μs. The average e.m.f. induced in the coil is ____V.

Answer (400)

Sol. Initial flux through inductor = LI

⇒

\(\begin{array}{l}\phi_i=20\times 10^{-3}\times\frac{20}{10}\end{array} \)

= 4 × 10^–2 weber

Final flux = 0

⇒ average emf

\(\begin{array}{l}=\frac{\left|\phi_i-\phi_f \right|}{100~\mu \textup{s}}\end{array} \)

\(\begin{array}{l}=\frac{4\times10^{-2}}{10^{-4}}=400~\text{V}\end{array} \)

6. A light ray is incident, at an incident angle θ₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ________ degree.

JEE Main 2022 June 26 Shift 1 Physics NQ6

Answer (210)

Sol.

On first reflection angel of deviation is 90° and on second reflection angle of deviation is 120°

so total deviation is δ = 90° +120° = 210°

7. In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be
____ ×10^–2 mm.

Answer (5)

Sol.

\(\begin{array}{l}LC=\frac{1MSD}{VSD}=\frac{\frac{1}{20}\textup{cm}}{10}\end{array} \)

\(\begin{array}{l}=\frac{1}{200}\textup{cm}\end{array} \)

= 5 × 10^–2 mm

8. As per the given circuit, the value of current through the battery will be ______ A.

JEE Main 2022 June 26 Shift 1 Physics NQ8

Answer (1)

Sol. Because of diode D₂ current will not flow through it so new circuit diagram is

JEE Main 2022 June 26 Shift 1 Physics NQ8 A so R_net =10 Ω

and

\(\begin{array}{l}i=\frac{V}{R_{net}}=1A\end{array} \)

9. A 110 V,50 Hz,AC source is connected in the circuit (as shown in figure). The current through the resistance 55Ω, at resonance in the circuit, will be _______ A.

JEE Main 2022 June 26 Shift 1 Physics NQ9

Answer (0)

Sol. At resonance

\(\begin{array}{l}\left ( \omega=\frac{1}{\sqrt{LC}} \right )\end{array} \)

, impedance of the circuit is infinite.

⇒Current through resistance = 0.

10. An ideal fluid of density 800kgm^–3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to

\(\begin{array}{l}\frac{a}{2}\end{array} \)

. The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is

\(\begin{array}{l}\frac{\sqrt{x}}{6}\end{array} \)

ms^–1 for x=_______. (Given g=10 ms^–2)

JEE Main 2022 June 26 Shift 1 Physics NQ10

Answer (363)

Sol. Applying Bernoulli’s theorem:

\(\begin{array}{l}P_1+\rho gh+\frac{1}{2}\rho v^2 = P_2+0 +\frac{1}{2}\rho\left ( 2v \right )^2\end{array} \)

Putting the values,

\(\begin{array}{l}4100=800\left\{ \frac{3}{2}v^2-10\right\}\end{array} \)

\(\begin{array}{l}\Rightarrow v=\frac{\sqrt{363}}{6}\ m/s\end{array} \)

JEE Main 2022 June 26 – Shift 1 Physics Question Paper with Solutions

JEE Main 2022 Physics Question Paper with Solutions June 26 – Shift 1

Comments

Leave a Comment Cancel reply

FREE TEXTBOOK SOLUTIONS