JEE Main 2022 June 25 – Shift 1 Physics Question Paper with Solutions

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JEE Main 2022 June 25 – Shift 1 Physics Question Paper with Solutions

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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. If

\(\begin{array}{l}Z=\frac{A^2B^3}{C^4}\end{array} \)

, then the relative error in Z will be

(A)

\(\begin{array}{l}\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}\end{array} \)

(B)

\(\begin{array}{l}\frac{2\Delta A}{A}+\frac{3\Delta B}{B}-\frac{4\Delta C}{C}\end{array} \)

(C)

\(\begin{array}{l}\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}\end{array} \)

(D)

\(\begin{array}{l}\frac{\Delta A}{A}+\frac{\Delta B}{B}-\frac{\Delta C}{C}\end{array} \)

Answer (C)

Sol.

\(\begin{array}{l}Z=\frac{A^{2}B^{3}}{C^{4}}\end{array} \)

∴

\(\begin{array}{l}\frac{\Delta Z}{Z}=\frac{2\Delta A}{A}+3\times\frac{\Delta B}{B}+ \frac{4\Delta C}{C}\end{array} \)

\(\begin{array}{l}\vec{A}\end{array} \)

is a vector quantity such that

\(\begin{array}{l}|\vec{A}| \end{array} \)

= non-zero constant. Which of the following expression is true for

\(\begin{array}{l}\vec{A}\end{array} \)

(A)

\(\begin{array}{l}\vec{A}\cdot\vec{A} = 0\end{array} \)

(B)

\(\begin{array}{l}\vec{A}\times\vec{A} < 0\end{array} \)

(C)

\(\begin{array}{l}\vec{A}\times\vec{A} = 0\end{array} \)

(D)

\(\begin{array}{l}\vec{A}\times\vec{A} > 0\end{array} \)

Answer (C)

Sol.

\(\begin{array}{l}\vec{A}\times\vec{A} = A\times A\times \sin 0^{\circ}\end{array} \)

= 0

3. Which of the following relations is true for two unit vector

\(\begin{array}{l}\hat{A}~\text{and}~\hat{B}\end{array} \)

making an angle θ to each other?

(A)

\(\begin{array}{l}|\hat{A}+\hat{B}|=|\hat{A}-\hat{B}|tan\frac{\theta}{2}\end{array} \)

(B)

\(\begin{array}{l}|\hat{A}-\hat{B}|=|\hat{A}+\hat{B}|tan\frac{\theta}{2}\end{array} \)

(C)

\(\begin{array}{l}\left| \hat{A}+\hat{B}\right|= \left|\hat{A}-\hat{B}\right|\cos\frac{\theta}{2}\end{array} \)

(D)

\(\begin{array}{l}\left| |\hat{A}-\hat{B}\right|=\left| \hat{A}+\hat{B}\right|\cos\frac{\theta}{2}\end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\left| \hat{A}-\hat{B}\right|=2\sin\left ( \frac{\theta}{2} \right )\end{array} \)

and,

\(\begin{array}{l}\left| \hat{A}+\hat{B}\right|=2\cos\left ( \frac{\theta}{2} \right )\end{array} \)

⇒

\(\begin{array}{l}\frac{\left| \hat{A}-\hat{B}\right|}{{\left| \hat{A}+\hat{B}\right|}}=\tan\left ( \frac{\theta}{2} \right )\end{array} \)

4. If force

\(\begin{array}{l}\vec{F}=3\hat{i}+4\hat{j}-2\hat{k} \end{array} \)

acts on a particle having position vector

\(\begin{array}{l}2\hat{i}+\hat{j}+2\hat{k} \end{array} \)

then, the torque about the origin will be

(A)

\(\begin{array}{l}3\hat{i}+4\hat{j}-2\hat{k} \end{array} \)

(B)

\(\begin{array}{l}-10\hat{i}+10\hat{j}+5\hat{k} \end{array} \)

(C)

\(\begin{array}{l}10\hat{i}+5\hat{j}-10\hat{k} \end{array} \)

(D)

\(\begin{array}{l}10\hat{i}+\hat{j}-5\hat{k} \end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\vec{\tau}=\vec{r}\times\vec{F}\end{array} \)

\(\begin{array}{l}=\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\times\left ( 3\hat{i}+4\hat{j}-2\hat{k} \right )\end{array} \)

\(\begin{array}{l}=-10\hat{i}+10\hat{j}+5\hat{k}\end{array} \)

5. The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be (Given g = acceleration due to gravity at the surface of earth).

(A)

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

(B)

\(\begin{array}{l}\frac{g}{4}\end{array} \)

(C)

\(\begin{array}{l}\frac{g}{3}\end{array} \)

(D)

\(\begin{array}{l}\frac{g}{9}\end{array} \)

Answer (D)

Sol.h = 2R

∴

\(\begin{array}{l}{g}\ ‘ =\frac{GM}{\left ( R+h \right )^2}\end{array} \)

\(\begin{array}{l}\frac{GM}{9R^2}\end{array} \)

\(\begin{array}{l}=\frac{g}{9}\end{array} \)

6. The terminal velocity (v_t) of the spherical rain drop depends on the radius (r) of the spherical rain drop as

(A) r^1/2 (B) r

Answer (C)

Sol.

\(\begin{array}{l}6\pi\eta v_tr=\frac{4}{3}\pi r^3\left ( \rho-\sigma \right )g\end{array} \)

⇒ v_t = Cr² where C is a constant

or v_t∝r²

7. The relation between root mean square speed (v_rms) and most probable speed (v_p) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be

(A)

\(\begin{array}{l}v_{rms}=\sqrt{\frac{2}{3}}v_p\end{array} \)

(B)

\(\begin{array}{l}v_{rms}=\sqrt{\frac{3}{2}}v_p\end{array} \)

(D)

\(\begin{array}{l}v_{rms}=\sqrt{\frac{1}{3}}v_p\end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}v_{rms}=\sqrt{\frac{3RT}{M}}\end{array} \)

\(\begin{array}{l}v_{p}=\sqrt{\frac{2RT}{M}}\end{array} \)

⇒

\(\begin{array}{l}v_{rms}=\sqrt{\frac{3}{2}}v_p\end{array} \)

⇒ option (B)

8. In the figure, a very large plane sheet of positive charge is shown. P₁ and P₂ are two points at distance l and 2l from the charge distribution. If σ is the surface charge density, then the magnitude of electric fields E₁ and E₂ and P₁ and P₂ respectively are

JEE Main 2022 June 25 Shift 1 Physics Q8

(A)

\(\begin{array}{l}E_1=\frac{\sigma}{\varepsilon_0},E_2=\frac{\sigma}{2\varepsilon_0}\end{array} \)

(B)

\(\begin{array}{l}E_1=\frac{2\sigma}{\varepsilon_0},E_2=\frac{\sigma}{\varepsilon_0}\end{array} \)

(C)

\(\begin{array}{l}E_1=E_2=\frac{\sigma}{2\varepsilon_0}\end{array} \)

(D)

\(\begin{array}{l}E_1=E_2=\frac{\sigma}{\varepsilon_0}\end{array} \)

Answer (C)

Sol. For an infinite charged plane

\(\begin{array}{l}E=\frac{\sigma}{2\varepsilon_0}\end{array} \)

for any value of l

⇒

\(\begin{array}{l}E_1=E_2=\frac{\sigma}{2\varepsilon_0}\end{array} \)

⇒ option (C)

9. Match List-I with List-II

List-I	List-II
(A)	AC generator	(I)	Detects the presence of current in the circuit
(B)	Galvanometer	(II)	Converts mechanical energy into electrical energy
(C)	Transformer	(III)	Works on the principle of resonance in AC circuit
(D)	Metal detector	(IV)	Changes an alternating voltage for smaller or greater value

Choose the correct answer from the options given below

(A) (A) – (II), (B) – (I), (C) – (IV), (D) – (III)

(B) (A) – (II), (B) – (I), (C) – (III), (D) – (IV)

(D) (A) – (III), (B) – (I), (C) – (II), (D) – (IV)

Answer (A)

Sol.

AC generator	→	Converts mechanical energy into electrical energy
Galvanometer	→	Detects the presence of current in the circuit
Transformer	→	Change AC voltage for smaller or greater value
Metal detector	→	Works on the principle of resonance in AC circuit

⇒ Option (A) is correct

10. A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r<R) from its centre will be

(A) B∝r²

(B) B∝r

(C)

\(\begin{array}{l}B\propto\frac{1}{r^2}\end{array} \)

(D)

\(\begin{array}{l}B\propto\frac{1}{r}\end{array} \)

Answer (B)

Sol.

\(\begin{array}{l}\int\overline{B}\cdot\overline{dl}=\mu_0l_{in} \end{array} \)

⇒

\(\begin{array}{l}B\times2\pi r=\frac{\mu_0l}{\pi R^2}\times\pi r^2\end{array} \)

⇒ B∝r

⇒ option (B) is correct

11. If wattless current flows in the AC circuit, then the circuit is :

(A) Purely Resistive circuit

(B) Purely Inductive circuit

(D) RC series circuit only

Answer (B)

Sol. For wattless current to flow in AC circuit the circuit will bePurely Inductive circuit

12. The electric field in an electromagnetic wave is given by E = 56.5 sinω(t – x/c) NC^–1. Find the intensity of the wave if it is propagating along x-axis in the free space.

(Given ∈₀ = 8.85 × 10^–12C²N^–1m^–2)

(A) 5.65 Wm^–2

(B) 4.24 Wm^–2

(D) 56.5 Wm^–2

Answer (B)

Sol.

\(\begin{array}{l}I=\frac{1}{2}\varepsilon_0 E_0^2c\end{array} \)

\(\begin{array}{l}=\frac{1}{2}8.5\times10^{-12}\times\left ( 56.5 \right )^2\times3\times10^8\end{array} \)

= 4.24 W/m²

13. The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point P and π at point Q. Then the difference between the resultant intensities at P and Q will be:

(A) 2I

(B) 6I

(D) 7I

Answer (B)

Sol.

\(\begin{array}{l}I_P=I+9I+2\sqrt{I\times9I}cos\frac{\pi}{2}=10I\end{array} \)

\(\begin{array}{l}I_Q=I+9I+2\sqrt{I\times9I}cos\pi=4I\end{array} \)

So, I_P – I_Q = 6I

14. A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :

(Given : relative permeability of medium μ_r= 1)

(A) 10° (B) 20°

Answer (D)

Sol.

\(\begin{array}{l}n=\sqrt{k\mu}=2\end{array} \)

(n⇒ refractive index)

So for TIR

\(\begin{array}{l}\theta>\sin^{-1}\left (\frac{1}{n}\right )\end{array} \)

θ> 30

Only option is 60°

15. Given below are two statements :

Statement I: Davisson-Germer experiment establishes the wave nature of electrons.

Statement II: If electrons have wave nature, they can interfere and show diffraction.

In the light of the above statements choose the correct answer from the option given below :

(A) Both statement I and statement II are true.

(B) Both statement I and statement II are false.

(D) Statement I is false but statement II is true.

Answer (A)

Sol. Davisson-Germer experiment is done and establishes the wave nature of electrons. Interference and diffraction establishes wave nature.

16. The ratio for the speed of the electron in the 3^rd orbit of He⁺ to the speed of the electron in the 3^rd orbit of hydrogen atom will be :

(A) 1 : 1

(B) 1 : 2

(D) 2 : 1

Answer (D)

Sol. We know that

\(\begin{array}{l}V\propto\frac{Z}{n}\end{array} \)

⇒ Required ratio

\(\begin{array}{l}=\frac{\frac{2}{3}}{\frac{1}{3}}\end{array} \)

= 2 : 1

17. The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because :

(A) fractional change in majority carriers produce higher forward bias current

(B) fractional change in majority carriers produce higher reverse bias current

(D) fractional change in minority carriers produce higher reverse bias current

Answer (D)

Sol. A photodiode is reverse biased. When light falling on it produces charge carriers, the fractional change, in minority carriers is high since the original current is very small.

18. A signal of 100 THz frequency can be transmitted with maximum efficiency by :

(A) Coaxial cable

(B) Optical fibre

(D) Water

Answer (B)

Sol. Optical fibres supports frequency of electromagnetic waves in the range 10¹⁴ Hz to 10¹⁵ Hz.

19. The difference of speed of light in the two media A and B(υ_A – υ_B) is 2.6 × 10⁷ m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is: (Given: speed of light in vacuum C = 3 × 10⁸ms^–1)

(A) 1.303

(B) 1.318

(D) 0.12

Answer (C)

Sol. Speed of light in a medium

\(\begin{array}{l}=\frac{c}{n}\end{array} \)

⇒ According to given information,

\(\begin{array}{l}\frac{c}{n_A}-\frac{c}{n_B}=2.6\times10^7\end{array} \)

⇒

\(\begin{array}{l}\frac{n_B}{n_A}-1=\frac{2.6\times10^7}{3\times10^8}\times n_B\end{array} \)

⇒

\(\begin{array}{l}\frac{n_B}{n_A}\simeq 1.13\end{array} \)

20. A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for