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JEE Advanced 2022 Physics Question Paper 2 with Solutions

SECTION – 1 (Maximum marks : 24)

∙ This section contains EIGHT (08) questions.

∙ The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 TO 9, BOTH INCLUSIVE.

∙ For each question, enter the correct integer corresponding to the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 If ONLY the correct integer is entered;

Zero Marks : 0 If the question is unanswered;

Negative Marks : −1 In all other cases.

1. A particle of mass 1 kg is subjected to a force which depends on the position as

$\begin{array}{l} \vec{F} = - k (x \hat{i} + y \hat{j}) k g m s^{- 2} \end{array}$

with k = 1 kg s^–2. At time t = 0, the particle’s position
$\begin{array}{l} \vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) m and its velocity \vec{v} = (- \sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{π} \hat{k}) m s^{- 1} . \end{array}$
Let v_x and v_y denote the x and the y components of the particle’s velocity, respectively. Ignore gravity. When z = 0.5 m, the value of (xv_y – yv_x) is _______ m²s^–1.

Answer (3)

Sol. F_x = – x = ma_x.

So

\begin{array}{l} a_{x} = \frac{d^{2} x}{d t^{2}} = - x \end{array}

\begin{array}{l} \Rightarrow x = A_{x} \sin (ω t + ϕ_{x}) \end{array}

(ω = 1 rad/s)

and

\begin{array}{l} v_{x} = A_{x} ω \cos (ω t + ϕ_{x}) \end{array}

at t = 0,

\begin{array}{l} x = \frac{1}{\sqrt{2}} m \end{array}

and

\begin{array}{l} v_{x} = - \sqrt{2} m / s \end{array}

So

\begin{array}{l} \frac{1}{\sqrt{2}} = A_{x} \sin ϕ_{x} \end{array}

and

\begin{array}{l} - \sqrt{2} = A_{x} \cos ϕ_{x} \end{array}

\begin{array}{l} \Rightarrow \tan ϕ_{x} = - \frac{1}{2} \dots . (1) \end{array}

and

\begin{array}{l} A_{x} = \sqrt{\frac{5}{2}} m \dots . (2) \end{array}

Similarly

F_y = – y = ma_y.

\begin{array}{l} \Rightarrow \frac{d^{2} y}{d t^{2}} = - y \end{array}

So, y = A_y sin(ωt + φ_y) (ω = 1 rad/s)

and

\begin{array}{l} v_{y} = A_{y} ω \cos (ω t + ϕ_{y}) \end{array}

at t = 0, y = √2 m and v_y = √2 m/s

So

\begin{array}{l} \sqrt{2} = A_{y} \sin ϕ \end{array}

and

\begin{array}{l} \sqrt{2} = A_{y} \cos ϕ \end{array}

\begin{array}{l} \Rightarrow ϕ = \frac{π}{4} \end{array}

and A_y = 2 (3 and 4).

So, (xv_y – yv_x)

\begin{array}{l} = \sqrt{\frac{5}{2}} \sin (ω t + ϕ_{x}) \times 2 \cos (ω t + ϕ_{y}) - 2 \sin (ω t + ϕ_{y}) \times \sqrt{\frac{5}{2}} \cos (ω t + ϕ_{x}) \end{array}

\begin{array}{l} = \sqrt{\frac{5}{2}} \times 2 (\sin (ω t + ϕ_{x}) \cos (ω t + ϕ_{y}) - \sin (ω t + ϕ_{y}) \times \cos (ω t + ϕ_{x}) \end{array}

\begin{array}{l} = \sqrt{10} \sin (ϕ_{x} - ϕ_{y}) \end{array}

\begin{array}{l} = \sqrt{10} (\sin ϕ_{x} \cos ϕ_{y} - \cos ϕ_{x} \sin ϕ_{y}) \end{array}

\begin{array}{l} = \sqrt{10} (\frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{2}} - (- \frac{2}{\sqrt{5}}) \times \frac{1}{\sqrt{2}}) \end{array}

= 3

2. In a radioactive decay chain reaction,

$\begin{array}{l} _{90}^{230} T h nucleus decays into_{84}^{214} P o nucleus . \end{array}$

The ratio of the number of α to number of β^– particles emitted in this process is _______.

Answer (2)

Sol. Let number of α particles are n_α and β particles are n_β so

4n_α = 230 – 214

⇒ n_α = 4

n_β = 84 – (90 – 2n_α)

n_β = 2

So

\begin{array}{l} \frac{n_{α}}{n_{β}} = 2 \end{array}

3. Two resistances R₁ = X Ω and R₂ = 1 Ω are connected to a wire AB of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 mm at A to 1 mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from each end along its axis, shows zero deflection when A and B are connected to a battery. The value of X is _______.

JEE Advanced Question Paper 2022 Physics Paper 2 Q3

Answer (5)

Sol. Resistance of frustum shaped conductor shown is

JEE Advanced Question Paper 2022 Physics Paper 2 A3 (i)

\begin{array}{l} R = ρ \frac{l}{π a b} \end{array}

For the shown conductor in the diagram.

JEE Advanced Question Paper 2022 Physics Paper 2 A3 (ii)

\begin{array}{l} r = \frac{a + b}{2} = \frac{0.2 + 1}{2} = 0.6 \end{array}

thus, the resistance of left half is

\begin{array}{l} P = \frac{ρ \times 0.5 \times 10^{6}}{π \times 0.2 \times 0.6} \end{array}

and the resistance of right half is

\begin{array}{l} Q = \frac{ρ \times 0.5 \times 10^{6}}{π \times 0.6 \times 1} \end{array}

for Wheatstone to be balanced

\begin{array}{l} \frac{R_{1}}{P} = \frac{R_{2}}{Q} \end{array}

\begin{array}{l} \frac{X π \times 0.2 \times 0.6}{ρ \times 0.5 \times 10^{6}} = \frac{1 π \times 0.6 \times 1}{ρ \times 0.5 \times 10^{6}} \end{array}

⇒ X = 5

4. In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, electron mass m_e, Planck’s constant h, and Coulomb’s constant

$\begin{array}{l} k = \frac{1}{4 π ε_{0}}, \end{array}$

where ε₀ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]^α [m_e]^β[h]^γ[k]^δ. The value of α + β + γ + δ is ______.

Answer (4)

Sol. [B] = [e]^α [m_e]^β [h^γ] [k]^δ

[M¹T^–2I^–1] = [IT]^α [M]^β [ML²T^–1]^γ [ML³T^–4]^δ

So, β + γ + δ = 1 …(i)

2γ + 3δ = 0 …(ii)

α – γ – 4δ = –2 …(iii)

α – 2δ = –1 …(iv)

On solving

so, α + β + γ + δ = 4

5. Consider a configuration of n identical units, each consisting of three layers. The first layer is a column of air of height h = ⅓ cm, and the second and third layers are of equal thickness

$\begin{array}{l} d = \frac{\sqrt{3} - 1}{2} c m, and refractive indices μ_{1} = \sqrt{\frac{3}{2}} a n d μ_{2} = \sqrt{3}, \end{array}$

respectively. A light source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is incident on the second layer of the first unit at an angle of θ = 60° to the normal. For a specific value of n, the ray of light emerges from the bottom of the configuration at a distance
$\begin{array}{l} l = \frac{8}{\sqrt{3}} c m, \end{array}$
as shown in the figure. The value of n is _______.

JEE Advanced Question Paper 2022 Physics Paper 2 Q5

Answer (4)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A5

\begin{array}{l} x_{1} = \frac{1}{3} \times \tan 60^{\circ} = \frac{1}{\sqrt{3}} c m \end{array}

and,

\begin{array}{l} 1 \times \frac{\sqrt{3}}{2} = \sqrt{\frac{3}{2}} \times \sin θ_{2} \end{array}

⇒ θ₂ = 45°

⇒ x₂ = d

and,

\begin{array}{l} 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \times \sin θ_{2} \end{array}

⇒ θ₃ = 30°

\begin{array}{l} \Rightarrow x_{3} = \frac{d}{\sqrt{3}} \end{array}

\begin{array}{l} ∴ x_{1} + x_{2} + x_{3} = \frac{1}{\sqrt{3}} + \frac{(\sqrt{3} - 1)}{2} (1 + \frac{1}{\sqrt{3}}) \end{array}

\begin{array}{l} = \frac{2}{\sqrt{3}} c m \end{array}

\begin{array}{l} ∴ n = \frac{l}{x_{1} + x_{2} + x_{3}} = \frac{8 / \sqrt{3}}{2 / \sqrt{3}} = 4 \end{array}

6. A charge q is surrounded by a closed surface consisting of an inverted cone of height h and base radius R, and a hemisphere of radius R as shown in the figure. The electric flux through the conical surface is

$\begin{array}{l} \frac{n q}{6 ε_{0}} (in SI units) . \end{array}$

The value of n is _________.

JEE Advanced Question Paper 2022 Physics Paper 2 Q6

Answer (3)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A6

φ through cone

\begin{array}{l} = \frac{q}{2 ε_{0}} \end{array}

∴ n = 3

7. On a frictionless horizontal plane, a bob of mass m = 0.1 kg is attached to a spring with natural length l₀ = 0.1 m. The spring constant is k₁ = 0.009 Nm^–1 when the length of the spring l > l₀ and is k₂ = 0.016 Nm^–1 when l < l₀. Initially the bob is released from l = 0.15 m. Assume that Hooke’s law remains valid throughout the motion. If the time period of the full oscillation is T = (n π) s, then the integer closest to n is _______.

Answer (6)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A7

\begin{array}{l} ω_{1} = \sqrt{\frac{k_{1}}{m}} a n d ω_{2} = \sqrt{\frac{k_{2}}{m}} \end{array}

\begin{array}{l} ∴ Time period = π \sqrt{\frac{m}{k_{1}}} + π \sqrt{\frac{m}{k_{2}}} \end{array}

\begin{array}{l} = π \sqrt{\frac{0.1}{0.009}} + π \sqrt{\frac{0.1}{0.016}} \end{array}

\begin{array}{l} = \frac{π}{0.3} + \frac{π}{0.4} \end{array}

\begin{array}{l} = π \times (\frac{4 + 3}{12}) \times 10 \end{array}

\begin{array}{l} = \frac{70}{12} π \end{array}

= 5.83π

8. An object and a concave mirror of focal length f = 10 cm both move along the principal axis of the mirror with constant speeds. The object moves with speed V₀ = 15 cm s^–1 towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by u. When u = 30 cm, the speed of the mirror V_m is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of V_m is _____ cm s^–1.

JEE Advanced Question Paper 2022 Physics Paper 2 Q8

Answer (3)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A8

\begin{array}{l} m = \frac{f}{u - f} \end{array}

\begin{array}{l} = \frac{10}{30 - 10} = \frac{1}{2} \end{array}

\begin{array}{l} ∴ (V_{0} - V_{m}) \times m^{2} - V_{m} = 0 \end{array}

\begin{array}{l} \Rightarrow (V_{0} - V_{m}) \times \frac{1}{4} = V_{m} \end{array}

\begin{array}{l} \Rightarrow V_{0} = 5 V_{m} \end{array}

\begin{array}{l} \Rightarrow V_{m} = \frac{V_{0}}{5} \end{array}

\begin{array}{l} = \frac{15}{5} \end{array}

= 3 cm/s

SECTION – 2 (Maximum marks : 24)

∙ This section contains SIX (06) questions.

∙ Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct answer(s).

∙ For each question, choose the option(s) corresponding to (all) the correct answer(s).

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 ONLY if (all) the correct option(s) is(are) chosen;

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;

Partial Marks : + 2 If three or more options are correct but ONLY two options are chosen, both of which are correct;

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct option;

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : –2 In all other cases.

9. In the figure, the inner (shaded) region A represents a sphere of radius r_A = 1, within which the electrostatic charge density varies with the radial distance r from the center as ρ_A = kr, where k is positive. In the spherical shell B of outer radius r_B, the electrostatic charge density varies as ρ_B =2k/r. Assume that dimensions are taken care of. All physical quantities are in their SI units.

JEE Advanced Question Paper 2022 Physics Paper 2 Q9

Which of the following statement(s) is/(are) correct?

(A) If

\begin{array}{l} r_{B} = \sqrt{\frac{3}{2},} \end{array}

then the electric field is zero everywhere outside B.

(B) If

\begin{array}{l} r_{B} = \frac{3}{2}, \end{array}

then the electric potential just outside B is

\begin{array}{l} \frac{k}{ε_{0}} . \end{array}

(C) If r_B = 2, then the total charge of the configuration is 15πk.

(D) lf

\begin{array}{l} r_{B} = \frac{5}{2}, \end{array}

then the magnitude of the electric field just outside B is

\begin{array}{l} \frac{13 π k}{ε_{0}} . \end{array}

Answer (B)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A9

\begin{array}{l} Q_{T o t a l} = \int_{0}^{r_{A}} k r (4 π r^{2}) d r + \int_{r_{A}}^{r_{B}} \frac{2 k}{r} (4 π r^{2}) d r \end{array}

\begin{array}{l} = \frac{4 π k}{4} r_{A}^{4} + \frac{8 π k}{2} (r_{B}^{2} - r_{A}^{2}) \end{array}

\begin{array}{l} = π k + 4 π k (r_{B}^{2} - r_{A}^{2}) \end{array}

If

\begin{array}{l} r_{B} = \sqrt{\frac{3}{2}} \end{array}

\begin{array}{l} Q_{T o t a l} = π k r_{A}^{4} + 4 π k (\frac{3}{2} - r_{A}^{2}) \end{array}

\begin{array}{l} = π k + 4 π k (\frac{3}{2} - 1) \end{array}

\begin{array}{l} = π k + 2 π k = 3 π k \end{array}

If

\begin{array}{l} r_{B} = \frac{3}{2} \end{array}

\begin{array}{l} Q_{T o t a l} = π k + 4 π k (\frac{9}{4} - 1) \end{array}

\begin{array}{l} = π k + 4 π k (\frac{5}{4}) = 6 π k \end{array}

\begin{array}{l} V = \frac{1}{4 π ε_{0}} \frac{6 π k}{r_{B}} = \frac{3 k}{2} \frac{2}{3 ε_{0}} = \frac{k}{ε_{0}} \end{array}

(B) is correct

If r_B = 2

\begin{array}{l} Q_{T o t a l} = π k + 4 π k (4 - 1) \end{array}

= 13πk

Option (C) is incorrect

If

\begin{array}{l} r_{B} = \frac{5}{2} \end{array}

\begin{array}{l} Q_{T o t a l} = π k + 4 π k (\frac{25}{4} - 1) \end{array}

\begin{array}{l} = π k + π k (21) \end{array}

\begin{array}{l} = 22 π k \end{array}

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{22 π k}{25} \times 4 \end{array}

\begin{array}{l} = \frac{22 k}{25 ε_{0}} \end{array}

10. In Circuit-1 and Circuit-2 shown in the figures, R₁ = 1 Ω, R₂ = 2 Ω and R₃ = 3 Ω.

P₁ and P₂ are the power dissipations in Circuit-1 and Circuit-2 when the switches S₁ and S₂ are in open conditions, respectively.

Q₁ and Q₂ are the power dissipations in Circuit-1 and Circuit-2 when the switches S₁ and S₂ are in closed conditions, respectively.

JEE Advanced Question Paper 2022 Physics Paper 2 Q10

Which of the following statement(s) is(are) correct?

(A) When a voltage source of 6 V is connected across A and B in both circuits, P₁ < P₂

(B) When a constant current source of 2 Amp is connected across A and B in both circuits, P₁ > P₂

(C) When a voltage source 6 V is connected across A and B in Circuit-1, Q₁ > P₁

(D) When a constant current source of 2 Amp is connected across A and B in both circuits, Q₂ < Q₁

Answer (A, B, C)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A10

When S₁ and S₂ are open

\begin{array}{l} {(R_{e q})}_{1} = 1 + \frac{5 \times \frac{1}{2}}{5 + \frac{1}{2}} = 1 + \frac{5}{11} = \frac{16}{11} \end{array}

\begin{array}{l} P_{1} = \frac{V^{2}}{R_{e q}} = \frac{{(6)}^{2}}{16} \times 11 = \frac{36 \times 11}{16} = 24.75 W \end{array}

\begin{array}{l} (R_{e q})_{2} = \frac{6}{11} Ω \end{array}

\begin{array}{l} P_{2} = \frac{V^{2}}{R_{e q}} = \frac{{(6)}^{2}}{6} \times 11 = \frac{36 \times 11}{6} = 66 W \end{array}

\begin{array}{l} P_{2} > P_{1} \end{array}

Option (A) is correct.

⇒ If 2 A source is used in both the cases.

\begin{array}{l} P_{1} = i^{2} (R_{e q})_{1} = (2)^{2} \times \frac{16}{11} = \frac{64}{11} = 5.818 W \end{array}

\begin{array}{l} P_{2} = i^{2} (R_{e q})_{2} = (2)^{2} \times \frac{6}{11} = \frac{24}{11} = 2.1818 W \end{array}

P₁ > P₂

Option (B) is correct

For Q₁

\begin{array}{l} R_{e q} = \frac{5}{11} Ω \end{array}

\begin{array}{l} Q_{1} = \frac{V^{2}}{R_{e q}} = \frac{{(6)}^{2}}{\frac{5}{11}} = \frac{36 \times 11}{5} = 79.2 W \end{array}

P₁ = 24.75 W

Q₁ > P₁

Option (C) is correct.

For option (D)

\begin{array}{l} Q_{1} = i^{2} R_{e q} = (2)^{2} \times \frac{5}{11} = \frac{20}{11} = 1.81 W \end{array}

\begin{array}{l} Q_{2} = i^{2} R_{e q} = (2)^{2} \times \frac{1}{2} = \frac{4}{2} = 2 W \end{array}

Q₂ > Q₁

Option (D) is incorrect.

11. A bubble has surface tension S. The ideal gas inside the bubble has ratio of specific heats γ = 5/3. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is P_a₁, the radius of the bubble is found to be r₁ and the temperature of the enclosed gas is T₁. When the atmospheric pressure is P_a₂, the radius of the bubble and the temperature of the enclosed gas are r₂ and T₂, respectively.

Which of the following statement(s) is(are) correct?

(A) If the surface of the bubble is a perfect heat insulator, then

\begin{array}{l} {(\frac{r_{1}}{r_{2}})}^{5} = \frac{P_{a 2} + \frac{2 S}{r_{2}}}{P_{a 1} + \frac{2 S}{r_{1}}} \end{array}

(B) If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.

(C) If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then

\begin{array}{l} {(\frac{r_{1}}{r_{2}})}^{3} = \frac{P_{a 2} + \frac{4 S}{r_{2}}}{P_{a 1} + \frac{4 S}{r_{1}}} \end{array}

(D) If the surface of the bubble is a perfect heat insulator, then

\begin{array}{l} {(\frac{T_{2}}{T_{1}})}^{\frac{5}{2}} = \frac{P_{a 2} + \frac{4 S}{r_{2}}}{P_{a 1} + \frac{4 S}{r_{1}}} \end{array}

Answer (C, D)

Sol. S : Surface tension

JEE Advanced Question Paper 2022 Physics Paper 2 A11

When,

JEE Advanced Question Paper 2022 Physics Paper 2 A11 (ii)

For adiabatic process

\begin{array}{l} P_{1} {V_{1}}^{γ} = P_{2} {V_{2}}^{γ} \end{array}

\begin{array}{l} (P_{a 1} + \frac{4 T}{r_{1}}) {(\frac{4}{3} π r_{1}^{3})}^{\frac{5}{3}} = (P_{a 2} + \frac{4 T}{r_{2}}) {(\frac{4}{3} π r_{2}^{3})}^{\frac{5}{3}} \end{array}

\begin{array}{l} {(\frac{r_{1}}{r_{2}})}^{5} = \frac{(P_{a 2} + \frac{4 T}{r_{2}})}{(P_{a 1} + \frac{4 T}{r_{1}})} \end{array}

___________________

\begin{array}{l} T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1} \end{array}

\begin{array}{l} \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} = {(\frac{r_{1}}{r_{2}})}^{3 (\frac{2}{3})} \end{array}

\begin{array}{l} (\frac{T_{2}}{T_{1}}) = {(\frac{P_{a 2} + \frac{4 T}{r_{2}}}{P_{a 1} + \frac{4 T}{r_{1}}})}^{\frac{2}{5}} \end{array}

For option (B) Total internal energy + surface energy will not be same as word done by gas will be there.

Option (B) is incorrect.

For option (C)

\begin{array}{l} P_{1} V_{1} = P_{2} V_{2} \end{array}

\begin{array}{l} (P_{a 1} + \frac{4 T}{r_{1}}) (\frac{4}{3} π r_{1}^{3}) = (P_{a 2} + \frac{4 T}{r_{2}}) (\frac{4}{3} π r_{2}^{3}) \end{array}

\begin{array}{l} {(\frac{r_{1}}{r_{2}})}^{3} = (\frac{P_{a 2} + \frac{4 T}{r_{2}}}{P_{a 1} + \frac{4 T}{r_{1}}}) \end{array}

Option (C) is correct

12. A disk of radius R with uniform positive charge density σ is placed on the xy plane with its center at the origin. The Coulomb potential along the z-axis is

$\begin{array}{l} V (z) = \frac{σ}{2 \in_{0}} (\sqrt{R^{2} + z^{2}} - z) \end{array}$

A particle of positive charge q is placed initially at rest at a point on the z-axis with z = z₀ and z₀ > 0. In addition to the Coulomb force, the particle experiences a vertical force

$\begin{array}{l} \vec{F} = - c \hat{k} with c > 0. Let β = \frac{2 c \in_{0}}{q σ} . \end{array}$

Which of the following statement(s) is(are) correct?

(A) For

\begin{array}{l} β = \frac{1}{4} a n d z_{0} = \frac{25}{7} R, \end{array}

the particle reaches the origin.

(B) For

\begin{array}{l} β = \frac{1}{4} a n d z_{0} = \frac{3}{7} R, \end{array}

the particle reaches the origin.

(C) For

\begin{array}{l} β = \frac{1}{4} a n d z_{0} = \frac{R}{\sqrt{3}}, \end{array}

the particle returns back to z = z₀

(D) For β > 1 and z₀ > 0, the particle always reaches the origin.

Answer (A, C, D)

Sol.

\begin{array}{l} V (z) = \frac{σ}{2 \in_{0}} (\sqrt{R^{2} + z^{2}} - z) \end{array}

U_z = cz

\begin{array}{l} \Rightarrow U (z)_{n e t} = \frac{σ q}{2 \in_{0}} (\sqrt{R^{2} + z^{2}} - z) + c z \end{array}

\begin{array}{l} = c [\frac{σ q}{2 c \in_{0}} (\sqrt{R^{2} + z^{2}} - z) + z] \end{array}

\begin{array}{l} = c [4 \sqrt{R^{2} + z^{2}} - 3 z] a t β = \frac{1}{4} \end{array}

At z = 0,

\begin{array}{l} β = \frac{1}{4} \end{array}

\begin{array}{l} U (z)_{n e t} = c [4 R] = 4 R c \dots . (i) \end{array}

At z = z₀ = 25/7 R, β = ¼

\begin{array}{l} U (z)_{n e t} = c [4 \times \frac{26 R}{7} - 3 \times \frac{25 R}{7}] = \frac{29}{7} R c \dots . (i i) \end{array}

At z = z₀ = 3/7 R, β = ¼

\begin{array}{l} U (z)_{n e t} = c [4 \times \frac{\sqrt{58}}{7} R - \frac{9 R}{7}] \approx 3 R c \dots . (i i i) \end{array}

\begin{array}{l} At z = \frac{R}{\sqrt{3}} β = \frac{1}{4} \end{array}

\begin{array}{l} U (z)_{n e t} = c [\frac{8 R}{\sqrt{3}} - \frac{3 R}{\sqrt{3}}] \approx 2.887 R c \end{array}

⇒ In option (A) particle reaches at origin with positive kinetic energy

\begin{array}{l} \frac{d U (z)}{d z} = 0 a t z = \frac{3 R}{\sqrt{7}} \end{array}

\begin{array}{l} a t β = \frac{1}{4} a n d z = \frac{3 R}{\sqrt{7}} \end{array}

\begin{array}{l} U (z)_{n e t} = \sqrt{7} R c = 2.645 \end{array}

⇒ In option B at U(z)_net

\begin{array}{l} \approx 3 R c \end{array}

⇒ The kinetic energy at origin will become negative

\begin{array}{l} a t z = \frac{R}{\sqrt{3}} \end{array}

⇒ In option (C),

\begin{array}{l} U (z)_{n e t} a t z = \frac{R}{\sqrt{3}} < U (z)_{n e t} \end{array}

at z = 0,

And

\begin{array}{l} \begin{matrix} U (z)_{n e t} a t z = \frac{R}{\sqrt{3}} > U (z)_{n e t} a t z = \frac{3 R}{\sqrt{7}} \end{matrix} \end{array}

⇒ Particle will return back to z₀.

In option (D) (β > 1 and z₀ > 0)

U(z)_netwill keep on increasing with z

⇒ Particle always reaches the origin.

⇒ Answer (A, C, D)

13. A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n₂. The other slit is at the interface of this medium with another medium 1 of refractive index n₁ (≠ n₂). The line joining the slits is perpendicular to the interface and the distance between the slits is d. The slit widths are much smaller than d. A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle θ from the line joining them, so that θ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

JEE Advanced Question Paper 2022 Physics Paper 2 Q13

Which of the following statement(s) is(are) correct?

(A) The phase difference between the two rays is independent of d.

(B) The two rays interfere constructively at the detector.

(C) The phase difference between the two rays depends on n₁ but is independent of n₂.

(D) The phase difference between the two rays vanishes only for certain values of d and the angle of incidence of the beam, with q being the corresponding angle of refraction.

Answer (A, B)

Sol.

JEE Advanced Question Paper 2022 Physics Paper 2 A13

AB = (d)(tanθ)

and BC = AB sinα = (d) (tanθ)(sinα)

Also, AD = AB sinθ

⇒ Path difference (in vaccum)

= n₁BC – n₂AD

= n₁(AB) sinα – n₂(AB sinθ)

= AB (n₁sinα – n₂sinθ) = 0

⇒ (A), (B) are correct

(C), (D) are incorrect.

14. In the given P–V diagram, a monoatomic gas (γ = 5/3) is first compressed adiabatically from state A state B. Then it expands isothermally from state B to state C. [Given :

$\begin{array}{l} {(\frac{1}{3})}^{0.6} ≃ 0.5, I n 2 ≃ 0.7 \end{array}$

].

JEE Advanced Question Paper 2022 Physics Paper 2 Q14

Which of the following statement(s) is(are) correct?

(A) The magnitude of the total work done in the process A → B → C is 144 kJ.

(B) The magnitude of the work done in the process B → C is 84 kJ.

(C) The magnitude of the work done in the process A → B is 60 kJ.

(D) The magnitude of the work done in the process C → A is zero.

Answer (B, C, D)

Sol.Pv^γ = c

\begin{array}{l} \Rightarrow 100 (0.8)^{5 / 3} = 300 (v)^{5 / 3} \end{array}

\begin{array}{l} \Rightarrow V_{B} = \frac{0.8}{3^{3 / 5}} \end{array}

\begin{array}{l} \Rightarrow W_{A B} = \frac{P_{A} V_{A} - P_{B} V_{B}}{\frac{5}{3} - 1} = \frac{80 - 300 \times \frac{0.8}{3^{3 / 5}}}{2 / 3} k J \end{array}

\begin{array}{l} = \frac{80 - 240 (0.5)}{2 / 3} k J \end{array}

= – 60 kJ

⇒ (C) is correct

C → A is isochoric ⇒ (D) is correct

BC : W_BC

\begin{array}{l} = n R T \ln \frac{V_{2}}{V_{1}} = P V \ln \frac{V_{2}}{V_{1}} \end{array}

\begin{array}{l} = 300 \times 0.8 \times 0.5 \ln (\frac{0.8}{\frac{0.8}{3^{3 / 5}}}) \end{array}

\begin{array}{l} = \frac{240}{3^{3 / 5}} \ln (3^{3 / 5}) \end{array}

\begin{array}{l} = 120 \times \frac{3}{5} \ln 3 k J \end{array}

= 79 kJ

⇒ (A) & (B) are not correct.

SECTION – 3 (Maximum marks : 12)

∙ This section contains FOUR (04) questions.

∙ Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.

∙ For each question, choose the option corresponding to the correct answer.

∙ Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 If ONLY the correct option is chosen;

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);

Negative Marks : −1 In all other cases.

15. A flat surface of a thin uniform disk A of radius R is glued to a horizontal table. Another thin uniform disk B of mass M and with the same radius R rolls without slipping on the circumference of A, as shown in the figure. A flat surface of B also lies on the plane of the table. The center of mass of B has fixed angular speed ω about the vertical axis passing through the center of A. The angular momentum of B is nMωR² with respect to the center of A. Which of the following is the value of n?

JEE Advanced Question Paper 2022 Physics Paper 2 Q15

(A) 2

(B) 5

(C) 7/2

(D) 9/2

Answer (B)

Sol. Angular momentum of B with respect to center of A

\begin{array}{l} \vec{L} = {\vec{L}}_{C M} + {\vec{L}}_{B o d y a b o u t C M} \end{array}

\begin{array}{l} = M (2 R)^{2} ω \hat{k} + \frac{M R^{2}}{2} (ω_{b o d y}) \hat{k} \end{array}

\begin{array}{l} = M (2 R)^{2} ω \hat{k} + \frac{M R^{2}}{2} (2 ω) \hat{k} \end{array}

\begin{array}{l} = 5 M R^{2} ω \hat{k} \end{array}

Comparing the magnitude with nMωR²

n = 5

16. When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take hc/e = 1.24 × 10^–6 J m C^–1.]

(A) 1.72 x 10^–7 m, 1.20 eV

(B) 1.72 x 10^–7 m, 5.60 eV

(C) 3.78 x 10^–7 m, 5.60 eV

(D) 3.78 x 10^–7 m, 1.20 eV

Answer (A)

Sol.

\begin{array}{l} h ν - ϕ = 6 e V \end{array}

\begin{array}{l} \frac{h c}{λ} - ϕ = 6 e V \dots . (i) \end{array}

\begin{array}{l} \frac{h c}{4 λ} - ϕ = 0.6 e V \end{array}

\begin{array}{l} \frac{3 h c}{4 λ} = 5.4 e V \end{array}

\begin{array}{l} ∴ λ = \frac{3 h c}{4 \times 5.4 e V} = \frac{3 \times 1.24 \times 10^{- 6}}{4 \times 5.4} \end{array}

= 1.72 × 10^–7 m

⇒ from equation (i)

\begin{array}{l} \frac{h c}{1.72 \times 10^{- 7}} \times \frac{1}{1.6 \times 10^{- 19}} - ϕ = 6 e V \end{array}

\begin{array}{l} \frac{2 \times 10^{- 25}}{2.75 \times 10^{- 26}} - ϕ = 6 \end{array}

\begin{array}{l} \Rightarrow ϕ = (7.27 - 6) ≅ 1.2 e V \end{array}

17. Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is 0.5 mm. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

Measurement condition	Main scale reading	Circular scale reading
Two arms of gauge touching each other without wire	0 division	4 divisions
Attempt-1: With wire	4 divisions	20 divisions
Attempt-2: With wire	4 divisions	16 divisions

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

(A) 2.22 ± 0.02 mm, π (1.23 ± 0.02) mm²

(B) 2.22 ± 0.01 mm, π (1.23 ± 0.01) mm²

(C) 2.14 ± 0.02 mm, π (1.14 ± 0.02) mm²

(D) 2.14 ± 0.01 mm, π (1.14 ± 0.01) mm²

Answer (C*) (Bonus declared by JEE)

Sol.Reading-1

MSR = 4 × 0.5 = 2 mm

\begin{array}{l} C S R = \frac{1}{100} \times 20 = 0.20 m m \end{array}

Zero error

\begin{array}{l} = \frac{1}{100} \times 4 = 0.04 m m \end{array}

R₁ = MSR + CSR – (Zero error)

= (2 + 0.20 – 0.04) mm

= 2.16 mm

Reading-2

MSR = 2 mm

CSR = 0.16 mm

Zero error = 0.04 mm

R₂ = (2 + 0.16 – 0.04) mm = 2.12 mm

Reading/average

\begin{array}{l} = \frac{R_{1} + R_{2}}{2} \end{array}

= 2.14 mm = R_m (say)

Average mean error

\begin{array}{l} = \frac{| R_{m} - R_{1} | + | R_{m} - R_{2} |}{2} \end{array}

= 0.02 mm

⇒ Diameter = (2.14 ± 0.02) mm = d (say)

Area

\begin{array}{l} = \frac{π d^{2}}{4} \end{array}

\begin{array}{l} \Rightarrow d A = \frac{2 π d}{4} (Δ d) \end{array}

\begin{array}{l} \Rightarrow Δ A r e a ≅ π (0.02) \end{array}

\begin{array}{l} \Rightarrow A r e a = \frac{π d^{2}}{4} \pm Δ A r e a = π (1.14 \pm 0.02) m m \end{array}

18. Which one of the following options represents the magnetic field

$\begin{array}{l} \vec{B} \end{array}$

at O due to the current flowing in the given wire segments lying on the xy plane?

JEE Advanced Question Paper 2022 Physics Paper 2 Q18

(A)

\begin{array}{l} \vec{B} = \frac{- μ_{0} I}{L} (\frac{3}{2} + \frac{1}{4 \sqrt{2} π}) \hat{k} \end{array}

(B)

\begin{array}{l} \vec{B} = \frac{- μ_{0} I}{L} (\frac{3}{2} + \frac{1}{2 \sqrt{2} π}) \hat{k} \end{array}

(C)

\begin{array}{l} \vec{B} = \frac{- μ_{0} I}{L} (1 + \frac{1}{4 \sqrt{2} π}) \hat{k} \end{array}

(D)

\begin{array}{l} \vec{B} = \frac{- μ_{0} I}{L} (1 + \frac{1}{4 π}) \hat{k} \end{array}

Answer (C)

Sol.B_net = B_semicircle + B_quarter + B_straight

\begin{array}{l} = \frac{μ_{0} I}{4 (\frac{L}{2})} + \frac{μ_{0} I}{8 \times (\frac{L}{4})} + \frac{μ_{0} I}{4 π \times L} (\frac{1}{\sqrt{2}}) (- \hat{k}) \end{array}

\begin{array}{l} = (\frac{μ_{0} I}{2 L} + \frac{μ_{0} I}{2 L} + \frac{μ_{0} I}{4 \sqrt{2} π L}) (- \hat{k}) \end{array}

\begin{array}{l} = \frac{μ_{0} I}{L} (1 + \frac{1}{4 \sqrt{2} π}) (- \hat{k}) \end{array}

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JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced Question Paper 2022 Physics Paper 2

JEE Advanced 2022 Paper 2 – Live Analysis and Solutions

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Frequently Asked Questions – FAQs

Q1

What was the Class-wise distribution of Physics questions in JEE Advanced 2022 Physics Paper 2?

In JEE Advanced 2022 Physics Paper 2, 6 questions were asked from class 11 and 12 from the class 12 syllabus.

Q2

Was the JEE Advanced 2022 Physics question paper 2 challenging?

The questions asked in the JEE Advanced 2022 Physics question paper 2 were easy to medium to answer.

Q3

What was the overall difficulty level of Physics Paper 2 in the JEE Advanced 2022 question paper?

The overall difficulty level of Physics Paper 2 in the JEE Advanced 2022 question paper can be stated as 2.11 on a scale of 3.

Q4

From which chapters are easy questions asked in the JEE Advanced 2022 Physics paper 2?

The chapters from which the easy questions asked in the JEE Advanced 2022 Physics question paper 2 are: Nuclei, System of Particles and Rotational Motion, Dual Nature of Radiation and Matter

Q5

Is it compulsory to attempt all the questions in the Chemistry part of JEE Advanced 2022 paper 2?

There are three sections in the Chemistry part of JEE Advanced 2022 paper 2, and candidates can attempt the questions based on their choice.

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