JEE Advanced 2023 — Paper 1 — PHY-IV solution

JEE Advanced2023Paper 1PHY-IV

A thin conducting rod MN of mass $20$ gm, length $25$ cm and resistance $10 Ω$ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field $B_{0} = 4$ T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time $t = 0$ and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option.

[Given: The acceleration due to gravity $g = 10 m s^{- 2}$ and $e^{- 1} = 0.4$ ]

List-I	List-II
(P) At $t = 0.2$ s, the magnitude of the induced emf in Volt	(1) 0.07
(Q) At $t = 0.2$ s, the magnitude of the magnetic force in Newton	(2) 0.14
(R) At $t = 0.2$ s, the power dissipated as heat in Watt	(3) 1.20
(S) The magnitude of terminal velocity of the rod in m s $^{- 1}$	(4) 0.12
	(5) 2.00

A
P $\to$ 5, Q $\to$ 2, R $\to$ 3, S $\to$ 1
B
P $\to$ 3, Q $\to$ 1, R $\to$ 4, S $\to$ 5
C
P $\to$ 4, Q $\to$ 3, R $\to$ 1, S $\to$ 2
D
P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 5

Solution

Equation of motion. The induced emf is $B ℓ v$ , the current is $B ℓ v / R$ , and the magnetic retarding force is $B^{2} ℓ^{2} v / R$ . Newton's second law: $m \frac{d v}{d t} = m g - \frac{B ^{2} ℓ ^{2} v}{R}$ .

Terminal velocity. Setting acceleration to zero: $v_{T} = \frac{m g R}{B ^{2} ℓ ^{2}} = \frac{( 20 \times 1 0 ^{- 3} ) ( 10 ) ( 10 )}{16 \cdot ( 1/16 )} = 2 m s^{- 1}$ . (S) $\to$ (5).

Time constant. Define $k = B^{2} ℓ^{2} / (m R) = 16 \cdot (1/16) / (20 \times 1 0^{- 3} \cdot 10) = 5 s^{- 1}$ , so $v (t) = v_{T} (1 - e^{- k t})$ .

Velocity at $t = 0.2$ s. $k t = 1$ , so $v = 2 (1 - e^{- 1}) = 2 (1 - 0.4) = 1.2 m s^{- 1}$ .

(P) Induced emf. $ε = B ℓ v = 4 \cdot 0.25 \cdot 1.2 = 1.20$ V. Match $\to$ (3).

(Q) Magnetic force. $F = \frac{B ^{2} ℓ ^{2} v}{R} = \frac{1 \cdot 1.2}{10} = 0.12$ N. Match $\to$ (4).

(R) Power dissipated. $P = ε^{2} / R = 1.44/10 = 0.144$ W $\approx 0.14$ W. Match $\to$ (2).

Mapping: P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 5. Answer: (D).

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