JEE Advanced 2023 · Paper 1 · Physics Q17 · Dronacharya Exams

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PhysicsList match+3 / -1Question 34 of 51

A thin conducting rod MN of mass 20 gm, length 25 cm and resistance $10\,\Omega$ 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 $g = 10\ \mathrm{m\,s^{-2}}$ and $e^{-1} = 0.4$ ]

Two long vertical conducting rails 25 cm apart, joined at the top by a horizontal conducting segment. A horizontal rod MN rests across the rails just below the top. A uniform magnetic field B₀ points out of the page (dot symbols) everywhere; gravity g points down.

List-I

P
At $t = 0.2$ s, the magnitude of the induced emf in Volt
Q
At $t = 0.2$ s, the magnitude of the magnetic force in Newton
R
At $t = 0.2$ s, the power dissipated as heat in Watt
S
The magnitude of terminal velocity of the rod in $\mathrm{m\,s^{-1}}$

List-II

1
$0.07$
2
$0.14$
3
$1.20$
4
$0.12$
5
$2.00$

A

P → 5; Q → 2; R → 3; S → 1

B

P → 3; Q → 1; R → 4; S → 5

C

P → 4; Q → 3; R → 1; S → 2

D

P → 3; Q → 4; R → 2; S → 5

Answer and worked solution

Official answer: D

Equation of motion. With $B\ell=4\times0.25=1$ T m:
$m\frac{dv}{dt}=mg-\frac{B^2\ell^2}{R}v.$

Terminal speed: $v_t=\dfrac{mgR}{B^2\ell^2}=\dfrac{0.02\times10\times10}{1}=2$ m/s → (S) 5.
Time constant: $\tau=\dfrac{mR}{B^2\ell^2}=0.2$ s.

At $t=0.2$ s $=\tau$ . $v=v_t(1-e^{-1})=2\times0.6=1.2$ m/s.

(P) emf $=B\ell v=1.2$ V → 3.
(Q) $I=0.12$ A, so the force is $BI\ell=0.12$ N → 4.
(R) Power $=\varepsilon^2/R=0.144\approx0.14$ W → 2.

P→3, Q→4, R→2, S→5.

Topics:Motional EMF Newton's laws

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