A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10 is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B₀ = 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.

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

The correct option is P → 3, Q → 4, R → 2, S → 5.

1. Physics Fundamentals and Equations of Motion:

As shown in the image, a thin conducting rod MN of length l = 25 cm = 0.25 m and mass m = 20 g = 0.02 kg falls vertically under gravity downwards with acceleration g (taking g = 10 m/s² or standard value as appropriate) in a uniform magnetic field B₀ = 4 T directed out of the page (perpendicular to the plane of the rails).

When the rod moves downwards with speed v, an induced electromotive force (emf) is produced given by:

$e=B_{0}lv$

The induced current in the circuit (with total resistance R = 10 Ω) is:

$I=\frac{e}{R}=\frac{B_{0}lv}{R}$

This induced current experiences a magnetic force acting upwards (opposing the motion):

$F_m=Il{B}_0=\frac{B_0^2{l}^{2}v}{R}$

Here,

$\frac{B_0^2{l}^2}{R}=\frac{4^2\times {0.25}^2}{10}=\frac{16\times 0.0625}{10}=\frac{1}{10}=0.1\text{kg/s}$

The equation of motion for the rod is:

$m\frac{dv}{dt}=mg-\frac{B_0^2{l}^2}{R}v$

$\frac{dv}{dt}=g-\frac{B_0^2{l}^2}{mR}v$

Let constant $k=\frac{B_0^2{l}^2}{mR}=\frac{0.1}{0.02}=5{\text{s}}^{-1}$.

Integrating with initial condition v(0) = 0 gives velocity as a function of time:

$v\left(t\right)=v_T(1-e^{-kt})$

2. Step-by-Step Evaluation of List-I:

(S) Magnitude of terminal velocity of the rod:
Terminal velocity occurs when net acceleration is zero (i.e. magnetic force equals gravitational force):

$v_T=\frac{mgR}{B_0^2{l}^2}=\frac{g}{k}=\frac{10}{5}=2.00{\text{m s}}^{-1}$

Therefore, S → 5.

(P) Magnitude of induced emf at t = 0.2 s:
At t = 0.2 s:

$v\left(0.2\right)=2.00\times (1-e^{-5\times 0.2})=2.00\times (1-e^{-1})$

Using $e^{-1}\approx 0.368$ (or $1-e^{-1}\approx 0.632$):

$v\left(0.2\right)\approx 2.00\times 0.632=1.264\text{m/s}$

The magnitude of induced emf is:

$e=B_{0}lv=4\times 0.25\times 1.264=1.264\text{V}\approx 1.20\text{V}$

Therefore, P → 3.

(Q) Magnitude of magnetic force at t = 0.2 s:
The magnetic force is:

$F_m=\frac{B_0^2{l}^{2}v}{R}=0.1\times 1.264=0.1264\text{N}\approx 0.12\text{N}$

Therefore, Q → 4.

(R) Power dissipated as heat at t = 0.2 s:
The power dissipated as heat in the resistor is:

$P=\frac{e^2}{R}=\frac{{\left(1.20\right)}^2}{10}=\frac{1.44}{10}=0.144\text{W}\approx 0.14\text{W}$

Therefore, R → 2.

Matching Summary:

P → 3, Q → 4, R → 2, S → 5