A thin conducting rod MN of mass 20gm , length 25cm and resistance...

Induced emf $ε = Blv$

$⇒$ Induced current $i = \dfrac{ε}{R} = \dfrac{Blv}{R}$

Direction of induced current would be from $N\text{to}M$.

Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write

$⇒ mg - ilB = ma$ [Applying ${2}^{nd}$ law]

$⇒ mg - \dfrac{{B}^{2}{l}^{2}v}{R} = m\dfrac{dv}{dt}$

$⇒ {∫}_{0}^{v}\dfrac{dv}{mg - \dfrac{{B}^{2}{l}^{2}v}{R}} = {∫}_{0}^{t}\dfrac{dt}{m} ⇒ \dfrac{ln{{\left[mg - \dfrac{{B}^{2}{l}^{2}v}{R}\right]}_{0}}^{v}}{\dfrac{- {B}^{2}{l}^{2}}{R}} = \dfrac{t}{m}$

$⇒ \dfrac{mg - \dfrac{{B}^{2}{l}^{2}v}{R}}{mg} = {e}^{\dfrac{- {B}^{2}{l}^{2}}{mR}t}$

Now, $mg = \left(20 \times {10}^{- 3}\right) \times 10 = 0 . 2$, $\dfrac{{B}^{2}{l}^{2}}{R} = \dfrac{{4}^{2} \times {\left(0 . 25\right)}^{2}}{10} = 0 . 1$ and $\dfrac{{B}^{2}{l}^{2}}{mR} = \dfrac{{4}^{2} \times {\left(0 . 25\right)}^{2}}{\left(20 \times {10}^{- 3}\right) \times 10} = 5$

Therefore,

$⇒ \dfrac{0 . 2 - 0 . 1v}{0 . 2} = {e}^{- 5t}$

$⇒ v = 2 [ 1 - {e}^{– 5t} ]$

$⇒$ At $t = 0 . 2s, v = 2\left[1 - \dfrac{1}{e}\right] = 2 \times \left(0 . 6\right) = 1 . 2m{s}^{- 1}$

$⇒ ε = Blv = 4 \times 0 . 25 \times 1 . 2 = 1 . 2V$ volts

Then, current at $t = 0 . 2s$ will be $i = \dfrac{4 \times 0 . 25 \times 1 . 2}{10} = 0 . 12A$

and magnetic force $= ilB = 0 . 12 \times 0 . 25 \times 4 = 0 . 12N$

and power dissipated $= Fv = 0 . 12 \times 1 . 2 = 0 . 144W$

also for maximum velocity(terminal velocity) $t \to \infty , {e}^{- 5t} \to 0$ and hence $v = 2\left[1 - 0\right] = 2m{s}^{- 1}$

$⇒$ Correct match is (D)