Induced emf $\epsilon =B\mathcal{l}v$

$\Rightarrow$ Induced current $i=\frac{\epsilon }{R}=\frac{B\mathcal{l}v}{R}$

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

$\Rightarrow mg-i\mathcal{l}B=ma$ [Applying $2^{\mathrm{nd}}$ law]

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

$$\begin{gathered} \Rightarrow {\int }_0^v \frac{dv}{mg-{\displaystyle \frac{B^2{\mathcal{l}}^{2}v}{R}}}={\int }_0^t\frac{dt}{m} \\ \Rightarrow \frac{\ln {{\left[mg-{\displaystyle \frac{B^2{\mathcal{l}}^{2}v}{R}}\right]}_0}^v}{{\displaystyle \frac{-B^2{\mathcal{l}}^2}{R}}}=\frac{t}{m} \end{gathered}$$

$\Rightarrow \frac{mg-{\displaystyle \frac{B^2{\mathcal{l}}^{2}v}{R}}}{mg}=e^{\frac{-B^2{\mathcal{l}}^2}{mR}t}$

Now, $mg=\left(20\times {10}^{-3}\right)\times 10=0.2$, $\frac{B^2{\mathcal{l}}^2}{R}=\frac{4^2\times {\left(0.25\right)}^2}{10}=0.1$ and $\frac{B^2{\mathcal{l}}^2}{mR}=\frac{4^2\times {\left(0.25\right)}^2}{\left(20\times {10}^{-3}\right)\times 10}=5$

Therefore,

$\Rightarrow \frac{0.2-{\displaystyle 0.1v}}{0.2}=e^{-5t}$

$\Rightarrow v=2[1-e^{–5t}]$

$\Rightarrow$ At $t=0.2 \mathrm{s}, v=2\left[1-\frac{1}{e}\right]=2\times \left(0.6\right)=1.2 \mathrm{m} {\mathrm{s}}^{-1}$

$\Rightarrow \epsilon =B\mathcal{l}v=4\times 0.25\times 1.2=1.2 \mathrm{V}$ volts

Then, current at $t=0.2 \mathrm{s}$ will be $i=\frac{4\times 0.25\times 1.2}{10}=0.12 \mathrm{A}$

and magnetic force $=i\mathcal{l}B=0.12\times 0.25\times 4=0.12 \mathrm{N}$

and power dissipated $=Fv=0.12\times 1.2=0.144 \mathrm{W}$

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

$\Rightarrow$ Correct match is (D)