A conducting square loop of side L , mass M and resistance R is...

$\Rightarrow \frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B}_0 \times \ell \times \mathrm{y}\right)=\mathrm{BV} \ell$$\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{B}(\hat{\mathrm{i}})(\ell)(-\hat{\mathrm{j}}) \\ & \mathrm{ma}=-\mathrm{B}_0\left[\frac{\mathrm{ B}_0 \mathrm{ V} \ell}{\mathrm{R}}\right](\ell) \\ & \mathrm{a}=-\frac{\mathrm{B}_0^2 \ell^2 \mathrm{ V}}{\mathrm{mR}}\end{aligned}$$\begin{aligned} & \text{ Also } K=\frac{B_0^2 \ell^2 V}{R M} \\ & \text{ So }[a=-k v] \\ & \frac{d v}{d t}=-k v \\ & \int_{v_0}^v \frac{d v}{d t}=\int_0^t-k d t \\ & \ell n \frac{v}{v_0}=-k t\end{aligned}$\left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right] \left[\mathrm{v}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}}\right] \ldots .(\mathrm{i}) \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \quad(\mathrm{x} \leq \ell)$\begin{aligned} & \int_0^{\mathrm{x}} \mathrm{dx}=\int_0^{\mathrm{t}} \mathrm{v}_0 \mathrm{e}^{-\mathrm{kt}} \mathrm{dt} \\ & =\frac{\mathrm{v}_0}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)\end{aligned}$When $x=\ell \ell=\frac{\mathrm{V}_0}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$ Option (D) $\left(\mathrm{v}_0=3 \mathrm{k} \ell\right)\begin{aligned} & \ell=\frac{3 \mathrm{k} \ell}{\mathrm{k}}\left(1-\mathrm{e}^{-\mathrm{kt}}\right) \\ & \frac{1}{3}=1-\mathrm{e}^{-\mathrm{kt}} \\ & \mathrm{f} \frac{2}{3}=2 \mathrm{e}^{-\mathrm{kt}} \\ & -\mathrm{kt}=8 \mathrm{n}\left(\frac{2}{3}\right) \\ & \mathrm{t}=\frac{1}{\mathrm{k}} \ell \mathrm{n}\left(\frac{2}{3}\right)\end{aligned}$$$Complete loop will enter at$\mathrm{t}=\frac{1}{\mathrm{k}} \ln \left(\frac{2}{3}\right)$Option (B)$\frac{\mathrm{d} \phi}{\mathrm{dt}}=0, \underline{\underline{\mathrm{e}}}=0, \mathrm{i}=0, \mathrm{ F}=0$ Ans. B,D)