$(A)\;2 \pi ( \frac{Ml}{2qE})^{1/2} \\ (B)\;2 \pi ( \frac{2Ml}{qE})^{1/2} \\ (C)\;2 \pi ( \frac{Ml}{qE})^{1/2} \\ (D)\;\frac{\pi}{2} ( \frac{Ml}{2qE})^{1/2} $

The force acting at both the charged constitute a couple.

z= Force $\times$ perpendicular disatnce

=>$z= qE \times AB \sin \theta$

$\qquad= qEl \sin \theta$

Since , $\theta$ is small

$\therefore \sin \theta =0$

=>$ z =qEl \theta$

Restoring torque $z= -qEL \theta$

If $\alpha $ is angular acceleration

$z=I \alpha$

$I= \large\frac{Ml^2}{2}$

=> $z= \large\frac{Ml^2}{2}$$ \alpha$

=> $\alpha =-w^2 \theta$

Now, $\alpha =-w^2 \theta$

=> $w^2=\bigg(\large\frac{2qE}{Ml}\bigg)$

=> $w=\bigg(\large\frac{2qE}{Ml}\bigg)^{1/2}$

Time period of oscillation $=\large\frac{2 \pi}{w}$

=> $ T= 2 \pi \bigg( \large\frac{Ml}{2qE}\bigg) ^{1/2}$

Rotating in clock wise direction, minimum time taken by rod to align itself parallel to electric field is time taken to complete $1/4\; th$ oscillation.

=> $ t= \large\frac{T}{4} =\frac{\pi}{2} \bigg( \large\frac{ Ml}{2qE}\bigg)^{1/2}$

Hence D is the correct answer.

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