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Two point particles of same mass M are attached to two ends of a massless non-conducting rigid rod of length l. The charge on particle are +q and -q. The rod is held in a region of uniform electric field E making an angle $\theta $( very small) with E. When rod is released , its angular frequency will be :


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

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The force acting at both the charged constitute a couple.
$\tau$= Force $\times$ perpendicular disatnce
=>$\tau= qE \times AB \sin \theta$
$\qquad= qEl \sin \theta$
Since , $\theta$ is small
$\therefore \sin \theta =0$
=>$ \tau =qEl \theta$
Restoring torque $\tau= -qEL \theta$
If $\alpha $ is angular acceleration
$\tau=I \alpha$
$I= \large\frac{Ml^2}{2}$
=> $\tau= \large\frac{Ml^2}{2}$$ \alpha$
=> $\alpha =-w^2 \theta$
Now, $\alpha =-w^2 \theta$
=> $w^2=\bigg(\large\frac{2qE}{Ml}\bigg)$
=> $w=\bigg(\large\frac{2qE}{Ml}\bigg)^{1/2}$
Hence B is the correct answer.


answered Jan 15, 2014 by meena.p
edited Aug 11, 2014 by thagee.vedartham

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