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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. The time t , after itself parallel to electric field after releasing it is :


$(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} $

1 Answer

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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.
answered Jan 15, 2014 by meena.p

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