Question

A dielectric rod of mass M and length L is kept in uniform electric field $E_0$ as shown in figure. Half of the rod is charged to +Q and other half to -Q uniformly. Frequency of SHM if rod is rotated slightly and released is

• A. $\frac{1}{2\lambda }\sqrt{\frac{6Q{E}_0}{ML}}$
• B. $\frac{1}{\lambda }\sqrt{\frac{Q{E}_0}{ML}}$
• C. $\frac{1}{\lambda }\sqrt{\frac{2Q{E}_0}{ML}}$
• D. $None$

Answer 1

The correct option is A $\frac{1}{2\lambda }\sqrt{\frac{6Q{E}_0}{ML}}$


$dp=2\times 2q=2x\frac{\frac{Q}{L}}{2}\times dx=4\frac{Q}{L}\times dx$
$p=\int dp=\frac{4Q}{L}{\int }_0^{\frac{L}{2}}xdx=\frac{4Q}{L}=(\frac{x^2}{2}{)}_0^{\frac{L}{2}}=\frac{QL}{2}$
$\tau =PE\theta =\frac{QL}{2}{E}_0\theta$ ....(1)
Also $\tau =I\propto =\frac{M{L}^2}{12}\propto$ ....(2)
$\frac{M{L}^2}{12}\propto =\frac{QL{E}_0}{2}\theta \Rightarrow \propto =\left(\frac{6Q{E}_0}{ML}\right)\theta$
$w=\sqrt{\frac{6Q{E}_0}{ML}}$ and $T=\frac{w}{2\pi }=\frac{1}{2\pi }\sqrt{\frac{6Q{E}_0}{ML}}$