$(a)\;2\pi \sqrt{\large\frac{I}{PE}}\qquad(b)\;\pi \sqrt{\large\frac{I}{PE}}\qquad(c)\;\pi \sqrt{\large\frac{PE}{I}}\qquad(d)\;2\pi \sqrt{\large\frac{PE}{I}}$

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Answer : (a) $\;2\pi \sqrt{\large\frac{I}{PE}}$

Explanation :

When displaced at an angle $\;theta$, from its mean position the magnitude of restoring torque is

$\tau=-P sin \theta$

For small angular displacement $\;sin \theta \approx \theta$

$\tau=-P E \theta$

$\alpha =\large\frac{\tau}{I}=-(\large\frac{PE}{I})\;\theta=-w^2 \theta$

$w^2=\large\frac{P E}{I}$

$T=2 \pi \sqrt{\large\frac{I}{PE}}\;.$

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