Question

A bob is suspended from an ideal string of length L. Now it is pulled to a side till the string makes an angle ${60}^o$ to the vertical and whirled along a horizontal circle then its period of revolution it:

• A. $\pi \sqrt{\frac{L}{g}}$
• B. $\pi \sqrt{\frac{L}{2g}}$
• C. $\pi \sqrt{\frac{3L}{2g}}$
• D. $\pi \sqrt{\frac{2L}{g}}$

Answer 1

The correct option is D $\pi \sqrt{\frac{2L}{g}}$

$\begin{array}{c}T=2\pi \sqrt{\frac{l\cos {60}^0}{g}}\\ =2\pi \sqrt{\frac{l}{2g}}\\ =\pi \sqrt{\frac{2l}{g}}\\ Hence,\\ optionDiscorrectanswer.\end{array}$