question_answer

A ring is suspended from its one end and oscillating then its time period for small oscillations will be :-

A)  \[2\pi \sqrt{\frac{R}{g}}\]            

B)  \[2\pi \sqrt{\frac{2R}{g}}\]

C)  \[2\pi \sqrt{\frac{R}{2g}}\]          

D)  \[2\pi \sqrt{\frac{3R}{2g}}\]

Correct Answer: B

Solution :

\[T=2\pi \sqrt{\frac{\ell +\frac{{{K}^{2}}}{\ell }}{g}}\] Here \[\ell =R,\]\[M{{K}^{2}}=M{{R}^{2}}\Rightarrow K=R\] \[\Rightarrow \]  \[T=2\pi \sqrt{\frac{R+R}{g}}=2\pi \sqrt{\frac{2R}{g}}\]