question_answer

A rod of length L and mass M is bent to form a semicircular ring as shown in figure. The moment of inertia about XY is                     

A)  \[\frac{1}{4}\frac{M{{L}^{2}}}{{{\pi }^{2}}}\]                                      

B) \[\frac{2M{{L}^{2}}}{3{{\pi }^{2}}}\]

C) \[\frac{M{{L}^{2}}}{2}\]                

D)        \[\frac{M{{L}^{2}}}{2}\]

Correct Answer: A

Solution :

Here,                 \[L=\pi R\]or\[R=\frac{L}{\pi }\] \[\therefore \]Moment of inertia about XY                         \[=\frac{1}{2}\left( \frac{1}{2}M{{R}^{2}} \right)\]                 \[=\frac{1}{4}\frac{M{{L}^{2}}}{{{\pi }^{2}}}\]