Question

A rod of length $l$ and mass $m$ is bent in shape of a ring. The moment of inertia of the ring about any of its diameter is:

• A. $\frac{m{l}^2}{{\pi }^2}$
• B. $\frac{m{l}^2}{4{\pi }^2}$
• C. $\frac{m{l}^2}{2{\pi }^2}$
• D. $\frac{m{l}^2}{8{\pi }^2}$

Answer 1

The correct option is D $\frac{m{l}^2}{8{\pi }^2}$

As per given,
Circumference of the ring = Length of the rod
$\Rightarrow 2\pi R=l$
$\Rightarrow R=\frac{l}{2\pi }$
Now, we know that MOI of ring about any of its diameter is,
$I=\frac{m{R}^2}{2}$
$=\frac{m{l}^2}{8{\pi }^2}$