Question

A wire of mass $m$ and length $I$ is bent in the form of a circular ring, the moment of inertia of the ring about its axis is

• A. $\left(\frac{1}{8{\pi }^2}\right)m{l}^2$
• B. $\left(\frac{1}{2{\pi }^2}\right)m{l}^2$
• C. $\left(\frac{1}{4{\pi }^2}\right)m{l}^2$
• D. $m{l}^2$

Answer 1

The correct option is C $\left(\frac{1}{4{\pi }^2}\right)m{l}^2$

Length of the wire $=l$.
Let it is bent in the form of a circular ring of radius r.
Thus, radius of the ring
$r=\frac{l}{2\pi }$
The moment of inertia of the ring about its axis,
$I=m{r}^2=m{\left(\frac{1}{2\pi }\right)}^2=m\times \frac{l^2}{4{\pi }^2}$
$=\left(\frac{1}{4{\pi }^2}\right)m{l}^2$.