Question

A wire of length 'I' is bent in the form of a circular ring, moment of inertia about its axis is

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

Answer 1

The correct option is C $\left(\frac{I}{4{\pi }^2}\right)M{I}^2$

Lets consider $M=$ mass of ring

$I=$ length of circular ring

$r=$ radius of circular ring

Length of wire is equal to the circumference of circular ring,

$I=2\pi r$

$r=\frac{I}{2\pi }$. . . .(1)

The moment of inertia about the axis of circular ring is given by

$M{r}^2=\frac{M{I}^2}{4{\pi }^2}=\left(\frac{1}{4{\pi }^2}\right)M{I}^2$

The correct option C.