Question

If $I_1$ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and $I_2$ is the moment of inertia of the ring (formed by bending the rod) about an axis perpendicular to the plane of ring and passing through its centre, then the ratio $\frac{I_1}{I_2}$ is

• A. $\frac{3}{{\pi }^2}$
• B. $\frac{2}{{\pi }^2}$
• C. $\frac{{\pi }^2}{2}$
• D. $\frac{{\pi }^2}{3}$

Answer 1

The correct option is D $\frac{{\pi }^2}{3}$

let $L$ be the length of rod,
$\Rightarrow I_1=\frac{m{L}^2}{12}$
Mass of the ring formed by bending the rod will be $m$ and for the radius of ring,
$2\pi r=L$
$r=\frac{L}{2\pi }$
The moment of inertia of ring about an axis $\perp$ to its plane and passing through its centre is,
$I_2=m{r}^2$
$I_2=\frac{m{L}^2}{4{\pi }^2}$
$\Rightarrow \frac{I_1}{I_2}=\frac{{\pi }^2}{3}$