Question

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is I. If the same rod is bent into a ring and its moment of inertia about its diameter is I’, then the ratio $\frac{I}{I^{\prime }}$ is

• A. $\frac{2}{3}{\pi }^2$
• B. $\frac{3}{2}{\pi }^2$
• C. $\frac{5}{3}{\pi }^2$
• D. $\frac{8}{3}{\pi }^2$

Answer 1

The correct option is A

$\frac{2}{3}{\pi }^2$

For a thin rod, $I=\frac{M{L}^2}{12}$
For a ring, about diameter, $I^{\prime }=\frac{M{R}^2}{2}$
$\frac{I}{I^{\prime }}=\frac{L^2}{12}\times \frac{2}{R^2}=\frac{L^2}{6R^2}[L=2\pi R\Rightarrow R=\frac{L}{2\pi }]$
$=\frac{L^2}{6\times L^2}\times 4{\pi }^2=\frac{2}{3}{\pi }^2$