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 $I^{\prime }$, 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

Answer: (A)

We know that, radius of ring,
$R=\frac{L}{2\pi }\dots (i)$
Moment of inertia of thin uniform rod,
$I=\frac{M{L}^2}{12}\dots (ii)$
and same rod is bent into a ring, then its moment of inertia,
$I^{\prime }=\frac{1}{2}M{R}^2$
From Eq. (i).
$I^{\prime }=\frac{1}{2}\frac{M{L}^2}{4{\pi }^2}$
$I^{\prime }=\frac{M{L}^2}{8{\pi }^2}$
On dividing Eq. (ii) by Eq. (iii), we get
$\frac{I}{I^{\prime }}=\frac{M{L}^2}{12}\times \frac{8{\pi }^2}{M{L}^2}$
$\frac{I}{I^{\prime }}=\frac{8{\pi }^2}{12}$
$\frac{I}{I^{\prime }}=\frac{2{\pi }^2}{3}$