Question

The moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through one end is $I$. The same rod is bent into a ring and its moment of inertia about the diameter is $I_1$. The ratio $\frac{I}{I_1}$ is

• A. $\frac{4\pi }{3}$
• B. $\frac{8{\pi }^2}{3}$
• C. $\frac{5\pi }{3}$
• D. $\frac{8{\pi }^2}{5}$

Answer 1

Answer: (B)

$\frac{8{\pi }^2}{3}$

Moment of inertial of a thin uniform rod about perpendicular axis through it one end
$I=\frac{1}{3}M{l}^2$ ....(i)
Same rod is bent into a ring
$\therefore l=2\pi r\Rightarrow \frac{l}{r}=2\pi$ ....(ii)
Moment of inertia of ring about diameter
$I_1=\frac{M{R}^2}{2}$ .....(iii)
Dividing equation (i) by (ii), we get
$\frac{I}{I_1}=\frac{2}{3}\frac{M{l}^2}{M{R}^2}=\frac{2}{3}\frac{l^2}{R^2}$
$\frac{I}{I_1}=\frac{2}{3}{\left(2\pi \right)}^2=\frac{8{\pi }^2}{3}$ [using equation (ii)]