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 (about central axis) of the ring formed by bending the rod, then

• A. $I_1:I_2=1:1$
• B. $I_1:I_2={\pi }^2:3$
• C. $I_1:I_2=\pi :4$
• D. $I_1:I_2=3:5$

Answer 1

The correct option is B $I_1:I_2={\pi }^2:3$

$I_1=\frac{M{l}^2}{12}$

$\Rightarrow I_2=M{R}^2$

$\Rightarrow l=2\pi R$

$\Rightarrow R=\frac{l}{2\pi }$

$\Rightarrow I=\frac{M{l}^2}{4{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{M{l}^2}{12}\times \frac{4{\pi }^2}{M{l}^2}=\frac{{\pi }^2}{3}$

$\Rightarrow I_1:I_2={\pi }^2:3.$

Hence, the answer is $I_1:I_2={\pi }^2:3.$