Question

If $I^{}$ 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

Answer: (B)

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

Moment of Inertia of rod = $I_1=\frac{M{L}^2}{12}$
Moment of Inertia of ring= $I_2=M{R}^2$
Also, $2\pi R=LR=\frac{L}{2\pi }$
Therefore $\frac{I_1}{I_2}=\frac{\frac{M{L}^2}{12}}{\frac{M{L}^2}{{\left(2\pi \right)}^2}}\frac{I_1}{I_2}=\frac{{\pi }^2}{3}$