Question

From a uniform circular disc of radius R and mass $9\text{M}$, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is. (Given that moment of inertia of removed portion about the given axis is $M{R}^2/2$)

• A. $4M{R}^2$
• B. $\frac{40}{9}M{R}^2$
• C. $10M{R}^2$
• D. $\frac{37}{9}M{R}^2$

Answer 1

The correct option is A $4M{R}^2$

Moment of inertia of remaining solid
= Moment of inertia of complete solid
- Moment of inertia of removed portion
$\therefore I=\frac{9M{R}^2}{2}-\frac{M{R}^2}{2}$
$\Rightarrow I=4M{R}^2$