Question

A disc has mass $9M$ . A small disc of radius $R/3$ is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre ${}^{\prime }{O}^{\prime }$ of the disc and perpendicular to the plane of the disc is :

• A. $8M{R}^2$
• B. $4M{R}^2$
• C. $\frac{40}{9}M{R}^2$
• D. $\frac{37}{9}M{R}^2$

Answer 1

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

Mass of the disc $=9M$
Mass of the removed portion of disc $=M$
The moment of inertia of the complete disc about an axis passing through its centre $O$ and perpendicular to its plane is
$I_1=\frac{9}{2}M{R}^2$
Now, the moment of inertia of the disc with removed portion
$I_2=\frac{1}{2}M{\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2=\frac{1}{18}M{R}^2+\frac{4}{9}M{R}^2=\frac{1}{2}M{R}^2$
Therefore, moment of inertia of the remaining portion of disc about $O$ is
$=9\frac{M{R}^2}{2}-\frac{M{R}^2}{2}=4M{R}^2$