Question

A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is :

• A. $\frac{219 \, MR^2}{256}$
• B. $\frac{237\, MR^2}{512}$
• C. $\frac{19 \, MR^2}{512}$
• D. $\frac{197\, MR^2}{256}$

Answer 1

Answer: (B)

$I_{D}=\frac{m r^{2}}{2}$
$I_{\text {removed }}=\frac{1}{2} \frac{m}{16} \frac{r^{2}}{16}+\frac{m}{16} \frac{9 r^{2}}{16}(I m+m d)$
$=\frac{m r^{2}+18 m r^{2}}{512}$
$=\frac{19 \,m r^{2}}{512}$
$I_{\text {remaining }}=\frac{m r^{2}}{2}-\frac{19}{512} m r^{2}$
$=\frac{237}{512} m r^{2}$