Question

A circular disc of radius $R$ and thickness $\frac{R}{6}$ has a moment inertia $I$ about an axis perpendicular to the plane of plate and passing through its centre. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is:

• A. $\frac{I}{5}$
• B. $\frac{2I}{5}$
• C. $\frac{4I}{5}$
• D. $\frac{I}{10}$

Answer 1

Answer: (A)

$\frac{I}{5}$

Moment of Inertia of disc $I=\frac{M{R}^2}{2}$
Volume of disc $=\left(\pi R^2\right)\frac{R}{6}=\frac{\pi R^3}{6}$
Now, $\frac{\pi R^3}{6}=\frac{4}{3}\pi R_1^3$
So, $R_1^3=\frac{R^3}{8}\Rightarrow R_1=\frac{R}{2}$
Hence, Moment of Inertia of solid sphere $=\frac{2}{5}M{R}_1^2=\frac{2}{5}\times \frac{M\times R^2}{4}=\frac{1}{5}\left(\frac{M{R}^2}{2}\right)=\frac{I}{5}$