Q. A circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is
A) $I\ $
B) $\frac{2I}{3}$
C) $\frac{I}{5}$
D) $\frac{I}{10}$
Explanation
(c) Moment of inertia of a disc
$I = \frac{1}{2}MR^{2}$
Disc is melted and recasted into a solid sphere.
$\therefore$ Volume of sphere=Volume of disc
$\frac{4}{3}\pi R_{1}^{3} = \pi R^{2} \times \frac{R}{6}$
$\frac{4}{3}R_{1}^{3} = \frac{R^{3}}{6}$
$R_{1}^{3} = \frac{R^{3}}{8}$ $\Longrightarrow$ $R_{1} = \frac{R}{2}$
$\therefore$ Moment of inertia of sphere
$I^{'} = \frac{2}{5}MR_{1}^{2} = \frac{2}{5}M\left( \frac{R}{2} \right)^{2} = \frac{2}{5}\frac{MR^{2}}{4} = \frac{1}{5}\left( \frac{1}{2}MR^{2} \right) = \frac{I}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.