Q.
A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to
A) $\frac{2}{\sqrt{15}}\ R$
B) $\frac{2}{\sqrt{5}}\ R$
C) $\frac{3}{\sqrt{15}}\ R$
D) $\frac{\sqrt{3}}{\sqrt{15}}\ R$
Explanation
(a) $\frac{2}{5}MR^{2} = \frac{1}{2}Mr^{2} + Mr^{2}$
or $\frac{2}{5}\ MR^{2} = \frac{3}{2}Mr^{2}$
$\therefore$ $r = \frac{2}{\sqrt{15}}R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.