Q81 / 120 · JEE Advanced 2006 · Physics
$$
\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }
$$
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
- Correct answerA$\frac{2}{\sqrt{15}} R$
- B$\frac{2}{\sqrt{5}} R$
- C$\frac{3}{\sqrt{15}} R$
- D$\frac{\sqrt{3}}{\sqrt{15}} R$
Solution
$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$
Hence, the correct answer is option: 'A'.