Question

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. $\frac{\sqrt{3}}{\sqrt{15}}R$

Answer 1

Answer: (A)

$\frac{2}{5}M{R}^2=\frac{1}{2}M{r}^2+M{r}^2$
or $\frac{2}{5}M{R}^2=\frac{3}{2}M{r}^2$
$\therefore$ $r=\frac{2}{\sqrt{15}}R$