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

The correct option is A $\frac{2}{\sqrt{15}}R$

lets say solid sphere has mass $=M$ and

It is known that moment of inertia of a solid sphere about its geometrical axis =$\frac{2}{5}M{R}^2$

$I=\frac{2}{5}M{R}^2$

$M=\frac{5I}{2R^2}$

moment of Inertia of disc of mass $M$ and radius $r$ about COM is given by

$I^{\prime }=\frac{1}{2}M{r}^2$.

from perpendicular axis theorem:

$I_t^{\prime }=I^{\prime }+m{d}^2$ here $d=r$

$I_t^{\prime }=\frac{1}{2}M{r}^2+M{r}^2$

$I_t^{\prime }=\frac{3}{2}M{r}^2$

according to question $I_t^{\prime }=I$

so $I=\frac{3}{2}M{r}^2$

putting value of $M=\frac{5I}{2R^2}$

$I=\frac{3}{2}\times \frac{5I}{2R^2}{r}^2$

$r=\frac{2}{\sqrt{15}}R$