A circular disc of radius ‘R’ and thickness $\text{R/6}$ 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 M.I of the sphere about its diameter as axis of rotation is

Radius = R

Thickness (t) $=\frac{\text{R}}{\text{6}}$

Volume = Area × thickness

$\text{v}=\text{A}\times \text{t}$

$=\pi {\text{R}}^2\times \frac{\text{R}}{\text{6}}=\frac{\pi {\text{R}}^3}{6}$

${\text{r}}_{\text{S}}=?$

${\text{v}}_{\text{s}}\to$ Volume of a sphere

${\text{v}}_{\text{s}}=\text{v}$

$\frac{4}{3}\pi {\text{r}}_{\text{s}}^{\text{3}}=\frac{\pi {\text{R}}^3}{6}$ ( $\because$ Disc method and recasted as a solid sphere)

${\text{r}}_{\text{s}}=\frac{\text{R}}{\text{2}}$

$I=\frac{{\text{MR}}^{\text{2}}}{\text{2}}$

$I_s=\frac{2}{5}{\text{Mr}}_{\text{s}}^{\text{2}}$

$=\frac{2}{5}{\text{Mr}}_{\text{s}}^{\text{2}}$

$=\frac{2}{5}\text{M}\times \frac{{\text{R}}^{\text{2}}}{\text{4}}$

$=\frac{{\text{MR}}^{\text{2}}}{\text{2}}\times \frac{1}{5}$

$I_{\text{s}}=\frac{\text{I}}{\text{5}}$