A circular disc of radius ' \(R\) ' and thickness \(\frac{R}{8}\) 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 then moment of inertia of sphere about an axis passing through diameter is

M.I. of disc, \(I=\frac{1}{2} M R_d^2\)
M.I. of sphere, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{MR}_{\mathrm{s}}^2\)
volume of disc \(=\) volume of sphere \(\ldots\) (given)
\(\begin{array}{ll}
\therefore & \pi \mathrm{R}_{\mathrm{d}}^2\left(\frac{\mathrm{R}_{\mathrm{d}}}{8}\right)=\frac{4}{3} \pi \mathrm{R}_{\mathrm{s}}^3 \\
\therefore & \mathrm{R}_{\mathrm{d}}^3=\frac{32 \mathrm{R}_{\mathrm{s}}^3}{3} \\
\therefore & \mathrm{R}_{\mathrm{s}}=\left(\frac{3}{32}\right)^{1 / 3} \mathrm{R}_{\mathrm{d}}...(i)
\end{array}\)
Substitute equation (iii) in equation (ii)
\(\begin{aligned}
\therefore \quad \mathrm{I}_{\text {sphere }} & =\frac{2}{5} \mathrm{M}\left(\frac{3}{32}\right)^{2 / 3} \mathrm{R}_{\mathrm{d}}{ }^2 \\
& \approx \frac{1}{5}\left(\frac{2}{5} \mathrm{MR}_{\mathrm{d}}^2\right)=\frac{1}{5}
\end{aligned}\)
[From (i)]