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A disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia I about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
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Verified Answer
The correct answer is:
$\frac{\mathrm{I}}{5}$
M.I. of disc, $I=\frac{1}{2} \mathrm{MR}_{\mathrm{d}}^2$... (i)
M.I. of sphere, $I_{\text {sphere }}=\frac{2}{5} \mathrm{MR}_{\mathrm{s}}^2$... (ii)
$\because \quad$ volume of disc $=$ volume of sphere
$\begin{array}{ll}
\therefore & \pi \mathrm{R}_{\mathrm{d}}^2\left(\frac{\mathrm{R}_{\mathrm{d}}}{6}\right)=\frac{4}{3} \pi \mathrm{R}_{\mathrm{s}}^3 \\
\therefore & \mathrm{R}_{\mathrm{d}}^3=8 \mathrm{R}_{\mathrm{s}}^3 \\
\therefore & \mathrm{R}_{\mathrm{S}}=\frac{\mathrm{R}_{\mathrm{d}}}{2} ... (iii)
\end{array}$
Substitute equation (iii) in equation (ii)
$\begin{aligned}
\therefore \quad \mathrm{I}_{\text {sphere }} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}_{\mathrm{d}}}{2}\right)^2=\frac{2}{5} \times \frac{1}{4} \mathrm{MR}_{\mathrm{d}}^2 \\
& =\frac{1}{5}\left(\frac{1}{2} \mathrm{MR}_{\mathrm{d}}^2\right)=\frac{\mathrm{I}}{5}....[from (i)]
\end{aligned}$
M.I. of sphere, $I_{\text {sphere }}=\frac{2}{5} \mathrm{MR}_{\mathrm{s}}^2$... (ii)
$\because \quad$ volume of disc $=$ volume of sphere
$\begin{array}{ll}
\therefore & \pi \mathrm{R}_{\mathrm{d}}^2\left(\frac{\mathrm{R}_{\mathrm{d}}}{6}\right)=\frac{4}{3} \pi \mathrm{R}_{\mathrm{s}}^3 \\
\therefore & \mathrm{R}_{\mathrm{d}}^3=8 \mathrm{R}_{\mathrm{s}}^3 \\
\therefore & \mathrm{R}_{\mathrm{S}}=\frac{\mathrm{R}_{\mathrm{d}}}{2} ... (iii)
\end{array}$
Substitute equation (iii) in equation (ii)
$\begin{aligned}
\therefore \quad \mathrm{I}_{\text {sphere }} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}_{\mathrm{d}}}{2}\right)^2=\frac{2}{5} \times \frac{1}{4} \mathrm{MR}_{\mathrm{d}}^2 \\
& =\frac{1}{5}\left(\frac{1}{2} \mathrm{MR}_{\mathrm{d}}^2\right)=\frac{\mathrm{I}}{5}....[from (i)]
\end{aligned}$
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