Question

A circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment inertia I about an axis passing through its centre perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter is

• A. $I$
• B. $\frac{2I}{8}$
• C. $\frac{I}{5}$
• D. $\frac{I}{10}$

Answer 1

Answer: (C)

According to problem disc is melted and recasted into a solid sphere so their volume will be same.
$V_{\text{Disc }}=V_{\text{Sphere }}$
$\Rightarrow \pi R_{\text{Disc }}^{2}t=\frac{4}{3}\pi R_{\text{Sphere }}^3$
$\Rightarrow \pi R_{\text{Disc }}^2\left(\frac{R_{\text{Disc }}}{6}\right)=\frac{4}{3}\pi R_{\text{Sphere }}^3\left[t=\frac{R_{\text{Disc }}}{6},\text{ given }\right]$
$\Rightarrow R_{\text{Disc }}^3=8R_{\text{Sphere }}^3$
$\Rightarrow R_{\text{Sphere }}=R_{\text{Sphere }}=\frac{R_{\text{Disc }}}{2}$
Moment of inertia of disc
$I_{\text{Sphere }}=\frac{1}{2}M{R}_{\text{Sphere }}^2=\frac{2}{5}M{\left(\frac{R_{\text{Disc }}}{2}\right)}^2$
$=\frac{M}{10}{\left(R_{\text{Disc }}\right)}^2=\frac{2I}{10}=\frac{I}{5}$.