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}=8 R_{\text {Sphere }}^{3} $
$\Rightarrow R_{\text {Sphere }}=R_{\text {Sphere }}=\frac{R_{\text {Disc }}}{2}$
Moment of inertia of disc
$I_{\text {Sphere }}=\frac{1}{2} M R_{\text {Sphere }}^{2}=\frac{2}{5} M\left(\frac{R_{\text {Disc }}}{2}\right)^{2}$
$=\frac{M}{10}\left(R_{\text {Disc }}\right)^{2}=\frac{2 I}{10}=\frac{I}{5}$.