Q125 / 125 · BITSAT 2016 · Physics
A circular disc of radius $\mathrm{R}$ 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$\frac{2 I}{8}$
- Correct answerC$\frac{I}{5}$
- D$\frac{I}{10}$
Solution
According to problem disc is melted and recasted into a solid sphere so their volume
$$
\begin{array}{c}
\text { will be same. } \\
V_{\text {Disc }}=V_{\text {Sphere }} \Rightarrow \pi R_{\text {Disc }}^{2} t=\frac{4}{3} \pi R_{\text {Sphere }}^{3} \\
\Rightarrow \pi R_{\text {Disc }}^{2}\left(\frac{R_{\text {Disc }}}{6}\right)=\frac{4}{3} \pi R_{\text {Sphere }}^{3}\left[t=\frac{R_{\text {Disc }}}{6}, \text { given }\right] \\
\Rightarrow R_{\text {Disc }}^{3}=8 R_{\text {Sphere }}^{3} \Rightarrow R_{\text {Sphere }}=\frac{R_{\text {Disc }}}{2}
\end{array}
$$
Moment of inertia of disc
$$
\begin{array}{l}
I_{\text {Disc }}=\frac{1}{2} M R_{\text {Disc }}^{2}=I \text { (given) } \\
\therefore \text { M }\left(R_{\text {Disc }}\right)^{2}=2 I
\end{array}
$$
Moment of inertia of sphere
$$
\begin{array}{l}
I_{\text {Sphere }}=\frac{1}{2} M R_{\text {Sphere }}^{2} \\
=\frac{2}{5} M\left(\frac{R_{\text {Disc }}}{2}\right)^{2}=\frac{M}{10}\left(R_{\text {Disc }}\right)^{2}=\frac{2 I}{10}=\frac{I}{5}
\end{array}
$$
Hence, the correct answer is option: 'C'.