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A circular 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. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is :
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The correct answer is:
$\frac{I}{5}$
Moment of inertia of a disc,
$I=\frac{1}{2} M R^2$
Disc is melted and recasted into a solid sphere.
$\therefore$ Volume of sphere $=$ Volume of disc
$\frac{4}{3} \pi R_1^3=\pi R^2 \times \frac{R}{6} \Rightarrow \frac{4}{3} R_1^3=\frac{R^3}{6}$
$R_1^3=\frac{R^3}{8} \Rightarrow R_1=\frac{R}{2}$
$\therefore \quad$ Moment of inertia of sphere,
$I^{\prime}=\frac{2}{5} M R_1^2=\frac{2}{5} M\left(\frac{R}{2}\right)^2=\frac{2}{5} \frac{M R^2}{4}$
$=\frac{1}{5}\left(\frac{1}{2} M R^2\right)=\frac{I}{5}$
$I=\frac{1}{2} M R^2$
Disc is melted and recasted into a solid sphere.
$\therefore$ Volume of sphere $=$ Volume of disc
$\frac{4}{3} \pi R_1^3=\pi R^2 \times \frac{R}{6} \Rightarrow \frac{4}{3} R_1^3=\frac{R^3}{6}$
$R_1^3=\frac{R^3}{8} \Rightarrow R_1=\frac{R}{2}$
$\therefore \quad$ Moment of inertia of sphere,
$I^{\prime}=\frac{2}{5} M R_1^2=\frac{2}{5} M\left(\frac{R}{2}\right)^2=\frac{2}{5} \frac{M R^2}{4}$
$=\frac{1}{5}\left(\frac{1}{2} M R^2\right)=\frac{I}{5}$
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