Q159 / 160 · TS EAMCET 2003 · Physics

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 :

  • A
    $I$
  • B
    $\frac{2 I}{3}$
  • Correct answer
    C
    $\frac{I}{5}$
  • D
    $\frac{I}{10}$

Solution

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, $I^{\prime}=\frac{2}{5} M R_1^2=\frac{2}{5} M\left(\frac{R}{2}\right)^2=\frac{2}{5} \frac{M R^2}{4}$ $=\frac{1}{5}\left(\frac{1}{2} M R^2\right)=\frac{I}{5}$

Hence, the correct answer is option: 'C'.

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