MCQ
A 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. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
  • $\frac {I}{5}$
  • B
    $\frac {I}{6}$
  • C
    $\frac {I}{32}$
  • D
    $\frac {I}{64}$

Answer

Correct option: A.
$\frac {I}{5}$
a
Moment of inertia of a disk $\mathrm{I}=\frac{\mathrm{MR}^{2}}{2}$ Moment of inertia of a solid sphere $I'$ $=$

$\frac{2}{5} \mathrm{M}^{\prime} \mathrm{R}^{\prime 2}$

$\mathrm{M}=\mathrm{M}^{\prime}$ and $\mathrm{V}=\mathrm{V}^{\prime}(\mathrm{Volume})$

$\pi \mathrm{R}^{2} \cdot \frac{\mathrm{R}}{6}=\frac{4 \pi}{3} \mathrm{R}^{3} \rightarrow \mathrm{R}^{\prime}=\frac{\mathrm{R}}{2}$

$I’=\frac{2}{5} M\left(\frac{R}{2}\right)^{2}=\frac{1}{5}\left(\frac{M R^{2}}{2}\right)=\frac{I}{5}$

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