A solid sphere of mass ' $m$ ', radius ' $R$ ', having moment of inertia about an axis passing through…

A solid sphere of mass ' $m$ ', radius ' $R$ ', having moment of inertia about an axis passing through center of mass as ' $I$ ' is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains ' $I$ '. Then the radius of disc is
  1. $\frac{2 \mathrm{R}}{\sqrt{15}}$
  2. $\left(\sqrt{\frac{2}{15}}\right) \mathrm{R}$
  3. $\frac{4 \mathrm{R}}{\sqrt{15}}$
  4. $\frac{\mathrm{R}}{4}$

Solution

The moment of inertia of a solid sphere $=\frac{2}{5} \mathrm{MR}^2$
Moment of inertia of a disc through its rim, $=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$
Since both the moment of inertias are equal, $\therefore \quad \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2$, where r is the radius of the disc $\therefore \quad r=\frac{2 R}{\sqrt{15}}$

Asked in: MHT CET 2024 (02 May Shift 2)

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