Question

A solid sphere of mass $M$, radius $R$ and having moment of inertia about an axis passing through the centre of mass as $I$, is recast into a disc of thickness $t$, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. Then, radius of the disc will be:

• A. $\frac{2R}{\sqrt{15}}$
• B. $R\sqrt{\frac{2}{15}}$
• C. $\frac{4R}{\sqrt{15}}$
• D. $\sqrt{\frac{2R}{15}}$

Answer 1

The correct option is A $\frac{2R}{\sqrt{15}}$

Let $R_1$ be the radius of the disc formed and also mass of the disc and the sphere must be same.

Moment of inertia of the disc about an axis passing through the centre and perpendicular to the plane, $I^{\prime }=\frac{1}{2}M{R}_1^2$

Moment of inertia of the disc about an axis passing through the edge and perpendicular to the plane, $I^{\prime \prime }=I^{\prime }+M{R}_1^2=\frac{3}{2}M{R}_1^2$

For sphere: $I=\frac{2}{5}M{R}^2$

Given: $I^{\prime \prime }=I$

$\frac{3}{2}M{R}_1^2=\frac{2}{5}M{R}^2⟹R_1=\frac{2R}{\sqrt{15}}$