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. $\frac{R}{4}$

Answer 1

Answer: (A)

$\frac{2R}{\sqrt{15}}$

Answer 2

Moment of inertia of solid sphere of mass M and radius R about an axis passing through the centre of mass is $I=\frac{2}{5}MR^{2}$. Let the radius of the disc is $r.$ Moment of inertia of circular disc of radius $r$ and mass $M$ about an axis passing through the centre of mass and perpendicular to its plane $=\frac{1}{2}Mr^{2}$ Using theorem of parallel axes, moment of inertia of disc about its edge is $I '=\frac{1}{2}Mr^{2}+Mr^{2}=\frac{3}{2}Mr^{2}$ Given, $I = I ?$\therefore \frac{2}{5}MR^{2}=\frac{3}{2}Mr^{2}$or$r^{2}=\frac{4}{15}R^{2}$or$r=\frac{2R}{\sqrt{15}}$