Q. 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) $\frac{R}{4}$
Explanation
(a) $\frac{2}{5}MR^{2} = \frac{3}{2}Mr^{2} \Rightarrow r = \frac{2R}{\sqrt{15}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.