Q. The moment of inertia of a sphere of mass $M$ and radius $R$ about an axis passing through its centre is $\frac{2}{5}MR^{2}$. The radius of gyration of the sphere about a parallel axis to the above and tangent to the sphere is
A) $\frac{7}{5}R$
B) $\frac{3}{5}R$
C) $\left( \sqrt{\frac{7}{5}} \right)R$
D) $\left( \sqrt{\frac{3}{5}} \right)R$
Explanation
(c) Given, $I = \frac{2}{5}MR^{2}$
Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangential to the sphere is
$I^{'} = I + MR^{2} = \frac{2}{5}MR^{2} + MR^{2} = \frac{7}{5}MR^{2}$
$\therefore\ I^{'} = MK^{2} = \frac{7}{5}MR^{2},\ \ K = \left( \sqrt{\frac{7}{5}} \right)R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.