Q. The moment of inertia of a solid sphere about an axis passing through centre of gravity is $\frac{2}{5}{MR}^{2}$ , then its radius of gyration about a parallel axis at a distance 2$R$ from first axis is
A) $5R$
B) $\sqrt{\frac{22}{5}}R$
C) $\frac{5}{2}\ R$
D) $\sqrt{\frac{12}{5}}R$
Explanation
(b) $I = MK^{2} = \sum_{}^{}{mR^{2}}\ $
where $M$ is the total mass of the body.
This means that
$K = \sqrt{\left( \frac{I}{M} \right)}$
According to thermo of parallel axis
$I = I_{CG} + M(2R)^{2}$
[image:7_image95.emf]
where, $I_{CG}$ is moment of inertia about an axis through centre of gravity.
$\therefore$ $I = \frac{2}{5}MR^{2} + 4MR^{2} = \frac{22}{5}MR^{2}$
or $MK^{2} = \frac{22}{5}MK^{2}$
$\therefore$ $K = \sqrt{\frac{22}{5}}R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.