Question

The moment of inertia of a sphere of mass $M$ and radius $R$ about an axis passing through its centre is $\frac{2}{5}M{R}^2.$ 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. $\left(\sqrt{\frac{3}{5}}\right)R$

Answer 1

Answer: (C)

Given, $I=\frac{2}{5}M{R}^2$
Using the theorem of parallel axes, moment of inertia of the sphere about a parallel axis tangential to the sphere is
$I^{\prime }=I+M{R}^2=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$
$\therefore I^{\prime }=M{K}^2=\frac{7}{5}M{R}^2,K=\left(\sqrt{\frac{7}{5}}\right)R$
(Here, $K$ is radius of gyrations)