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
$I'=I+M R^{2}=\frac{2}{5} M R^{2}+M R^{2}=\frac{7}{5} M R^{2}$
$\therefore I'=M K^{2}=\frac{7}{5} M R^{2}, K=\left(\sqrt{\frac{7}{5}}\right) R$
(Here, $K$ is radius of gyrations)