Question

The moment of inertia of a solid sphere about an axis passing through the centre of gravity is $2/5M{R}^2$, then its radius of gyration about a parallel axis at a distance $2R$ from first axis is

• A. $5R$
• B. $\sqrt{\frac{22}{5}}R$
• C. $\frac{5}{2}R$
• D. $\sqrt{\frac{12}{5}}R$

Answer 1

The correct option is B $\sqrt{\frac{22}{5}}R$

By, Parallel axis theorm

$I$ at $2R$ is $2M{R}^2/5+M(2R{)}^2=22M{R}^2/5$

Radius of gyration, $M{k}^2=22M{R}^2/5\Rightarrow k=\sqrt{22/5}\times R$.