Question

The moment of inertia of a solid sphere about an axis passing through the centre of gravity is $\frac{2}{5}M{R}^2$; 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 A $5R$

Moment of inertia of sphere= $\frac{2}{5}M{R}^2$

Moment of inertia w.r.t. $y{y}^{\prime }$ aixs:
$I_{y{y}^{\prime }}=I_{cm}+M(2R{)}^2$
$I_{y{y}^{\prime }}=\frac{2}{5}M{R}^2+M(2R{)}^2{I}_{y{y}^{\prime }}=\frac{22}{5}M{R}^2$

On considering the point mass, having radius of gyratio = $k$
$M{k}^2=\frac{22}{5}M{R}^{2}k=\sqrt{\frac{22}{5}}R$
Hence, option A is correct.