Question

The moment of inertia of a solid sphere about an axis passing through 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

Answer: (B)

$I=M{K}^2=\mathbf{\Sigma }M{R}^2$
$\Rightarrow K=\sqrt{\left(\frac{I}{M}\right)}$
where M is the total mass of the body.
According to theorem of parallel axis
$I=I_{CG}+M(2R{)}^2$
where, $I_{CG}$ is moment of inertia about an axis through centre of gravity.
$\therefore I=\frac{2}{5}M{R}^2+4M{R}^2=\frac{22}{5}M{R}^2$
or $M{K}^2=\frac{22}{5}M{R}^2$
$\therefore K=\sqrt{\frac{22}{5}}R$