Physics - Rotation Question with Solution | TestHub

The moment of inertia of a solid sphere about an axis passing through its center of mass is given by $I_{cm}=\frac{2}{5}M{R}^2$.

Using the parallel axis theorem, the moment of inertia about the given axis of rotation is:

$I=I_{cm}+M{d}^2$

where $d$ is the distance from the axis of rotation to the center of mass of the sphere.

Given $R=5$ cm and $d=10$ cm, we have:

$I=\frac{2}{5}M(5{)}^2+M(10{)}^2$

$I=10M+100M=110M$

The radius of gyration $k$ is defined by $I=M{k}^2$.

$M{k}^2=110M$

$k^2=110$

$k=\sqrt{110}$ cm

Comparing this with $\sqrt{x}$ cm, we get $x=110$.