Physics - Rotation Question with Solution | TestHub

Moment of inertia of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is
$I=\frac{M{L}^2}{12} \mathrm{a}\mathrm{n}\mathrm{d} I=M{K}_1^2$(where${\text{K}}_1$radius of gravitation)
$\Rightarrow K_1=\frac{L}{2\sqrt{3}}$..... (i)
When axis of rotation of rod is passing through one end of rod, then
$I=M{K}_2^2=\frac{M{L}^2}{3}\Rightarrow K_2=\frac{L}{\sqrt{3}}$..... (ii)
Taking ratios of (i) and (ii) we get
$\frac{K_1}{K_2}=\frac{L}{2\sqrt{3 }}\times \frac{\sqrt{3}}{L}=\frac{1}{2}$
$\Rightarrow \frac{K_1}{K_2}=\frac{1}{2}$