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(B) The moment of inertia $(I)$ of a thin uniform rod rotating about an axis passing through its center and perpendicular to its length is given by $I

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$\frac{1}{2}$

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(B) The moment of inertia $(I)$ of a thin uniform rod rotating about an axis passing through its center and perpendicular to its length is given by $I_1 = \frac{ML^2}{12}$.Since $I = MK^2$,where $K$ is the radius of gyration,we have $MK_1^2 = \frac{ML^2}{12}$,which gives $K_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.In the second case,the axis of rotation passes through one end and is perpendicular to the length of the rod. The moment of inertia is $I_2 = \frac{ML^2}{3}$.Using $MK_2^2 = \frac{ML^2}{3}$,we get $K_2 = \frac{L}{\sqrt{3}}$.The ratio of the radius of gyration in the first case to the second case is $\frac{K_1}{K_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{1}{2}$.

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