Let M and L be the mass and length of a thin | vedclass

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(B) The radius of gyration $k$ is given by the formula $I = Mk^2$, where $I$ is the moment of inertia and $M$ is the mass of the body.Case $1$: Axis p

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$1/2$

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(B) The radius of gyration $k$ is given by the formula $I = Mk^2$, where $I$ is the moment of inertia and $M$ is the mass of the body.Case $1$: Axis passes through the centre and is perpendicular to the length.The moment of inertia is $I_1 = \frac{ML^2}{12}$.Thus, $Mk_1^2 = \frac{ML^2}{12} \implies k_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.Case $2$: Axis passes through one end and is perpendicular to the length.The moment of inertia is $I_2 = \frac{ML^2}{3}$.Thus, $Mk_2^2 = \frac{ML^2}{3} \implies k_2 = \frac{L}{\sqrt{3}}$.The ratio of the radius of gyration in the first case to the second case is:$\frac{k_1}{k_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$.

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