The radius of gyration of a solid sphere of r | vedclass

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(C) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is $I_d = (2/5)MR^2$. Since $I = MK^2$, we have $MK_d^2 = (2

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$(2/7)^{1/2}$

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(C) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is $I_d = (2/5)MR^2$. Since $I = MK^2$, we have $MK_d^2 = (2/5)MR^2$, which gives $K_d = R \sqrt{2/5}$. The moment of inertia of a solid sphere about its tangent is given by the parallel axis theorem: $I_t = I_{cm} + MR^2 = (2/5)MR^2 + MR^2 = (7/5)MR^2$. Since $MK_t^2 = (7/5)MR^2$, we have $K_t = R \sqrt{7/5}$. The ratio $K_d/K_t$ is $\frac{R \sqrt{2/5}}{R \sqrt{7/5}} = \sqrt{\frac{2/5}{7/5}} = \sqrt{2/7} = (2/7)^{1/2}$.

The radius of gyration of a solid sphere of radius $R$ and mass $M$ about its diameter is $K_d$ and that about a tangent of a solid sphere is $K_t$. The ratio of $K_d$ to $K_t$ is

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