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(B) The moment of inertia of a solid sphere about its diameter is $I_s = \frac{2}{5} mR^2$. The moment of inertia of a solid cylinder about its geomet

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$1: 5$

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(B) The moment of inertia of a solid sphere about its diameter is $I_s = \frac{2}{5} mR^2$. The moment of inertia of a solid cylinder about its geometrical axis is $I_c = \frac{1}{2} mR^2$. Let the angular speed of the sphere be $\omega$. Then the angular speed of the cylinder is $2\omega$. The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$. Kinetic energy of the sphere: $K_s = \frac{1}{2} I_s \omega^2 = \frac{1}{2} (\frac{2}{5} mR^2) \omega^2 = \frac{1}{5} mR^2 \omega^2$. Kinetic energy of the cylinder: $K_c = \frac{1}{2} I_c (2\omega)^2 = \frac{1}{2} (\frac{1}{2} mR^2) (4\omega^2) = mR^2 \omega^2$. The ratio of kinetic energy of the sphere to the cylinder is $\frac{K_s}{K_c} = \frac{\frac{1}{5} mR^2 \omega^2}{mR^2 \omega^2} = \frac{1}{5}$. Thus,the ratio is $1: 5$.

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