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(D) The total kinetic energy $(K_{total})$ of a rolling sphere is the sum of its linear kinetic energy $(K_{linear})$ and rotational kinetic energy $(

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$2 : 7$

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(D) The total kinetic energy $(K_{total})$ of a rolling sphere is the sum of its linear kinetic energy $(K_{linear})$ and rotational kinetic energy $(K_{rotational})$.$K_{total} = K_{linear} + K_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the angular velocity is $\omega = \frac{v}{r}$.Substituting these values into the rotational kinetic energy expression:$K_{rotational} = \frac{1}{2} \left(\frac{2}{5}mr^2ight) \left(\frac{v}{r}ight)^2 = \frac{1}{5}mv^2$.Now,calculating the total kinetic energy:$K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left(\frac{5+2}{10}ight)mv^2 = \frac{7}{10}mv^2$.The ratio of rotational kinetic energy to total kinetic energy is:$\frac{K_{rotational}}{K_{total}} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{1}{5} 	imes \frac{10}{7} = \frac{2}{7}$.Thus,the ratio is $2:7$.

If a sphere is rolling,the ratio of its rotational energy to the total kinetic energy is given by

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