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(D) The rotational kinetic energy of a body is given by $K_{	ext{rot}} = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is t

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$2$

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(D) The rotational kinetic energy of a body is given by $K_{	ext{rot}} = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.For a spherical shell,the moment of inertia about its center is $I = \frac{2}{3} mR^2$.For pure rolling,$\omega = \frac{v}{R}$,so $K_{	ext{rot}} = \frac{1}{2} (\frac{2}{3} mR^2) (\frac{v}{R})^2 = \frac{1}{3} mv^2$.The total kinetic energy is $K_{	ext{total}} = K_{	ext{trans}} + K_{	ext{rot}} = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 = \frac{5}{6} mv^2$.The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{	ext{rot}}}{K_{	ext{total}}} = \frac{\frac{1}{3} mv^2}{\frac{5}{6} mv^2} = \frac{1}{3} 	imes \frac{6}{5} = \frac{2}{5}$.Given that the ratio is $\frac{x}{5}$,we have $\frac{x}{5} = \frac{2}{5}$,which implies $x = 2$.

For a rolling spherical shell,the ratio of rotational kinetic energy and total kinetic energy is $\frac{x}{5}$. The value of $x$ is ................

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