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(D) The velocity $V$ of a body rolling down an inclined plane of height $H$ without slipping is given by the formula: $V = \sqrt{\frac{2gH}{1 + k^2/R^

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$\sqrt{\frac{14}{15}}$

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(D) The velocity $V$ of a body rolling down an inclined plane of height $H$ without slipping is given by the formula: $V = \sqrt{\frac{2gH}{1 + k^2/R^2}}$,where $k$ is the radius of gyration.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2/R^2 = 1/2$.For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $k^2/R^2 = 2/5$.The ratio of velocities is given by: $\frac{V_{	ext{cylinder}}}{V_{	ext{sphere}}} = \sqrt{\frac{1 + k_{	ext{sphere}}^2/R^2}{1 + k_{	ext{cylinder}}^2/R^2}}$.Substituting the values: $\frac{V_{	ext{cylinder}}}{V_{	ext{sphere}}} = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} 	imes \frac{2}{3}} = \sqrt{\frac{14}{15}}$.

$A$ solid cylinder and a solid sphere,having same mass $M$ and radius $R$,roll down the same inclined plane from the top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere,with which they reach the ground,will be:

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