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(B) The total kinetic energy of a rolling body is given by $K E = \frac{1}{2} m v^2 (1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.For a r

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$\sqrt{3}: \sqrt{4}$

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(B) The total kinetic energy of a rolling body is given by $K E = \frac{1}{2} m v^2 (1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.For a ring,the moment of inertia $I = mR^2$,so $k^2 = R^2$. Thus,$K E_{	ext{ring}} = \frac{1}{2} m v_R^2 (1 + 1) = m v_R^2$.For a disc,the moment of inertia $I = \frac{1}{2} mR^2$,so $k^2 = \frac{1}{2} R^2$. Thus,$K E_{	ext{disc}} = \frac{1}{2} m v_d^2 (1 + \frac{1}{2}) = \frac{3}{4} m v_d^2$.Given $K E_{	ext{ring}} = K E_{	ext{disc}}$,we have $m v_R^2 = \frac{3}{4} m v_d^2$.Therefore,$\frac{v_R^2}{v_d^2} = \frac{3}{4}$,which implies $\frac{v_R}{v_d} = \frac{\sqrt{3}}{\sqrt{4}}$.Thus,the ratio of the velocity of the ring to the disc is $\sqrt{3}: \sqrt{4}$.

$A$ disc and a ring of the same mass are rolling. If their kinetic energies are equal,then the ratio of their velocities will be

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