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(B) For a body rolling down an inclined plane of height $h$,the conservation of energy gives $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $v

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$3$

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(B) For a body rolling down an inclined plane of height $h$,the conservation of energy gives $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $v = R\omega$,we have $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{I}{mR^2})$.For a ring,$I = mR^2$,so $mgh = \frac{1}{2}mv_R^2(1 + 1) = mv_R^2$. Thus,$v_R = \sqrt{gh}$.For a solid cylinder,$I = \frac{1}{2}mR^2$,so $mgh = \frac{1}{2}mv_c^2(1 + \frac{1}{2}) = \frac{3}{4}mv_c^2$. Thus,$v_c = \sqrt{\frac{4gh}{3}}$.The ratio of velocities is $\frac{v_R}{v_c} = \frac{\sqrt{gh}}{\sqrt{4gh/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.Comparing this to $\frac{\sqrt{x}}{2}$,we get $x = 3$.

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