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(D) For a solid sphere rolling down an inclined plane without slipping, the conservation of energy gives: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omeg

Vedclass · Accepted Answer

$\sqrt{5} : \sqrt{7}$

Vedclass · Answer

(D) For a solid sphere rolling down an inclined plane without slipping, the conservation of energy gives: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omega^2$. Since $I = \frac{2}{5}mr^2$ and $\omega = \frac{v_1}{r}$, we have: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_1}{r})^2 = \frac{1}{2}mv_1^2 + \frac{1}{5}mv_1^2 = \frac{7}{10}mv_1^2$. Thus, $v_1 = \sqrt{\frac{10gh}{7}}$. When the sphere slides down (no friction), the potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv_2^2$. Thus, $v_2 = \sqrt{2gh}$. The ratio $V_1 : V_2 = \sqrt{\frac{10gh}{7}} : \sqrt{2gh} = \sqrt{\frac{10}{7}} : \sqrt{2} = \sqrt{\frac{5}{7}} = \sqrt{5} : \sqrt{7}$.

$A$ solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '$V_1$'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '$V_2$'. The ratio $V_1 : V_2$ is (neglect friction).

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