A solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^\circ$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^\circ$ while keeping $L$ constant, then the new speed of the sphere at the bottom of the plane is $v_2$. The ratio $v_1^2 : v_2^2$ is
Answer: (A) $1 : \sqrt2$
For rolling without slipping from height $h$:
$$mgh = \frac{1}{2}mv^2\left(1 + \frac{k^2}{r^2}\right) \Rightarrow v^2 = \frac{2gh}{1 + k^2/r^2}$$
With the same sphere, $v^2 \propto h = L\sin\theta$:
$$\frac{v_1^2}{v_2^2} = \frac{\sin30^\circ}{\sin45^\circ} = \frac{1/2}{1/\sqrt2} = \frac{1}{\sqrt2}$$
Solution by Sreeraj P, M.Sc Physics