A block of mass M is released from the top of an inclined plane. Its velocity on reaching the bottom of the plane is v. A circular disc of the same mass M rolls down the incline plane from the top. Its velocity on reaching the bottom is $v^{\text{'}}$. The ratio $v^{\text{'}}/v$ will be

The acceleration of the block sliding down the plane is
$a=g\sin \theta$
where $\theta$ is the angle of inclination. If l is the length of the inclined plane, the velocity of the block on reaching the bottom is given by
$$\begin{gathered} v^2=2al=2g\sin \theta \times l \\ \text{ or } v=\sqrt{2gl\sin \theta } \end{gathered}$$
The acceleration of the disc rolling down the plane is 
$a^{\text{'}}=\frac{2}{3}g\sin \theta$
Therefore, the velocity of the disc on reaching the bottom is given by
$$\begin{gathered} v^{\text{'}2}=2a^{\text{'}}l=\frac{4}{3}gl\sin \theta \text{ or }{v}^{\text{'}}=2\sqrt{\frac{gl\sin \theta }{3}} \\ \frac{v^{\text{'}}}{v}=\sqrt{\frac{2}{3}} \end{gathered}$$