A body of mass \(m\) slides down an incline and reaches the bottom with a velocity \(V\). If the same mass were in the form of a disc which rolls down this incline, the velocity of the disc at bottom would have been

Case 1: \(\frac{1}{2} m V^2=m g h\)
Case 2: \(\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{\prime}\right)^2+\frac{1}{2} \mathrm{I} \omega^2=\mathrm{mgh}\)
For disc, \(\mathrm{I}=\frac{1}{2} \mathrm{mR}^2\)
\(\begin{aligned}
\therefore \quad & \frac{1}{2} m\left(v^{\prime}\right)^2+\frac{1}{2}\left(\frac{1}{2} m R^2\right) \frac{\left(v^{\prime}\right)^2}{R^2}=m g h \\
& \frac{1}{2} m\left(v^{\prime}\right)^2+\frac{1}{4} m\left(v^{\prime}\right)^2=m g h \\
& \frac{3}{4} m\left(v^{\prime}\right)^2=m g h \quad \text {....(ii) }
\end{aligned}\)
From (i) and (ii),
\(\begin{aligned}
& \frac{1}{2} m V^2=\frac{3}{4} m v^{\prime 2} \\
\therefore \quad & v^{\prime}=\sqrt{\frac{2}{3}} V
\end{aligned}\)