A solid cylinder of radius R is at rest at a height h on an inclined plane. If it rolls down then its velocity on reaching the ground is

The given situation is shown below
 
Let v be the velocity of solid cylinder on reaching the ground.
According to conservative of energy,
Total energy at point A = Total energy at point B
i.e,  $mgh+0=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2+0\Rightarrow mgh=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2+\frac{1}{2}m{v}^2$
$\Rightarrow (where,I=\frac{1}{2}M{R}^{2}and\omega =\frac{v}{R})\Rightarrow gh=\frac{v^2}{4}+\frac{v^2}{2}\Rightarrow gh=\frac{3v^2}{4}$

$\Rightarrow v^2=\frac{4gh}{3}\Rightarrow v=\sqrt{\frac{4gh}{3}}$