Let a plane be inclined at an angle $\mathrm{\theta }$ and a cylinder rolls down then the acceleration of the cylinder of mass $m$, radius $R$, and $\mathrm{I }$as moment of inertia is given by

$a=\frac{\mathrm{g}\sin \mathrm{\theta }}{\left(1+\frac{I}{m{R}^2}\right)}$

Moment of inertia $\left(I\right)$ of a cylinder =$\frac{m{R}^2}{2}$

$\therefore$ $a=\frac{\mathrm{g}\sin \mathrm{\theta }}{\left(\frac{m{R}^2}{1+\frac{2}{m{R}^2}}\right)}=\frac{2}{3}\mathrm{g}\sin 30°$

$\Rightarrow$ $a=\frac{\mathrm{g}}{3}$