Explanation

(a) Let a plane be inclined at an angle $\theta$ and a cylinder rolls down then the acceleration of the cylinder of mass $m$, radius $R$, and $I$
$I$ as moment of inertia is given by
$a = \frac{g\sin\theta}{\left( 1 + \frac{I}{mR^{2}} \right)}$
Moment of inertia $(I)$ of a cylinder =$\frac{mR^{2}}{2}$
$\therefore$ $a = \frac{g\sin\theta}{\left( \frac{mR^{2}}{1 + \frac{2}{mR^{2}}} \right)} = \frac{2}{3}g\sin{30{^\circ}}$
$\Rightarrow$ $a = \frac{g}{3}$