A cylinder is rolling down on an inclined plane of inclination 60 °...

For rolling motion, the acceleration of the cylinder along the inclined plane can be expressed as

$a = \dfrac{gsin\theta}{1 + \dfrac{{I}_{cm}}{M{R}^{2}}} . . . \left(1\right)$

The moment of inertia of the cylinder can be written as

${I}_{cm} = \dfrac{1}{2}M{R}^{2} . . . \left(2\right)$

From equations (1) and (2), it follows that

$\begin{cases}a & = & \dfrac{gsin\theta}{1 + \dfrac{1}{2}} \\ & = & \dfrac{2 \times 10 \times \dfrac{\sqrt{3}}{2}}{3} \\ & = & \dfrac{10}{\sqrt{3}}m{s}^{- 2}\end{cases}$

Therefore, $x = 10$.