A cylinder of mass 2kg is released from rest from the top of an...

Potential energy of cylinder at top position

$U = mg\left(AB\right)$

= 2 \times 10\left[1 . \sin 30 °\right]\left[$\begin{array}{ll}∵ \sin 30 ° = \dfrac{AB}{BC} \\ ∴ AB = 1\sin 30 °\end{array}$\right]

$= 10J$

If $v$ is the linear speed of cylinder when it reaches at bottom, then at bottom, total potential energy of cylinder is converted into kinetic energy,

i.e. $K = U$

$\dfrac{1}{2}I{\omega}^{2} + \dfrac{1}{2}m{v}^{2} = 10$

$\dfrac{1}{2} \cdot \dfrac{m{r}^{2}}{2} \cdot \dfrac{{v}^{2}}{{r}^{2}} + \dfrac{1}{2}m{v}^{2} = 10\left[∵ \omega = \dfrac{v}{r}\right]$

$⇒ \dfrac{3}{4}m{v}^{2} = 10$

$v = \sqrt{\dfrac{40}{3m}} = \sqrt{\dfrac{40}{3 \times 2}} = \sqrt{\dfrac{20}{3}}m{s}^{- 1}$