A cylinder of mass $2 \mathrm{kg}$ is released from rest from the top of an inclined plane of inclination $30°$ and length $1 \mathrm{m}$. If the cylinder rolls without slipping, then its speed when it reaches the bottom is [$g=10 \mathrm{m} {\mathrm{s}}^{-2}$]

Potential energy of cylinder at top position

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

$=2\times 10\left[1.\sin 30°\right] \left[\begin{array}{l}\because \sin 30°=\frac{AB}{BC}\\ \therefore AB=1\sin 30°\end{array}\right]$

$=10 \mathrm{J}$

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. $K=U$

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

$\frac{1}{2}·\frac{m{r}^2}{2}·\frac{v^2}{r^2}+\frac{1}{2}m{v}^2=10\left[\because \omega =\frac{v}{r}\right]$

$\Rightarrow \frac{3}{4}m{v}^2=10$

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