A solid cylinder is released from rest from the top of an inclined...

Let speed of cylinder upon reaching the bottom of the inclined plane be $v$.

Here, gain in kinetic energy=loss in potential energy

So, $\dfrac{1}{2}I{\omega}^{2} + \dfrac{1}{2}m{v}^{2} = mglsin\theta$

$⇒ \dfrac{1}{2}\left(\dfrac{m{r}^{2}}{2}\right){\omega}^{2} + \dfrac{1}{2}m{v}^{2} = mglsin30 °$

For pure rolling, $v = r\omega$

$⇒ \dfrac{1}{2}\left(\dfrac{m{r}^{2}}{2}\right){\left(\dfrac{v}{r}\right)}^{2} + \dfrac{1}{2}m{v}^{2} = mglsin30 °$

$⇒ \dfrac{1}{4}m{v}^{2} + \dfrac{1}{2}m{v}^{2} = mglsin30 ° ⇒ \dfrac{3}{4}m{v}^{2} = \dfrac{mgl}{2} ⇒ v = \sqrt{\dfrac{4gl}{6}}$

Putting the values, we have

$⇒ v = \sqrt{\dfrac{4 \times 10 \times 0 . 6}{6}} = 2m{s}^{- 1}$