Answer: Option B
:
B
Let the mass of the cylinder be m and its radius r. Suppose the linear speed of the cylinder when it reaches the bottom is v. As the cylinder rolls without slipping, its angular speed about its axis is $\omega =\frac{v}{r}$. The kinetic energy at the bottom will be= $\frac{3}{4}m{v}^2$
Note: Try reducing this result on your own for practice.
Thus, $\frac{3}{4}m{v}^2=mglsin\theta$
or, v=$\sqrt{\frac{4}{3}glsin\theta }$