A cylinder, released from the top of an inclined plane, rolls without sliding and reaches the bottom with speed v_r. Another identical cylinder, released from the top of the same inclined plane, slides without rolling and reaches the bottom with speed v_s. Then

When the cylinder rolls without sliding, the acceleration down the plane is

 ${\mathrm{a}}_{\mathrm{r}}=\frac{2}{3}\mathrm{gsin}\mathrm{\theta } \left(\because \mathrm{a}=\frac{\mathrm{gsin\theta }}{1+\mathrm{\beta }} \mathrm{and} \mathrm{\beta }=\frac{1}{2}\right)$

When the cylinder slides without rolling, the acceleration is

${\mathrm{a}}_{\mathrm{s}}=\mathrm{gsin}\mathrm{\theta }$

where $\mathrm{\theta }$ is the inclination of the plane.

If h is the height of the inclined plane, their speeds on reaching the bottom are given by 

${\mathrm{v}}_{\mathrm{r}}=\sqrt{2{\mathrm{a}}_{\mathrm{r}}\mathrm{h}}\text{ and }{\mathrm{v}}_{\mathrm{s}}=\sqrt{2{\mathrm{a}}_{\mathrm{s}}\mathrm{h}}$

Since ${\mathrm{a}}_{\mathrm{s}}>{\mathrm{a}}_{\mathrm{r}}$, it follows that ${\mathrm{v}}_{\mathrm{s}}>{\mathrm{v}}_{\mathrm{r}}$, which is choice (b).