A uniform solid cylinder rolls without slipping with centre speed 4 m s^-1 along a rough horizontal floor and then, without any loss of speed, passes…

On the smooth incline there is no friction, so no torque acts about the centre of mass and the angular velocity of the cylinder stays constant. The rotational kinetic energy $\tfrac12 I\omega^2=\tfrac14 mv^2$ therefore remains locked in the spin and cannot be converted into height. Only the translational energy $\tfrac12 mv^2$ is available to climb: $mgh=\tfrac12 mv^2$, so $h=v^2/2g=16/20=0.8\ \mathrm{m}$. The distance along the incline is $s=h/\sin30^\circ=1.6\ \mathrm{m}$. The common mistake is to use $\tfrac34 mv^2$ (rolling on a rough incline), which gives $h=1.2\ \mathrm{m}$ and $s=2.4\ \mathrm{m}$; that applies only when friction converts spin energy. Using $h$ itself as the distance gives $0.8\ \mathrm{m}$. Hence option (B).