$A$ solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '$V_1$'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '$V_2$'. The ratio $V_1 : V_2$ is (neglect friction).

$A$ ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \,s$, then the height of the top of the inclined plane, in metres, is. . . . . . Take $g=10 \,m \,s^{-2}$.

$A$ solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
$(a)$ Will it reach the bottom with the same speed in each case?
$(b)$ Will it take longer to roll down one plane than the other?
$(c)$ If so,which one and why?

$A$ hollow cylinder of mass $m$ and radius $R$ is spinned to a clockwise angular velocity $\omega_0$ and then gently placed on an inclined plane for which the coefficient of friction is $\mu = \tan \theta$,where $\theta$ is the angle of the inclined plane with the horizontal. The centre of mass of the cylinder will remain stationary for time:

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \,cm$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is (in $\,m/s$)

The rotational kinetic energy of a solid sphere of mass $3 \; kg$ and radius $0.2 \; m$ rolling down an inclined plane of height $7 \; m$ is (in $; J$)