$A$ ring,a solid sphere,and a disc are rolled down an inclined plane from the same height. The order in which they reach the bottom is:

$A$ solid sphere rolls down two different inclined planes of the same height,but of different inclinations. In both cases:

An object of mass $m$ slides down an inclined plane and reaches the bottom with a velocity $v$. If the same object were in the form of a ring and rolled down the same inclined plane,its velocity at the bottom would be:

$A$ 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 $g=10 \ m \ s^{-2}$.

Derive expressions for the kinetic energy and velocity of a body rolling without sliding down an inclined plane of inclination $\theta$.

$A$ disc of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be