An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolls down from the top of the inclined plane from rest without slipping. Its linear acceleration along the plane is equal to (where $g$ is acceleration due to gravity and $\sin 30^{\circ} = 0.5$):

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively,as shown in the figure. The velocities of the two discs at the points $A$ and $B$ at the bottom of the inclined planes are related as:

$A$ thin metal disc of radius $0.25\,m$ and mass $2\,kg$ starts from rest and rolls down an inclined plane. If its rotational kinetic energy is $4\,J$ at the foot of the inclined plane,then the linear velocity (in $m/s$) at the same point is:

$A$ cylinder rolls down two different inclined planes of the same height but of different inclinations.

$A$ thin uniform circular ring is rolling down an inclined plane of inclination $30^o$ without slipping. Its linear acceleration along the inclined plane will be

$A$ solid spherical ball rolls on a horizontal surface at $10 \ m \ s^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \ kg$ and frictional losses are negligible,the value of $h$,where the ball stops and starts rolling down the inclination is $($Assume $g = 10 \ m \ s^{-2} )$ (in $m$)