An inclined plane makes an angle of $30^o$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

$A$ uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is $v$. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is

$A$ solid sphere is rolling on a surface as shown in the figure,with a translational velocity $v \, ms^{-1}$. If it is to climb the inclined surface continuing to roll without slipping,then the minimum velocity for this to happen is:

$A$ disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$,from rest without sliding,is

Two discs having masses in the ratio $1:2$ and radii in the ratio $1:8$ roll down without slipping one by one from an inclined plane of height $h$. The ratio of their linear velocities on reaching the ground is ........