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:

$A$ 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,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

An inclined plane makes an angle of $30^\circ$ with the horizontal. $A$ solid sphere starts rolling down from rest without slipping. Its linear acceleration will be:

$A$ solid ball of mass $m$ and radius $r$ rolls without slipping along the track shown in the figure. The radius of the circular part of the track is $R$. The ball starts rolling down the track from rest from a height of $8R$ from the ground level. When the ball reaches the point $P$,then its velocity will be

$A$ solid uniform sphere having a mass $M$,radius $R$ and moment of inertia of $\frac{2}{5} M R^2$ rolls down a plane inclined at an angle $\theta$ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is $\mu_s$. Then,

$A$ disc rolls without slipping on an inclined plane. What fraction of its total energy is in the form of rotational kinetic energy?