$A$ circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane,it takes $t \ s$. When it rolls down the plane,it takes $\left(\frac{\alpha}{2}\right)^{1/2} t \ s$,where $\alpha$ is:

$A$ round uniform body of radius $R$,mass $M$,and moment of inertia $I$ rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

$A$ solid sphere at rest rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of the plane is ($g=$ acceleration due to gravity).

$A$ 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:

$A$ sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping,how far up the sphere will travel?

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. When these bodies roll down to the foot of the inclined plane,then