$A$ body slides down a smooth inclined plane having angle $\theta$ and reaches the bottom with velocity $v$. If the body is a solid sphere rolling down the same plane,then its linear velocity at the bottom of the plane is

$A$ solid cylinder of radius $R$ and mass $M$ rolls down an inclined plane of height $h$. When it reaches the bottom of the plane,its rotational kinetic energy is ($g =$ acceleration due to gravity).

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)

$A$ solid cylinder rolls without slipping down an inclined plane of height $h$. The velocity of the cylinder when it reaches the bottom is

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).

$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is: