$A$ small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $3v^2/4g$ with respect to the initial position. The object is

$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

$A$ body of mass $m$ slides down an incline and reaches the bottom with a velocity $V$. If the same mass were in the form of a disc which rolls down this same incline,the velocity of the disc at the bottom would be

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

Three bodies: a ring $(R)$,a solid cylinder $(C)$,and a solid sphere $(S)$ having the same mass and same radius roll down an inclined plane without slipping. They start from rest. If $v_{R}$,$v_{C}$,and $v_{S}$ are the velocities of the respective bodies on reaching the bottom of the plane,then:

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 $\sin 30^{\circ} = 0.5$):