$A$ hollow sphere of mass $m$ and radius $R$ is rolling downward on a rough inclined plane of inclination $\theta$. If the coefficient of friction between the hollow sphere and incline is $\mu$,then ........

$A$ small sphere rolls without slipping from the top of a vertical track. The track has an inclined part and a horizontal part. The horizontal part is $1.0 \ m$ above the ground,and the top of the track is $2.4 \ m$ above the ground. The sphere falls to the ground at point $E$. The horizontal distance from the point directly below $C$ to point $E$ is $R$. Find the value of $R$ in meters.

Starting from rest,at the same time,a ring,a coin (disc),and a solid ball of the same mass roll down an incline without slipping. The ratio of their translational kinetic energies at the bottom will be:

$Assertion$ : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane,compared to when it rolls down the same plane.
$Reason$ : In rolling down,a body acquires both kinetic energy of translation and rotation.

$A$ solid sphere of mass $2M$ and a thin hollow spherical shell of mass $M$ of the same radius roll down an inclined plane simultaneously. Then,

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].