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:

$A$ 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.

$A$ 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:

The acceleration of a body rolling down on an inclined plane does not depend upon

$A$ solid sphere of mass $4 \ kg$ and radius $28 \ cm$ is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is $3.5 \ m \ s^{-2}$,then the acceleration of the sphere when it slides down without rolling is (in $m \ s^{-2}$)

$A$ ring and a disc are initially at rest,side by side,at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \ s$,then the height of the top of the inclined plane,in metres,is. . . . . . . . Take $g=10 \ m \ s^{-2}$.