$A$ ring takes time $t_1$ in slipping down an inclined plane of length $L$ and takes time $t_2$ in rolling down the same plane. The ratio $\frac{t_1}{t_2}$ is

$A$ solid sphere and a hollow cylinder roll up without slipping on the same inclined plane with the same initial speed $v$. The sphere and the cylinder reach maximum heights $h_1$ and $h_2$,respectively,above the initial level. The ratio $h_1: h_2$ is $\frac{n}{10}$. The value of $n$ is . . . . . . .

$A$ solid spherical ball rolls on a horizontal surface at $10 \ m \ s^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \ kg$ and frictional losses are negligible, the value of $h$, where the ball stops and starts rolling down the inclination is $($Assume $g = 10 \ m \ s^{-2} )$ (in $m$)

If a sphere is rolling,the ratio of its rotational energy to the total kinetic energy is given by

$Assertion$ : 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 cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)