$A$ solid sphere rolls down without slipping on an inclined plane,then the percentage of rotational kinetic energy of the total energy will be ........ $\%.$

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

$A$ disc of mass $m$ and radius $r$ 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).

$A$ body starts rolling down an inclined plane of length $L$ and height $h$. This body reaches the bottom of the plane in time $t$. The relation between $L$ and $t$ is:

$A$ drum of radius $R$ and mass $M$ rolls down without slipping along an inclined plane of angle $\theta$. The frictional force:

$A$ 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 . . . . . . .