$A$ ring is rolling on an inclined plane. The ratio of the linear and rotational kinetic energies will be

$A$ solid cylinder and a hollow cylinder,both of the same mass and same external diameter,are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

$A$ 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 ........

Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (see figure) starting at $A$ and $C$ with linear speeds $v_1$ and $v_2$,respectively,and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_1 = 3 \ m/s$,then $v_2$ in $m/s$ is $(g = 10 \ m/s^2)$.

$A$ 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 $m$)

$A$ solid cylinder of mass $M$ and radius $R$ rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity $v$. What is the height of the inclined plane? (Assume $g$ is the acceleration due to gravity.)