$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

$A$ hollow cylinder and a solid cylinder, initially at rest at the top of an inclined plane, are rolling down without slipping. If the time taken by the hollow cylinder to reach the bottom of the inclined plane is $2 \ s$, the time taken by the solid cylinder to reach the bottom of the inclined plane is: (in $s$)

$A$ solid sphere is rolling on a surface as shown in the figure,with a translational velocity $v \, ms^{-1}$. If it is to climb the inclined surface continuing to roll without slipping,then the minimum velocity for this to happen is:

$A$ 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 $g=10 \ m \ s^{-2}$.

$A$ body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \%$ of its translational kinetic energy. The body is :

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. Identify the correct statement$(s)$.