$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

$A$ 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 cylinder rolls without slipping down an inclined plane of height $h$. The velocity of the cylinder when it reaches the bottom is

$A$ cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}} \ m/s^2$,where $x=$ . . . . . . . (use $g=10 \ m/s^2$)

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

$A$ ring,a solid sphere,and a disc are rolled down an inclined plane from the same height. The order in which they reach the bottom is: