$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

$A$ solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$,respectively,then

$A$ solid uniform disk of mass $m$ rolls without slipping down a fixed inclined plane with an acceleration $a$. The frictional force on the disk due to the surface of the plane 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 $g=10 \ m/s^2$)

$A$ 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 $m$)

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively,as shown in the figure. The velocities of the two discs at the points $A$ and $B$ at the bottom of the inclined planes are related as: