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

$A$ hollow sphere of mass $m$ filled with a non-viscous liquid of same mass $m$ is released on a slope inclined at angle $\theta$ with the horizontal. The friction between the sphere and the slope is sufficient to prevent sliding,and frictional forces between the inner surface of the sphere and the liquid are negligible. After descending a certain height,the ratio of translational and rotational kinetic energies is found to be $x:y$. Find the numerical value of the expression $(x+y)_{min}$.

$A$ cylinder rolls down two different inclined planes of the same height but of different inclinations.

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

$A$ hollow cylinder of mass $m$ and radius $R$ is spinned to a clockwise angular velocity $\omega_0$ and then gently placed on an inclined plane for which the coefficient of friction is $\mu = \tan \theta$,where $\theta$ is the angle of the inclined plane with the horizontal. The centre of mass of the cylinder will remain stationary for time:

$A$ solid sphere is rolling on a frictionless surface with a linear velocity $v \; m/s$,as shown in the figure. If the sphere climbs up to a height $h$,then the value of $v$ will be