The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

$A$ 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 cylinder and a solid sphere,having same mass $M$ and radius $R$,roll down the same inclined plane from the top without slipping. They start from rest. The ratio of the velocity of the solid cylinder to that of the solid sphere,with which they reach the ground,will be:

$A$ circular disc reaches from top to bottom of an inclined plane of length $l$. When it slips down the plane,it takes $t \ s$. When it rolls down the plane,it takes $\left(\frac{\alpha}{2}\right)^{1/2} t \ s$,where $\alpha$ is:

$A$ 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 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, then the speed when it reaches the bottom is (in $\,m/s$)