$A$ 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

$A$ 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 $m$)

$A$ circular ring and a solid sphere having the same radius roll down an inclined plane from rest without slipping. The ratio of their velocities when they reach the bottom of the plane is $\sqrt{\frac{x}{5}}$,where $x=$ . . . . . . .

An inclined plane makes an angle of $30^o$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

$A$ 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 is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is