$A$ sphere of mass $0.5 \ kg$ and diameter $1 \ m$ is rolling without slipping with a constant velocity of $5 \ m/s$. The ratio of its rotational kinetic energy to its total kinetic energy is:

$A$ disc of radius $r$ is rotating about its centre with an angular speed $\omega_0$. It is gently placed on a rough horizontal surface. After what time will it be in pure rolling?

Two spheres are rolling with the same velocity (for their $C.M.$). Their ratio of kinetic energy is $2:1$ and the ratio of their radii is $2:1$. Their mass ratio will be:

$A$ wheel of radius $1 \text{ m}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

$A$ wheel is rolling on the ground with a speed of $2 \ m/s$. What is the velocity of the endpoints of the horizontal diameter of the wheel?

At time $t=0$,a disk of radius $1 \text{ m}$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha = \frac{2}{3} \text{ rad s}^{-2}$. $A$ small stone is stuck to the disk. At $t=0$,it is at the contact point of the disk and the plane. Later,at time $t=\sqrt{\pi} \text{ s}$,the stone detaches itself and flies off tangentially from the disk. The maximum height (in $\text{m}$) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$. The value of $x$ is. . . . . . . [Take $g=10 \text{ m s}^{-2}$.]