If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is

$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $4 \ J$,then the total kinetic energy of the disc is: (in $J$)

$A$ uniform non-deformable cylinder of mass $m$ and radius $R$ is rolling without slipping on a horizontal rough surface. The force of friction is

$A$ disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points on the same horizontal line passing through $C$,such that $Q$ is at a distance $r$ from $C$ and $P$ is at a distance $r$ from $C$ on the opposite side. Let $V_P, V_Q$ and $V_C$ be the magnitudes of velocities of points $P, Q$ and $C$ respectively,then:

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 $g=10 \text{ m s}^{-2}$.]