$A$ solid sphere of mass $2\,kg$ is making pure rolling on a horizontal surface with kinetic energy $2240\,J$. The velocity of the centre of mass of the sphere will be $..........\,m/s$.

$A$ cylinder rolls without slipping down an inclined plane. The number of degrees of freedom it has is:

$A$ uniform solid sphere of radius $R$ and radius of gyration $K$ about an axis passing through the centre of mass,is rolling without slipping. Then the fraction of total energy associated with its rotation will be

$Assertion$: $A$ rigid disc rolls without slipping on a fixed rough horizontal surface with uniform angular velocity. Then the acceleration of the lowest point on the disc is zero.
$Reason$: For a rigid disc rolling without slipping on a fixed rough horizontal surface,the velocity of the lowest point on the disc is always zero.

$A$ uniform solid cylinder of mass $m$ and radius $R$ is set in rotation about its axis with an angular velocity $\omega_0$,then lowered with its lateral surface onto a horizontal plane and released. The coefficient of friction between the cylinder and plane is equal to $\mu$. The time after which the cylinder starts rolling without slipping 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: