If a sphere is rolling,the ratio of its rotational energy to the total kinetic energy is given by

$A$ solid sphere rolls on a surface with a translational velocity $v \ m/s$. It climbs up a curved surface without slipping. The minimum value of $v$ required to reach the height $h$ is:

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

$Assertion$ : The velocity of a body at the bottom of an inclined plane of given height is more when it slides down the plane,compared to when it rolls down the same plane.
$Reason$ : In rolling down,a body acquires both kinetic energy of translation and rotation.

$A$ solid cylinder of mass $m$ is wrapped with an inextensible light string and is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is: [The coefficient of static friction,$\mu_{s}$,is $0.4$]

$A$ solid sphere (mass $2M$) and a thin hollow spherical shell (mass $M$) both of the same size,roll down an inclined plane,then