$A$ 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 $t_2$?

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane,then the linear velocity at the bottom of the plane is:

$A$ solid sphere,a solid cylinder,a disc,and a ring are rolling down an inclined plane. Which of these bodies will reach the bottom simultaneously?

$A$ solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^{\circ}$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant,the new speed of the sphere at the bottom of the plane is $v_2$. The ratio of $v_1^2 : v_2^2$ is

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

One solid sphere and one disc of the same radius are rolling down an inclined plane without slipping. One reaches the bottom earlier than the other due to: