$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

$A$ solid cylinder $P$ rolls without slipping from rest down an inclined plane,attaining a speed $v_p$ at the bottom. Another smooth solid cylinder $Q$ of the same mass and dimensions slides without friction from rest down the inclined plane,attaining a speed $v_q$ at the bottom. The ratio of the speeds $\frac{v_q}{v_p}$ is

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

$A$ small ring is rolling without slipping on the circumference of a large bowl as shown in the figure. The ring is moving down at $P_{1}$,comes down to the lowest point $P_{2}$,and is climbing up at $P_{3}$. Let $v_{CM}$ denote the velocity of the centre of mass of the ring. Choose the correct statement regarding the frictional force on the ring.

$A$ solid spherical ball rolls on a horizontal surface at $10 \ m \ s^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \ kg$ and frictional losses are negligible, the value of $h$, where the ball stops and starts rolling down the inclination is $($Assume $g = 10 \ m \ s^{-2} )$ (in $m$)

$A$ solid sphere rolls down without slipping on a $30^{\circ}$ inclined plane. If $g = 10\,m/s^2$,the acceleration of the rolling sphere is