$A$ ring is rolling on an inclined plane. The ratio of the linear and rotational kinetic energies will be

$A$ 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

$A$ solid homogeneous sphere rolls without slipping down an inclined plane of height $h$. The velocity of the sphere at the bottom is:

$A$ solid uniform disk of mass $m$ rolls without slipping down a fixed inclined plane with an acceleration $a$. The frictional force on the disk due to the surface of the plane 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, then the speed when it reaches the bottom is (in $\,m/s$)

$A$ solid uniform sphere having a mass $M$,radius $R$ and moment of inertia of $\frac{2}{5} M R^2$ rolls down a plane inclined at an angle $\theta$ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is $\mu_s$. Then,