$A$ solid cylinder and a disc of same radii are allowed to roll down a rough inclined plane from the top of a plane. The ratio of their times taken to reach the bottom of the inclined plane is

$A$ 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,

$A$ uniform disk of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $l$ and inclination $\theta$. Initially,the disk was at rest at the top of the inclined plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to:

$A$ thin uniform circular ring is rolling down an inclined plane of inclination $30^o$ without slipping. Its linear acceleration along the inclined plane will be

$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?

An object of mass $m$ slides down an inclined plane and reaches the bottom with a velocity $v$. If the same object were in the form of a ring and rolled down the same inclined plane,its velocity at the bottom would be: