$A$ ring,a solid sphere,and a disc are rolling down from the top of an inclined plane of the same height. What is the sequence in which they reach the surface?

$A$ 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:

The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal place: $(i)$ a ring of radius $R$,$(ii)$ a solid cylinder of radius $\frac{R}{2}$,and $(iii)$ a solid sphere of radius $\frac{R}{4}$. If,in each case,the speed of the center of mass at the bottom of the incline is the same,the ratio of the maximum heights they climb 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 sphere is rolling without slipping on a semi-circular track of radius $R = 10 \ m$ as shown in the figure. The radius of the solid sphere is much smaller than the radius of the semi-circular track. At the lowest point, it has a velocity $v = 10 \ m/s$. To what maximum angle $\theta$ from the vertical will the sphere travel before it comes back down? Neglect the rolling friction between the sphere and the track. (Take $g = 10 \ m/s^2$)

Three bodies: a ring $(R)$,a solid cylinder $(C)$,and a solid sphere $(S)$ having the same mass and same radius roll down an inclined plane without slipping. They start from rest. If $v_{R}$,$v_{C}$,and $v_{S}$ are the velocities of the respective bodies on reaching the bottom of the plane,then: