$A$ hollow sphere of mass $m$ and radius $R$ is rolling downward on a rough inclined plane of inclination $\theta$. If the coefficient of friction between the hollow sphere and incline is $\mu$,then ........

$A$ solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

$A$ disc and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

$A$ ring and a disc are initially at rest,side by side,at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \ s$,then the height of the top of the inclined plane,in metres,is. . . . . . . . Take $g=10 \ m \ s^{-2}$.

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:

$A$ cylinder is pure rolling up an incline plane. It stops momentarily and then rolls back. The force of friction