$A$ solid spherical ball is rolled up an inclined plane of angle of inclination $30^{\circ}$ with an initial speed of $4 \ m/s$ at the bottom of the inclination. How far will the ball go up the plane (in $cm$)? (Use $g=10 \ m/s^2$)

$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $30 \ m$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $[$use $g=10 \ m \ s^{-2}]$ (in $m \ s^{-1}$)

The following bodies,
$(1)$ a ring
$(2)$ a disc
$(3)$ a solid cylinder
$(4)$ a solid sphere,
of same mass $m$ and radius $R$ are allowed to roll down without slipping simultaneously from the top of an inclined plane. The body which will reach first at the bottom of the inclined plane is ...........
[Mark the body as per their respective numbering given in the question]

$A$ solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
$(a)$ Will it reach the bottom with the same speed in each case?
$(b)$ Will it take longer to roll down one plane than the other?
$(c)$ If so,which one and why?

$A$ uniform ring of radius $R$ is moving on a horizontal surface with speed $v$,then climbs up a ramp of inclination $30^{\circ}$ to a height $h$. There is no slipping in the entire motion. Then,$h$ is

$A$ solid sphere rolls down from the top of an inclined plane. On reaching the bottom of the plane, its velocity is '$V_1$'. When the same sphere slides down from the top of the same plane of same height, its velocity on reaching the bottom is '$V_2$'. The ratio $V_1 : V_2$ is (neglect friction).