$A$ 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?

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

$A$ solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

$A$ ring,a solid sphere,and a disc are rolled down an inclined plane from the same height. The order in which they reach the bottom is:

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)

$A$ uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is $v$. If the ball is now thrown vertically upwards with the same velocity $v$,the maximum height to which the ball will rise is