$A$ 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 $g=10 \,m/s^2$].

An inclined plane makes an angle of $30^o$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

$A$ thick-walled hollow sphere has outside radius $R_0$. It rolls down an incline without slipping and its speed at the bottom is $v_0$. Now the incline is waxed,so that it is practically frictionless and the sphere is observed to slide down (without any rolling). Its speed at the bottom is observed to be $5v_0/4$. The radius of gyration of the hollow sphere about an axis through its centre is

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane, then the linear velocity at the bottom of the plane is:

$A$ disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v,$ the height to which the disc will rise will be

$A$ 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 $m \ s^{-1}$)