An object of mass $m$ slides down an inclined plane and reaches the bottom with a velocity $v$. If the same object were in the form of a ring and rolled down the same inclined plane,its velocity at the bottom would be:

$A$ solid uniform sphere having a mass $M$,radius $R$ and moment of inertia of $\frac{2}{5} M R^2$ rolls down a plane inclined at an angle $\theta$ to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is $\mu_s$. Then,

$A$ sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping,how far up the sphere will travel?

$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

$A$ horizontal force $F$ is applied at the center of mass of a cylindrical object of mass $m$ and radius $R$,perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is $\mu$. The center of mass of the object has an acceleration $a$. The acceleration due to gravity is $g$. Given that the object rolls without slipping,which of the following statement$(s)$ is(are) correct?
$(A)$ For the same $F$,the value of $a$ does not depend on whether the cylinder is solid or hollow
$(B)$ For a solid cylinder,the maximum possible value of $a$ is $2 \mu g$
$(C)$ The magnitude of the frictional force on the object due to the ground is always $\mu m g$
$(D)$ For a thin-walled hollow cylinder,$a = \frac{F}{2m}$

$A$ disc of mass $m$ and radius $r$ rolls down an inclined plane of height $h$. When it reaches the bottom of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).