$A$ body of mass $m$ slides down an incline and reaches the bottom with a velocity $V$. If the same mass were in the form of a disc which rolls down this same incline,the velocity of the disc at the bottom would be

$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

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,

$A$ solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^{\circ}$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant,the new speed of the sphere at the bottom of the plane is $v_2$. The ratio of $v_1^2 : v_2^2$ is

$A$ body slides down a smooth inclined plane having angle $\theta$ and reaches the bottom with velocity $v$. If the body is a solid sphere rolling down the same plane,then its linear velocity at the bottom of the plane is