$A$ solid cylinder of mass $M$ and radius $R$ rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity $v$. What is the height of the inclined plane? (Assume $g$ is the acceleration due to gravity.)

$A$ 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 is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

$A$ 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

$A$ 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

Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface,while $Q$ has most of its mass concentrated near the axis. Which statement$(s)$ is (are) correct?