$A$ body of mass $1\,kg$ is rotating in a vertical circle of radius $1\,m$. What will be the difference in its kinetic energy at the top and bottom of the circle? $(g = 10\,m/s^2)$

$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)

$A$ small body slides down a smooth uneven surface from a height $H$,which eventually emerges into a circular loop of radius $R (< H)$. What should be the value of $H$,so that the force on the body at $A$ is $\sqrt{2}$ times its weight?

$A$ bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and is set to perform vertical circular motion as shown in the figure. Initially,by applying a horizontal velocity $v_0$ at point $A$,the string becomes slack when the bob reaches point $D$. The ratio of the kinetic energy of the bob at points $B$ and $C$ is:

$A$ body of mass $2 \, kg$ is tied to a string of length $2 \, m$ and rotated in a vertical circle. What is the difference between the maximum and minimum kinetic energy in $J$?

$A$ stone of mass $m$ $kg$ is tied to a string of length $L$ $m$ and moved in a vertical circle of radius $49$ $cm$ in a vertical plane. If it completes $30$ revolutions per minute,the tension in the string when it is at the lowermost point is nearly. [Take $\pi^2=10$ and acceleration due to gravity,$g=10$ $m/s^2$]