$A$ body tied to a string of length $L$ is revolved in a vertical circle with minimum velocity. When the body reaches the uppermost point,the string breaks and the body moves under the influence of the gravitational field of the Earth along a parabolic path. The horizontal range $AC$ of the body will be

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a horizontal velocity $v_{o}$ at the lowest point $A$ such that it completes a circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point,$C$. This is shown in the figure. Obtain an expression for: $(i) v_{o}$; $(ii)$ the speeds at points $B$ and $C$; $(iii)$ the ratio of the kinetic energies $(K_{B} / K_{C})$ at $B$ and $C$. Comment on the nature of the trajectory of the bob after it reaches the point $C$.

$A$ bob of mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in the figure. If the string becomes slack at some point $P$ making an angle $\theta$ with the horizontal,the ratio of the speed $v_p$ of the bob at point $P$ to its initial speed $v_0$ is:

$A$ mass attached to one end of a string crosses the topmost point on a vertical circle with critical speed. Its centripetal acceleration when the string becomes horizontal will be ($g =$ gravitational acceleration).

$A$ stone of mass $m$ is moving in a vertical circle of radius $20 \ cm$. What is the difference in kinetic energy between the lowest and highest points?