If a particle completes a vertical circular motion,what is the ratio of its velocities at points $A$,$B$,and $C$?

$A$ ball of mass $0.1 \ kg$ is suspended by a string. It is displaced through an angle of $60^\circ$ and released. When the ball passes through the mean position,the tension in the string is ........ $N$.

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes a full vertical circle,reaching the topmost position $B$. The ratio of kinetic energies $\frac{(\text{K.E.})_A}{(\text{K.E.})_B}$ is:

$A$ point mass '$m$' attached at one end of a massless,inextensible string of length '$\ell$' performs a vertical circular motion and the string rotates in a vertical plane,as shown in the diagram. The increase in the centripetal acceleration of the point mass when it moves from point $A$ to point $C$ is $(g = \text{acceleration due to gravity})$:

$A$ particle is tied to a $20 \ cm$ long string. It performs circular motion in a vertical plane. What is the angular speed of the particle when the tension in the string at the top is zero? (Take $g = 9.8 \ m/s^2$)

$A$ stone of mass $1 \ kg$ is tied to a string $2 \ m$ long and is rotated at a constant speed of $40 \ ms^{-1}$ in a vertical circle. The ratio of the tension at the top and the bottom is [Take $g = 10 \ ms^{-2}$].