The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

In a vertical circle of radius $r$,at what point in its path does a particle have tension equal to zero if it is just able to complete the vertical circle?

$A$ 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:

In order to just complete the vertical circular motion,the ratio of kinetic energy of a particle at the highest point to that at the lowest point is

$A$ simple pendulum with a bob of mass $m = 1 \ kg$, charge $q = 5 \ \mu C$ and string length $l = 1 \ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 \times 10^6 \ V/m$ at its bottom-most point $A$, as shown in the figure. It is given a speed $u$ such that the particle leaves the circular path at its topmost point $C$. Find the speed $u$. (Take $g = 10 \ m/s^2$)

In the given figure,for an initial velocity $u = u_0/3$,find the height from the ground at which the block leaves the hemisphere,where $u_0 = \sqrt{gr}$.