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?

$A$ bucket full of water is revolved in a vertical circle of radius $2\,m$. What should be the maximum time-period of revolution so that the water does not fall off the bucket? (Take $g = 10\,m/s^2$) ......... $\sec$.

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the highest and lowest positions of the circular path will be:

Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

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 $u_0 = \sqrt{gr}$.

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).