$A$ bucket filled with water is revolved in a vertical circle of radius $4 \ m$ and the water just does not fall down. The time period of revolution will be ........ $s$.

$A$ simple pendulum consisting of a light inextensible string of length $\ell$ attached to a heavy small bob of mass $m$ is at rest. The bob is imparted a horizontal impulsive force which gives it a speed of $u = \sqrt{4 g \ell}$. The speed of the bob at its highest point is ($g$ is the acceleration due to gravity).

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 $u_0 = \sqrt{gr}$.

$A$ body of mass $m$ hangs at one end of a string of length $l$,the other end of which is fixed. It is given a horizontal velocity such that the string just reaches a position where it makes an angle of $60^{\circ}$ with the vertical. The tension in the string at the mean position is:

$A$ rod of length $L$ is hung from one of its ends,and a mass $m$ is attached to its free end. What tangential velocity must be imparted to $m$ so that it reaches the top of the vertical circle? ($g$ = acceleration due to gravity)

The bob of a simple pendulum is of mass $10 \, g$. It is suspended with a thread of $1 \, m$. If we hold the bob so as to stretch the string horizontally and release it,what will be the tension at the lowest position? (Take $g = 10 \, m/s^2$)