A glass tube of circular cross section is closed at one end. This end is weighted and the tube floats vertically in water, heavy end down. How far below the water surface is the end of the tube? Given: outer radius of the tube is `0.14 cm`, mass of weighted tube is `0.2 g`, surface tension of water `73 dyn//cm` and `g = 980 cms^(-12)`.

Explanation

Let `I` be the length of the tbe inside water. The forces acting on the tube are:
a. Upthrust of water acting upward
`=pirho^(2)lxx1xx980`
`=22/7x(0.14)^(2)lxx980=60.368l`dyne
b. Weight of the system acting downwards
`=mg=0.2xx980=196`dyne
c. Force of surface tension acting downward
`=2pirhoT=64.24`dyne
Since the tube is in eqilibrium the upward force is balanced by the downward force. That is
`60.368l=196+64.24=260.24`
`l=4.31cm`