A glass tube 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 `l` be the length of the tube inside water. The forces acting on the tube are :

`(i)` buoyancive force of water acting upward
`B = pir^(2)l xx 1 xx 980 = (22)/(7) xx (0.14)^(2)l xx 980 = 60.368 l "dyne"`.
`(ii)` Weight of the system acting downward
`= mg = 0.2 xx 980 = 196 "dyne"`.
`(iii)` Force of surface tension acting downward
`= 2pirT`
`= 2 xx (22)/(7) xx 0.14 xx 73 = 64.24 "dyne"`
Since the tube is in equilibrium, the upward force is balanced by the downward forces. That is,
`60.368 l = 196 + 64.24 = 260.24`.
`:. l = (260.24)/(60.368) = 4.31 cm`.