Do you mean this-
from capillary rise,
H=2Tcos(theta)/rÏg, cos(theta)=1 for water
so H=2T/rÏg, P=2T/r ?
or from excess pressure in a water droplet, P=2T/r?
Suppose there is an air bubble at the bottom of a beaker filled with water to a height 'h'. What is the pressure inside it? I guess it will be Patm + hÏg + 2T/r
When a capillary tube of radius 0.5mm is dipped inside water (T=0.075N/m), what is the pressure in the tube (say at a height 5cm) below the surface? I am getting confused in this one. First of all, the water will rise to a height given by the capillary rise formula (0.03m). Next, the total height of water at a depth 5cm below the surface will be 5 + 0.03m.
So total pressure = (5 + 0.03)Ïg + Patm + 2T/r
Is this correct? thanks alot in advance