What is the pressure at the bottom of a capillary tube dipped in a liquid? is it hÏg + 2T/R
or hÏg - 2T/R? One more thing, this 2T/R is the excess pressure due to what?? at the bottom there is no bubble or droplet.
What about at the meniscus i.e. just below the liquid surface.?
Any ideas appreciated. Thanks alot in advance.
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2 Answers
62This is due to surface tension...
There is surface tension and depending on the meniscus and its direction the + or - sign exists...
The best way to solve this problem is through the FBD like most things in mechanics..
The fact that you get a expression 2T/R comes from the definition of surface tension (force per unit length) and again using the FBD to find surface tension...
1Do 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