
When one end of the capillary tube of radius r is immersed into a liquid of density ρ then the shape of the liquid meniscus in the tube becomes concave upwards.
The liquid can be water and it should wet the sides of the capillary tube.
Let R be the radius of curvature of the liquid meniscus
P be the atmospheric pressure
T be the surface tension of the liquid
Pressure at point A – Above the liquid meniscus is atmospheric pressure – P
Pressure at point B – Below the liquid meniscus= P -2T/R
Pressure at point C – atmospheric pressure – P
Pressure at point D – atmospheric pressure – P
Though the points B (in the capillary tube) and D are on the same horizontal plane, the pressure will not be the same in these two points.
Hence, in order to maintain an equilibrium, the liquid level rises in the capillary tube upto height h
Now, the pressure at E = pressure at B + pressure due to height h (=BE) of the liquid column = ( P – 2T/R) + hρg
As equilibrium, pressure at E = pressure at D which means P = ( P – 2T/R) + hρg
hρg = 2T/R which implies h = 2T/Rρg
In the above expression, as R the radius of curvature of the meniscus is proportional to r the radius of the capillary tube
Actually, R = r/cos θ where θ is the angle of contact between the liquid and glass
h = 2T cosθ /rρg is ascent formula