When one end of capillary tube of radius r is immersed into a liquid of density ρ which wets the sides of capillary tube (say water and capillary tube of glass), the shape of the liquid is in the tube becomes concave upwards in figure.
Let R = Radius of curvature of liquid meniscus
P = Atmospheric pressure
S = Surface tension of the liquid,
The pressure at point A just above the liquid meniscus in the capillary tube is atmospheric pressure = P
The pressure at point B, just below the liquid meniscus (on convex side).
= P -2S/R.
Pressure at point C and D, just above and below the place surface of liquid in the vessel is also P (i.e. atmospheric pressure). The point B and D are in the same horizontal plane in liquid but the pressure at these point is different. Hence there will be an equilibrium.
In order to maintain an equilibrium the liquid level rises in the capillary tube upto a height h, so that the pressure at point D and E which are in the same level in liquid may become equal, from figure.
Now, pressure at E = pressure at B + pressure due to height h (= BE) of the liquid column
= (P − \(\frac{2S}{R}\)) + hρg
As there is equilibrium, therefore
Pressure at E = Pressure at D
i.e., P – \(\frac{2S}{R}\) + hρg = P
or hρg = \(\frac{2S}{R}\)
or h = \(\frac{2S}{Rρg}\) …(1)
Calculation of R. Let I be the centre of curvature of liquid meniscus GXY in the tube and GS be the tangent to the liquid surface at point G.
GI = R, GO = r
∠IGO = θ = angle of contact
In ∆IGO,
cos θ = \(\frac{GO}{GI}\) = r/R
R = \(\frac{r}{cos\, θ}\)
Putting this value in (1) eqn.
h = \(\frac{2Scos\, θ}{rρg}\)
