Consider a capillary tube of a radius r and opened at both ends. Let it be dipped in the liquid of surface tension T (The liquid wets the glass). Liquid rises in the tube to height h as shown in Figure.

Surface tension T acts along the tangents of the spherical meniscus of the liquid at point A and B. The reaction R = T acts just opposite to the surface tension.

Resolve R = T into two components.

(i) T cosθ which acts at every point of the meniscus in the upward direction and is responsible for the rise of liquid in the capillary tube.

(ii) T sinθ acts at right angle to the length of the capillary tube. It has no effect on the rise of liquid in the tube.

The total vertical force acting on the circular meniscus is given by

F = T cosθ x 2πr   ...(i)

Due to this force liquid rises in the tube. As the liquid rises, weight of the liquid in the tube increases. Weight of the liquid acts vertically downwards. When the weight of the liquid in the tube becomes equal to the vertical force due to surface tension, liquid stops rising further in the tube.

Volume of liquid in the capillary tube is

V = Volume of cylinder of length h and radius r + (Volume of cylinder ABCD of radius r and length r) - (Volume of hemisphere of radius r)

= πr²h + (πr²r 1/2 x 4/3πr³)

= πr²h + (πr³r - 2/3 πr³) = πr³h + π/3 r³

V = πr²(h + r/3)

If ρ be the density of the liquid, then mass of the liquid raised in the capillary tube is

m = Vρ = πr²(h + r/3)ρ

Weight of the liquid raised in the capillary tube is

W = mg = πr²(h + r/3)ρg    ....(ii)

In equilibrium W = F

or, πr²(h + r/3)ρg = T cosθ x 2πr

(h + r/3) = {2T cosθ}/{rρg}

If r << h, then r/3 can be neglected as compare θ to h.

h = {2T cosθ}/{rρg}