Chapter: Mechanical Properties of Fluids· Physics
Answer: Capillary rise \(h = \dfrac{2T\cos\theta}{r\rho g}\)
Consider a capillary tube of radius \(r\) dipped vertically into a wetting liquid of density \(\rho\). The liquid rises to a height \(h\) and forms a concave meniscus. Let \(\theta\) be the angle of contact between the meniscus and the glass, and \(R\) the radius of curvature of the meniscus (taken as part of a sphere).
From the geometry of the meniscus (the radius of curvature at the edge makes angle \(\theta\) with the horizontal radius of the tube):
\(r = R\cos\theta\)
Surface tension \(T\) acts along the line of contact, tangential to the meniscus, at angle \(\theta\) with the wall. Neglect the small volume of liquid in the meniscus.
The free liquid surface outside is at atmospheric pressure \(P_0\). The pressure in the liquid at depth \(h\) below the meniscus (i.e. at the level of the outside free surface) is \(P\). The gauge pressure there is
\(P - P_0 = h\rho g\) ...(1)
By Laplace's law for a spherical surface, the excess pressure across the curved meniscus is
\(P - P_0 = \dfrac{2T}{R}\) ...(2)
From (1) and (2), using \(R = \dfrac{r}{\cos\theta}\):
\(h\rho g = \dfrac{2T}{R} = \dfrac{2T\cos\theta}{r}\)
\(\therefore h = \dfrac{2T\cos\theta}{r\rho g}\) ...(3)
Thus, the narrower the capillary tube, the greater the capillary rise (\(h \propto \dfrac{1}{r}\)).
Rearranging (3), the surface tension is
\(T = \dfrac{hr\rho g}{2\cos\theta}\) ...(4)
Equations (3) and (4) also hold for capillary depression of a non-wetting liquid (e.g. mercury in glass). Then the meniscus is convex and \(\theta\) is obtuse, so \(\cos\theta\) is negative and \(h\) comes out negative, showing a fall of the liquid in the tube. \(T\) is positive in both cases.
For capillary rise, Eq. (3) is also called the ascent formula.
