What is capillary action? Derive the formula for the rise of liquid in a capillary tube immersed vertically in a liquid.

(N/A) The phenomenon of the rise or fall of a liquid in a capillary tube (held vertically in a liquid) is called capillarity. In this phenomenon,the surface tension of the liquid plays an important role.
The word 'capilla' means hair in Latin. If the tube is as thin as a hair,the rise would be very large for a liquid with an acute angle of contact. This type of tube is known as a capillary tube.
Consider a vertical capillary tube of circular cross-section (radius $a$) inserted into an open vessel of water. The contact angle $\theta$ between water and glass is acute $(\theta < 90^{\circ})$. Thus,the surface of the water in the capillary is concave. This implies a pressure difference between the two sides of the top surface.
$P_{i} - P_{0} = \frac{2S}{r}$,where $r$ is the radius of the meniscus ...$(1)$
From the geometry,$\cos \theta = \frac{a}{r}$,therefore $r = \frac{a}{\cos \theta}$ ...$(2)$
Substituting equation $(2)$ into $(1)$:
$P_{i} - P_{0} = \frac{2S \cos \theta}{a}$ ...$(3)$
At the same horizontal level as the free surface of the liquid in the vessel,the pressure inside the tube must be equal to the atmospheric pressure $P_{0}$.
Thus,$P_{0} = P_{i} + h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
$P_{0} - P_{i} = h \rho g$
Comparing this with equation $(3)$,we get:
$h \rho g = \frac{2S \cos \theta}{a}$
Therefore,the height of the liquid rise is:
$h = \frac{2S \cos \theta}{a \rho g}$