Rise of a liquid in a capillary tube:

Let us consider a capillary tube of radius r is dipped in a liquid vertically. The liquid within the tube rises until it stands at certain a height.

The surface column at the top inside the capillary tube is hemispherical and concave upwards. Let \( h \) be the height of the meniscus inside the tube from the level of the free surface of the liquid outside the tube. A force acts inwards along the tangent to the liquid surface at each point of contact of the liquid surface with wall of the tube due to surface tension. According to the Newton’s third law a reaction force acts along the opposite direction. The vertical component of this reaction is \( T\cos\theta \), which acts vertically upward at each point of contact. Let, \( \theta \) be the angle of contact.

Since the liquid inside the capillary tube meets the tube in a circle of periphery \( 2\pi{r} \), so the total upward force is given by, \( 2\pi{r}\times{T}\cos\theta \).
Due to this force the liquid inside the capillary tube rises until it is balanced by the weight of the liquid cylinder of height \( h \) and radius \( r \) plus the weight of the liquid in the meniscus.

Now the weight of the liquid cylinder of height \( h \) and radius \( r \) is given by, \( \pi{r^2}h\rho{g} \), where \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity.

Again the weight of the liquid inside the meniscus = (weight of the liquid cylinder of radius \( r \) and height \( r \))-( weight of the liquid hemisphere of radius \( r \))

So the weight of the liquid inside the meniscus is given by,

\( \pi{r^3}\rho{g}-\frac{2}{3}\pi{r^3}\rho{g}\\=\rho{g}(\pi{r^3}-\frac{2}{3}\pi{r^3})\\=\frac{1}{3}\rho{g}\pi{r^3} \)

Hemce for equilibrium condistion,

\( 2\pi{r}{T}\cos\theta =\pi{r^2}h\rho{g}+\frac{1}{3}\rho{g}\pi{r^3}\\or,\ \displaystyle{T=\frac{\rho{rg}(h+\frac{r}{3})}{2\cos\theta}} \)

If the capillary is very small then \( r \) is very small and \( \frac{r}{3} \) can be neglected with respect to \( h \). In such case, the previous equation will be

\( \displaystyle{T=\frac{\rho{r}gh}{2\cos\theta}\\or,\ h=\frac{2T\cos\theta}{\rho{r}g}} \)

In case of the liquid for which \( \theta=0 \), \( \cos\theta=1 \),

Therefore, \( \displaystyle{h=\frac{2T}{\rho{r}g}} \)

(i) In case of a liquid which does not wet the glass, such as in case of mercury, \( \theta>90^\circ \) so \( \cos\theta \) is negative. Hence \( h \) is negative, this means that the liquid inside the capillary tube is depressed below the surface outside the capillary tube.

(ii) If a capillary tube with its inside coated with paraffin, is dipped vertically into water, the surface of the water column inside the tube also lies below that in the outer surface since \( \theta>90^\circ \).

(iii) Two different liquids having the same values of the surface tension \( T \) but different values of angle of contact ( \( \theta \)), rise to different heights when the same capillary tube is introduced into them.