Question

If the surface tension of water is $0.06\,N{m^{ - 1}}$ ,what is the capillary rise in the tube of diameter $1\,mm$? $(\theta = {0^ \circ })$
A. $3.86\,cm$
B. $3.12\,cm$
C. $2.4\,cm$
D. $1.22\,cm$

Answer 1

Hint: Capillary rise is the elevation of the liquid level above zero pressure level in a capillary tube. The height to which the liquid rises in a capillary tube is directly related to the surface tension of the liquid and inversely related to the diameter.
The capillary rise formula can be used to solve this question. It is given as
$h = \dfrac{{2T\cos \theta }}{{\rho rg}}$
Where $h$ is the elevation of the liquid, $T$ is the surface tension, $\theta $ is the angle of contact of the liquid with the capillary tube, $\rho $ is the density of the liquid, $g$ is the acceleration due to gravity and $r$ is the radius of the capillary tube.

Complete step by step answer:
Capillary rise is the rise of a liquid above the level of zero pressure due to an upward force produced by the attraction of molecules in liquid to the surface of the capillary tube.
This happens when the adhesion of the liquid to solid is greater than the cohesion of liquid to itself
Cohesion is a property by which like molecules stick together and adhesion is the force of attraction between dissimilar molecules.
The surface tension of a liquid is the tendency of a liquid to reduce the surface area to the minimum possible value.
At the interface between liquid and air, the surface tension is a result of the greater attraction of liquid molecules to each other than the attraction between the liquid molecules and the air molecules.
Given,
Diameter, $d = 1\,mm$
Therefore,
Radius is half the diameter.
$\Rightarrow r = \dfrac{1}{2}\,mm $
$\Rightarrow r = \dfrac{1}{2} \times {10^{ - 3}}\,m $
Tension, $T = 0.06\,N{m^{ - 1}}$
The relation between capillary rise and surface tension is given by the capillary rise formula given below.
$h = \dfrac{{2T\cos \theta }}{{\rho rg}}$
Where $h$ is the elevation of the liquid, $T$is the surface tension, $\theta $ is the angle of contact of the liquid with the capillary tube, $\rho $ is the density of the liquid, $g$ is the acceleration due to gravity and $r$ is the radius of the capillary tube.
We know that density of water, $\rho = 1000\,kg/{m^3}$
Acceleration due to gravity, $g = 10\,m/{s^2}$
Now let us substitute the given values.
$h = \dfrac{{2 \times 0.06 \times 1}}{{1000 \times \dfrac{1}{2} \times {{10}^{ - 3}} \times 10}}$
Since, $\cos {0^ \circ } = 1$
$ \Rightarrow h = \dfrac{{0.12}}{5}\,m$
$ \Rightarrow h = 0.024\,m$
Therefore,
$h = 2.4\,cm$

Therefore, option C is the correct answer.

Note:
The angle of contact $\theta $ is an important factor that determines whether a capillary rise or capillary depression occurs. If this angle is acute such as in case of water capillary rise takes place. In the case where this angle is obtuse such as in the case of mercury capillary depression takes place. The diameter of the tube is also an important factor. Both capillary rise and depression are observed only in tubes of small diameter.