Capillary Action

Capillary action is a fascinating phenomenon where liquid rises in narrow tubes or porous materials against the force of gravity. This process occurs due to the cohesive forces within the liquid and the adhesive forces between the liquid and the surface of the material. For instance, capillary action is crucial in plants, where water is transported from roots to leaves through tiny vessels. In everyday life, this principle is observed in action when a paper towel absorbs spilt liquid or when ink climbs up a fountain pen nib. Understanding capillary action not only unveils the intricate workings of natural systems but also highlights the role of fundamental physics in daily life, from the way our morning coffee percolates to the efficiency of various absorbent materials.

Capillary Action

Capillary action is a unique physical phenomenon where liquid rises or moves through narrow spaces or porous materials against the force of gravity. This occurs due to the interplay of cohesive forces within the liquid and adhesive forces between the liquid and the walls of the material. In nature, this action is vital for the movement of water and nutrients in plants, allowing them to thrive even in a challenging environment.

Capillarity

Capillarity, or capillary action, refers to the ability of a liquid to flow in narrow spaces without the assistance of external forces, such as gravity. This phenomenon arises from the interplay between cohesive forces within the liquid and adhesive forces between the liquid and the walls of the container or material. A common example is how water rises in the thin tubes of plants, facilitating the transport of essential nutrients from the roots to the leaves. If a capillary tube is dipped in a liquid, it is found that the liquid in the capillary either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarity.

Examples of capillarity

A towel soaks water.

Ascent Formula

When one end of the capillary tube of radius r is immersed into a liquid of density (For example- water and a capillary tube of glass), the shape of the liquid meniscus in the tube becomes concave upwards as shown in the figure.

Then the height h up to which the liquid level rises in the capillary tube is given by Ascent Formula

which says $h=\frac{2 T \cos \theta}{\rho g r}$

where

$T$ - sur face Tension
$r$ - radius of capillary tube
$\rho$ - liquid density
$\theta$ - Angle of contact

The capillary rise depends on the nature of liquid and solid both. I.e on $T, \rho, \theta, r$

For a given liquid and solid pair as $T, \rho, \theta, r$ is constant then. i.e., the greater the radius of capillary greater the rise and vice-versa.

Capillary action for various liquid-solid pair

For $\theta<90^{\circ}$ (1.e for water and capillary tube of glass)
So Meniscus will take Concave shape and liquid in the capillary will rise/ascend.
For $\theta>90^{\circ}$ (l.e for Mercury and glass capillary tube)
So Meniscus will take Convex shape and liquid in the capillary will fall/descend.
For $\theta=90^{\circ}$ (i.e. for Pure water and silver-coated capillary tube.)

So Meniscus will take a Plane shape and the liquid in the capillary will show No rise and no fall.

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Solved Examples Based on Capillary Action

Example 1: A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = 0.05 Nm^-1, density = 667 kg m^-3 ) Which rises to height h in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary ) make an angle of 60^o with one another. Then h is close to (g=10 ms^-2).

1)0.172 m

2)0.049 m

3)0.087 m

4)0.137 m

Solution:

So Angle of contact=30
So using

$
h=\frac{2 T \cos \theta}{r \rho g}=\frac{2 \times 0.05 \times\left(\frac{\sqrt{3}}{2}\right)}{0.15 \times 10^{-3} \times 667 \times 10}=0.087 \mathrm{~m}
$
Hence, the answer is the option (3).

Example 2: Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of the two limbs of the tube?
[ Take surface tension of water $T=7 \cdot 3 \times 10^{-2} \mathrm{Nm}^{-1}$, angel of contact $=0, g=10 \mathrm{~ms}^{-2}$ and density of water $=1 \cdot 0 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ ]

1) $5 \cdot 34 \mathrm{~mm}$
2) $3 \cdot 62 \mathrm{~mm}$
3) $2 \cdot 19 \mathrm{~mm}$
4) 4.97 mm

Solution:

$\begin{aligned} r_1 & =\frac{5}{2} \times 10^{-3} \mathrm{~m} \\ r_2 & =\frac{8}{2} \times 10^{-3} \mathrm{~m} \\ h_1 & =\frac{2 T \cos \theta}{r_1 \rho g} \quad \frac{h_1-h_2}{h_2}=\frac{r_2-r_1}{r_1} \\ h_2 & =\frac{2 T \cos \theta}{r_2 \rho g}=\frac{1 \cdot 5}{\frac{5}{2}} \\ \frac{h_1}{h_2} & =\frac{r_2}{r_1} \frac{\Delta h}{h_2}=\frac{3}{5} \\ h_2 & =\frac{2 \times 7 \cdot 3 \times 10^{-2} \times \cos 0^{\circ}}{4 \times 10^{-3} \times 10^3 \times 10} \\ h_2 & =3 \cdot 65 \mathrm{~mm} \\ \Delta h & =\frac{3}{5} \times 3 \cdot 65 \\ & =\frac{10 \cdot 95}{5} \\ \Delta h & =2 \cdot 19\end{aligned}$

Hence, the answer is the option (3).

Example 3: If M is the mass of water that rises in a capillary tube of radius r, then the mass of water which will rise in a capillary tube of radius 2r will be xM, then find the value of x :

1) 1

2) 0.5

3) 4

4) 2

Solution:

$\begin{aligned} & \frac{2 T \cos \Theta}{\rho g r} \\ & T-\text { wherein face } T \text { tension } \\ & r-\text { radius of capillary tube } \\ & \rho-\text { liquid density } \\ & \theta-\text { Angle of contact }\end{aligned}$

$\begin{aligned} m & =\rho A h \\ m & =(\rho)\left(\pi r^2\right) \frac{(2 T \cos \theta)}{\rho g r} \\ m & \propto r \\ \frac{m_1}{m_2} & =\frac{r_1}{r_2} \\ \frac{M}{m_2} & =\frac{r}{2 r} \\ M_2 & =2 M\end{aligned}$

Hence, the answer is the option (4).

Example 5: Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, then the rate of flow becomes:

1) $4 Q$
2) $Q$
3) $\frac{Q}{4}$
4) $\frac{Q}{8}$

Solution:

Use, $V=\frac{\pi P r^4}{8 \eta l}$

$
\begin{array}{ll}
\therefore & V \propto P r^4 \\
\therefore & \frac{V_2}{V_1}=\left(\frac{P_2}{P_1}\right)\left(\frac{r_2}{r_1}\right)^4=2 \times\left(\frac{1}{2}\right)^4=\frac{1}{8} \\
\therefore & V_2=\frac{Q}{8}
\end{array}
$
Hence, the answer is the option (4).

Summary

Capillary action, also known as capillarity, is the process where liquids move through narrow spaces or porous materials against gravity due to cohesive and adhesive forces. This principle is crucial in various natural and practical contexts, such as water transport in plants and the absorption of liquids by paper towels. The height to which a liquid rises in a capillary tube is determined by factors like surface tension, liquid density, and the tube's radius, as described by the ascent formula. Practical examples, such as the rise of liquids in capillary tubes or the effect of tube diameter on liquid height, illustrate the diverse applications and significance of capillary action in both scientific and everyday scenarios.