Contents
Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What is Capillary Action? Definition and Examples
The word ‘capillary’ has originated from capillus, a Latin word which means ‘hair’. So a tube with a uniform fine small bore like a hair is known as a capillary tube.
When a capillary tube is inserted vertically upright into water, the water rises within the tube and its level inside the tube stands higher than the water level outside. Moreover, the surface of water inside the tube takes a concave shape instead of a horizontal one [Fig.(a)]. Liquids like water, alcohol, copper sulphate solution etc. can wet glass. If a
capillary tube made of glass is (lipped into these liquids, then the same phenomenon occurs. On the other hand, if a capillary glass tube is dipped into mercury (or into some other liquid that does not wet glass), then the level of mercury in the tube falls slightly, i.e., the mercury stands at a lower level inside the capillary tube than that outside it. Moreover, the upper meniscus of mercury takes a convex shape [Fig.(b)].
The narrower the bore of the capillary tube, the greater the ascent or descent of the liquid level.
This type of rise or fall of a liquid in a capillary tube is called capillary action. The curvature of a liquid surface in contact with a solid surface is also due to this action. The surface tension of a liquid is the origin of these phenomena.
Examples of capillarity: Due to capillary action, a blotting paper absorbs ink, a sponge absorbs water, oil rises through a wick and a towel absorbs water droplets from our body. These objects contain small pores and when water or any liquid comes in contact with them, the liquid enters these pores.
Underground water rises through the small pores of soil and travels upwards so that the soil remains moist. The stems of a tree sends water from the moist soil to the leaves through capillary action. The pores of sandy soil are quite large and hence water cannot rise up to a great height through it. Therefore the upper layer of sandy soil remains dry.
Angle of contact:
Definition: When a liquid is in contact with a solid, the angle between the solid surface and the tangent to the free surface of the liquid at the point of contact, measured from inside the liquid is called the angle of contact for that specific pair of solid and liquid.
In Fig., the upper surface of the liquid touches the solid at the point A. The tangent drawn on the liquid surface is AB and this tangent makes the angle ∠BAC with the solid inside the liquid. This angle ∠BAC is the angle of contact (θ) between the solid and the liquid. The angle of contact for various liquid-solid pairs ranges from 0° to 180°. The angle of contact does not depend upon the inclination of the solid with the liquid surface. This angle between a glass plate and water remains the same irrespective of whether the plate is immersed straight or in an oblique manner.
The rise or fall of a liquid inside a capillary tube depends on the angle of contact (θ). If the value of θ is less than 90°, then the liquid rises through a capillary tube (for example, water rises in a capillary glass tube) and the tipper meniscus is concave [Fig. (a)l. On the other hand, if the value of θ is more than 90°, then a liquid is depressed inside a capillary tube (for example. mercury level is lower inside a capillary tube) and the upper meniscus is convex [Fig.(b)].
Factors on which the angle of contact depends:
1. Nature of solid and liquid: If the adhesive force, between a liquid and the material of its solid container, is stronger than the cohesive force among the liquid molecules, then the liquid wets the solid. For example, water wets glass because the water-glass adhesive force is stronger than the cohesive force of water molecules. On the other hand, mercury-glass adhesive force is weaker than the cohesive force among mercury molecules; so mercury does not wet its glass container.
If a drop of pure water is poured on a clean piece of glass, the water spreads quickly on the glass surface. In this case, the angle of contact is very small— almost zero, i.e., θ ≈ 0. On the other hand, if a drop of water falls on a lotus petal, the drop does not spread. Instead, it rests on the petal like a dewdrop. Here, the angle of contact, θ > 90°. So, if θ < 90° the liquid wets the solid in contact with it, but, if θ > 90°, then it does not wet the solid. For example, a drop of mercury on a piece of glass does not wet it as θ ≈ 140°. Again, the angle of contact (θ) of water and silver is 90°, so, when water is kept in a vertical silver container, the surface of water remains horizontal.
2. Medium in contact with the exposed liquid surface: If the medium in contact with the surface of mercury is air, then the value of the angle of contact between mercury and glass is not the same as when air is replaced by water as the medium in contact with mercury
3. Impurities present in the liquid: For instance, the angle of contact between clear glass and very pure water is nearly zero, but that between ordinary water and glass is nearly 8°.
Rise of liquid in a capillary tube: Suppose, a capillary tube of glass having a uniform bore of radius r is dipped vertically in water or any liquid which wets it. The liquid immediately rises up into it up to the height ‘h’ (say) and the shape of the liquid meniscus is spherical and it concaves upwards [Fig.]. The liquid meniscus in the tube is along a circle of circumference 2πr which is in contact with the glass tube. Let T be the surface tension of the liquid and θ be the angle of contact for the liquid and the glass of the tube.
The surface tension T of the liquid acts inwards along the tangent to the liquid meniscus at every point of its contact with the inner surface of the tube, making an angle θ with the wall of the tube. Thus an inward pull is exerted on the glass in this direction at all these points. Since in accordance with Newton’s third law of motion, action and reaction are equal and opposite, the wall of the tube also exerts an equal (reaction) force T on the liquid. This reaction force T may be resolved into two perpendicular components
1. Tcosθ acting vertically upwards and
2. Tsinθ acting horizontally outwards. Taking the whole meniscus into consideration, for each horizontal component Tsinθ, there is an equal and opposite component so that the two neutralise each other. The vertical components being in the same direction get added up. It is this force which supports the weight of the liquid column so raised.
Hence, total upward force on the liquid in the tube
= 2πr ᐧ Tcosθ
It is this force which supports the weight of the column h of the liquid in the tube plus the weight of a volume V of the liquid in the meniscus itself. h ¡s the length of the liquid column from the horizontal surface of the liquid in the container to the bottom of the meniscus at E.
So, the weight of the total volume of the liquid
= (πr2h + V)ρg
or, T = \(\left(\frac{\pi r^2 h+V}{2 \pi r \cos \theta}\right) \rho g\) ………. (1)
If the volume of the liquid in the meniscus is negligible in comparison with that in the column h i.e., if the tube is of a
very fine bore we have,
T = \(\frac{\pi r^2 h \rho g}{2 \pi r \cos \theta}\) = \(\frac{r h \rho g}{2 \cos \theta}\)
or, h = \(\frac{2 T \cos \theta}{r \rho g}\) ………. (2)
However if V is not negligible and θ = 0, the meniscus of the liquid in the tube is hemispherical in shape of radius equal to r. So the volume of the liquid in the meniscus is equal to the difference between the volumes of a cylinder of radius r and length r and a hemisphere of radius r.
∴ V = volume of the cylinder ABCD – volume of the hemisphere AEB
[taking cosθ = 1]
If the temperature of the liquid remains unchanged, T, ρ and θ are constants. In that case it can be said from equation (2),
hr = constant ……. (4)
This relation is known as Jurin’s law. The graph between the radius (r) and the rise in height (h) of the liquid is a rectangular hyperbola. The same relation holds good for depression of a liquid in a capillary tube but with a negative sign.
Jurin’s law: The ascent or descent of a liquid inside a capillary tube is inversely proportional to the radius of the tube.
From this law, it is concluded that smaller the radius (r) of the tube, higher is the rise (h) of the liquid level inside the tube. But, if the radius of the tube is large, i.e., if the capillarity of the tube is not maintained, then this law becomes erroneous.
If the angle of contact is less than 90°, then cos θ is positive and hence the value of h as indicated by equation (1) is also positive; in this case, the liquid rises upwards in the capillary tube. On the other hand, if the angle of contact is more than 90°, then cos θ becomes negative and hence h will also be negative. In this case, the liquid level falls downwards in the capillary tube.
In the case of clean glass and pure water, the angle of contact, θ ≈ 0°; so cos θ ≈ 1 and hence from equation (2), h = \(\frac{2 T}{r \rho g}\) or, T = \(\frac{1}{2} r h \rho g\)
Measuring the rise of water in a capillary tube, its surface tension can be determined from this equation.
Rise of a liquid in a capillary tub of insuffficient height: The expression for capillary rise h of a liquid in a tube is
h = \(\frac{2 T \cos \theta}{r \rho g}\),
where, r = radius of the tube;
ρ, T = density and surface tension of the liquid, respectively;
θ = angle of contact;
g = acceleration due to gravity.
When the height of the capillary tube is more than h, the liquid rises to the full height h to attain equilibrium [Fig.(a)]. If R be the radius of the concave upper surface, then
r = Rcosθ
So, h = \(\frac{2 T \cos \theta}{(R \cos \theta) \rho g}\) = \(\frac{2 T}{R \rho g}\)
or, hR = \(\frac{2 T}{\rho g}\) = constant
Now, if a capillary tube of the same radius has a shorter height h’ [Fig.(b)] above the external liquid surface, i.e., if h’ < h, then the liquid can rise only upto that height h’. In this case, the liquid adjusts the curvature of its upper concave surface in such a way that the radius of curvature R’ satisfies the relation, h’R’ = hR, because hR is a constant. Naturally, R’ = R\(\frac{h}{h^{\prime}}\) > R, as h’ < h. So the liquid in the capillary attains equilibrium by increasing the radius of curvature of its upper concave surface. This means that the curvature of the surface becomes less. It is to be noted that, the liquid will not spill over the top due to the insufficient height of the capillary tube.
Numerical Examples
Example 1.
A liquid of density 830 kg ᐧ m-3 rises through 0.0893 ni in a capillary tube of diameter 1.68 × 10-4 m. Determine the surface tension of the liquid. Take the angle of contact as 0°.
Solution:
We know that, T = \(\frac{r h \rho g}{2 \cos \theta}\)
Here, r = \(\frac{r h \rho g}{2 \cos \theta}\) = 8.4 × 10-5m,
h = 0.0893 m, ρ = 830 kg ᐧ m-3 and θ = 0°
∴ T = \(\frac{8.4 \times 10^{-5} \times 0.0893 \times 830 \times 9.8}{2 \cos 0^{\circ}}\) = 0.0305 N ᐧ m-1
Example 2.
A capillary tube with a bore diameter of 0.5 mm is dipped upright in a liquid having surface tension 0.03 N ᐧ m-1. The specific gravity of the liquid is 0.8 and the liquid wets the surface of the tube completely. Up to what height will the liquid rise in the tube?
Solution:
If a liquid wets the wall of the tube completely, then the angle of contact becomes approximately zero.
∴ T = \(\frac{r h \rho g}{2}\) [As θ is very small, cosθ \(\simeq\) 1]
or, h = \(\frac{2 T}{r \rho g}\) [Here, T = 0.03 N ᐧ m-1, r = 0.25 × 10-3 m, ρ = 0.8 × 103 kg ᐧ m-3]
= \(\frac{2 \times 0.03}{0.25 \times 10^{-3} \times 0.8 \times 10^3 \times 9.8}\) = 0.0306 m.
Example 3.
The diameter of a barometer tube is 3 mm. What will be the error in the barometer reading due to surface tension? For mercury in the glass tube, the surface tension = 0.647 N ᐧ m-1; the angle of contact = 128° and the density of mercury = 13500 kg ᐧ m-3.
Solution:
We know that h = \(\frac{2 T \cos \theta}{r \rho g}\)
Here, T = 0.647 N ᐧ m-1, θ = 128°,
r = \(\frac{3 \times 10^{-3}}{2}\) = 1.5 × 10-3m,
ρ = 13500 kg ᐧ m-1, g = 9.8 m ᐧ s-2,
cosθ = cos 128° = cos(90° + 38°)
= -sin38° = -0.6157
∴ h = –\(\frac{2 \times 0.647 \times 0.6157}{1.5 \times 10^{-3} \times 13500 \times 9.8}\)
= -0.004 m
So, to obtain the actual atmospheric pressure, 0.004 m has to be added to the barometric reading.
Example 4.
The diameters of the two arms of a U-tube are 1 mm and 2 mm. The tube is filled partly with water and is kept vertical. If the surface tension of water is 70 dyn ᐧ cm-1, then find the difference in the level of water in the two arms of the U-tube. [Angle of contact = 8°]
Solution:
Suppose due to capillary action, the water level rises in one arm of the U-tube to h1 and in the other arm to h2.
Example 5.
There are two parallel glass plates having some water between them [Fig.]. The surface tension of water is 0.07 N/m. When the two plates are moved apart from each other by 0.2 mm a water layer is formed between the plates. In this case what is the minimum force to be applied to separate the plates. Consider, the angle of contact between water and glass plate is zero and area of contact is 4 cm2.
Solution:
In two dimensional picture, the shape of the freé surface of the water layer formed between two parallel plates separated by the distance d is concave [Fig.].
If the radius of curvature of the concave surface is r, then
d = 2 rcosθ [where, θ = the angle of contact between water and glass]
The pressure difference between point A (outside the water layer) and point B (inside the water layer) is
pA – pB = \(\frac{T}{r}\) [where T = surface tension of water]
Butt pA = atmospheric pressure (p0)
i.e., p0 – pB = \(\frac{T}{r}\)
Atmospheric pressure is also exerted on the two glass plates.
Therefore, the minimum force required to separate the two plates is
Example 6.
Water reaches upto a height 2 cm along a capillary tube. Calculate the height upto which water can reach along another capillary tube of radius \(\frac{1}{3}\)rd of the previous one.
Solution:
Let the rise of liquid in the capillary tube of radius \(\frac{r}{3}\) be h1.
According to Jurin’s law, hr = constant.
Hence, h1(\(\frac{r}{3}\)) = hr
or, h1 = 3h = 3 × 2 = 6 cm
∴ The rise of the liquid surface in the second tube
= 6 cm.