According to the definition of surface tension, it is the phenomenon that occurs when the surface of a liquid is in contact with another phase (it can be a liquid as well). Liquids tend to acquire the least surface area possible. The surface of the liquid behaves like an elastic sheet.

Surface tension is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimise surface area.

Cohesive force is the force of attraction acting between the molecules of the same substance. These forces between molecules in a liquid are shared with all neighbouring molecules. Those on the surface have no adjacent molecules above.

Hence, these molecules have an unbalanced force acting on them compared to the molecules present in bulk.  The unbalanced forces acting on the surface molecules tend to pull them inwards, which is why a liquid’s surface behaves like a stretched membrane.

Surface Tension Definition

Surface tension is a force per unit length acting in the plane of the interface between the plane of the liquid and any other substance. 

Let \(F\) be the common magnitude of the forces exerted on each other by the two parts of the surface across a line of length \(l\). We define the surface tension \(T\) of the liquid as
\(
T=F / l
\)

The value of the surface tension of a liquid depends on the liquid’s temperature and the medium on the other side of the surface. It decreases with temperature rise and becomes zero at the critical temperature.

\(\text { The SI unit of surface tension is } \mathrm{N} \mathrm{m}^{-1}\) and
\(
\text { the dimensional formula of surface tension is } \mathbf{M T}^{-2}
\)

Example 1: Water is kept in a beaker of radius \(5.0 \mathrm{~cm}\). Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. Surface tension of water \(=0.075 \mathrm{~N} \mathrm{~m}^{-1}\).

Solution:

The length of the diameter is
\(
\begin{aligned}
l=2 r & =10 \mathrm{~cm} \\
& =0.1 \mathrm{~m} .
\end{aligned}
\)
The surface tension is \(T=F / l\). Thus,
\(
\begin{aligned}
F & =T l \\
=\left(0.075 \mathrm{~N} \mathrm{~m}^{-1}\right) \times(0.1 \mathrm{~m}) & =7.5 \times 10^{-3} \mathrm{~N} .
\end{aligned}
\)

What Causes Surface Tension?

Surface tension has been well-explained by the molecular theory of matter, and Laplace explained this phenomenon based on intermolecular forces. For example, we know that if the distance between two molecules is less than the molecular range ” \(d “\left(\approx 10^{-9}\right.\) meter \()\), they attract each other. Still, if the distance is more than this, they attract each other, but the attraction force decreases considerably with distance.
Therefore, if we draw a sphere of radius \(d\) with a molecule at its centre, only those enclosed within this sphere can attract, or be attracted by, the molecule placed at the sphere’s centre. This is called the ‘sphere of molecular activity.

To understand the tension acting in a liquid’s free surface, let us consider four liquid molecules like \(A, B, C\) and \(D\) along with their spheres of molecular activity. The molecule \(D\) is well inside the liquid, and so it is attracted equally in all directions. Hence the resultant force acting on it is zero. The sphere of molecule \(C\) is just below the liquid surface, and the resultant force on it is also zero.

Molecule \(B\), which is a little below the liquid surface, has its sphere of molecular activity partly outside the liquid. The number of liquid molecules in the upper half (attracting downward) is less than the upward attraction. Hence the molecule \(B\) experiences a resultant downward force. Molecule \(A\) is on the surface of the liquid so that its sphere of molecular activity is half outside the liquid and half inside. As such, it experiences a maximum downward force. Thus all the molecules situated between the surface and a plane \(X Y\), distance \(d\) below the surface, experience a downward cohesive force.

When the surface area of liquid is increased, molecules from the interior of the liquid rise to the surface; as these molecules reach near the surface, work is done against the (downward) cohesive force. This work is stored in the molecule in the form of potential energy. Thus the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid. But a system is in stable equilibrium when its potential energy is minimum. Hence, to have minimum potential energy, the liquid surface tends to have a minimum number of molecules. In other words, the surface tends to be in contact with the minimum possible area. This tendency is exhibited as surface tension.

Relation Between Surface Energy and Surface Tension

When the surface area of a liquid is increased, the molecules from the interior rise to the surface. This requires work against the force of attraction of the molecules just below the surface. This work is stored in the form of potential energy in the newly formed surface. Thus the molecules on the surface have some additional energy due to their position. This extra energy per unit area of the surface is called surface energy.
Consider a liquid film trapped in a metal frame, with a moveable bar as shown in figure 10.17.

\(
W=\mathbf{F} \cdot \mathbf{d}=F d .
\)

From the conservation of energy, this is stored as additional energy in the film. If the surface energy of the film is \(S\) per unit area, the extra area is \(2 dl\). A film has two sides and the liquid in between, so there are two surfaces and the extra energy is
\(
\begin{aligned}
& S(2 d l)=F d \dots(10.21) \\
& \text { Or, } S=F d / 2 d l=F / 2 l \dots(10.22)
\end{aligned}
\)

This quantity \(S\) is the magnitude of surface tension. It is equal to the surface energy per unit area of the liquid interface and is also equal to the force per unit length exerted by the fluid on the movable bar.

Now suppose the wire is slowly pulled out by the external force through a distance \(x\) so that the area of the frame is increased by \(l x\). As there are two surfaces of the solution, a new surface area \(2 l x\) is created. The liquid from the inside is brought to create the new surface.

The work done by the external force in the displacement is
\(
W=F x=2 S l x=S(2 l x) .
\)
As there is no change in kinetic energy, the work done by the external force is stored as the potential energy of the new surface.
The increase in surface energy is
Thus, \(\quad \frac{U}{(2 l x)}=S\)
or, \(\quad \frac{U}{A}=S\).
We see that the surface tension of a liquid is equal to the surface energy per unit surface area.

In this interpretation, the SI unit of surface tension may be written as \(\mathrm{J} \mathrm{m}^{-2}\). It may be verified that \(\mathrm{N} \mathrm{m}^{-1}\) is equivalent to \(\mathrm{J} \mathrm{m}^{-2}\).

Example 2: A water drop of radius \(10^{-2} \mathrm{~m}\) is broken into 1000 equal droplets. Calculate the gain in surface energy. Surface tension of water is \(0.075 \mathrm{~N} \mathrm{~m}^{-1}\).

Solution:

The volume of the original drop is
\(
V=\frac{4}{3} \pi R^3 \text {, where } R=10^{-2} \mathrm{~m} .
\)
If \(r\) is the radius of each broken droplet, the volume is also
\(
V=1000 \times \frac{4}{3} \pi r^3 \text {. }
\)
Thus, \(\quad 1000 r^3=R^3\)
or, \(\quad r=R / 10\).
The surface area of the original drop is \(A_1=4 \pi R^2\) and the surface area of the 1000 droplets is
\(
A_2=1000 \times 4 \pi r^2=40 \pi R^2 .
\)
The increase in area is
\(
\Delta A=A_2-A_1=40 \pi R^2-4 \pi R^2=36 \pi R^2 .
\)
The gain in surface energy is
\(
\begin{aligned}
\Delta U & =(\Delta A) S=36 \pi R^2 S \\
& =36 \times 3.14 \times\left(10^{-4} \mathrm{~m}^2\right) \times\left(0.075 \mathrm{~N} \mathrm{~m}^{-1}\right) \\
& =8.5 \times 10^{-4} \mathrm{~J} .
\end{aligned}
\)

Surface Tension Example

1. The shape of droplets: Liquid droplets acquire their shape because of the surface tension. Water drops come together due to their forces of attraction, and because of the surface tension, they take up the spherical shape to have the least possible area.
2. Soaps and detergents: commonly used for washing dishes or clothes are also based on surface tension. When these are added to water, they lower the surface tension of the water so that it more readily soaks into pores and soiled areas.

Measuring Surface Tension

A fluid will stick to a solid surface if the surface energy between the fluid and the solid is smaller than the sum of surface energies between solid air and fluid air. Now there is an attraction between the solid surface and the liquid. It can be directly measured experimentally as schematically shown in Fig.10.18. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side, with its horizontal edge just over the water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears the water.

Suppose the additional weight required is \(W\). Then from Eq. \(10.22\) and the discussion given there, the surface tension of the liquid-air interface is
\(
S_{\mathrm{la}}=(W / 2 l)=(\mathrm{mg} / 2 l) \dots(10.23)
\)
where \(\mathrm{m}\) is the extra mass and \(l\) is the length of the plate edge. The subscript (la) emphasizes the fact that the liquid-air interface tension is involved.

Angle of Contact

The surface of liquid near the plane of contact, with another medium, is in general curved. The angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid is termed as the angle of contact. It is denoted by \(\theta\). It is different at interfaces of different pairs of liquids and solids. The value of \(\theta\) determines whether a liquid will spread on the surface of a solid or it will form droplets on it. For example, water forms droplets on the lotus leaf as shown in Fig. 10.19 (a) while spreads over a clean plastic plate as shown in Fig. 10.19(b).

We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air, and solid-liquid denoted by \(S_{\mathrm{la}}, S_{\mathrm{sa}}\) and \(S_{\text {sl }}\), respectively as given in Fig. 10.19 (a) and (b). At the line of contact, the surface forces between the three media must be in equilibrium. From the Fig. 10.19(b) the following relation is easily derived.
\(
S_{\mathrm{la}} \cos \theta+S_{\mathrm{sl}}=S_{\mathrm{sa}} \dots(10.24)
\)
The angle of contact is an obtuse angle if \(S_{\text {sl }}>S_{\text {la }}\) as in the case of water-leaf interface while it is an acute angle if \(S_{\mathrm{sl}}<S_{\mathrm{la}}\) as in the case of the water-plastic interface.

Drops and Bubbles

One consequence of surface tension is that free liquid drops and bubbles are spherical if effects of gravity can be neglected.

Another interesting consequence of surface tension is that the pressure inside a spherical drop Fig.10.20 (a) is more than the pressure outside. Suppose a spherical drop of radius \(r\) is in equilibrium. If its radius increase by \(\Delta r\). The extra surface energy is
\(
\left[4 \pi(r+\Delta r)^{2-} 4 \pi r^2\right] S_{\text {la }}=8 \pi \pi \Delta r S_{\text {la }} \quad(10.25)
\)
If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference \(\left(P_i-P_0\right)\) between the inside of the bubble and the outside. The work done is
\(
W=\left(P_i-P_0\right) 4 \pi r^2 \Delta r \dots(10.26)
\)
so that
\(
\left(P_i-P_{\mathrm{o}}\right)=\left(2 S_{1 \mathrm{a}} / r\right) \dots(10.27)
\)

Note: The pressure inside the surface is greater than the pressure outside the surface by an amount \(2 S / R\).

In general, for a liquid-gas interface, the convex side has a higher pressure than the concave side. For example, an air bubble in a liquid would have higher pressure inside it. See Fig 10.20 (b).

A bubble Fig 10.20 (c) differs from a drop and a cavity; in this, it has two interfaces. Applying the above argument we have for a bubble
\(
\left(P_i-P_{\mathrm{o}}\right)=\left(4 \mathrm{~S}_{\mathrm{la}} / r\right) \dots(10.28)
\)
This is probably why you have to blow hard, but not too hard, to form a soap bubble. A little extra air pressure is needed inside!

Example 3: Find the excess pressure inside a mercury drop of radius \(2.0 \mathrm{~mm}\). The surface tension of mercury \(=0 \cdot 464 \mathrm{~N} \mathrm{~m}^{-1}\).

Solution:

The excess pressure inside the drop is
\(
\begin{gathered}
P_2-P_1=2 S / R \\
=\frac{2 \times 0.464 \mathrm{~N} \mathrm{~m}^{-1}}{2.0 \times 10^{-3} \mathrm{~m}}=464 \mathrm{~N} \mathrm{~m}^{-2} .
\end{gathered}
\)

Excess Pressure in a Soap Bubble

Let the pressure of the air outside the bubble be \(P_1\), that within the soap solution be \(P^{\prime}\) and that of the air inside the bubble be \(P_2\). Looking at the outer surface, the solution is on the concave side of the surface, hence
\(
P^{\prime}-P_1=2 S / R
\)
where \(R\) is the radius of the bubble. As the thickness of the bubble is small on a macroscopic scale, the difference in the radii of the two surfaces will be negligible.

Similarly, looking at the inner surface, the air is on the concave side of the surface, hence
\(
P_2-P^{\prime}=2 S / R \text {. }
\)
Adding the two equations
\(
P_2-P_1=4 \mathrm{~S} / R .
\)
The pressure inside a bubble is greater than the pressure outside by an amount \(4 S / R\).

Example 4: A \(0.02 \mathrm{~cm}\) liquid column balances the excess pressure inside a soap bubble of radius \(7.5 \mathrm{~mm}\). Determine the density of the liquid. Surface tension of soap solution \(=0.03 \mathrm{~N} \mathrm{~m}^{-1}\).

Solution:

The excess pressure inside a soap bubble is
\(
\Delta P=4 S / R=\frac{4 \times 0.03 \mathrm{~N} \mathrm{~m}^{-1}}{7.5 \times 10^{-3} \mathrm{~m}}=16 \mathrm{~N} \mathrm{~m}^{-2} .
\)
The pressure due to \(0.02 \mathrm{~cm}\) of the liquid column is
\(
\begin{aligned}
\Delta P & =h \rho g \\
& =\left(0.02 \times 10^{-2} \mathrm{~m}\right) \rho\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) . \\
\text { Thus, } 16 \mathrm{~N} \mathrm{~m}^{-2} & =\left(0.02 \times 10^{-2} \mathrm{~m}\right) \rho\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right) \\
\text { or, } \quad \quad \quad \quad \quad & =8.2 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} .
\end{aligned}
\)

Capillary Rise

One consequence of the pressure difference across a curved liquid-air interface is the well-known effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin; if the tube were hair thin, the rise would be very large. To see this, consider a
vertical capillary tube of circular cross-section (radius a) inserted into an open vessel of water (Fig. 10.21). The contact angle between water and glass is acute.

Thus the surface of water in the capillary is concave. This means that there is a pressure difference between the two sides of the top surface. This is given by

\(
\begin{aligned}
& \left(P_i-P_o\right)=(2 S / r)=2 S /(a \sec \theta) \\
& =(2 S / a) \cos \theta \dots(10.29)
\end{aligned}
\)

Thus the pressure of the water inside the tube, just at the meniscus (air-water interface) is less than the atmospheric pressure. Consider the two points A and B in Fig. 10.21(a). They must be at the same pressure, namely
\(
P_0+h \rho g=P_i=P_A \dots(10.30)
\)
where \(\rho\) is the density of water and \(h\) is called the capillary rise [Fig. \(10.21(\mathrm{a})\) ]. Using Eq. (10.29) and (10.30) we have
\(
h \rho g=\left(P_i-P_o\right)=(2 S \cos \theta) / a \quad \text { (10.31) }
\)
The discussion here, and the Eqs. (10.26) and (10.27) make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few \(\mathrm{cm}\) for fine capillaries. For example, if \(a=0.05 \mathrm{~cm}\), using the value of surface tension for water (Table 10.3), we find that
\(
\begin{aligned}
h & =2 S /(\rho g a) \\
& =\frac{2 \times\left(0.073 \mathrm{~N} \mathrm{~m}^{-1}\right)}{\left(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(5 \times 10^{-4} \mathrm{~m}\right)} \\
& =2.98 \times 10^{-2} \mathrm{~m}=2.98 \mathrm{~cm}
\end{aligned}
\)
Notice that if the liquid meniscus is convex, as for mercury, 1.e., if \(\cos \theta\) is negative then from Eq. (10.30) for example, it is clear that the liquid will be lower in the capillary !

Example 5: A capillary tube of radius \(0.20 \mathrm{~mm}\) is dipped vertically in water. Find the height of the water column raised in the tube. Surface tension of water \(=0.075 \mathrm{~N} \mathrm{~m}^{-1}\) and density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\). Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

We have,
\(
\begin{aligned}
h & =\frac{2 S \cos \theta}{r \rho g} \\
& =\frac{2 \times 0.075 \mathrm{~N} \mathrm{~m}^{-1} \times 1}{\left(0.20 \times 10^{-3} \mathrm{~m}\right) \times\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)} \\
& =0.075 \mathrm{~m}=7.5 \mathrm{~cm} .
\end{aligned}
\)

Note: \(a = r\)

Detergents and Surface Tension

We clean dirty clothes containing grease and oil stains sticking to cotton or other fabrics by adding detergents or soap to water, soaking clothes in it, and shaking. Washing with water does not remove grease stains. This is because water does not wet greasy dirt; i.e., there is very little area of contact between them. If water could wet grease, the flow of water could carry some grease away. Something of this sort is achieved through detergents. The molecules of detergents are hairpin shaped, with one end attracted to water and the other to molecules of grease, oil, or wax, thus tending to form water-oil interfaces.

In our language, we would say that the addition of detergents, whose molecules attract at one end and say, oil on the other, reduces drastically the surface tension S (water-oil). It may even become energetically favourable to form such interfaces, i.e., globs of dirt surrounded by detergents and then by water. This kind of process using surface active detergents or surfactants is important not only for cleaning, but also in recovering oil, mineral ores, etc.

Example 6: The lower end of a capillary tube of diameter \(2.00 \mathrm{~mm}\) is dipped \(8.00\) \(\mathrm{cm}\) below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at the temperature of the experiments is \(7.30 \times 10^{-2} \mathrm{Nm}^{-1} .1\) atmospheric pressure \(=\) \(1.01 \times 10^5 \mathrm{~Pa}\), the density of water \(=1000 \mathrm{~kg} / \mathrm{m}^3\), \(\mathrm{g}=9.80 \mathrm{~m} \mathrm{~s}^2\). Also, calculate the excess pressure.

Solution:

The excess pressure in a bubble of gas in a liquid is given by \(2 S / r\), where \(S\) is the surface tension of the liquid-gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure, in that case, is \(4 S / r\).) The radius of the bubble is \(\mathrm{r}\). Now the pressure outside the bubble \(P_0\) equals atmospheric pressure plus the pressure due to \(8.00 \mathrm{~cm}\) of water column. That is
\(
\begin{aligned}
& P_{\mathrm{o}}=\left(1.01 \times 10^5 \mathrm{~Pa}+0.08 \mathrm{~m} \times 1000 \mathrm{~kg} \mathrm{~m}^{-3}\right. \\
& =1.01784 \times 10^5 \mathrm{~Pa} \\
& \times 9.80 \mathrm{~m} \mathrm{~s}^{-2} \text { ) } \\
&
\end{aligned}
\)
Therefore, the pressure inside the bubble is
\(
\begin{aligned}
P_i & =P_0+2 S / r \\
& =1.01784 \times 10^5 \mathrm{~Pa}+\left(2 \times 7.3 \times 10^2 \mathrm{Pam} / 10^{-3} \mathrm{~m}\right) \\
& =(1.01784+0.00146) \times 10^5 \mathrm{~Pa} \\
& =1.02 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical! (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is \(146 \mathrm{~Pa}\).

Example 7: There is an air bubble of radius \(1.0 \mathrm{~mm}\) in a liquid of surface tension \(0.075 \mathrm{~N} \mathrm{~m}^{-1}\) and density \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\). The bubble is at a depth of \(10 \mathrm{~cm}\) below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? Take \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

Let the atmospheric pressure be \(P_a\). The pressure of the liquid just outside the bubble is (figure above)
\(
P=P_a+h \rho g \text {. }
\)
The pressure inside the bubble is
\(
P^{\prime}=P+\frac{2 S}{r}=P_a+h \rho g+\frac{2 S}{r}
\)
or, \(P^{\prime}-P_a\)
\(
\begin{aligned}
& =(10 \mathrm{~cm})\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)+\frac{2 \times 0.075 \mathrm{~N} \mathrm{~m}^{-1}}{1.0 \times 10^{-3} \mathrm{~m}} \\
& =980 \mathrm{~N} \mathrm{~m}^{-2}+150 \mathrm{~N} \mathrm{~m}^{-2} \\
& =1130 \mathrm{~Pa} .
\end{aligned}
\)

Example 8: A light wire \(A B\) of length \(10 \mathrm{~cm}\) can slide on a vertical frame as shown in the figure below. There is a film of soap solution trapped between the frame and the wire. Find the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soap solution \(=25\) dyne \(\mathrm{cm}^{-1}\). Take \(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

Soap solution film will be formed on both sides of the frame. Each film is in contact with the wire along a distance of \(10 \mathrm{~cm}\). The force exerted by the film on the wire
\(
\begin{aligned}
& =2 \times(10 \mathrm{~cm}) \times\left(25 \text { dyne } \mathrm{cm}^{-1}\right) \\
& =500 \text { dyne }=5 \times 10^{-3} \mathrm{~N} .
\end{aligned}
\)
This force acts vertically upward and should be balanced by the load. Hence the load that should be suspended is \(5 \times 10^{-3} \mathrm{~N}\). The mass of the load should be \(\frac{5 \times 10^{-3} \mathrm{~N}}{10 \mathrm{~m} \mathrm{~s}^{-2}}=5 \times 10^{-4} \mathrm{~kg}=0.5 \mathrm{~g}\)

Example 9: The lower end of a capillary tube is dipped into water and it is seen that the water rises through \(7.5 \mathrm{~cm}\) in the capillary. Find the radius of the capillary. Surface tension of water \(=7.5 \times 10^{-2} \mathrm{~N} \mathrm{~m}^{-1}\). The contact angle between water and glass \(=0^{\circ}\). Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

We have,
\(
\begin{aligned}
h & =\frac{2 S \cos \theta}{r \rho g} \\
\text { or, } \quad \quad r & =\frac{2 S \cos \theta}{h \rho g} \\
& =\frac{2 \times\left(7.5 \times 10^{-2} \mathrm{~N} \mathrm{~m}^{-1}\right) \times 1}{(0.075 \mathrm{~m}) \times\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \times\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)} \\
& =2 \times 10^{-4} \mathrm{~m}=0.2 \mathrm{~mm} .
\end{aligned}
\)

Example 10: Two mercury drops each of radius \(r\) merge to form a bigger drop. Calculate the surface energy released.

Solution:

The surface area of one drop before merging \(=4 \pi r^2\).
The total surface area of both the drops \(\quad=8 \pi r^2\).
Hence, the surface energy before merging \(=8 \pi r^2 S\).
When the drops merge, the volume of the bigger drop
\(
=2 \times \frac{4}{3} \pi r^3=\frac{8}{3} \pi r^3 \text {. }
\)
If the radius of this new drop is \(R\),
or
\(
\frac{4}{3} \pi R^3=\frac{8}{3} \pi r^3
\)
or, \(\quad 4 \pi R^2=4 \times 2^{2 / 3} \times \pi r^2\).
Hence, the surface energy \(=4 \times 2^{2 / 3} \times \pi r^2 S\).
The released surface energy \(=8 \pi r^2 S-4 \times 2^{2 / 3} \pi r^2 S\)
\(\approx 1 \cdot 65 \pi r^2 S\).

Example 11: A large wooden plate of area \(10 \mathrm{~m}^2\) floating on the surface of a river is made to move horizontally with a speed of \(2 \mathrm{~m} \mathrm{~s}^{-1}\) by applying a tangential force. If the river is \(1 \mathrm{~m}\) deep and the water in contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river \(=10^{-2}\) poise.

Solution:

The velocity decreases from \(2 \mathrm{~m} \mathrm{~s}^{-1}\) to zero in \(1 \mathrm{~m}\) of perpendicular length. Hence, the velocity gradient
\(
=d v / d x=2 \mathrm{~s}^{-1} \text {. }
\)
Now,
or,
\(
\eta=\left|\frac{F / A}{d v / d x}\right|
\)
or,
\(
\begin{aligned}
10^{-3} \frac{\mathrm{N}-\mathrm{s}}{\mathrm{m}^2} & =\frac{F}{\left(10 \mathrm{~m}^2\right)\left(2 \mathrm{~s}^{-1}\right)} \\
F & =0.02 \mathrm{~N}
\end{aligned}
\)

Example 12: The velocity of water in a river is \(18 \mathrm{~km} \mathrm{~h}^{-1}\) near the surface. If the river is \(5 \mathrm{~m}\) deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water \(=10^{-2}\) poise.

Solution:

The velocity gradient in the vertical direction is

\(
\frac{d v}{d x}=\frac{18 \mathrm{~km} \mathrm{~h}^{-1}}{5 \mathrm{~m}}=1.0 \mathrm{~s}^{-1}
\)
The magnitude of the force of viscosity is
\(
F=\eta A \frac{d v}{d x}
\)
The shearing stress is
\(
F / A=\eta \frac{d v}{d x}=\left(10^{-2} \text { poise }\right)\left(1 \cdot 0 \mathrm{~s}^{-1}\right)=10^{-3} \mathrm{~N} \mathrm{~m}^{-2} \text {. }
\)

Example 13: Find the terminal velocity of a raindrop of radius \(0.01 \mathrm{~mm}\). The coefficient of viscosity of air is \(1.8 \times 10^{-5}\) \(\mathrm{N}-\mathrm{s} \mathrm{m}^{-2}\) and its density is \(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\). Density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\). Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

The forces on the rain drop are
(a) the weight \(\frac{4}{3} \pi r^3 \rho g\) downward,
(b) the force of buoyancy \(\frac{4}{3} \pi r^3 \sigma g\) upward,
(c) the force of viscosity \(6 \pi \eta r v\) upward.
Here \(\rho\) is the density of water and \(\sigma\) is the density of air. At terminal velocity the net force is zero. As the density of air is much smaller than the density of water, the force of buoyance may be neglected.
Thus, at terminal velocity
\(
\begin{aligned}
6 \pi \eta r v & =\frac{4}{3} \pi r^3 \rho g \\
\text { or, } \quad v & =\frac{2 r^2 \rho g}{9 \eta} . \\
& =\frac{2 \times(0.01 \mathrm{~mm})^2 \times\left(1000 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}{9 \times\left(1.8 \times 10^{-5} \mathrm{~N}-\mathrm{s} \mathrm{m}^{-2}\right)} \\
& \approx 1.2 \mathrm{~cm} \mathrm{~s}^{-2} .
\end{aligned}
\)