Chapter 10.17 - Pressure Inside Drops and Bubbles

In the previous section, we saw contact angle and capillarity. In this section we will see pressure inside a drop. Some basic details can be written in 21 steps:

1. Consider a drop of a liquid
    ♦ A molecule in the interior of the drop will not experience any net force
    ♦ But a molecule at the surface will experience a net inward force
• We saw this same situation when we first saw the basics about surface tension (See the first fig.10.48 of section 10.15)
2. So all the surface molecules of the drop will experience a net inward force
• As a result, the drop will roll into a spherical shape
3. Also we know that, the surface will act like  stretched membrane
    ♦ That means, the surface of the drop tries to shrink to the least possible area
    ♦ That means, the drop will be trying to attain the least possible surface area
4. Which shape has the least possible surface area?
• We can find the answer in 2 steps:
(i) Let there be two samples of any substance
• Let both have the same volume V
(ii) Let us mold them into two different shapes:
    ♦ The first sample is made into a cubical shape
    ♦ The second sample is made into a spherical shape
• Then, the second sample will be having lesser surface area
• In fact, we can compare a sphere with any other shape with the same volume
■ The surface area of the sphere will always be lesser
■ So naturally, the liquid drop will be spherical
5. The force which tries to shrink the surface is the surface tension S
• This force will be tangential to the surface of the drop
6. We can draw infinite number of tangents on the surface of a sphere
• Three of them are shown in fig.10.61(a) below:

Fig.10.61

• Surface tension will be acting along all those tangents
7. For our present case, we are going to draw tangents at a few specific points
This can be explained in steps:
(i) Consider a sphere
(ii) Cut the sphere using a horizontal plane
    ♦ This horizontal plane should be passing through the centre of the sphere
    ♦ This is shown in fig.10.61(b) above
(iii) The horizontal plane will intersect the ‘surface of the sphere’ along a horizontal circle
    ♦ Draw vertical tangents of the sphere at the points on that circle
    ♦ These are the tangents that we need for our present problem
8. Surface tension will be acting along those tangents
    ♦ To indicate those surface tension forces, we give arrow heads to the tangents
    ♦ This is shown in fig.c
9. Using the horizontal plane, cut the sphere into two halves. So we have two hemispheres
    ♦ Separate the top hemisphere
    ♦ This is shown in fig.d
10. We see that the surface tension forces are acting downwards
• We can easily calculate the total downward force in just 2 steps:
(i) S is the surface tension force per unit length
(ii) So total surface tension force = (S × circumference of the circle) = 2πrS
    ♦ Where r is the radius of the drop
11. So a force of 2πrS is acting in the downward direction
• But the hemisphere is in equilibrium
• That means, there are some vertical upward forces also acting on the hemisphere
• The next four steps from (12) to (15) will explain those vertical upward forces
12. Fig.10.62(a) below shows the 2D view of the hemisphere

Fig.10.62

• The red arrows indicate the pressure P inside the drop
    ♦ Some of those arrows will be vertical. Some will be horizontal
13. Consider those arrows which are neither vertical nor horizontal
• Such red arrows will have a vertical component and a horizontal component
    ♦ This is shown in fig.b
          ✰ Red arrows which are closer to vertical will have lesser 𝜃 values
          ✰ Red arrows which are closer to horizontal will have greater 𝜃 values
• The horizontal components will cancel each other
    ♦ So the net horizontal force will be zero
14. But there will be a net vertical upward force
• Let F_y(1), F_y(2), F_y(3), . . . be the vertical components
• Then the total vertical force will be given by:
F_y = F_y(1) + F_y(2) + F_y(3) + . . .
15. Since the hemisphere is in equilibrium, we have: F_y = 2πrS
• Thus we successfully obtain the net upward internal force F_y
16. If we divide this F_y by 'area on which Fy acts', we will get the internal pressure exerted by F_y
■ But by definition of pressure,
    ♦ the area on which the force acts
    ♦ must be perpendicular to
    ♦ the force
• In our present case, F_y and it's components are vertical
    ♦ So the area on which they act must be horizontal
17. But in our present case, the components of F_y acts on the surface of the hemisphere
• Surface of the hemisphere is not horizontal. It is curved
18. So we consider the 'projected area' of the curved surface
• That is., we project the curved surface of the hemisphere onto a horizontal plane
• Obviously, the 'projected area' thus obtained, will be the base of the hemisphere
• So the projected area = Base area of the hemisphere = πr^₂
19. Now we can find the pressure:
• Pressure in the vertical direction = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{F_y}{\pi r^2}}}$
20. But from (15), we have: F_y = 2πrS
• Substituting this in (19), we get:
Pressure in the vertical direction = $\mathbf\small{\rm{\frac{F_y}{\pi r^2}=\frac{2 \pi rS}{\pi r^2}=\frac{2S}{r}}}$
21. But inside a fluid, pressure at a point will be the same in all directions
(See fig.10.5 in section 10.1)
■ So we can write Eq.10.22: Pressure inside a liquid drop = $\mathbf\small{\rm{\frac{2S}{r}}}$

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Thus we successfully calculated the pressure inside a liquid drop. Based on that, we can write the details about the pressure inside a bubble. It can be written in 6 steps:

1. We know that a bubble is hollow inside. Also, the outer skin has two surfaces
• So the upper hemisphere will be as shown in fig.10.63(a) below:

Fig.10.63

• There are two rings of arrows
    ♦ The arrows in the outer ring are shown in magenta color
          ✰ They indicate the surface tension in the outer surface
    ♦ The arrows in the inner ring are shown in yellow color
          ✰ They indicate the surface tension in the inner surface
2. So the total downward force due to surface tension = 2×2πrS = 4πrS
• This is balanced by the force F_y arising due to the pressure inside the bubble
• So we get: F_y = 4πrS
3. F_y is the resultant of the vertical components of the red forces in fig.10.63(b)
• This F_y acts on a 'projected area' which is equal to
4. So the pressure = $\mathbf\small{\rm{\frac{Force}{Area}=\frac{F_y}{\pi r^2}}}$
5. But from (2), we have: F_y = 4πrS
• Substituting this in (4), we get:
Pressure in the vertical direction = $\mathbf\small{\rm{\frac{F_y}{\pi r^2}=\frac{4 \pi rS}{\pi r^2}=\frac{4S}{r}}}$
6. But inside a fluid, pressure at a point will be the same in all directions
■ So we can write Eq.10.23: Pressure inside a bubble = $\mathbf\small{\rm{\frac{4S}{r}}}$

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Detergents and Surface tension

We know that detergents, when used with water, help to remove grease and oil from fabric. We can write an explanation in 7 steps:
1. First let us try to clean the oily fabric using ordinary water (with out any detergents)
• In such a situation, we have the following three interfaces:
    ♦ Oil-air. This creates S_oa
    ♦ Oil-water. This creates S_ow
    ♦ Air-water. This creates S_aw
2. We have seen the details about contact angle 𝜃 (previous section) when three different materials meet
• In our present case, 𝜃 will be such that, the water rolls into spherical drops
    ♦ So it will not wet the oil
    ♦ That is., there will not be any attraction between oil and water
• So, neither running water, nor rinsing action, will remove the oil
3. Now we add some detergent to the water
• Each detergent molecule has a special shape as shown in fig.10.64(a) below:

Fig.10.64

• The rounded end has attraction towards water
    ♦ But this rounded end has repulsion towards oil
• The pointed end has attraction towards oil
    ♦ But this pointed end has repulsion towards water
4. Thus the pointed ends of the detergent stick to the oil particles
• They soon form spheres around the oil particles
    ♦ This is shown in fig.b
5. The spheres thus formed are favorable because:
• The pointed ends are protected from water as they are inside the spheres
• The rounded ends are in contact with the water
    ♦ The rounded ends do have attraction towards water
    ♦ So water need not repel away from the spheres
6. The spheres thus formed can be washed away by running water or rinsing
• Thus the detergents help to clean oily fabric
7. If we use hot water along with detergents, cleaning will become even more easy
• This is because, oil will melt into smaller particles and so more sphere will be formed

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Now we will see some solved examples
Solved example 10.35
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10_^-1 N m_^-1 . The atmospheric pressure is 1.01 × 10_⁵ Pa. Also give the excess pressure inside the drop
Solution:
1. Pressure inside a liquid drop is given by Eq.10.22: $\mathbf\small{\rm{P=\frac{2S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{P=\frac{2(4.65\times 10^{-1})}{(3\times 10^{-3})}}}$ = 310 Pa
• This pressure is exerted by the surface towards the interior of the drop
    ♦ The surface exerts such a pressure because, it is trying to shrink
3. The drop is already subjected to atmospheric pressure P_a
• This P_a is compressing the drop from all directions
• So the total pressure in the interior of the drop
= (1.01 × 10_⁵ Pa + 310 Pa) = 1.0131 × 10_⁵ Pa
4. But in the given data, atmospheric pressure has only three significant figures
• So the result must have only three significant figures
• Thus the result in (3) becomes: 1.01 × 10_⁵ Pa 

Solved example 10.36
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm,
given that the surface tension of soap solution at the temperature (20 °C) is 2.50  × 10_^-2 N m_^-1 ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 10_⁵ Pa).
Solution:
Part(a):
1. Excess pressure inside a bubble is given by Eq.10.22: $\mathbf\small{\rm{P=\frac{4S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{p=\frac{4(2.5\times 10^{-2})}{(5\times 10^{-3})}}}$ = 20.0 Pa
Part (b):
1. When air bubble is formed inside, there is only one surface
• Then excess pressure inside the bubble will be given by:  $\mathbf\small{\rm{P=\frac{2S}{r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{p=\frac{2(2.5\times 10^{-2})}{(5\times 10^{-3})}}}$ = 10.0 Pa
3. Total pressure will be equal to the sum of three items:
(i) Atmospheric pressure (ii) Pressure due to the soap solution above the bubble (iii) Excess pressure inside the bubble
4. Pressure due to soap solution above the bubble = ρgh = (1.2 × 10_³ × 9.8 × 0.40)
5. So we get:
Total pressure = 1.01 × 10_⁵ + (1.2 × 10_³ × 9.8 × 0.40) + 10.0 Pa
= 1.05714 × 10_⁵ Pa
6. But in the given data, atmospheric pressure has only three significant figures
• So the result must have only three significant figures
• Thus the result in (5) becomes: 1.06 × 10_⁵ Pa 

Solved example 10.37
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m_^-1 . Density of mercury = 13.6 × 10_³ kg m_^-3
Solution:
1. Capillary rise is given by: $\mathbf\small{\rm{h=\frac{2S \cos \theta}{\rho g r}}}$
2. Substituting the known values, we get: $\mathbf\small{\rm{h=\frac{(2)(0.465) \cos 140}{(13.6\times 10^{3})(9.8)(1\times 10^{-3})}}}$ = -0.00534 m
3. So the fall of mercury = 5.34 mm

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In the next chapter, we will see thermal properties of matter



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Chapter 10.16 - Contact Angle and Capillarity

In the previous section, we saw the basics about surface tension. In this section we will see contact angle and capillarity. Some basic details about contact angle can be written in 9 steps:

1. Fig.10.53 below shows a drop of liquid on the surface of a solid

Fig.10.53
• In the fig., three different substances are interacting with each other. They are:
    ♦ The liquid substance of the drop
    ♦ The solid substance on which the liquid rests
    ♦ The air which surrounds both the drop and the solid
2. Consider a chain of liquid molecules in the periphery of the drop
• Let the length of the chain be k meters
    ♦ If k is very small, the chain can be considered to be straight
    ♦ We can then indicate the chain by a small straight line
    ♦ This is the yellow line shown in the fig.10.53
3. The molecules in the yellow line are liquid molecules
• They are acted upon by a 'force of attraction'
    ♦ This force of attraction is from the solid molecules
• This force will be tangential to the solid surface
    ♦ In other words, it will be parallel to the solid surface
• This force of attraction is indicated by the green arrow
    ♦ The green arrow is parallel to the solid surface
    ♦ This will be clear if we look from the side
    ♦ The side view is shown in fig.10.54 below:

Fig.10.54
• Let S_sl be the magnitude of this force per unit length
    ♦ Then the total force experienced by the line will be equal to (S_sl × k)
    ♦ The subscript ‘sl’ indicates that:
          ✰ It is the force due to the interaction between solid and liquid
4. The liquid molecules in the yellow line are acted upon by another force also
• It is the force of attraction from the air molecules
• This force will be tangential to the liquid surface
    ♦ The tangent should be drawn at:
          ✰ The point of contact between the yellow line and the liquid
• This force is denoted by the magenta arrow
• Let S_la be the magnitude of this force per unit length
    ♦ Then the total force experienced by the line will be equal to (S_la × k)
    ♦ The subscript ‘la’ indicates that:
          ✰ it is the force due to the interaction between liquid and air
■ It may be noted that:
    ♦ S_la is the ordinary surface tension which we saw earlier based on fig.10.48
    ♦ If the liquid is water, and the surrounding gas is atmospheric air, S_la will be equal to 0.0727 N m^_-1
5. There is one more force that we have to consider
• It is the force acting on the air molecules
    ♦ The air molecules close to the yellow line need to be considered
    ♦ Such a line of air molecules close to the yellow line is denoted by the grey line in figs.10.53 and 10.54
    ♦ It's length is also k
• The air molecules in the grey line are acted upon by the force of attraction
♦ It is the force of attraction by the solid molecules
• This force will be tangential to the solid surface
♦ The tangent should be drawn at:
          ✰ The point of contact between the grey line and the solid
          ✰ So it will be parallel to the solid surface
• This force is denoted by the red arrow
• Let S_sa be the magnitude of this force per unit length
    ♦ Then the total force experienced by the grey line will be equal to (S_sa × l)
    ♦ The subscript ‘sa’ indicates that
          ✰ It is the force due to the interaction between solid and air
6. So now we know the three forces. Let us see how they interact with each other
• Two different cases are possible:
Case 1. This can be explained in 3 steps:
(i) If the adhesion between air and solid is high, the air will try to spread over more and more solid surface
• The grey line will then push the yellow line more and more towards the right
(ii) In such a situation, if the liquid has greater surface tension, the magenta arrow will have a high magnitude
So the drop will roll upwards into a spherical shape
(iii) This will reduce the contact surface between the drop and the solid
• The area vacated by the liquid will be occupied by air
    ♦ This is shown in fig.10.55(a) below:

Fig.10.55

• In such a situation, we say that: the liquid is not able to wet the solid properly
    ♦ Action of water on wax is an example
Case 2. This can be explained in 2 steps:
(i) If the liquid has high adhesion to solid, the drop will spread out
• The yellow line will push the grey line more and more towards the left
• The liquid will wet more and more solid surface
(ii) The direction of S_la will change as shown in fig.10.55(b)
    ♦ The action of water on plastic is an example
7. So we see that, there is a competition between the three forces
• We can form a mathematical model to represent this competition
• For that, we introduce a new quantity: the angle of contact
    ♦ It is denoted by 𝜃. This is shown in fig.10.55
8. The significance of 𝜃 can be written in 4 steps:
(i) In fig.10.55(a), the drop is in equilibrium
• So forces towards the left must be equal to the forces towards the right
• Thus we get: $\mathbf\small{\rm{S_{sa}k+S_{la}k \cos \theta=S_{sl}k}}$
⇒ $\mathbf\small{\rm{S_{sa}+S_{la} \cos \theta=S_{sl}}}$
• But cosine of any obtuse angle is negative
    ♦ So the S_la term will be added to the right side in the final result
(ii) In fig.10.53(b), the drop is in equilibrium
• So forces towards the left must be equal to the forces towards the right
• Thus we get: $\mathbf\small{\rm{S_{sa}k=S_{sl}k+S_{la}k \cos \theta}}$
⇒ $\mathbf\small{\rm{S_{sa}=S_{sl}+S_{la} \cos \theta}}$
    ♦ Here the S_la term is already on the right side
(iii) So the results in (i) and (ii) are basically, the same. We need not worry about the directions. The cosine of 𝜃 will take care of the sign
■ We can write the general equation:
Eq.10.19: $\mathbf\small{\rm{S_{sa}=S_{sl}+S_{la} \cos \theta}}$
(iv) The angle will depend on the properties of the liquid and the solid
• Comparison has to be made between two items:
(i) The cohesion between liquid molecules
(ii) The adhesion between liquid molecules and solid molecules
• If the first item is larger, the liquid molecules will group together to form a spherical shape. It will not wet the solid
    ♦ In this case, 𝜃 will be obtuse
• If the first item is smaller, the liquid molecules will spread out on the solid surface. It will wet the solid
    ♦ In this case, 𝜃 will be acute
• So a manufacturer of water proof materials will choose the solid material in such a way that, 𝜃 is obtuse
9. Now we can write the definition for 𝜃. It can be written in 5 steps:
(i) Mark the boundary between the solid and the liquid
    ♦ In fig.10.55(a), the boundary is just a point P
    ♦ In fig.10.55(b), the boundary is the line PQ
(ii) What ever be the shape of the boundary, within that boundary, there will be no contact between liquid and air
• So we take the point just outside the boundary
    ♦ This is the first point at which the liquid makes contact with the air
    ♦ Recall that, the grey line in fig. corresponds to this point
(iv) Draw a tangent at this point
• It must be tangential to the liquid surface at that point
    ♦ In fig.10.55, this tangent is the force S_la k
(v) Mark the angle between this tangent and the solid surface
■ This angle is the angle of contact
■ It is important to always mark the angle inside the liquid

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• Next we will see some practical applications of contact angle
• First we will see capillary rise
• In fig.10.56(a) below, a capillary tube is inserted vertically into an open vessel of water

Fig.10.56

• Water rises up to a height h inside the tube. This is known as capillary rise
• The word 'capilla' in Latin means hair
• If the inner diameter of the capillary tube is very small as in the case of a hair strand, the height of water h will be large
• The explanation for capillary rise can be written in 10 steps:
1. The angle of contact 𝜃 between glass and water is acute
• So the surface of water will be curved downwards
    ♦ When looked from above, the surface will have a concave shape
■ The curved surface of a liquid is called meniscus
• If 𝜃 is 90_^o, the surface will be flat
• If 𝜃 is obtuse, the surface will be convex
2. Fig.10.56(b) shows an enlarged view of the point of contact
• The surface tension of water (S_la) makes an angle 𝜃 with the glass surface
    ♦ For convenience, we will denote 'S_la' as simply 'S'
    ♦ This S is indicated in magenta color
    ♦ Recall that, we used the same magenta color to indicate 'S_la' in the earlier figs.10.53, 10.54 and 10.55
• The S indicates the tendency of water to spread over the glass surface
    ♦ This tendency is due to the adhesion between water and glass molecules
3. We see two 𝜃 angles in the fig.10.56(b)
(i) The lower 𝜃 is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper 𝜃 will be the same because, they are opposite angles
4. Since S is inclined, it can be resolved into horizontal and vertical components
    ♦ The horizontal component is S sin 𝜃
    ♦ The vertical component is S cos 𝜃
    ♦ These are shown in fig.10.56(c)
5. The vertical component will lift the water upwards
• We see that, the vertical component acts at the tip of the meniscus
• But what appear as a tip is actually a ring. It is the periphery of the meniscus
    ♦ This is clear from the 3D view shown below:

Fig.10.57

• The horizontal components will all cancel each other
6. S is the force per unit length
• So S cos 𝜃 is also the force per unit length
• So the total upward force = (S cos 𝜃 × Peripheral length of the meniscus) = (S cos 𝜃 × 2𝞹r)
    ♦ Where r is the radius of the capillary tube
7. This total upward force will support the weight of the water column
• We have: weight of the water column = (ρ × 𝞹r_²hg)
    ♦ Where ρ is the density of water
8. Equating the results in (6) and (7), we get:
(S cos 𝜃 × 2𝞹r) = (ρ × 𝞹r_²hg)
■ Thus we get Eq.10.20: $\mathbf\small{\rm{h=\frac{2S\cos \theta}{\rho gr}}}$
9. When water is in contact with glass, the angle 𝜃 is very small. It is nearly equal to zero
• When 𝜃 is zero, cos 𝜃 is 1
■ So, when the liquid inside the capillary tube is water, we have:
Eq.10.21: $\mathbf\small{\rm{h=\frac{2S}{\rho gr}}}$
10. From Eqs.10.20 and 10.21, it is clear that, more water will rise when radius of the tube decreases
• Let us calculate the rise of water in a glass tube of radius 0.05 cm:
    ♦ Using Eq.10.21, we get: $\mathbf\small{\rm{h=\frac{2S}{\rho gr}=\frac{2(0.073)}{(10^3)(9.8) (5\times 10^{-4})}}}$ = 2.98 cm
          ✰ Where S of water is 0.073 N m_^-1, which is obtained from the data book

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• Next we will see capillary depression

• In fig.10.58(a) below, a capillary tube is inserted vertically into an open vessel of mercury

Fig.10.58

• Mercury falls by a height h inside the tube. This is known as capillary depression
• If the inner diameter of the capillary tube is very small as in the case of a hair strand, the depression h will be large
• The explanation for capillary depression can be written in 10 steps:
1. The angle of contact 𝜃 between glass and mercury is obtuse
• So the surface of mercury will be curved downwards
    ♦ When looked from above, the surface will have a convex shape
■ The curved surface of a liquid is called meniscus
• If 𝜃 is 90_^o, the surface will be flat
• If 𝜃 is acute the surface will be concave
2. Fig.10.58(b) shows an enlarged view of the point of contact
• The surface tension of mercury (S_la) makes an angle 𝜃 with the glass surface
    ♦ For convenience, we will denote 'S_la' as simply 'S'
    ♦ This S is indicated in magenta color
    ♦ Recall that, we used the same magenta color to indicate 'S_la' in the earlier figs.10.53, 10.54 and 10.55
• The S indicates the tendency of mercury to repel away from the glass surface
    ♦ This tendency arises because:
          ✰ The cohesive force between mercury molecules
          ✰ is greater than
          ✰ the adhesion between mercury and glass molecules
          ✰ So the cohesive forces will try to pull mercury molecules into a sphere
3. We see two 𝜃 angles in the fig.10.58(b)
(i) The lower 𝜃 is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper 𝜃 will be the same because, they are opposite angles
4. Since S is inclined, it can be resolved into horizontal and vertical components
Note that, the angle with the vertical is (180-𝜃)
    ♦ The horizontal component = S sin (180-𝜃) = S sin 𝜃
    ♦ The vertical component is S cos (180-𝜃) = -S cos 𝜃
    ♦ These are shown in fig.10.58(c)
          ✰ The -ve sign indicates that the vertical component acts downwards
5. The vertical component will pull the mercury downwards
• We see that, the vertical component acts at the tip of the meniscus
• But what appear as a tip is actually a ring. It is the periphery of the meniscus
    ♦ This is clear from the 3D view shown in fig.10.57(b) above
• The horizontal components will all cancel each other
6. S is the force per unit length
• So S cos 𝜃 is also the force per unit length
• So the total downward force = (S cos 𝜃 × Peripheral length of the meniscus) = (S cos 𝜃 × 2𝞹r)
    ♦ Where r is the radius of the capillary tube
7. This total downward force will keep the weight of the mercury column suppressed
• We have: weight of the mercury column = (ρ × 𝞹r_²hg)
    ♦ Where ρ is the density of mercury
8. Equating the results in (6) and (7), we get:
(S cos 𝜃 × 2𝞹r) = (ρ × 𝞹r_²hg)
■ Thus we get the same Eq.10.20: $\mathbf\small{\rm{h=\frac{2S\cos \theta}{\rho gr}}}$
9. When mercury is in contact with glass, the angle 𝜃 is 130_^o
10. From Eqs.10.20, it is clear that, more mercury will fall when radius of the tube decreases
• Let us calculate the fall of mercury in a tube of radius 1.25 mm:
    ♦ Using Eq.10.20, we get: $\mathbf\small{\rm{h=\frac{2S \cos \theta}{\rho gr}=\frac{2(0.4355)}{(13.6 \times 10^3)(9.8) (1.25\times 10^{-3})}}}$ = -0.00336 m = -3.36 mm
          ✰ Where S of mercury is 0.4355 N m_^-1, which is obtained from the data book

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Next we will see a simple problem based on contact angle

Solved example 10.34
What is the maximum mass of a needle which can rest on the surface of water. The length of the needle is 4cm. Surface tension of water = 0.073 N m_^-1
Solution:
1. Fig.10.59(a) shows the 3D view of a needle resting on the surface of water

Fig.10.59

2. The weight of the needle acts on the surface of water
    ♦ Due to the weight of the needle, there will be a small depression
    ♦ Due to the depression, there will be a contact angle
    ♦ This is shown in fig.10.60(a) below:

Fig.10.60

3. If there is no depression, there will not be a contact angle
    ♦ This is clear from fig.10.60(b)
4. We see two 𝜃 angles in the fig.10.60(a)
(i) The lower 𝜃 is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper 𝜃 will be the same because, they are opposite angles
5. Since S is inclined, it can be resolved into horizontal and vertical components
    ♦ The horizontal component is S sin 𝜃
    ♦ The vertical component is S cos 𝜃
6. The vertical component will support the weight of the needle
    ♦ We see that, the vertical component acts at the tip of the meniscus
    ♦ But what appear as a tip is actually a line. It is the line of the needle
    ♦ In fact, there are two lines. One on either side of the needle
    ♦ This is shown in the earlier fig.10.59(b)
• So the total upward force = 2Sl cos 𝜃
    ♦ Where l is the length of the needle
7. When the mass of the needle increases, the water surface become more and more depressed as shown in fig.10.60(c)
• As a result, the angle 𝜃 becomes smaller and smaller
• The angle 𝜃 can decrease to such a level that, it is very close to zero
• But 𝜃 cannot become equal to zero because, it would then mean that:
    ♦ S is vertical and
    ♦ there is no horizontal component
• Surface tension cannot become vertical
8. However, for the extreme case (maximum possible mass of the needle), we can take 𝜃 to be very close to zero and hence cos 𝜃 as 1
• So result in (6) becomes:
    ♦ Total upward force = 2Sl
9. Equating this to the maximum possible weight, we get:
2Sl = m_(max)g
⇒ $\mathbf\small{\rm{m_{max}=\frac{2Sl}{g}=\frac{(2)(0.073)(0.04)}{(9.8)}}}$ = 0.0006 kg = 0.6 grams

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In the next section, we will see pressure inside a drop of liquid



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