In the previous section, we saw Stokes law and Reynolds number. In this section we will see surface tension. Some basic details about the surface tension can be written in 8 steps:
1. Fig.10.48 below shows a mass of liquid
• Three molecules A, B and C are marked
♦ A is in the interior
♦ B is near the surface
♦ C is at the surface
Fig.10.48
2. All three molecules will experience attractions from their surrounding molecules
• A is attracted in all directions
♦ Magnitudes of all those attractions are equal
• B is attracted in all directions
♦ Magnitudes of all those attractions are not equal
• C is not attracted in all directions
♦ There are no attractions from above
3. It is clear that:
♦ On A, there is no net attraction
♦ On B, there is a small net attraction towards the bottom
♦ On C, there is a large net attraction towards the bottom
4. Like the molecule C, all molecules at the surface, will experience a large downward attraction
• But those molecules cannot move downwards because, the molecules already present in the interior are incompressible
5. So the molecules at the surface are experiencing attraction
• Due to this attraction, there will be a potential energy
(This is just like a body of mass m and at a height h, possessing a potential energy of mgh. This potential energy is due to the gravitational force of attraction from the earth)
6. The potential energy at the surface of the liquid is called surface energy
• For attaining greater stability, the liquid will try to reduce the surface energy
• This can be achieved if the ‘number of molecules at the surface’ is reduced
• To reduce this number of molecules, the area of the surface must decrease
7. So we can write:
• The liquid will always try to achieve minimum possible surface area
■ That means, the surface will be always trying to shrink
8. Let us try to understand the behaviour of a ‘surface which tries to shrink’
• It can be written in 4 steps:
(i) In fig.10.49(a) below, a rubber sheet (shown in yellow color) is stretched between two supports and is pinned down
• Since the rubber sheet is stretched, it will be able to take some load as shown in fig.b
Fig.10.49
(ii) In fig.c, the rubber sheet is pinned down but it is not in a stretched position
• Placing a load on such an unstretched rubber sheet does not serve any purpose
♦ It is shown in fig.d
(iii) The stretched sheet in fig.a trying to regain it’s original shape
• In other words, the sheet is trying to shrink
(iv) In the same way, the liquid surface in fig.10.48 is also trying to shrink
• Such a liquid surface will be able to support small loads
• If we carefully place a small needle or a paper clip on the surface of water, they will float
• Remember that, though light in weight, needle and paper clip have higher densities than water. We expect ‘bodies having higher densities than the liquid’ to sink
1. In fig.10.50(a) below, two vertical channels are shown in yellow color
Fig.10.50
• Two horizontal red bars slide freely inside the channels
♦ The top red bar is fixed in position
♦ The bottom red bar is free to slide
• The length of both the red bars is l
• A liquid film is present in between the red bars
♦ It is shown in orange color
• In fig.a, the system is in equilibrium
2. Now, a small mass m is attached to the bottom red bar
• As a result, the bottom red bar moves downward by a distance d
• So the film is stretched by a distance d
• This is shown in fig.10.50(b)
• A 3D view is shown in fig.10.51(a) below
 |
| Fig.10.51 |
• The system attains a new equilibrium
• The film is able to support the weight mg because, it has the tendency to shrink
3. Let the force exerted by the film be S newton per meter length of the bottom red bar
• Then the total force exerted on the bar appears to be: (S × l) newton
• But this is not the correct value
♦ The film has two surfaces. Both the surfaces will pull on the red bar
♦ This is clear from fig.10.51(b). A portion of the film is removed
✰ So the presence of two surfaces becomes clear
• So the total force exerted by the film on the red bar is equal to (2 ×s × l) = 2sl
4. Thus at the new equilibrium, we have: mg = 2sl
• From this, we get: Eq.10.17: $\mathbf\small{\rm{S=\frac{mg}{2l}}}$
■ Thus we can write:
♦ The force S exerted by
♦ one surface of the film
♦ on each meter length
♦ of the red bar
♦ is equal to $\mathbf\small{\rm{\frac{mg}{2l}}}$
■ So S is the force per meter length. It is called surface tension. It’s unit is N m-1
• We will see the actual definition of surface tension in a later section
• The surface tension of some liquids are given below:
♦ Water at 20 oC: 0.0727 N m-1
♦ Mercury at 20 oC: 0.4355 N m-1
♦ Ethanol at 20 oC: 0.0227 N m-1
5. The above steps help us to understand the effects of surface tension on the red bar
• Let us see the effect of surface tension on the 'molecules on the surface of the film'. It can be written in steps:
(i) Draw a line on the surface of the film, as shown in green color in fig.10.50(c)
• Let it be a ‘line of molecules’
♦ All the molecules in that line are attracted upwards by molecules above that line
♦ All the molecules in that line are attracted downwards by molecules below that line
(ii) But the line is in equilibrium. It does not move
• That means:
♦ upward force per unit length
♦ is equal to
♦ downward force per unit length
(iii) If the green line is close to the bottom red bar, then the molecules in the line will experience only upward force
• In the above discussion, we have used the force method
• Let us now use the energy method. It can be written in 4 steps:
1. The work done by the newly added mass m will be equal to mgd
• This work will be stored as extra surface energy of the film
2. Let S’ be the surface energy per unit area
♦ Then extra energy in one surface of the film = S’ld
♦ Total extra energy from the two surfaces = 2S’ld
3. This extra energy must be equal to the work done by mg
• So we get: 2S’ld = mgd
• From this, we get: Eq.10.18: $\mathbf\small{\rm{S'=\frac{mg}{2l}}}$
■ Thus we can write:
♦ The surface energy S'
♦ per unit area of the surface
♦ is equal to $\mathbf\small{\rm{\frac{mg}{2l}}}$
■ So S' is the energy per unit area of the surface
4. We see that, the right side is same in both Eqs.10.17 and 10.18
■ That means:
♦ Surface tension S (which is the force per unit length)
♦ is same as
♦ Surface energy S' per unit area
• Let us do a dimensional analysis:
♦ The dimensions of 'Force per unit length' can be obtained as:
✰ $\mathbf\small{\rm{\frac{[MLT^{-2}]}{[L]}=[MT^{-2}]}}$
♦ The dimensions of 'Energy per unit area' can be obtained as:
✰ $\mathbf\small{\rm{\frac{[ML^2T^{-2}]}{[L^2]}=[MT^{-2}]}}$
• We see that, both dimensions are the same
Solved example 10.32
A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10-2 N
(which includes the small weight of the slider). The length of the slider is 30 cm.
What is the surface tension of the film ?
Solution:
1. The arrangement is shown in fig.10.52(a) below:
 |
| Fig.10.52 |
2. We have Eq.10.17: $\mathbf\small{\rm{S=\frac{mg}{2l}}}$
• Substituting the known values, we get:
$\mathbf\small{\rm{S=\frac{1.5\times 10^{-2}}{2 \times 0.30}=2.5\times 10^{-2}\,N\,m^{-1}}}$
Solved example 10.33
Figure 10.52(b) shows a thin liquid film supporting a small weight 4.5 × 10-2 N.
What is the weight supported by a film of the same liquid at the same temperature
in Fig. (c) and (d) ? Explain your answer physically
Solution:
1. The weight and lengths are given. Using those, we can calculate the surface tension S
2. In figs. (c) and(d), the liquid is the same. Temperature also is the same
• So S will be the same
3. In figs. (c) and(d), the length of the red bar is also same as that in fig.(b)
• So the weight will also be the same
• The value of S does not depend on the shape or area of the film
In the next section, we will see angle of contact
PREVIOUS
CONTENTS
NEXT
Copyright©2020 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment