Summary of Surface Tension Part - II by Ednexa

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Surface Tension (Part - II)
Problems
1. A needle of length 5 cm can just rest on the
surface of water of surface tension 0.073 N / m.
Find the vertical force required to detach this
floating needle from the surface of water.
Sol :
L = 5 cm = 5 × 10-2m
T = 0.073 N/m
F=?
The force due to surface tension is given as,
F = TL
The total length of the needle in contact with
water = 2 L
∴F=T×2L
= 0.073 × 2 × 5 × 10-2 = 0.073 × 10-1
∴ F = 7.3 × 10-3 N

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Note : This force is the weight of needle.
2. A horizontal circular loop of wire of diameter 0.08
m is lowered in to a oil. The force due to surface
tension required to pull the loop out of the liquid is
0.0226 N. Calculate the surface tension of the oil.
Sol:
d = 0.08 n
∴ r = 0.04 m
F = 0.0226N, T = ?
The force due to Surface tension is F = TL
T
F
L
F
2 2 r
0.0226
4 r
4
0.0226
0.16
3.14
T
0.0449 N / m
0.0226
3.14
0.04
0.0226
0.5024

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Angle of Contact
When a liquid is in contact with a solid, the angle
between the surface of the liquid and the tangent
drawn to the surface of the liquid, at the point of
contact, on the side of the liquid is called “angle of
contact” of the given solid liquid pair.
Features of Angle of Contact
1. For a given solid liquid pair the angle of contact is
constant.
2. If the liquid partially wets the solid, the angle of
contact is acute.
3. If the liquid doesn‟t wets the solid at all, the angle
of contact is obtuse.

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4. If the liquid wets the solid completely, the angle of
contact is approximately zero.
5. Any small contamination of the liquid can change
the angle of contact largely.
6. The angle of contact depends on the magnitudes
of adhesive forces between solid and liquid
molecules and cohesive forces between liquid
molecules.
Explanation of Angle of Contact

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For a liquid which completely wets the solid, the
adhesive forces are so strong, as compared to the
cohesive forces, that the resultant AR of these forces
is along AP. So, the tangent at the point of contact is
along the wall of the container. So, the liquid surface
remains plane.

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Shape of a liquid drop
T1 = Force due to surface tension at the liquid - solid
interface,
T2 = Force due to surface tension at the air - solid
interface,
T3 = Force due to surface tension at the air - liquid
interface,
For the equilibrium of the drop
T2
T1
i.e. cos
T 3 cos
T 2 T1
T3

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From this equation we get following cases:
1. If T2 > T1, and T2 - T1 < T3, cos θ is positive and
angle of contact θ is acute.
2. If T2 < T1, and T2 - T2 < T3, cos θ is negative and
angle of contact θ is obtuse.
3. If T2 - T1 = T3, cos θ = 1 and „θ‟ is nearly equal to
zero.
4. If T2 - T1 > T3 or T2 > T1 + T3, cos θ > 1 which is
impossible, liquid is spread over the solid surface and
drop shall not be formed.

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Multiple choice Questions
1. A water drop of radius R is split into n smaller
drops, each of radius r. If T is the surface tension
of water, then the work done in this process is
4 3 1 1
R T
(a)
3
r R
(c) 4 R3 T
1 1
r R
3 3 1
R T
(b)
4
R
1
r
1
R
1
r
(d) 6 R2 T
2. One thousand small water droplets of equal size
combine to form a big drop. The ratio of the final
surface energy to the initial surface energy is of
water drops is
(a) 1:1000
(b) 10:1
(c) 1:10
(d) 1000:1
3. What is the potential energy of a soap film formed
on a frame of area 5 10 3 m2 ? The surface
tension of soap solution is 30 10 3 N / m .
(a) 2 × 10-4J
(b) 2.5 × 10-4J

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(c) 3 × 10-4J
(d) 5 × 10-4J

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Ans:
(c) 4 R3 T
1 1
r R
In this process, the mass & hence the volume of
the water drop, remains constant but the surface
area is increased.
∴ Volume of the surface area is increased one
drop = Volume of n droplets
4 3
4 3
R n.
r
3
3
R3 nr 3
Change in area
4
R3
r
R2
4 R2
4 (nr 2
R2 )
R2
4
dA
nr 3
r
n.4 r 2
Work done
dW
4 R3
1 1
r R
(S.T.) (Change in area) T dA
4 R3 T
1 1
r R

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Ans:
(c) 1:10
4
4 3
R3 1000
r
3
3
R3 1000 r 3
R 10 r
Energy
or E T A
area
Energy of 1000 droplets (E1 ) 1000 T (4 r 2 )
But surface tension T
and Energy of the sin gle big drop
E2
T 4 (R2 )
T 4 (10 r)2
4 T(100 r 2 )
E2
E1
100 r 2
1000 r 2
1
10

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Ans:
(c) 3 × 10-4J
For a soap film, we have to consider double the
area, as there are two free surfaces.
P.E work done
2 30 10
3 10 4 J
3
T(2A)
5 10
3

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Drops and Bubbles
As it is spherical in shape, the inside pressure will
be greater than that of the outside. Let the outside
pressure be P0 and inside pressure be Pi, so that the
excess pressure is Pi – P0.
the radius of the drop increases from r to r + Δr,
where Δr is very small, so that the inside pressure
remains almost constant.
Initial surface area (A1) = 4πr2
Final surface area (A2) = 4π(r + Δr)2

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A2 = 4π (r2 + 2r Δr + Δr2)
A2 = 4πr2 + 8πr Δr + 4πΔr2
As Δr is very small, Δr2 is neglected (i.e. 4 π Δr2 ≅ 0)
Increase in surface area (dA)
= A2 – A1 = 4πr2 + 8πr Δr - 4πr2
Increase in surface area (dA) = 8 πr Δr ….. (1)
Work done to increase the surface area 8 πr Δr is
extra surface energy.
∴ dW = TdA
dW = T (8πr Δr) ….. (2)
This work done is also equal to product of force and
the distance Δr.
Let dW = dF Δr

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But dF = Excess pressure × area
dF = (Pi – P0) 4πr2
…… (3)
dW = (Pi – P0) 4πr2 Δr….. (4)
By comparing equation (2) and (4) we get
T (8πrΔr) = (Pi – P0) 4πr2 Δr
Here P i
P0
2T
…… (5)
r
This is called Laplace‟s law of a spherical membrane.
In case of soap bubble
Pi
P0
4T
r
....... 6
The threads of rain – coat are coated with water
proofing agents like resin etc. Which have very small
force of adhesion with water so rain coats become
water proof.