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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 × 2 L

= 0.073 × 2 × 5 × 10-2 = 0.073 × 10-1

∴ F = 7.3 × 10-3 N

Note : This force is the weight of needle.

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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

Angle of Contact

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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.

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3. If the liquid doesn’t wets the solid at all, the angle

of contact is obtuse.

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

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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

(a) (b)

(c) (d)

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 ? The surface

tension of soap solution is .

(a) 2 × 10-4J (b) 2.5 × 10-4J

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(c) 3 × 10-4J (d) 5 × 10-4J

Ans:

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(c)

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

Ans:

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(c) 1:10

Ans:

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(c) 3 × 10-4J

For a soap film, we have to consider double the

area, as there are two free surfaces.

Drops and Bubbles

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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

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)

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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

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 …… (5)

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This is called Laplace’s law of a spherical membrane.

In case of soap bubble

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.