**Chapter – 15, Surface Tension**

Above pictures are the phenomena related to surface tension. Let us understand that what internal force is before knowing about the surface tension.

* Inter Molecular Force : *The force of interaction between the molecules of a substance, is called as inter molecular force. There are two types of inter molecular force, one is adhesive force and other is cohesive force.

* Adhesive Force : *The force of interaction between the molecules of different substances, is called as Adhesive Force.

* Cohesive Force : *The force of interaction between the molecules of the same substance, is called as Cohesive Force.

* Molecular Range :* The maximum distance up to which a molecule can exert some appreciable force of attraction for another molecules, is called as the molecular range.

* Sphere Of Molecular Activity / Sphere Of Molecular Influence :* The sphere drawn around a molecule as center and molecular range as radius, is called as the sphere of influence. All the other molecules lying in this sphere are attracted towards the molecule at the center.

* Surface Tension : *The property of a liquid by virtue of which its free surface behaves like a stretched elastic membrane tending to contract to possess minimum surface area, is called as the surface tension.

It is a joint property of the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid one.

σ = F / l

where F = tangential force

l = length of the imaginary line

* Units :* Its CGS unit is dyn / cm and SI unit is N / m

* Dimension :* [ Surface tension ] = [ ML

^{0}T

^{-2}]

* Surface Film : *The thin film of liquid near its surface and having thickness equal to the molecular range for that liquid, is called as the Surface Film.

* Surface Energy :* The potential energy per unit area of the surface film is called as the surface energy. It is the amount of work done in increasing the surface area of the surface film through unity under isothermal conditions.

Its SI unit is J / m^{2} and CGS unit is dyne / cm which is equivalent to erg / cm^{2}.

Its dimension is given by

[ E ] = [ ML^{0}T^{-2} ]

* Angle Of Contact :* The angle between the tangent to the liquid surface at the point of contact and solid surface inside the liquid is known as Angle of contact for a given pair of solid and liquid.

- It does not depend on the manner of contact.
- For a liquid having concave meniscus, angle of contact is acute angle .
- For a liquid having convex meniscus, angle of contact is obtuse angle.
- For water and perfectly clean glass, angle of contact is zero degree.
- For ordinary water and glass, its value lies between 8
^{0}– 18^{0} - For pure water and pure silver its value is equal to 90
^{0}

**Factors influencing the value of angle of contact :**

1 ) The nature of the solid and the liquid in contact.

2 ) The cleanliness of the surface in contact

3 ) The medium above the free surface the liquid.

4 ) The temperature of the liquid that is θ ∝ t

5 ) On soluble impurity to a liquid decreases the angle of contact.

* Excess Pressure : *The difference of pressure of two sides of the liquid surface is called as the Excess Pressure.

We will discuss the excess pressure in the following three cases in this chapter.

1 ) Inside a liquid Drop

2 ) Inside a air bubble in a liquid

3 ) Inside a soap bubble

**1 ) Excess Pressure Inside A Liquid Drop :**

Let R = Radius of the spherical drop

σ = surface tension of the liquid surface

P =Outside pressure

p = excess pressure

dR = increase in radius

S = initial surface area = 4πR^{2}

S’ = Final surface area = 4π( R + dR )^{2}

dS = increase in surface area

⇒dS = S’ – S = 4πR^{2} – 4π( R + dR )^{2}

⇒ dS = 4πR^{2} – 4πR^{2} -8πRdR – 4π( dR )^{2}

⇒ dS = ( 8πdR )R + 4πR ( dR )^{2}

As ( dR )^{2} is negligible, so ignoring the term ( dR )^{2}, we get

⇒dS = ( 8πdR )R

dE = Increase in surface energy due to the work done by excess pressure

⇒dE = increase in surface area * surface tension

⇒dW = dS * σ

⇒dE = ( 8πdR )Rσ —————————— ( 1 )

Work done by excess pressure is given by

⇒dw = force * displacement

⇒dw = ( excess pressure * surface area ) * increase in radius

⇒dw = p * 4πR^{2} * dR

⇒dw = 4πpR^{2}dR ———————————- ( 2 )

By comparing equation ( 1 ) and ( 2 ) we get

dw = dE

⇒4πpR^{2}dR = ( 8πdR )Rσ

**p = 2 σ / R **

**2 ) Excess Pressure Inside An Air Bubble In A Liquid Drop :**

It is as same as the excess pressure inside a liquid drop

Therefore, the excess pressure inside an air bubble in a liquid drop is given by

p = 2σ / R

**3 ) Excess Pressure Inside A Soap Bubble :**

[caption id="attachment_1727" align="alignnone" width="300"] ** Soap Bubble **[/caption]

Let R = Radius of the spherical drop

σ = surface tension of the liquid surface

P =Outside pressure

p = excess pressure

dR = increase in radius

S = initial surface area = 4πR^{2}

S’ = Final surface area = 4π( R + dR )^{2}

dw = work done by the excess pressure

⇒dw = force * displacement

⇒dw = ( excess pressure * surface area ) * increase in radius

⇒dw = p * 4πR^{2} * dR

⇒dw = 4πpR^{2}dR ———————————- ( 1 )

The soap bubble has two free surfaces

Increase in surface area ( dS ) = 2 ( S’ – S )

⇒dS = 8πR^{2} – 8π( R + dR )^{2}

⇒ dS = 8πR^{2} – 8πR^{2} -8πRdR – 8π( dR )^{2}

⇒ dS = ( 8πdR )R + 4πR ( dR )^{2}

As ( dR )^{2} is negligible, so ignoring the term ( dR )^{2}, we get

⇒dS = ( 16πdR )R

Increase in surface energy ( dE ) = dS * σ

dE = ( 16πdR )Rσ ———————— ( 2 )

On comparing equation ( 1 ) and ( 2 ), we get

dw = dE

4πpR^{2}dR = ( 16πdR )Rσ

**p = 4σ / R **

**Important points :**

**Important points :**

1 ) Surface Tension = F / l

2 ) If a big dropper is splitted into ‘n’ small drops, then

i ) the radius of a small drop ( r ) = n^{-1/3}R

ii ) the increase in surface area ( dS ) = ( n^{1/3} – 1 ) *4πR^{2}

iii ) the energy required to split the bigger drop ( dU ) = ( n^{-1/3} – 1 ) *4πR^{2}σ

iv ) the decrease in temperature ( dT ) = ( 3σ / ρs ) { ( 1 / r ) – ( 1 / R ) }

where ρ = density of the liquid

σ = surface tension of the liquid surface

s = specific heat of the liquid

R = radius of the bigger drop

3 ) The excess pressure on the concave side of a curved surface is given by

p = σ { ( 1 / R_{1} ) + ( 1 / R_{2} ) }

a ) for spherical surface, R_{1} = R_{2} = R

p = 2 σ / R

b ) For cylindrical surface, R_{1} =R and R_{2} = ∞

p = σ / R

Special case : If the liquid is between the two plates of thickness t

then R = 2t i.e

p = σ / 2t

Work done to take away from each other = σA / R

where A = area of the plate

**Question – Answer zone**

**Question – Answer zone**

**1 ) What is the nature of the inter molecular force ?**

Ans : The nature of the inter molecular force is electrical.

**2 ) Is inter molecular forces obey the inverse square law ?**

Ans : No, inter molecular forces do not obey the inverse square law.

**3 ) Surface tension is what type of phenomenon ?**

Ans : Surface tension is molecular phenomenon.

**4 ) Why a rain drop is spherical in shape ?**

Ans : The sphere has minimum surface area among all shapes. Due to surface tension, a rain drop tends to occupy minimum surface area, it is spherical in shape.

**5 ) What is the relation between the surface tension of a liquid with its temperature ?**

Ans : The surface tension decreases with the increase in temperature.

**6 ) When a number of small drops coalesce to form a bigger drop, why there is liberation of energy ?**

Ans : When a number of small drops coalesce to form a bigger drop, the surface area decreases. Therefore, there is liberation of energy.

**7 ) What is the condition require for a curved surface of the liquid to be in equilibrium ?**

Ans : If there is excess pressure on the concave side of the curved surface of a liquid, then the curved surface area will be in equilibrium.

**8 ) What is the relation between the pressure on the concave side of a curved surface and the pressure on the convex surface of the liquid ?**

Ans : The pressure on the on the concave side of a curved surface area is always greater than the pressure on the convex surface of the liquid.

**9 ) Define capillary tube.**

Ans : A tube of very fine bore is called as capillary tube.

**10 ) What do you mean by capillary angle ?**

Ans : The angle between the tangent to the liquid surface at the point of contact and solid surface inside the liquid is known as Angle of contact for a given pair of solid and liquid.

**11 ) When will a liquid wet the solid ?**

Ans : For a liquid – solid interface, if angle of contact is acute then the liquid will wet the solid.

**12 ) What will be the shape of the meniscus of a liquid having acute angle of contact ?**

Ans : The shape of the meniscus of a liquid having acute angle of contact is concave.

**13 ) When will be a liquid rise in a capillary tube ?**

Ans : When the angle of contact is acute then a liquid will rise in a capillary tube.

**14 ) What is meant by capillarity ?**

Ans : When a tube of very fine bore is dipped in a liquid, the rise or fall of liquid in the tube is called as capillarity.

**15 ) At what height to which a liquid having surface tension σ rises in a tube of radius r ?**

Ans : h = ( 2σ cosθ ) / ( rgρ ) – ( r / 3 )

where h = height upto which a liquid rises ,θ is the angle of contact, g is the acceleration due to gravity and ρ is the density of liquid.

**16 ) What is the called the temperature at which surface tension of a liquid surface becomes zero ?**

Ans : The temperature at which surface tension of a liquid surface becomes zero, is called as critical temperature.

**17 ) If a bigger drop of radirs R is splitted in to n number of smaller drops of radius r, what will be the value of r ?**

Ans : r = n^{-1/3}R

**18 ) A U – shaped wire loop having a slider ( of negligible mass ) is dipped in a soap solution and then removed . A thin soap film is formed between the wire and the light slider, It is found that the film can support a weight of 0.006 N, before it will break, If the length of the slider is 0.1 m, what is the surface tension of the film ?**

Solution : weight ( F ) 0.006 N

Length of the slider ( l ) = 0.1 m

As this film has two free surfaces, the surface tension is given by

σ = F / 2l

σ = 0.006 / ( 2 * 0.1 ) = 0.03 N / m

**19 ) A wire ring of 3 cm radius is rested on the surface of a liquid and then raided. The pull required is 3.03 g more before the film breaks than it is after. Find the surface tension of the liquid.**

Solution :

Given that, mass ( m ) = 3.03 g

radius ( r ) = 3 cm

This film has two surface area.

The surface tension is given by

σ = force / ( 2 * circumference )

σ = mg / ( 2 * 2 πr )

σ = ( 3.03 * 980 ) / ( 2 * 2 * 3.14 * 3 )

σ = 78.80 dyne / cm

**20 ) A liquid drop of diameter D breaks up into 27 tiny drops. Find the resulting change in energy. Take surface tension of the liquid as T.**

Ans : Given that,

Diameter = D

Radius ( R ) = D / 2

Number of drops ( n ) = 27

Surface tension = T

Required change in energy is given by

dU = ( n^{1/3} – 1 ) *4πR^{2}T

dU = ( 27^{1/3} -1 ) * 4π ( D / 2)^{2}T

dU = ( 3 -1 ) * πD^{2}T

dU = 2πD^{2}T

**21 ) A glass tube of 1 mm bore is dipped vertically into a container of mercury, with its lower end 2 cm below the mercury surface. What must be the gauge pressure of air in the tube to blow a hemispherical bubble at its lower end ? Given that density of mercury = 13600 kg / m ^{3} and surface tension of mercury = 0.465 N / m.**

Solution :

Given that, Diameter of glass tube ( d ) = 1 mm

radius of the tube ( r ) = d / 2 = 1 / 2 = 0.5 mm = 0.5 * 10^{-3} m

the density of the mercury ( ρ ) = 13600 kg / m^{3}

surface tension of the liquid ( σ ) = 0.465 N / m

Excess pressure inside an air bubble inside the mercury is given by

p = 2 σ / r

p = ( 0.465 * 2 ) / ( 0.5 * 10^{-3 } ) = 1860 pascal

p = 1860 / ( ρg ) = 1860 / ( 13600 * 9.8 ) = 0.01395 m of Hg = 1.39 cm of Hg

The guage pressure is given by

P = p + 2 cm = 1.39 + 2 = 3.39 cm of Hg

**22 ) What is the excess pressure inside a bubble of soap solution of radius 3 mm, given that the surface tension of soap solution is 25 dyn / cm ? Also find the pressure inside the bubble, if the atmospheric pressure is 1.013 * 10**^{6}** dyne / cm ^{2}.**

Solution :

Given that, radius of the soap bubble ( r ) = 3 mm = 3 * 10^{-1} cm

Surface tension ( σ ) = 25 dyne / cm

Atmospheric pressure ( P ) = 1.013 * 10^{6} dyne / cm^{2}

Excess pressure inside the pressure ( p ) = 4σ / r = ( 4 * 25 ) / ( 3 * 10^{-1} )

p = 333.3

Required pressure =P + p = ( 1.013 * 10^{6} ) + (333.3 ) = 1.0133 * 10^{6} dyne / cm^{2}

**23 ) What would be the gauge pressure inside an air bubble of 0.2 mm, radius situated just below the surface of water ? Surface tension of water is 0.07 N/m**

Ans : Given that radius ( r ) = 0.2 mm = 0.2 * 10^{-3} m = 2 * 10^{-4} m

Surface tension ( T ) = 0.07 N / m

The gauge pressure = The excess pressure inside the bubble = 2 T / r = ( 2 * 0.07 ) / ( 2 * 10^{-4} ) = 700 N / m^{2}

**24 ) A small hollow sphere which has a small hole in it is immersed in water to a depth 40 cm before any water penetrates into it. If the surface tension of water is 75 dyne /cm, find the radius of the hole.**

Ans : Given that surface tension ( T ) = 75 dyne / cm

depth ( h ) = 40 cm

density of water ( d ) = 1 gm / cubic cm

Let radius of the bubble = r

acceleration due to gravity ( g ) = 980 dyne / cm^{2}

As the water start to enter into the hollow sphere through the small hole, at the hole, there is formation of an air bubble of the size of the hole. The water will enter if the excess pressure inside the bubble equals the gauge pressure but in opposite direction.

Therefore,

dgh = 2T / r

r = 2T / dgh

r = ( 2 * 75 ) / ( 1 * 980 * 40 )

r = 3.83 * 10^{-3}

**25 ) Calculate the force required to take away a flat plate of radius 5 cm from the surface of water. Given, surface tension of water = 72 dyn /cm**

Ans : Given that surface tension of water ( T ) = 72 dyn / cm

radius of plate ( r ) = 5 cm

The required force = T * 2πr = 72 * 5 * 3.14 * 5 = 2261.95 dyne

**26 ) A soap film is formed in a rectangular frame of length 7 cm dipping into soap solution. The framework hangs from the arm of a balance.. An extra wight of 0.38 g must be placed in the opposite arm to balance the pull of the film, Calculate the surface tension of soap solution.**

Ans : Given that the mass ( m ) 0.38 g

length of the rectangular soap film ( l ) = 7 cm

Force ( F ) = mg = 0.38 * 980 dyne

Surface tension ( T ) = F / 2l = 26.6 dyne

**27 ) A glass plate of length 10 cm, breadth 1.54 cm, and thickness 0.2 cm weighs 8.2 g in air, If it is held vertically with long side horizontal and the plate half immersed in water, what will be its apparent wight ? Surface tension of water = 73 dyne / cm.**

Ans : Given that the surface tension of water ( T ) = 73 dyne / cm

Length of the plate ( l ) = 10 cm

breadth of the plate ( b ) = 1.54 cm

thickness of the plate ( t ) = 0.2 cm

density of the water ( d ) = 1 gm /cc

Volume of the plate ( v ) = l* b* t = 10 * 1.54 * 0.2 = 3.08 c.c

weight of the plate ( mg ) = 8.2 gf = 9.2 * 980 = 8036 dyne

Force due to surface tension ( F ) = T * 2 ( l + t ) = 1489.2 dyne

Upthrust ( U ) = ( l * t * b/2 ) dg = 1509.2 dyne

The glass plate is under the action of three different forces i. e wight ( mg ) acting vertically downward, force due to surface tension ( F ) acting vertically downward and upthrust experienced due to the half immersed par of the plate ( U )

Hence, the apparent wight = mg + F – U

mg’ = 8036 + 1489.2 – 1509.2 = 8016 dyne

mg’ = 8016 / 980 = 8.18 gf

**28 ) Calculate the work done in blowing a soap bubble from a radius of 2 cm to 3 cm, the surface tension of the soap solution is 30 dyne / cm.**

Ans : Given that the surface tension ( T ) = 30 dyne / cm

Initial radius ( r ) = 2 cm

Final radius ( R ) = 3 cm

As the soap solution has two free surface, the change in surface area is given by

ds = 2 * 4π( R^{2} – r^{2} )

Required work done = ds * T = 8π( R^{2} – r^{2} ) * T = 3769.91 erg

**29 ) Calculate the amount of energy evolved, when 8 droplets of water ( surface tension 72 dyne / cm ) of radius 0.5 mm each combine into one.**

Ans : Give that number of droplets of water ( n ) = 8

surface tension ( T ) = 72 dyne / cm

radius of each drop ( r ) = 0.5 mm = 0.05 cm

Let radius of big drop = R

The volume of the big drop = the volume of 8 drops

R^{3} = 8r^{3}

R = 2r = 2 * 0.05 = 0.1 cm

Change in surface area ( ds ) = initial surface area – final surface area

ds = 8 * ( 4πr^{2} ) – 4πR^{2}

ds = ( 0.08 – 0.4 )π = 0.4 π

The required energy = ds * T = 0.04π * 72 = 9.047 erg

**30 ) Calculate the energy spent in spraying a drop of mercury of 1 cm radius into 10**^{6}** droplets all of same size, Surface tension of mercury is 35 * 10 ^{-3} N / m.**

**31 ) Find the work done in breaking a spherical drop of 1 mm radius of water into one million droplets . Surface tension of water is 72 dyn / cm.**

**32 ) A film of water is formed between two straight – parallel wires each 10 cm long and at separation 0.5 cm. Calculate the work required to increase 1 mm distance between the wires, Surface tension of water = 72 * 10 ^{-3} N / m.**

**33 ) A soap bubble is blown to a diameter of 7 cm. If 36,960 erg of work is done in blowing it further, find the mew radius, The surface tension of soap solution is 40 dyen / cm.**

**34 ) Calculate the excess of pressure inside ( a ) a drop of liquid of radius 0.2 cm and ( b ) a bubble of liquid of radius 0.3 cm. Given that surface tension 24 dyne / cm.**

**35 ) The air pressure inside a soap bubble of diameter 7 mm is 0.8 cm of water column above the atmosphere, Calculate surface tension of soap solution.**

**36 ) If the excess pressure inside a soap bubble of radius 1.0 cm is balanced by a column of oil 3 mm high, find the surface tension of the bubble, ( specific gravity of liquid = 1.8 )**

**37 ) What should be the pressure inside a small air bubble of 0.1 mm radius situated just below the free surface of water ? Surface tension of water = 72 dyn / cm and atmospheric pressure = 1.013 * 10 ^{6} dyne / cm^{-2}.**

**38 ) Two soap bubbles have radii in the ratio 1 : 2. Compare the excess of pressure inside these bubbles. Also compare the work done in blowing these bubbles.**