# Surface Tension: Derivations and Diagrams

#### Question – 1:

• Derive the relation between surface tension and surface energy per unit area. (MHSB March – 13-2 M)  OR  Show that the surface tension of the liquid is numerically equal to the surface energy per unit area. (MHSB October – 13-2 M)

• Consider a rectangular frame ABCD in which side CD is made of loose write and other three sides are fixed. The frame is immersed in a soap solution and taken out and held horizontally. A film of soap solution will be formed on the frame and it will at once try to shrink and pull the loose wire CD towards AB. If the length of loose wire CD is  ‘l’ and the film is of finite thickness, therefore the film will be in contact with the wire both on the upper surface as well as along its lower surface.  Hence the length of the wire in contact with the film is ‘2l‘.  The force acting on the wire is directed towards AB, per unit length of the contact line is surface tension (T). By definition of surface tension, we have

T = F / 2l

∴    F  =  T . 2 l     …  (1)

• Imagine an external force is applied on CD which is equal and opposite to force F Let the wire at CD moves to C’D’ through small distance dx.  Then the work done against the force of surface tension is given by

dW  =  F.dx         …  (2)

From equations (1) and (2),

dW = T.2l. dx

But,  2l . dx = dA = increase in Area of both the surface of the film.

∴  dW   =  T.dA

This work done is stored inside the films as potential energy dU.

∴   dU  =  T.dA

• If, initially CD is very close to AB, initial energy and initial area of the film can be taken as zero and dU and dA can be treated as total energy and the total area of the film respectively.

∴  T  =  dU / dA

This expression indicates that surface tension is equal to surface energy per unit area of the surface film.

#### Question – 2:

• Derive an expression for the excess pressure inside a drop of liquid.  (MHSB March – 15-3 M)
• Due to the spherical shape, the inside pressure Pis always greater than the outside pressure Po. The excess of pressure is Pi– Po.

• Let the radius of the drop increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Initial surface area = A = 4 π r²

Final surface area = A = 4 π (r + Δr)² = 4 π (r² + 2r.Δr + Δr²) =  4 πr² + 8 πr.Δr + 4 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 4 πΔr² can be neglected.

Final surface area = A =  4 πr² + 8 πr.Δr

Hence Change in area = A – A =  4 πr² + 8 πr.Δr  –  4 πr²

Change in area = dA =  8 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  8 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force ∴ displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (Pi– Po) × 4 πr²

Substituting in equation (2) we have

dW = (Pi– Po) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(Pi– Po) × 4 πr².Δr  = T.  8 πr.Δr

(Pi– Po) = 2T / r

This relation is known as Laplace’s law for the spherical membrane for a liquid drop.

#### Question – 3:

• Derive Laplace’s law of spherical membrane of the bubble due to surface tension. (MHSB March – 16-3 M)
• Due to the spherical shape, the inside pressure Pis always greater than the outside pressure Po. The excess of pressure is Pi– Po.

• Let the radius of the bubble increases from r to r + Δr, where Δr is very very small, hence the inside pressure is assumed to be constant.

Bubble has two free surfaces, one inside and another outside

Initial surface area = A = 2 × 4 π r²  = 8 π r²

Final surface area = A =  2 × 4 π (r + Δr)² = 8 π (r² + 2r.Δr + Δr²) =  8 πr² + 16 πr.Δr + 8 πΔr²

Δr is very very small, hence Δr² still smsll hence the term 8 πΔr² can be neglected.

Final surface area = A =  8 πr² + 16 πr.Δr

Hence Change in area = A – A =  8 πr² + 16 πr.Δr  –  8 πr²

Change in area = dA =  16 πr.Δr

Now, work done in increasing the surface area is given by

dW = T. dA = T.  16 πr.Δr    …………… (1)

By definition of work in mechanics we have

dW = Force ∴ displacement = F .Δr  …………… (2)

But P = F /A, Hence F = Excess pressure × Area

F = (Pi– Po) × 4 πr²

Substituting in equation (2) we have

dW = (Pi– Po) × 4 πr².Δr  …………… (3)

From equations (3) and (4) we have

(Pi– Po) × 4 πr².Δr  = T.  16 πr.Δr

(Pi– Po) = 4T / r

This relation is known as Laplace’s law for spherical membrane for a liquid bubble.

#### Question – 4:

• Obtain an expression for the rise of wetting liquid in a capillary tube.(Model Paper – 1 – 3M) OR  Derive an expression for the height of the liquid column in a capillary tube when dipped vertically in the liquid.    (Model Paper – 6 – 2M)
• If a glass tube of a smaller bore (capillary tube) is immersed in a liquid which wets the glass (water), then the liquid level inside the tube rises. If the tube is immersed in a liquid which does not wet the glass (mercury), then the liquid level inside the tube decreases. This phenomenon of the rise or fall of liquid in a capillary tube is called capillary action or capillarity.
• Consider a capillary tube immersed in a liquid that wets it. The liquid will rise in the capillary tube. The surface of the liquid will be concave.

• The surface tension ‘T’ acts along the tangent to the liquid surface at the point of contact as shown. Let q be the angle of contact. The force of surface tension is resolved into two components vertical T cos θ and horizontal T sin θ. The components T sinθ cancel each other as they are equal in magnitude and radially outward (opposite to each other). The unbalanced component T cos θ will push the liquid up into the capillary tube. This explains the rise in the liquid layer in the capillary tube.
1. If ‘r’ is the radius of the bore of the capillary tube, the length along which the force of surface tension acts is 2πr. Hence total upward force is  2πr T cos θ.
2. Due to this force the liquid rise up in the tube. The weight of liquid acts vertically downward. The liquid goes on rising till the force of surface tension is balanced by the weight of the liquid column.

Total upward force  =  Weight of liquid in the capillary tube.

2πr T cos θ  =    mg

Where ‘m’ is the mass of liquid in the capillary tube.

2πr T cos θ   =    V ρ g

Where ‘V’ is the volume of liquid in capillary and ρ is the density of the liquid in the capillary tube.

2πr T cos θ   =    π r²h ρ g

Where ‘h’ is the height of the liquid column in the capillary tube. then

This is an expression for the rise in the liquid in the capillary tube.

• Draw a neat labelled diagram showing the forces acting on the meniscus of water in the capillary tube.  (MHSB October – 15-2 M) OR  Draw a neat labelled diagram of the rise of liquid in capillary tube showing different components of tension (force). (MHSB July – 16-2 M)

Where, T = Surface tension

h = height of liquid column in capillary tube

θ = Angle of contact

r = Radius of cross-section of Capillary Tube

• Draw a neat labelled diagram of a mercury drop on glass showing all interfacial tensions.(Model Paper – 2 – 2M)

Where, θ  = the angle of contact of given solid-liquid pair.

T1 = Surface tension at the liquid-solid interface

T2 = Surface tension at the liquid-solid interface

T3 = Surface tension at the liquid-solid interface