### Angle of Contact:

- When the liquid is in contact with solid, the angle between the solid surface and the tangent to the free surface of liquid at the point of contact, measured from inside the liquid is called the angle of contact.

- When the liquid surface is curved concave upwards, the angle of contact is acute and when the liquid surface is curved convex upwards, the angle of contact is obtuse.

** Characteristics of the Angle of Contact:**

- The angle of contact is constant for a given liquid-solid pair.
- If there are impurities in liquid, then they alter the values of angle of contact.
- For a liquid which completely wets the solid, the angle of contact is equal to zero.
- When the angle of contact between the liquid and a solid surface is small (acute), the liquid is said to wet the surface. Thus water wets glass.
- If the angle of contact is large the surface is not wetted. Mercury does not wet glass.
- The angle of contact decreases with increase in temperature.

**Shape of a Liquid Drop on Solid Surface:**

- Consider a liquid drop put on solid surface.

Let, 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

For equilibrium of the drop

T_{2} = T_{1} + T_{3} cos θ

**Case -1:**T_{2}> T_{1}, and T_{2}-T_{1}< T_{3}, then cos θ is positive and the angle of contact is acute. e.g. kerosene on glass surface**Case – 2:**T_{2}< T_{1}, and T_{1}-T_{2}< T_{3}, then cos θ is negative and the angle of contact is obtuse. e.g. mercury on the glass surface.**Case – 3:**If T_{2}-T_{1}= T_{3}, then cos θ is 1 and the angle of contact is zero. For pure water, cos θ is almost 1 and liquid spreads on the surface. i.e. pure water on glass surface.**Case – 4:**T_{2}-T_{1}> T_{3}, then cos θ > 1 which is not possible. thus liquid will spread over the solid and drop shall not be formed.

**The Formation of Concave Surface of Liquid (Explanation of Acute Angle of Contact):**

- When impure water or kerosene is taken in a glass vessel, it is found that the surface near the walls is curved concave upwards.

- Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The magnitude of adhesive force A is greater than the magnitude of cohesive force C and resultant of the two molecular forces of attraction R is or outside the liquid. Hence the molecule A is attracted towards the walls of glass vessel.
- The free surface of water adjusts itself at right angles to the resultant R. Therefore molecules like M creep upward on the solid surface. Thus the water surface is curved concave upwards and the angle of contact is acute.

**The Formation of Convex Surface of Liquid ****(Explanation of Obtuse Angle of Contact)****:**

- When mercury is taken in a glass vessel, it is found that the surface near the walls is curved convex upwards.

- Consider a molecule of mercury M on the free surface very close to the wall of the glass vessel. The magnitude of adhesive force A is very less than the magnitude of cohesive force C and resultant of the two molecular forces of attraction R is directed inside the liquid. Hence the molecule A is attracted towards other molecules of mercury.
- The free surface adjusts itself at right angles to the resultant R. The molecule A creeps downwards on the glass surface. Thus the surface of the mercury in the glass is curved convex upwards and the angle of contact is obtuse.

**The Formation of Flat Surface of Liquid:**

- When pure is taken in a glass vessel, it is found that the surface of the water is perfectly flat.

- Consider a molecule of water M on the free surface very close to the wall of the glass vessel. The magnitude of adhesive force A is large hence the ersultant force R is along the direction of the adhesive force. Hence the molecule A is attracted towards glass molecules.
- The free surface adjusts itself at right angles to the wall. Thus the surface of the pure water is perfectly flat.

### Laplace’s Law of Spherical Membrane:

- Due to the spherical shape of drops and bubbles, the inside pressure P
_{i}is always greater than the outside pressure P_{o}. The excess of pressure is P_{i}– P_{o}.

**Laplace’s Law of a Spherical Membrane for a Liquid Drop:**

- 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.

Final surface area = A_{2 } = 4 πr² + 8 πr.Δr

Change in area = A_{2 } – A_{1 } = 4 π (r + Δr)² – 4 π r²

Change in area = A_{2 } – A_{1 } = 4 π (r² + 2r.Δr + Δr²) – 4 π r²

Change in area = A_{2 } – A_{1 } = 4 πr² + 8 πr.Δr + 4 πΔr² – 4 π r²

Change in area = A_{2 } – A_{1 } = 8 πr.Δr + 4 πΔr²

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

Change in area = dA = 8 πr.Δr

Now, work done = dW = T. dA = T. 8 πr.Δr …………… (1)

By definition of work in mechanics, we have dW = F .Δr …………… (2)

But P = F /A, Hence F =P × A

F = (P_{i}– P_{o}) × 4 πr²

Substituting in equation (2) we have

dW = (P_{i}– P_{o}) × 4 πr².Δr …………… (3)

From equations (3) and (4) we have

(P_{i}– P_{o}) × 4 πr².Δr = T. 8 πr.Δr

(P_{i}– P_{o}) = 2T / r

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

- For liquid bubble, there are two free surfaces, then the Laplace’s law for a spherical membrane for a liquid bubble is written as (P
_{i}– P_{o}) = 4T / r

**Problems Based on Surface Energy**

- A raindrop of diameter 4 mm is about to fall on the ground. Calculate the pressure inside the raindrop. Surface tension of water = 0.072 N/m. Atmospheric pressure = 1.013 x 10
^{5}N/m². (**Ans:**1.01372 × 10^{5 }N/m²) - Find the excess pressure inside the soap bubble of diameter 3 cm. The surface tension of the soap solution is 3 × 10
^{-2}N/m.**(Ans:**8 N/m²) - What should be the diameter of a soap bubble, in order that the excess pressure inside it is 51.2 N/m²? The surface tension of the soap solution is 3.2 × 10
^{-2}N/m.**Ans:**5 mm)

### For Solutions to Problems Click Here

### Capillarity:

- A capillary tube is a tube with a very fine bore. The phenomenon of a rise and fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillary action or capillarity.
- Example rise of water in capillary tube or drop of mercury in capillary tube.

**Examples of Capillarity:**

- Oil rises through wicks of lamps.
- Water rises through sap of trees.
- Ink is absorbed by blotting paper.
- Liquids are absorbed by sponges.

**Capillary Action in Water:**

**Capillary Action in case of Mercury:**

- In the case of glass capillary immersed in mercury, the meniscus is convex upwards. The pressure on concave side (liquid side) of the free surface is greater than that on the convex side (atmospheric pressure). It means that the pressure just below the meniscus i.e. on the concave side is more than atmospheric pressure.

** Expression for the Rise in the Liquid in a Capillary Tube:**

- Consider a capillary tube of internal radius of cross-section ‘r’. 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 θ 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..

- The length along which the force of surface tension acts is 2πr. Hence total upward force is 2πr T cos θ. Hence 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.

**Problems Based on Surface Energy**

- A capillary tube of a uniform bore is dipped vertically in water which rises by 7 cm in the tube. Find the radius of the capillary if the surface tension of is 70 dynes/cm g= 9.8 m/s
^{2}. (**Ans:**0.2 mm) - Water rises to a height of 4.5 cm in a capillary tube of radius r. Find r assuming the surface tension of water is 72 dynes/cm. Take angle of contact of water in the glass as 15°. Density of water=1000 kg/m³ , g = 9.8 m/s². (
**Ans:**0.315 mm) - A liquid rises to a height of 9 cm in a glass capillary of radius 0.02 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.03 cm? (
**Ans:**6 cm) - A liquid rises to a height of 4.5 cm in a glass capillary of radius 0.01 cm. What will be the height of the liquid column in a similar glass capillary of radius 0.02 cm? (
**Ans:**2.25 cm) - A liquid of density 900 kg/m³ rises to a height of 7 mm in a capillary tube of 2 mm internal diameter. If the angle of contact of the liquid in the glass is 25°, find the surface tension of the liquid. g=9.8 m/s². (
**Ans:**0.034 N/m) - When a capillary tube of radius 0.45 mm is dipped into water, the level inside the capillary tube rises to height 3 cm above the surface of the water. calculate the surface tension of water, if its angle of contact is zero and its density is 1 g/cm³, g = 9.8 m/s². (
**Ans:**0.066615 N/m) - A mercury barometer tube is 2.5 mm in internal radius. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=0.540 N/m. The density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 135°. (
**Ans:**2.29 mm should be added to the height of mercury) measurement.) - The tube of a mercury barometer is 3 mm in diameter. What error is introduced in the reading because of surface tension? Angle of contact of mercury in glass = 135° and S.T. of mercury = 0.460 N/m. Density of mercury = 13.6 g/cc, g=9.8 m/s². (
**Ans:**3.25 mm should be added to the height of mercury measurement.) - A mercury barometer tube is 1 cm in diameter. Find the error introduced in the observed reading because of surface tension. g = 9.8 m/s², T=435.5 dyne/cm . The density of mercury = 13600 kg /m³, angle of contact of mercury in glass = 140°. (
**Ans:**1.01 mm should be added to the height of mercury) - A capillary tube of radius 0.5 mm is dipped vertically in a liquid of surface tension 0.04 N/m and density 0.8 g/cc. Calculate the height of capillary rise, if the angle of contact is 10°. (
**Ans:**2.01 cm) - A capillary tube of radius 0.5 mm is immersed in a beaker of mercury. The mercury level inside the tube is found to be 0.80 cm below the level of the reservoir. Determine the angle of contact between mercury and glass. Surface tension of mercury = 0.465 N/m and density is 13.6 × 10
^{3 }kg/m³, g = 9.8 m/s^{2.}(**Ans:**124°58′) - Calculate the density of paraffin oil, if glass capillary of diameter 0.25 mm dipped in a paraffin oil of the surface tension 0.0245 N/m rises to a height of 4 cm. Angle of contact of paraffin oil with glass is 28°, g = 9.8 m/s
^{2}, (**Ans:**882.9 kg/m³) - Water rises to a height of 4 cm in a certain capillary tube. If the same capillary tube is dipped in a mercury, the level of mercury decreases to 3 cm. Compare the surface tension of water and mercury, if densities of mercury and water are 13.6 × 10
^{3 }kg/m³ and 10^{3 }kg/m³ respectively. The angle of contact for water and mercury are 0° and 135° respectively. (**Ans:**0.06932:1) - Water rises to height 3.2 cm in glass capillary tube. Find height to which same water rises in another capillary having half area of cross-section. (Ans: 4.525 cm)
- A U tube has one limb of radius 0.5 cm and the other limb of radius 0.1 cm. Water is poured in the tube. Find the difference in the level of water in the two tubes. S.T. of water = 0.070 N/m. g = 9.8 m/s
^{2}. (Ans: 1.142 cm)

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