Surface Tension and Capillarity

Master surface tension, capillary action, and surface energy for JEE Physics

Prerequisites

Before studying surface tension, make sure you understand:

What is Surface Tension?

Surface tension is the property of a liquid surface that makes it behave like a stretched elastic membrane.

It’s the tendency of liquid surfaces to minimize their surface area.

Real Life: Water Striders Walk on Water

Have you seen insects walking on water? Water striders can do this because of surface tension!

The liquid surface acts like a stretched rubber sheet. The insect’s weight creates a small depression, but surface tension provides an upward force that balances the weight — keeping the insect afloat.

Same reason you can carefully float a needle on water, even though steel is denser than water!

Molecular explanation:

Inside liquid:

  • Molecule experiences equal attractions from all sides
  • Net force = 0

At surface:

  • Molecule has more neighbors below than above
  • Net inward force pulls molecules into bulk
  • Surface tries to contract to minimum area

This creates tension in the surface layer!

Definition and Formula

Surface tension (γ or σ) is defined as:

Definition 1: Force per unit length

$$\boxed{\gamma = \frac{F}{L}}$$

Force acting perpendicular to an imaginary line of length L in the surface.

SI Unit: N/m (Newton per meter)

Definition 2: Energy per unit area

$$\boxed{\gamma = \frac{W}{A}}$$

Work done in increasing surface area by unit amount.

SI Unit: J/m² (Joule per meter squared)

Note: N/m = J/m² (both are equivalent!)

$$\text{N/m} = \frac{\text{N} \cdot \text{m}}{\text{m}^2} = \frac{\text{J}}{\text{m}^2}$$

Dimensions: [MT⁻²] (same as spring constant!)

Two Equivalent Definitions

Force definition: Surface tension is force per unit length

Energy definition: Surface tension is energy per unit area

Both give same numerical value and units!

Typical Surface Tension Values (at 20°C)

LiquidSurface Tension (N/m)Surface Tension (dyne/cm)
Water0.07373
Mercury0.465465
Ethanol0.02222
Glycerin0.06363
Soap solution0.02525
Blood0.05858
Olive oil0.03232
Memory Trick

Mercury has highest surface tension (≈ 0.47 N/m)

Water is moderate (≈ 0.073 N/m)

Alcohol, soap have low surface tension (≈ 0.02 N/m)

Order: Mercury > Water > Soap → “My Wife is Sweet”

Effect of Temperature

Surface tension decreases with increasing temperature.

Why? Higher temperature → greater molecular kinetic energy → weaker intermolecular forces → lower surface tension

At critical temperature: Surface tension becomes zero (liquid-gas distinction vanishes)

Surface Energy

Surface energy is the extra potential energy of molecules at the surface compared to those in the bulk.

Work done to create new surface:

$$\boxed{W = \gamma \cdot \Delta A}$$

where ΔA = increase in surface area

Surface energy per unit area = γ

Example: Soap Bubble Energy

A soap bubble has radius 2 cm. Find work done in blowing the bubble. (γ_soap = 0.03 N/m)

Solution:

Soap bubble has two surfaces (inner and outer)!

Total surface area = 2 × 4πr² = 8πr²

$$W = \gamma \times A = 0.03 \times 8\pi \times (0.02)^2$$ $$= 0.03 \times 8 \times 3.14 \times 0.0004$$ $$= 0.03 \times 0.01 = 3 \times 10^{-4} \text{ J}$$

W = 0.3 mJ

Excess Pressure in Drops and Bubbles

Due to surface tension, pressure inside a curved liquid surface is greater than outside.

1. Liquid Drop (one surface)

$$\boxed{P_{inside} - P_{outside} = \frac{2\gamma}{r}}$$ $$\boxed{\Delta P = \frac{2\gamma}{r}}$$

2. Soap Bubble (two surfaces)

$$\boxed{\Delta P = \frac{4\gamma}{r}}$$

Factor of 4 because bubble has inner and outer surfaces!

3. Air Bubble in Liquid (one surface, like drop)

$$\boxed{\Delta P = \frac{2\gamma}{r}}$$

Interactive Demo: Visualize Surface Tension

See how surface tension creates excess pressure in drops and bubbles.

Key Observations
  1. Smaller radius → Higher excess pressure

    • Tiny drops have enormous internal pressure!
  2. Pressure is same throughout inside

    • Uniform pressure inside drop/bubble
  3. Shape is spherical

    • Sphere has minimum surface area for given volume
Real Life: Soap Bubbles Burst

Ever notice how small bubbles burst faster than large ones?

From ΔP = 4γ/r:

  • Small bubble: High pressure, bursts easily
  • Large bubble: Lower pressure, more stable

Also, when two bubbles of different sizes connect, air flows from smaller to larger (higher to lower pressure), and small bubble shrinks while large one grows!

Derivation: Excess Pressure in Drop

Method 1: Force balance

Consider hemisphere of drop, radius r.

Inward force due to excess pressure:

$$F_{pressure} = \Delta P \times \pi r^2$$

Outward force due to surface tension:

$$F_{tension} = \gamma \times 2\pi r$$

(γ acts along circumference of circle)

At equilibrium:

$$\Delta P \times \pi r^2 = \gamma \times 2\pi r$$ $$\boxed{\Delta P = \frac{2\gamma}{r}}$$

Method 2: Energy consideration

Increase radius from r to r + dr.

Work done by pressure:

$$W_P = \Delta P \times 4\pi r^2 \times dr$$

Work done against surface tension:

$$W_\gamma = \gamma \times d(4\pi r^2) = \gamma \times 8\pi r \, dr$$

These must be equal:

$$\Delta P \times 4\pi r^2 \times dr = \gamma \times 8\pi r \, dr$$ $$\boxed{\Delta P = \frac{2\gamma}{r}}$$
Example: Pressure Inside Water Drop

Find excess pressure inside a water drop of radius 1 mm. (γ_water = 0.073 N/m)

Solution:

$$\Delta P = \frac{2\gamma}{r} = \frac{2 \times 0.073}{10^{-3}}$$ $$= \frac{0.146}{10^{-3}} = 146 \text{ Pa}$$

About 0.0014 atm — small but measurable!

For a drop of radius 1 μm (micrometer):

$$\Delta P = \frac{2 \times 0.073}{10^{-6}} = 1.46 \times 10^5 \text{ Pa} \approx 1.4 \text{ atm}$$

Enormous pressure in tiny drops!

Angle of Contact (θ)

Angle of contact is the angle between the solid surface and tangent to liquid surface at the point of contact, measured through the liquid.

Depends on:

  1. Nature of liquid
  2. Nature of solid
  3. Cleanliness of surfaces

Two cases:

1. Acute Angle of Contact (θ < 90°)

  • Liquid wets the surface
  • Concave meniscus (curved downward in tube)
  • Adhesive force > Cohesive force
  • Examples: Water on clean glass, alcohol on glass

2. Obtuse Angle of Contact (θ > 90°)

  • Liquid doesn’t wet the surface
  • Convex meniscus (curved upward in tube)
  • Cohesive force > Adhesive force
  • Examples: Mercury on glass, water on waxy surface

Special cases:

  • θ = 0°: Perfect wetting (spreads completely)
  • θ = 180°: No wetting at all (complete repulsion)
Contact Angles

Water on glass: θ ≈ 0° to 30° (wets)

Mercury on glass: θ ≈ 135° to 140° (doesn’t wet)

Water on Teflon: θ ≈ 108° (doesn’t wet — that’s why food doesn’t stick!)

Water on leaf (lotus effect): θ ≈ 160° (super-hydrophobic)

Capillarity (Capillary Action)

Capillary action is the rise or fall of liquid in a narrow tube (capillary) due to surface tension.

Capillary Rise (θ < 90°, liquid wets tube)

When a capillary tube is dipped in a liquid that wets it:

$$\boxed{h = \frac{2\gamma \cos\theta}{\rho g r}}$$

where:

  • h = height of rise (or depression)
  • γ = surface tension
  • θ = angle of contact
  • ρ = density of liquid
  • g = acceleration due to gravity
  • r = radius of capillary tube

For perfect wetting (θ = 0°, cos θ = 1):

$$\boxed{h = \frac{2\gamma}{\rho g r}}$$
Key Observations
  1. h ∝ 1/r — Thinner tube → greater rise
  2. h ∝ γ — Higher surface tension → greater rise
  3. h ∝ 1/ρ — Lighter liquid → greater rise
  4. In zero gravity: h → ∞ (liquid rises indefinitely!)

Capillary Depression (θ > 90°, liquid doesn’t wet tube)

For mercury in glass:

$$\boxed{h = -\frac{2\gamma \cos\theta}{\rho g r}}$$

Negative h means depression (liquid surface goes down).

Real Life: How Plants Drink Water

Capillary action is crucial for plant survival!

Water rises through xylem vessels (thin tubes) in plant stems from roots to leaves. Combined with:

  • Capillary rise (surface tension pulls water up)
  • Transpiration pull (evaporation from leaves creates suction)
  • Root pressure (osmotic pressure pushes water up)

Trees can transport water 100 meters high using these mechanisms!

The xylem tubes are microscopic (≈ 0.02 mm diameter), so capillary rise can be significant:

$$h = \frac{2 \times 0.073}{1000 \times 10 \times 10^{-5}} = 1.46 \text{ m}$$

Combined with other effects, water reaches the top of tallest trees!

Derivation of Capillary Rise

Forces on liquid column in capillary:

Upward force (due to surface tension):

$$F_{up} = \gamma \times 2\pi r \times \cos\theta$$

(γ acts along the contact line, vertical component is γ cos θ)

Downward force (weight of liquid column):

$$F_{down} = \rho \times \pi r^2 \times h \times g$$

At equilibrium:

$$\gamma \times 2\pi r \times \cos\theta = \rho \times \pi r^2 \times h \times g$$ $$2\gamma \cos\theta = \rho g r h$$ $$\boxed{h = \frac{2\gamma \cos\theta}{\rho g r}}$$
Example: Water in Capillary

Find the height water rises in a capillary tube of radius 0.5 mm. (γ = 0.073 N/m, ρ = 1000 kg/m³, θ = 0°, g = 10 m/s²)

Solution:

$$h = \frac{2\gamma \cos\theta}{\rho g r}$$ $$= \frac{2 \times 0.073 \times 1}{1000 \times 10 \times 0.5 \times 10^{-3}}$$ $$= \frac{0.146}{5} = 0.0292 \text{ m}$$

h ≈ 3 cm

In a tube of radius 0.05 mm (10 times thinner):

$$h = 10 \times 3 = 30 \text{ cm}$$

Much higher rise in thinner tube!

Applications of Surface Tension

1. Detergents and Soaps

Problem: Water has high surface tension → doesn’t wet greasy/oily surfaces well

Solution: Add soap/detergent

  • Reduces surface tension of water (from 0.073 to ~0.025 N/m)
  • Reduces contact angle with fabric
  • Water penetrates fabric better
  • Emulsifies oil/grease (breaks into tiny droplets)

2. Waterproofing

Hydrophobic coatings (Teflon, wax) create large contact angle (> 90°):

  • Water doesn’t wet surface
  • Forms droplets and rolls off
  • Surface stays dry

Lotus effect: Super-hydrophobic surface (θ ≈ 160°)

  • Water beads up and rolls off, carrying dirt
  • Self-cleaning surfaces

3. Oil Spreads on Water

Why oil spreads?

  • Surface tension of oil (≈ 0.03 N/m) < surface tension of water (0.073 N/m)
  • Oil film reduces total surface energy
  • Minimizes overall surface area

Application: Oil spills spread rapidly on ocean surface, causing environmental damage.

4. Hot Soup Tastes Better

Why?

  • Hot soup has lower surface tension than cold soup
  • Spreads more easily on tongue
  • Greater contact area with taste buds
  • Enhanced flavor!

5. Antiseptic Efficiency

Dettol/antiseptics have low surface tension:

  • Spread easily on skin
  • Penetrate wounds better
  • More effective killing bacteria

Common Mistakes to Avoid

Mistake 1: Forgetting Two Surfaces of Soap Bubble

Wrong: Excess pressure in soap bubble = 2γ/r

Correct: Soap bubble has inner and outer surfaces!

$$\Delta P = \frac{4\gamma}{r}$$

(Not 2γ/r like a drop)

Mistake 2: Contact Angle Measured from Solid

Wrong: Measuring angle from solid surface to liquid

Correct: Angle of contact is measured through the liquid

  • Water on glass: θ ≈ 10° (acute, measured inside water)
  • Mercury on glass: θ ≈ 140° (obtuse, measured inside mercury)
Mistake 3: Capillary Rise in Wide Tubes

Wrong: Using h = 2γ/(ρgr) for tubes of radius > 1 cm

Correct: Capillary formula is valid for narrow tubes only (r < few mm). For wide tubes, meniscus curvature changes, formula doesn’t apply.

Mistake 4: Surface Energy = Surface Tension

Wrong: They are the same thing.

Correct:

  • Surface tension (γ): Force per unit length (N/m) OR Energy per unit area (J/m²)
  • Surface energy (U): Total energy = γ × Area (Joules)

γ is intensive property, U is extensive.

Practice Problems

Level 1: JEE Main Basics

Problem 1

Find the excess pressure inside a water drop of radius 2 mm. (γ_water = 0.073 N/m)

Solution:

$$\Delta P = \frac{2\gamma}{r} = \frac{2 \times 0.073}{2 \times 10^{-3}}$$ $$= \frac{0.146}{0.002} = \boxed{73 \text{ Pa}}$$
Problem 2

Find work done in blowing a soap bubble of radius 5 cm. (γ_soap = 0.03 N/m)

Solution:

Surface area of bubble (two surfaces) = 2 × 4πr² = 8πr²

$$W = \gamma \times A = 0.03 \times 8\pi \times (0.05)^2$$ $$= 0.03 \times 8 \times 3.14 \times 0.0025$$ $$= 0.03 \times 0.0628 = \boxed{1.88 \times 10^{-3} \text{ J} = 1.88 \text{ mJ}}$$

Level 2: JEE Main/Advanced

Problem 3

Water rises to height 6 cm in a capillary tube of radius r. If radius is doubled, find new height of rise.

Solution:

From h = 2γ/(ρgr):

$$h \propto \frac{1}{r}$$ $$\frac{h_2}{h_1} = \frac{r_1}{r_2} = \frac{r}{2r} = \frac{1}{2}$$ $$h_2 = \frac{h_1}{2} = \frac{6}{2} = \boxed{3 \text{ cm}}$$

Doubling radius halves the rise.

Problem 4

Two soap bubbles of radii r₁ = 2 cm and r₂ = 4 cm coalesce to form a single bubble. Find radius of new bubble (assume isothermal process).

Solution:

Surface energy conserved (isothermal):

$$\gamma \times 8\pi r_1^2 + \gamma \times 8\pi r_2^2 = \gamma \times 8\pi R^2$$ $$r_1^2 + r_2^2 = R^2$$ $$4 + 16 = R^2$$ $$R^2 = 20$$ $$R = \sqrt{20} = 2\sqrt{5} \approx \boxed{4.47 \text{ cm}}$$

Level 3: JEE Advanced

Problem 5

A capillary tube of radius 0.5 mm is dipped vertically in water. Find: (a) Height of water rise (b) If tube is tilted at 60° to vertical, find length of water column in tube

(γ = 0.073 N/m, ρ = 1000 kg/m³, θ = 0°, g = 10 m/s²)

Solution:

(a) Vertical rise:

$$h = \frac{2\gamma}{\rho g r} = \frac{2 \times 0.073}{1000 \times 10 \times 0.5 \times 10^{-3}}$$ $$= \frac{0.146}{5} = 0.0292 \text{ m} = \boxed{2.92 \text{ cm}}$$

(b) Tilted tube:

Vertical height remains same (h = 2.92 cm)

But water column is now along the inclined tube!

If L = length of column:

$$L \cos 60° = h$$ $$L \times 0.5 = 2.92$$ $$L = \frac{2.92}{0.5} = \boxed{5.84 \text{ cm}}$$

Length doubles when tube is at 60° to vertical.

Problem 6

A soap bubble of radius 3 cm is blown at the end of a tube. By how much must the pressure inside the tube be increased to blow a bubble of radius 4 cm? (γ_soap = 0.03 N/m)

Solution:

Initial excess pressure:

$$\Delta P_1 = \frac{4\gamma}{r_1} = \frac{4 \times 0.03}{0.03} = 4 \text{ Pa}$$

Final excess pressure:

$$\Delta P_2 = \frac{4\gamma}{r_2} = \frac{4 \times 0.03}{0.04} = 3 \text{ Pa}$$

Change in pressure:

$$\Delta P_2 - \Delta P_1 = 3 - 4 = -1 \text{ Pa}$$

Pressure must be DECREASED by 1 Pa!

(Larger bubble needs less excess pressure)

Actually, question asks by how much to increase:

Initial: P_tube = P_atm + 4 Pa Final: P_tube = P_atm + 3 Pa

Decrease by 1 Pa (or increase by -1 Pa)

Problem 7

Mercury has surface tension 0.465 N/m, density 13600 kg/m³, and angle of contact 140° with glass. Find depression of mercury in a capillary tube of radius 1 mm. (g = 10 m/s²)

Solution:

$$h = \frac{2\gamma \cos\theta}{\rho g r}$$ $$= \frac{2 \times 0.465 \times \cos 140°}{13600 \times 10 \times 10^{-3}}$$

cos 140° = -cos 40° ≈ -0.766

$$= \frac{2 \times 0.465 \times (-0.766)}{136}$$ $$= \frac{-0.712}{136} = -0.00524 \text{ m}$$

h = -5.2 mm (depression)

Mercury depresses by about 5 mm in this tube.

Summary Table

ConceptFormulaUnits
Surface Tensionγ = F/L or W/AN/m or J/m²
Excess Pressure (Drop)ΔP = 2γ/rPa
Excess Pressure (Bubble)ΔP = 4γ/rPa
Capillary Riseh = 2γcosθ/(ρgr)m
Surface EnergyU = γAJ

Within Properties of Matter

Connected Chapters