Mechanical Properties of Fluids • Topic 3 of 3

Surface Tension

Why does a small water drop form a sphere, an insect walk on water, and a needle float if placed gently? All these are due to surface tension — the tendency of a liquid surface to behave like a stretched elastic membrane.

Surface tension ($T$) is the force per unit length acting along the surface of a liquid, perpendicular to an imaginary line drawn on the surface:

  • $T=\frac{F}{L}$, with SI unit $\text{N/m}$.
  • It arises because molecules at the surface have fewer neighbours than those inside, so they are pulled inward, making the surface contract to the smallest possible area. A sphere has the least surface area for a given volume, so free drops are spherical.

Cohesion and adhesion. Cohesive forces act between molecules of the same substance (water–water); adhesive forces act between molecules of different substances (water–glass). Their relative strength decides whether a liquid wets a surface or beads up.

Surface energy. Increasing the surface area of a liquid requires work against surface tension. The surface energy stored is the work done per unit increase of area, and numerically the surface energy per unit area equals the surface tension $T$. So $T$ can be defined either as force per unit length or as energy per unit area ($\text{J/m}^2=\text{N/m}$).

Angle of contact ($\theta$). The angle between the tangent to the liquid surface and the solid surface, measured inside the liquid, is the angle of contact. For water on clean glass $\theta$ is small (acute) — water wets glass and its meniscus is concave. For mercury on glass $\theta$ is obtuse — mercury does not wet glass and its meniscus is convex.

Capillary rise. When a narrow tube (capillary) is dipped in a wetting liquid, the liquid rises in the tube against gravity until the upward pull of surface tension balances the weight of the raised column. The height of rise is:

  • $h=\frac{2T\cos\theta}{\rho g r}$, where $r$ is the tube radius, $\theta$ the angle of contact, $\rho$ the density and $g$ gravity.
  • The rise is greater in a narrower tube ($h\propto\frac{1}{r}$). For a non-wetting liquid like mercury, $\cos\theta$ is negative, so the liquid is depressed instead of rising. Capillarity explains the rise of oil in a wick and water moving up through soil.

Excess pressure (Laplace's law). The curved surface of a drop or bubble has a higher pressure inside than outside:

  • For a liquid drop (one surface): $P_{excess}=\frac{2T}{R}$.
  • For a soap bubble (two surfaces — inner and outer): $P_{excess}=\frac{4T}{R}$.
  • The excess pressure is larger for a smaller radius, which is why a smaller bubble has a higher internal pressure than a larger one.
Capillary rise in a narrow tubeh2rh = 2T cos(theta) / (rho g r)water reservoir
Solved Examples
1
Worked Example
A liquid film of surface tension $0.05\ \text{N/m}$ supports a wire of length 0.2 m. Find the force due to surface tension on the wire (single surface).
Solution
  1. Step 1: Use $F=T\,L$ for one surface.
  2. Step 2: Substitute: $F=0.05\times0.2$.
  3. Step 3: Compute: $F=0.01\ \text{N}$.

Answer: $F=0.01\ \text{N}$.

2
Worked Example
Water rises to a height of 5 cm in a capillary tube. To what height will it rise in a tube of half the radius?
Solution
  1. Step 1: Since $h\propto\frac{1}{r}$, halving $r$ doubles $h$.
  2. Step 2: $h_2=h_1\frac{r_1}{r_2}=5\times\frac{r}{r/2}=5\times2$.
  3. Step 3: Compute: $h_2=10\ \text{cm}$.

Answer: $h_2=10\ \text{cm}$.

3
Worked Example
Find the height to which water ($T=0.072\ \text{N/m}$, $\rho=1000\ \text{kg/m}^3$) rises in a capillary of radius $0.5\ \text{mm}$. Assume $\theta=0^\circ$ and $g=9.8\ \text{m/s}^2$.
Solution
  1. Step 1: Use $h=\frac{2T\cos\theta}{\rho g r}$ with $\cos0^\circ=1$ and $r=5\times10^{-4}\ \text{m}$.
  2. Step 2: Numerator $=2\times0.072\times1=0.144$; denominator $=1000\times9.8\times5\times10^{-4}=4.9$.
  3. Step 3: Compute: $h=\frac{0.144}{4.9}\approx0.0294\ \text{m}\approx2.94\ \text{cm}$.

Answer: $h\approx2.94\ \text{cm}$.

4
Worked Example
Calculate the excess pressure inside a water drop of radius $1\ \text{mm}$. (Take $T=0.072\ \text{N/m}$.)
Solution
  1. Step 1: For a drop (one surface), $P_{excess}=\frac{2T}{R}$ with $R=10^{-3}\ \text{m}$.
  2. Step 2: Substitute: $P=\frac{2\times0.072}{10^{-3}}$.
  3. Step 3: Compute: $P=\frac{0.144}{10^{-3}}=144\ \text{Pa}$.

Answer: $P_{excess}=144\ \text{Pa}$.

5
Worked Example
Find the excess pressure inside a soap bubble of radius $2\ \text{cm}$. (Take $T=0.025\ \text{N/m}$.)
Solution
  1. Step 1: A soap bubble has two surfaces, so $P_{excess}=\frac{4T}{R}$ with $R=0.02\ \text{m}$.
  2. Step 2: Substitute: $P=\frac{4\times0.025}{0.02}$.
  3. Step 3: Compute: $P=\frac{0.1}{0.02}=5\ \text{Pa}$.

Answer: $P_{excess}=5\ \text{Pa}$.

6
Worked Example
How much work is done in increasing the radius of a soap bubble from 1 cm to 2 cm? (Take $T=0.03\ \text{N/m}$.)
Solution
  1. Step 1: A soap bubble has two surfaces, so work $=T\times\Delta A=T\times2\times4\pi(R_2^2-R_1^2)$.
  2. Step 2: $R_2^2-R_1^2=(0.02)^2-(0.01)^2=4\times10^{-4}-1\times10^{-4}=3\times10^{-4}\ \text{m}^2$.
  3. Step 3: $W=0.03\times8\pi\times3\times10^{-4}=0.03\times8\times3.1416\times3\times10^{-4}\approx2.26\times10^{-4}\ \text{J}$.

Answer: $W\approx2.26\times10^{-4}\ \text{J}$.

Key Points

  • Surface tension $T=\frac{F}{L}$ (N/m); it makes a liquid surface act like a stretched membrane and minimise its area (drops are spherical).
  • Surface energy per unit area equals $T$; increasing surface area needs work against surface tension.
  • Angle of contact is acute for wetting liquids (water on glass, concave meniscus) and obtuse for non-wetting ones (mercury on glass).
  • Capillary rise: $h=\frac{2T\cos\theta}{\rho g r}$; rise is greater in narrower tubes ($h\propto\frac{1}{r}$).
  • Excess pressure: $\frac{2T}{R}$ for a drop and $\frac{4T}{R}$ for a soap bubble; smaller radius means greater excess pressure.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The SI unit of surface tension is:
Explanation: Surface tension is force per unit length, $T=\frac{F}{L}$, measured in N/m.
Q2.The capillary rise of a liquid in a tube is given by:
Explanation: Balancing the surface-tension pull against the weight of the column gives $h=\frac{2T\cos\theta}{\rho g r}$.
Q3.The excess pressure inside a soap bubble of radius $R$ is:
Explanation: A soap bubble has two surfaces, so the excess pressure is $\frac{4T}{R}$ (double that of a single-surface drop).
Q4.Water rises in a glass capillary because the angle of contact for water on glass is:
Explanation: For water on clean glass the angle of contact is acute (adhesion exceeds cohesion), so $\cos\theta>0$ and the liquid rises.
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