Chapter 9 · Mechanical Properties of Fluids

9.4 Surface Tension and Capillarity

The surface of a liquid behaves like a stretched skin: it carries a surface tension $S$ (force per unit length, or energy per unit area). It makes drops round, raises the pressure inside drops and bubbles by $2S/r$ and $4S/r$, and lifts water up narrow tubes.

A pond skater walks on water without getting its feet wet; a steel needle laid gently on the surface floats although steel is eight times denser than water. Raindrops and drops of oil in water are almost perfect spheres. A soap film stretched across a wire loop pulls a loose thread into a perfect circle. Paper towels, the wick of an oil lamp and the soil around a plant's roots draw water upwards against gravity. None of these can be explained by pressure, flow or viscosity alone: they come from the special properties of a liquid's surface.

This section explains where those properties come from — the attraction between molecules — and measures them with one number, the surface tension, which is both a force per unit length along the surface and an energy per unit area of surface. You will meet the angle of contact, which decides whether a liquid wets a solid; find the extra pressure inside drops and bubbles; and derive the height to which a liquid rises (or sinks) in a narrow tube, the capillary rise.

Surface tension and surface energy

Molecules of a liquid attract one another strongly at short range (about a molecular diameter or two). A molecule deep inside is pulled equally in all directions by its neighbours, so the net force on it is zero. A molecule at the surface has neighbours only on the liquid side; it is pulled inwards, and has fewer neighbours to share attractive (negative) potential energy with. Bringing a molecule from the inside to the surface therefore needs work: the surface has extra energy. Like any system, a liquid left to itself settles towards the lowest energy it can reach — so it pulls its surface into the smallest area it can. For a given volume the shape of least area is a sphere, which is why small drops are round.

📘 Definition 9.8 — Surface energy

The surface energy of a liquid is the extra energy of the molecules in its surface compared with those inside. The work needed to increase the free surface area of a liquid by $\Delta A$ at constant temperature is proportional to $\Delta A$; the surface energy per unit area is measured in $\text{J/m}^2$.

Now stretch a surface and measure the force. Bend a wire into a U and lay a light straight wire of length $l$ across its arms so that it can slide. Dip the frame in soap solution: a film forms in the rectangle, and the film pulls the sliding wire inwards. To hold it still you must pull outwards with a force $F$. Pull it out a further distance $d$ slowly: you do work $Fd$, and the film's area increases by $ld$ on each of its two faces, $2ld$ in all. If the energy per unit area is $S$, then $Fd=S(2ld)$, so $F=2Sl$: the force is proportional to the length of the edge it acts on.

📘 Definition 9.9 — Surface tension

The surface tension $S$ of a liquid is the force per unit length acting across any line drawn in its surface, in the plane of the surface and perpendicular to the line:

$$S=\frac{F}{l}.$$

SI unit $\text{N/m}$; dimensions $[\text{M T}^{-2}]$. It depends on the liquid, on what is on the other side of the surface (usually air), and on the temperature.

📐 Result 9.11 — Surface tension equals surface energy per unit area

The surface tension $S$ (in $\text{N/m}$) is numerically equal to the surface energy per unit area (in $\text{J/m}^2$, the same unit). The work needed to increase a liquid surface's area by $\Delta A$ at constant temperature is

$$W=S\,\Delta A .$$

A film (soap film) has two surfaces, so a film of area $A$ has surface energy $2SA$, and pulls on a wire of length $l$ with force $2Sl$.

Liquid (against air)Temperature$S$ (N/m)
water$20\ ^\circ\text{C}$ / $100\ ^\circ\text{C}$$0.073$ / $0.059$
soap solution$20\ ^\circ\text{C}$about $0.025$–$0.030$
ethanol$20\ ^\circ\text{C}$$0.022$
olive oil$20\ ^\circ\text{C}$$0.032$
mercury$20\ ^\circ\text{C}$$0.465$

Surface tension falls as the temperature rises (hot water cleans better partly for this reason) and becomes zero at the critical temperature. Dissolved soaps and detergents lower it strongly; their molecules crowd into the surface with their water-loving ends in the water.

✏️ Example 1 — A soap film on a wire frame
A U-shaped frame holds a soap film ($S=0.030\ \text{N/m}$) closed by a sliding wire $5.0\ \text{cm}$ long. (a) What force holds the wire in equilibrium (neglect its weight)? (b) How much work is done in pulling the wire out a further $2.0\ \text{cm}$ slowly? (c) Check (b) using the surface energy.
(a) Two surfaces pull on the wire: $F=2Sl=2(0.030)(0.050)=3.0\times10^{-3}\ \text{N}$.
(b) $F$ stays the same as the film stretches (unlike a rubber sheet): $W=Fd=3.0\times10^{-3}\times0.020=6.0\times10^{-5}\ \text{J}$.
(c) New area of each face $0.050\times0.020=1.0\times10^{-3}\ \text{m}^2$; two faces: $\Delta A=2.0\times10^{-3}\ \text{m}^2$; $W=S\Delta A=6.0\times10^{-5}\ \text{J}$ ✓.
Check Units: $\text{N/m}\times\text{m}^2=\text{N m}=\text{J}$ ✓. A film's tension does not grow when stretched because new molecules simply move into the surface from the liquid inside the film.
✏️ Example 2 — The energy cost of a spray
A water drop of radius $1.0\ \text{mm}$ is broken up into $1000$ equal droplets. Taking $S=0.073\ \text{N/m}$, find the radius of each droplet and the work needed.
Radius Volume is conserved: $1000\times\tfrac43\pi r^3=\tfrac43\pi R^3$, so $r=\dfrac{R}{10}=0.10\ \text{mm}$.
Area Before: $4\pi R^2=4\pi(10^{-3})^2=1.26\times10^{-5}\ \text{m}^2$. After: $1000\times4\pi(10^{-4})^2=1.26\times10^{-4}\ \text{m}^2$ — ten times more. $\Delta A=1.13\times10^{-4}\ \text{m}^2$.
Work $W=S\Delta A=0.073\times1.13\times10^{-4}=8.3\times10^{-6}\ \text{J}$.
Check In general, splitting into $n$ droplets multiplies the area by $n^{1/3}$. The energy is small, but sprays and atomisers produce billions of droplets per second; going the other way, droplets that merge release surface energy as heat.

Angle of contact

Put a drop of water on clean glass and it spreads out; put a drop of mercury on glass and it stays as a bead; a drop of water on a waxy lotus leaf rolls about as a nearly perfect ball. What decides this is a competition between cohesion (the attraction of the liquid's molecules for each other) and adhesion (their attraction for the solid's molecules).

📘 Definition 9.10 — Angle of contact

Where a liquid surface meets a solid, the angle of contact $\theta$ is the angle between the solid surface and the tangent to the liquid surface at the line of contact, measured through the liquid.

  • $\theta\lt90^\circ$: the liquid wets the solid (adhesion wins); the liquid surface in a tube curves up at the walls (a concave meniscus). Water on clean glass, $\theta\approx0^\circ$; kerosene on glass, about $0^\circ$.
  • $\theta\gt90^\circ$: the liquid does not wet the solid (cohesion wins); it forms beads and its surface in a tube curves down at the walls (a convex meniscus). Mercury on glass, $\theta\approx140^\circ$; water on wax, about $105^\circ$.

The angle can be found from a balance of surface tensions at the edge of a drop resting on a solid: if $S_{la}$, $S_{sa}$ and $S_{sl}$ are the surface tensions of the liquid–air, solid–air and solid–liquid boundaries, the horizontal forces at the edge balance when $S_{la}\cos\theta+S_{sl}=S_{sa}$, so $\cos\theta=\dfrac{S_{sa}-S_{sl}}{S_{la}}$. Detergents and wetting agents lower the liquid's surface tension and make $\theta$ small, so water spreads into the fibres of cloth and carries grease away (the detergent molecules surround grease particles). Waterproofing agents do the opposite: they make $\theta$ large, so water beads up and rolls off a raincoat or umbrella.

Drops and bubbles: excess pressure

Because its surface is stretched like a skin, a drop squeezes the liquid inside it. The pressure inside a drop must therefore be greater than outside. To find by how much, use energy. Take a drop of radius $r$ with pressure $p_i$ inside and $p_o$ outside, and imagine it grows slightly to $r+\Delta r$ (by adding a little liquid). The extra surface energy needed is

$$S\left[4\pi(r+\Delta r)^2-4\pi r^2\right]\approx S\,8\pi r\,\Delta r$$

(neglecting the tiny term in $\Delta r^2$). It is supplied by the work done by the pressure difference pushing the surface out — pressure difference × area × distance — which is $(p_i-p_o)\,4\pi r^2\,\Delta r$. Equating the two gives $p_i-p_o=\dfrac{2S}{r}$. A soap bubble is a thin film with two surfaces, an inner and an outer one of almost equal radius, so the surface energy and the excess pressure are doubled.

📐 Result 9.12 — Excess pressure inside drops and bubbles

For a curved liquid surface of radius $r$ and surface tension $S$, the pressure on the concave (inner) side exceeds that on the convex side by

for a liquid drop, or an air bubble inside a liquid (one surface),

$$\Delta p=\frac{2S}{r};$$

for a soap bubble in air (two surfaces),

$$\Delta p=\frac{4S}{r}.$$

The smaller the drop or bubble, the greater the excess pressure.

✏️ Example 3 — Drops, bubbles and a soap bubble
Find the excess pressure inside (a) a water drop of radius $1.0\ \text{mm}$, (b) a soap bubble of radius $2.0\ \text{cm}$ ($S=0.025\ \text{N/m}$), (c) an air bubble of radius $0.10\ \text{mm}$ just below the surface of water. Take $S_\text{water}=0.073\ \text{N/m}$.
(a) One surface: $\Delta p=\dfrac{2(0.073)}{1.0\times10^{-3}}=146\ \text{Pa}\approx1.5\times10^2\ \text{Pa}$.
(b) Two surfaces: $\Delta p=\dfrac{4(0.025)}{0.020}=5.0\ \text{Pa}$ — tiny, which is why a soap bubble is so easy to blow.
(c) One surface (water–air): $\Delta p=\dfrac{2(0.073)}{1.0\times10^{-4}}=1.46\times10^3\ \text{Pa}$.
Check $\dfrac{2S}{r}$: ten times smaller radius, ten times larger excess pressure — (c) is ten times (a). Interactive 1 shows all three cases.
Interactive 1 — Pressure Inside Drops and Bubbles

Choose a water drop, a soap bubble or an air bubble in water, and change its radius (the slider is logarithmic: each step multiplies $r$). The surface pulls inwards all round (purple arrows), so the pressure inside is higher by $2S/r$ or $4S/r$. On the log–log graph $\Delta p\propto1/r$ is a straight line falling one decade for every decade of radius. Press Play to make the drop grow and shrink.

1.0 mm
radius $r$—
excess pressure $\Delta p$—
surface energy—
$\Delta p$ in mm of water—
✏️ Example 4 — A bubble under water
A tiny air bubble of radius $0.050\ \text{mm}$ is $10\ \text{cm}$ below the surface of water ($S=0.073\ \text{N/m}$), and the atmospheric pressure is $1.013\times10^5\ \text{Pa}$. Find the pressure of the air inside the bubble.
Outside The water just outside the bubble is at $p_a+\rho gh=1.013\times10^5+1000(9.8)(0.10)=1.013\times10^5+980\ \text{Pa}$.
Excess One water surface: $\dfrac{2S}{r}=\dfrac{2(0.073)}{5.0\times10^{-5}}=2.92\times10^3\ \text{Pa}$.
Inside $p=1.013\times10^5+980+2920=1.052\times10^5\ \text{Pa}$.
Check For a bubble this small, surface tension adds three times as much pressure as the $10\ \text{cm}$ of water above it. As a rising bubble grows, both effects shrink.

Capillary rise

Dip a narrow glass tube — a capillary, from the Latin for "hair" — into water. The water rises inside it, well above the level outside, and the narrower the tube the higher it climbs. In mercury, a glass capillary does the opposite: the mercury inside stands lower than outside.

Here is why. Water wets glass ($\theta\approx0^\circ$), so its surface in the tube is curved, concave upwards — a hemisphere if $\theta=0$, of the same radius $r$ as the tube. More generally the meniscus is part of a sphere of radius $R=\dfrac{r}{\cos\theta}$. By Result 9.12, the pressure just below the meniscus (on its convex side) is lower than the atmospheric pressure above it by $\dfrac{2S}{R}=\dfrac{2S\cos\theta}{r}$. Liquid at the level of the outside surface is at $p_a$ (Law 9.1), so water is pushed up the tube until the column of height $h$ supplies the difference:

$$\rho gh=\frac{2S\cos\theta}{r}.$$

The same result comes from a force balance: the surface tension pulls up round the rim of the meniscus with a vertical component $(2\pi r)S\cos\theta$, which holds up the weight of the column, $\pi r^2h\rho g$ (neglecting the small amount of liquid in the curved meniscus itself).

📐 Result 9.13 — Capillary rise

A liquid of density $\rho$, surface tension $S$ and angle of contact $\theta$ rises in a clean vertical capillary tube of radius $r$ to a height

$$h=\frac{2S\cos\theta}{\rho gr}$$

above the outside level (for a narrow tube, so that the meniscus is spherical). If $\theta\gt90^\circ$, $\cos\theta\lt0$ and the liquid is depressed below the outside level by $|h|$ — mercury in glass. Note $h\propto\dfrac1r$: halve the radius, double the rise.

Capillarity draws water up the threads of a towel, ink into the nib of a fountain pen, oil up a lamp wick and water through the tiny pores between soil particles to plant roots; bricks without a damp-proof layer let ground water creep up walls. Blotting paper works because its fibres form a maze of fine capillaries.

✏️ Example 5 — Water in a glass capillary
A clean glass capillary of internal radius $0.20\ \text{mm}$ stands vertically in water ($S=0.073\ \text{N/m}$, $\theta=0^\circ$). (a) How high does the water rise? (b) Check the answer with a force balance. (c) How high would it rise in a tube of radius $0.10\ \text{mm}$?
(a) $h=\dfrac{2(0.073)(1)}{1000(9.8)(2.0\times10^{-4})}=\dfrac{0.146}{1.96}=0.074\ \text{m}=7.4\ \text{cm}$.
(b) Upward pull: $2\pi rS=2\pi(2.0\times10^{-4})(0.073)=9.2\times10^{-5}\ \text{N}$. Column weight: $\pi r^2h\rho g=\pi(2.0\times10^{-4})^2(0.0745)(1000)(9.8)=9.2\times10^{-5}\ \text{N}$ ✓.
(c) Half the radius: $14.9\ \text{cm}\approx15\ \text{cm}$.
Check $hr=1.49\times10^{-5}\ \text{m}^2$ for water in clean glass, whatever the tube — a handy number. This is the default of Interactive 2.
Interactive 2 — Capillary Rise and Tube Radius

Four fixed glass tubes (radii $0.10$, $0.20$, $0.40$ and $0.80\ \text{mm}$, drawn wider than scale) and one highlighted test tube whose radius you set stand in a dish. Pick the liquid: water and ethanol wet glass and rise, mercury does not and is pushed down, and water in a waxed tube ($\theta=105^\circ$) is depressed too. The graph shows $h=\dfrac{2S\cos\theta}{\rho gr}$ — a hyperbola. Press Play to sweep the test tube's radius.

0.20 mm
height $h$ (test tube)—
$h\times r$—
pull $2\pi rS\cos\theta$—
meniscus radius $r/\cos\theta$—
✏️ Example 6 — Mercury in a glass tube
A glass tube of internal radius $0.50\ \text{mm}$ dips vertically into mercury ($S=0.465\ \text{N/m}$, $\rho=13.6\times10^3\ \text{kg/m}^3$, $\theta=140^\circ$). By how much does the mercury in the tube stand above or below the outside level?
Formula $h=\dfrac{2S\cos\theta}{\rho gr}$ with $\cos140^\circ=-0.766$.
Numbers $h=\dfrac{2(0.465)(-0.766)}{13\,600(9.8)(5.0\times10^{-4})}=\dfrac{-0.712}{66.6}=-0.011\ \text{m}$.
Answer The mercury is depressed by about $1.1\ \text{cm}$, with a convex meniscus.
Check This is why the reading of a narrow mercury barometer must be corrected upward, and why barometer tubes are made wide (about $1\ \text{cm}$ across), where the depression is only about a millimetre.
⚠️ Common mistakes

1. One surface or two? A drop, and an air bubble inside a liquid, have one surface: $2S/r$. A soap bubble and a soap film have two: $4S/r$ and $2Sl$.

2. Radius or diameter? Every formula here uses the radius. A tube "of diameter $0.40\ \text{mm}$" has $r=0.20\ \text{mm}$.

3. Units. $S$ in $\text{N/m}$, $r$ in metres. With $r$ in millimetres you get heights a thousand times too small.

4. Forgetting $\cos\theta$ — or its sign. For $\theta\gt90^\circ$ the "rise" is a depression.

5. "The liquid overflows from a short tube." It cannot: in a tube shorter than $h$, the liquid reaches the top and the meniscus flattens (its radius grows) until $\rho gh'=\dfrac{2S}{R'}$. There is no fountain — that would be perpetual motion.

6. Surface tension is not a force. It is a force per unit length; multiply by the length of the edge it acts along.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — two soap bubbles

In JEE Main, NEET and JEE Advanced (as part of excess pressure in drops and bubbles); not required by the CBSE syllabus.

JEE problems often combine two soap bubbles of radii $r_1\lt r_2$ (surface tension $S$).

Joined by a tube. The smaller bubble has the larger excess pressure, $\dfrac{4S}{r_1}\gt\dfrac{4S}{r_2}$, so air flows from the small bubble into the large one: the small bubble shrinks and the large one grows, the opposite of what most people expect. (The same instability is why the lungs need a surfactant, as the Research–Teaching Nexus below explores.)

Stuck together. Two bubbles in contact share a common film. The pressure in the smaller bubble exceeds the outside by $\dfrac{4S}{r_1}$, in the larger by $\dfrac{4S}{r_2}$; the common film must support the difference, so its radius $r$ satisfies $\dfrac{4S}{r}=\dfrac{4S}{r_1}-\dfrac{4S}{r_2}$:

$$r=\frac{r_1r_2}{r_2-r_1},$$

curving into the larger bubble. Example. Bubbles of radii $3.0\ \text{cm}$ and $6.0\ \text{cm}$: $r=\dfrac{3.0\times6.0}{3.0}=6.0\ \text{cm}$. Equal bubbles share a flat film ($r\to\infty$).

Exercises

Take $g=9.8\ \text{m/s}^2$, $\rho_\text{water}=1.00\times10^3\ \text{kg/m}^3$, $S_\text{water}=0.073\ \text{N/m}$ (angle of contact with clean glass $0^\circ$), and for mercury $S=0.465\ \text{N/m}$, $\rho=13.6\times10^3\ \text{kg/m}^3$, $\theta=140^\circ$ with glass.

I
Surface tension and surface energy.
  • 1 Find the dimensions of surface tension and show that $\text{N/m}$ and $\text{J/m}^2$ are the same unit.
  • 2 A steel needle $4.0\ \text{cm}$ long floats on water. Find the largest upward force surface tension can exert on it (the surface pulls along both sides), and the largest mass of needle it could support, ignoring buoyancy.
  • 3 The sliding wire of a soap-film frame is $6.0\ \text{cm}$ long and $S=0.025\ \text{N/m}$. Find the force needed to hold it, and the work done in pulling it out $1.5\ \text{cm}$.
  • 4 A thin wire ring of radius $2.0\ \text{cm}$ lies flat on a water surface. Estimate the extra force, beyond its weight, needed to pull it free (the water surface clings to both the inside and the outside of the ring).
  • 5 Explain, in terms of energy, why small drops of liquid are spherical.
  • 6 Find the work needed to blow a soap bubble of radius $5.0\ \text{cm}$ ($S=0.030\ \text{N/m}$).
  • 7 A mercury drop of radius $3.0\ \text{mm}$ is split into $27$ equal droplets. Find the energy needed.
1 $[S]=\dfrac{[\text{M L T}^{-2}]}{[\text{L}]}=[\text{M T}^{-2}]$; $\dfrac{\text{J}}{\text{m}^2}=\dfrac{\text{N m}}{\text{m}^2}=\dfrac{\text{N}}{\text{m}}$ ✓.
2 $F=2Sl=2(0.073)(0.040)=5.8\times10^{-3}\ \text{N}$; $m=\dfrac{F}{g}=6.0\times10^{-4}\ \text{kg}=0.60\ \text{g}$.
3 $F=2Sl=2(0.025)(0.060)=3.0\times10^{-3}\ \text{N}$; $W=Fd=4.5\times10^{-5}\ \text{J}$.
4 Total edge length $2\times2\pi r$: $F=4\pi rS=4\pi(0.020)(0.073)=1.8\times10^{-2}\ \text{N}$.
5 The surface energy is $S\times$area, and a liquid settles into the state of least energy. For a fixed volume, the sphere has the least surface area; for small drops gravity is negligible compared with surface effects.
6 Two surfaces: $W=S\times2\times4\pi r^2=0.030\times8\pi(0.050)^2=1.9\times10^{-3}\ \text{J}$.
7 Droplet radius $\dfrac{3.0}{27^{1/3}}=1.0\ \text{mm}$. $\Delta A=4\pi[27(1.0\times10^{-3})^2-(3.0\times10^{-3})^2]=2.26\times10^{-4}\ \text{m}^2$; $W=0.465\times2.26\times10^{-4}=1.1\times10^{-4}\ \text{J}$.
II
Angle of contact and excess pressure.
  • 1 Define the angle of contact. Give one liquid–solid pair with an acute angle and one with an obtuse angle.
  • 2 Why does water wet clean glass while mercury does not?
  • 3 Explain how detergents help to clean greasy clothes.
  • 4 Find the excess pressure inside a water drop of radius $0.50\ \text{mm}$.
  • 5 Find the excess pressure inside a soap bubble of radius $1.5\ \text{cm}$ ($S=0.030\ \text{N/m}$).
  • 6 Two soap bubbles of the same solution have radii $1.0\ \text{cm}$ and $2.0\ \text{cm}$. Which has the greater excess pressure, and by what factor?
  • 7 An air bubble of radius $0.10\ \text{mm}$ is $20\ \text{cm}$ below the surface of water. Find the pressure inside it ($p_a=1.013\times10^5\ \text{Pa}$).
  • 8 The excess pressure inside a soap bubble ($S=0.025\ \text{N/m}$) is $4.0\ \text{Pa}$. Find its radius.
1 The angle between the solid surface and the tangent to the liquid surface at the line of contact, measured through the liquid. Acute: water on clean glass ($\approx0^\circ$). Obtuse: mercury on glass ($\approx140^\circ$).
2 Water molecules are attracted to glass more strongly than to each other (adhesion exceeds cohesion), so water spreads; mercury atoms attract each other much more strongly than they attract glass, so mercury pulls itself into beads.
3 Detergent lowers the surface tension of water and its angle of contact with fabric, so the water spreads into the fine spaces between fibres; detergent molecules also attach to grease (one end) and to water (the other), so grease is lifted off in small particles and rinsed away.
4 $\dfrac{2S}{r}=\dfrac{2(0.073)}{5.0\times10^{-4}}=2.9\times10^2\ \text{Pa}$ ($292$).
5 $\dfrac{4S}{r}=\dfrac{4(0.030)}{0.015}=8.0\ \text{Pa}$.
6 The smaller one, by a factor of $2$ ($\Delta p\propto\tfrac1r$).
7 $1.013\times10^5+1000(9.8)(0.20)+\dfrac{2(0.073)}{1.0\times10^{-4}}=1.013\times10^5+1960+1460=1.047\times10^5\ \text{Pa}$.
8 $r=\dfrac{4S}{\Delta p}=\dfrac{0.10}{4.0}=0.025\ \text{m}=2.5\ \text{cm}$.
III
Capillary rise.
  • 1 How high does water rise in a clean glass capillary of internal diameter $0.50\ \text{mm}$?
  • 2 Water rises $3.0\ \text{cm}$ in a certain capillary. How high would it rise in a tube of half the radius? Of twice the radius?
  • 3 Find the depression of mercury in a glass tube of radius $1.0\ \text{mm}$.
  • 4 A liquid of density $800\ \text{kg/m}^3$ that wets glass ($\theta=0^\circ$) rises $3.0\ \text{cm}$ in a capillary of radius $0.25\ \text{mm}$. Find its surface tension.
  • 5 Explain why a towel soaks up water and why the oil in a lamp rises up the wick.
  • 6 Water rises $3.0\ \text{cm}$ in a capillary on the Earth. How high would it rise in the same tube on the Moon, where $g$ is one-sixth as large?
  • 7 For the tube of Example 5, show that the upward surface-tension force on the column equals its weight, and find that force.
1 $r=0.25\ \text{mm}$: $h=\dfrac{2(0.073)}{1000(9.8)(2.5\times10^{-4})}=0.060\ \text{m}=6.0\ \text{cm}$.
2 $h\propto\tfrac1r$: $6.0\ \text{cm}$ and $1.5\ \text{cm}$.
3 $h=\dfrac{2(0.465)(-0.766)}{13\,600(9.8)(1.0\times10^{-3})}=-5.3\times10^{-3}\ \text{m}$: depressed by $5.3\ \text{mm}$.
4 $S=\dfrac{h\rho gr}{2\cos\theta}=\dfrac{0.030(800)(9.8)(2.5\times10^{-4})}{2}=2.9\times10^{-2}\ \text{N/m}$.
5 The gaps between the fibres of the towel or the wick act as very fine capillary tubes that the liquid wets, so surface tension draws the liquid up into them, against gravity.
6 $h\propto\tfrac1g$: $6\times3.0=18\ \text{cm}$.
7 $2\pi rS=\pi r^2h\rho g$ is exactly $h=\dfrac{2S}{\rho gr}$ rearranged. Force $=2\pi(2.0\times10^{-4})(0.073)=9.2\times10^{-5}\ \text{N}$.
IV
Applications and harder problems.
  • 1 (Modelling) The water-conducting vessels (xylem) of a tree have radii of about $10\ \mu\text{m}$. How high could capillarity alone raise water in them? Can it explain how water reaches the top of a $30\ \text{m}$ tree?
  • 2 Two clean glass capillaries of radii $0.30\ \text{mm}$ and $0.60\ \text{mm}$ stand side by side in water. Find the difference between the water levels in them.
  • 3 A soap bubble of radius $4.0\ \text{cm}$ ($S=0.030\ \text{N/m}$) and a water drop have the same excess pressure. Find the radius of the water drop.
  • 4 A soap bubble ($S=0.030\ \text{N/m}$) is blown up from a radius of $2.0\ \text{cm}$ to $4.0\ \text{cm}$. Find the work done against surface tension.
  • 5 A capillary tube is $5.0\ \text{cm}$ long above the water surface, but water would rise $7.4\ \text{cm}$ in it. Describe what happens and find the radius of curvature of the meniscus if the tube's radius is $0.20\ \text{mm}$.
1 $h=\dfrac{2(0.073)}{1000(9.8)(1.0\times10^{-5})}=1.5\ \text{m}$ — far short of $30\ \text{m}$. Trees rely mainly on the "pull" of water evaporating from the leaves, transmitted down unbroken columns of water.
2 $h_1=\dfrac{0.146}{9800\times3.0\times10^{-4}}=4.97\ \text{cm}$, $h_2=2.48\ \text{cm}$; difference $2.5\ \text{cm}$, higher in the narrower tube.
3 $\dfrac{4(0.030)}{0.040}=3.0\ \text{Pa}=\dfrac{2(0.073)}{r}$, so $r=4.9\ \text{cm}$.
4 $W=S\times2\times4\pi(r_2^2-r_1^2)=0.030\times8\pi(0.0016-0.0004)=9.0\times10^{-4}\ \text{J}$.
5 The water rises to the top of the tube and stops; it does not overflow. The meniscus flattens so that $\rho g(0.050)=\dfrac{2S}{R}$: $R=\dfrac{2(0.073)}{1000(9.8)(0.050)}=3.0\times10^{-4}\ \text{m}=0.30\ \text{mm}$, larger than the tube radius ($hR$ is constant: $7.45\times0.20=5.0\times R$).
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E1. Board · 3 marks Derive an expression for the excess pressure inside a liquid drop of radius $r$ and surface tension $S$. Hence write the excess pressure inside a soap bubble, giving the reason for the difference.

Set-up Let the radius increase from $r$ to $r+\Delta r$ under the excess pressure $p_i-p_o$. Increase in surface area $8\pi r\Delta r$, so increase in surface energy $S\,8\pi r\Delta r$. [1]
Work Work done by the excess pressure $=(p_i-p_o)4\pi r^2\Delta r$. Equating: $p_i-p_o=\dfrac{2S}{r}$. [1]
Bubble A soap bubble has two surfaces (inner and outer), so the surface energy and the excess pressure are doubled: $\dfrac{4S}{r}$. [1]
Exam tip State clearly that the work done by the pressure difference is stored as surface energy (temperature constant).

E2. NEET Water rises to a height of $6.0\ \text{cm}$ in a capillary tube. The tube is pushed down until only $4.0\ \text{cm}$ of it stands above the water surface. The water in the tube will (a) overflow from the top as a small fountain (b) rise to the top and stay there, its meniscus becoming flatter (c) fall to $2.0\ \text{cm}$ (d) not rise at all

Idea The rise is set by $\rho gh=\dfrac{2S}{R}$, with $R$ the meniscus radius. With only $4.0\ \text{cm}$ available, the meniscus adjusts to a larger radius $R'=\dfrac{6.0}{4.0}R=1.5R$ and the column stops at the top.
Answer (b).
Exam tip A capillary fountain would deliver energy for free, forever — reject (a) on energy grounds alone.

E3. JEE Main (Numerical value) A drop of liquid of radius $1.0\ \text{cm}$ and surface tension $0.070\ \text{N/m}$ breaks up into $1000$ identical droplets at constant temperature. The work done, in units of $10^{-4}\ \text{J}$, is closest to which integer?

Set-up $r=\dfrac{R}{1000^{1/3}}=1.0\ \text{mm}$. $\Delta A=4\pi\left[1000(10^{-3})^2-(10^{-2})^2\right]=4\pi(9.0\times10^{-4})=1.13\times10^{-2}\ \text{m}^2$.
Answer $W=S\Delta A=0.070\times1.13\times10^{-2}=7.9\times10^{-4}\ \text{J}$: $8$.
Exam tip Shortcut: $W=4\pi R^2S\left(n^{1/3}-1\right)=4\pi(10^{-4})(0.070)(9)$.

E4. JEE Advanced (One or more correct) A clean glass capillary tube of radius $r$ stands vertically in water ($\theta=0^\circ$), and the water rises to a height $h$. Which statements are correct? (A) In a tube of radius $2r$ the rise would be $\dfrac h2$. (B) If the tube is tilted to make $60^\circ$ with the vertical (and is long enough), the length of the water column in it becomes $2h$. (C) In a freely orbiting space station, water would creep to the top of any length of such a tube. (D) The rise $h$ is the same whatever the temperature of the water.

(A) $h\propto\tfrac1r$ ✓.
(B) The vertical height is still $h$ (it is fixed by $\rho gh=\tfrac{2S}{r}$), so the length along the tube is $\dfrac{h}{\cos60^\circ}=2h$ ✓.
(C) With effective $g=0$ there is no weight to balance the surface-tension pull, so the water rises until it reaches the top of the tube ✓ (it still does not overflow).
(D) $S$ falls with temperature, so hot water rises less ✗.
Answer (A), (B), (C).
Exam tip In tilted-tube questions, height and length are different things — the same trap as the tilted barometer of Section 9.1.

E5. JEE Main NEET The air pressures inside two soap bubbles in the open air are $1.01\ \text{atm}$ and $1.02\ \text{atm}$ (atmospheric pressure $1.00\ \text{atm}$). The ratio of their volumes (first : second) is (a) $2:1$ (b) $8:1$ (c) $1:8$ (d) $4:1$

Idea Excess pressures $0.01$ and $0.02\ \text{atm}$. $\Delta p=\dfrac{4S}{r}$, so $r\propto\dfrac{1}{\Delta p}$: $r_1:r_2=2:1$.
Answer $V\propto r^3$: $8:1$, (b).
Exam tip Use the excess pressures, not $1.01:1.02$. (c) inverts the ratio — the bubble with lower pressure is the bigger one.

E6. Board · Case study A student dips a clean glass capillary of internal radius $0.30\ \text{mm}$ vertically into water ($S=0.072\ \text{N/m}$, $\theta=0^\circ$). (i) Calculate the height to which the water rises. [2] (ii) The inside of the tube is now coated with wax, for which the angle of contact of water is greater than $90^\circ$. What happens? [1] (iii) Why does the water rise less when the experiment is repeated with hot water? [1]

(i) $h=\dfrac{2S\cos\theta}{\rho gr}=\dfrac{2(0.072)}{1000(9.8)(3.0\times10^{-4})}=0.049\ \text{m}=4.9\ \text{cm}$. [2]
(ii) $\cos\theta\lt0$: the meniscus becomes convex and the water level in the tube falls below the outside level. [1]
(iii) Surface tension decreases as temperature rises, and $h\propto S$. [1]
Exam tip Write the formula with $\cos\theta$ even when $\theta=0$: it earns the method mark and sets up part (ii).
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Why do the lungs need a special "surfactant" — and why are babies born very early so often short of breath?

Why it matters. The lungs contain about $300$ million tiny air sacs (alveoli), each lined with a thin film of watery liquid. Surface tension in that film tends to collapse them, and by Result 9.12 the effect is strongest in the smallest sacs. Healthy lungs secrete a surfactant that lowers the surface tension. Babies born before about $32$ weeks often lack it and develop respiratory distress syndrome; giving them artificial surfactant, introduced in the 1990s, has saved many lives in neonatal units around the world.

What to do. Model an alveolus as a sphere of radius $0.10\ \text{mm}$ lined with liquid, with a single liquid–air surface on the inside, so that $\Delta p=\dfrac{2S}{r}$.

  1. Find the pressure difference needed to keep an alveolus open if its lining had the surface tension of water, $0.070\ \text{N/m}$. Express it in Pa and in mm of mercury ($1\ \text{mm Hg}=133\ \text{Pa}$).
  2. The breathing muscles can comfortably produce a pressure difference of only a few mm Hg in normal breathing. Compare.
  3. With surfactant, $S$ falls to about $0.005\ \text{N/m}$ when the sac is small. Repeat part 1.
  4. Two connected alveoli of radii $0.05\ \text{mm}$ and $0.10\ \text{mm}$ have the same lining. Without surfactant, which way would air flow? Why would this be dangerous? (See the extension box on bubbles.)
  5. Surfactant lowers $S$ more as the sac shrinks (the surfactant molecules crowd together). Explain how this keeps small and large alveoli stable together.

Tools: Calculator; two soap bubbles blown on the ends of a T-shaped straw with a clip in the middle, to watch the small bubble empty into the large one; an introductory physiology text or a reliable medical website for the lung data.

Hand in: Your calculations, a labelled sketch of the two-bubble demonstration with an explanation, and a half-page note for parents explaining in plain language what surfactant treatment does.

Going further. Find out how surfactant is produced in the lungs, why its production starts late in pregnancy, and how antenatal steroid injections help.

Instructor note: water-like lining: $\Delta p=\dfrac{2(0.070)}{1.0\times10^{-4}}=1.4\times10^3\ \text{Pa}\approx10.5\ \text{mm Hg}$ — several times the pressure differences of quiet breathing (about $1$–$3\ \text{mm Hg}$). With $S=0.005$: $100\ \text{Pa}\approx0.75\ \text{mm Hg}$, easily managed. Connected sacs: the $0.05\ \text{mm}$ sac has twice the excess pressure, so it empties into the larger one and collapses — the instability surfactant prevents, because a shrinking sac's $S$ falls faster than its radius.

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of surface tension is

Q2. The excess pressure inside a soap bubble of radius $r$ and surface tension $S$ is

Q3. Water rises $4.0\ \text{cm}$ in a capillary tube. In a tube of the same glass with half the radius it rises

Q4. When a glass capillary is dipped in mercury, the mercury in the tube

Q5. Adding detergent to water

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