Drops and bubbles

Blow a soap bubble and watch it drift, perfectly round, catching rainbow colours on its skin. Nobody pushes it into that shape — surface tension does, pulling the film tight like a stretched balloon on every side at once. A stretched balloon skin also squeezes whatever is inside it to a pressure a little higher than outside. That "little higher" is not a vague idea; it has an exact formula, and the formula for a plain drop is not the same as the formula for a soap bubble, even though both look like simple spheres of liquid film. The difference comes down to counting surfaces. A raindrop has liquid inside and air outside — one curved liquid–air boundary. A soap bubble is a thin sheet of soapy water with air on both sides — two curved boundaries, each squeezing independently. Get that counting wrong in an exam and you use $4S/r$ where $2S/r$ was needed, or the reverse, and the numeric answer collapses even though every other step was correct. This article builds the excess-pressure formulas for a drop, a bubble, and an air bubble trapped inside a liquid, from a single force-balance argument on a hemisphere — no new physics beyond surface tension and simple mechanics. Along the way you will see why smaller drops and bubbles always carry higher excess pressure than larger ones: connect two soap bubbles of different sizes with a tube and the small one empties into the big one, never the other way round. By the end you should be able to derive both formulas, tell instantly which one a problem needs, and combine excess pressure with atmospheric or hydrostatic pressure to get an absolute pressure — the calculation CBSE actually asks for.

Key takeaways

A curved liquid surface, unlike a flat one, sustains a pressure difference across it because surface tension has a net inward pull toward the centre of curvature.
A liquid drop (or an air bubble inside a liquid) has one curved surface: excess pressure $\Delta P = \dfrac{2S}{r}$ .
A soap bubble in air has two curved surfaces (outer film face and inner film face): excess pressure $\Delta P = \dfrac{4S}{r}$ — double the drop value for the same $r$ and $S$ .
Excess pressure is inversely proportional to radius, $\Delta P \propto 1/r$ , so smaller drops and bubbles have higher internal excess pressure than larger ones of the same liquid.
$\Delta P$ is only the excess; the absolute pressure inside always needs atmospheric pressure (and hydrostatic pressure, if the bubble sits underwater) added on top.
Connecting two unequal soap bubbles by a tube makes the smaller one shrink and empty into the larger one, because the smaller bubble is at the higher pressure.
Key idea: a curved liquid surface squeezes its contents to an excess pressure of 2S/r for one surface (a drop, or a bubble inside a liquid) or 4S/r for two surfaces (a soap bubble in air), and this excess must be added to the surrounding atmospheric or hydrostatic pressure to get the absolute pressure inside.

The core insight

Surface tension curves a liquid–air surface into a state of excess internal pressure equal to $2S/r$ per curved surface, so a one-surface drop or submerged bubble has $\Delta P = 2S/r$ while a two-surface soap bubble in air has $\Delta P = 4S/r$ , and smaller radii always mean larger excess pressure.

Where it fits in the syllabus

This is part of the CBSE Class 11 Physics unit Mechanical Properties of Fluids, sitting inside the surface-tension section on "excess pressure across a curved surface." The board expects you to state and apply $\Delta P = 2S/r$ for a drop and $\Delta P = 4S/r$ for a soap bubble, explain the factor-of-two difference between them physically (not just recite it), and solve numerical problems that combine this excess pressure with atmospheric or hydrostatic pressure — including the classic case of a bubble formed inside a liquid rather than in open air. Only the algebraic force-balance derivation is prescribed; nothing here needs calculus or the Young–Laplace equation by name.

The physical picture

A perfectly flat liquid surface carries surface tension forces that all lie in the plane of the surface, pulling sideways on any small patch but never pushing it inward or outward — so a flat surface sustains no pressure difference across it. Curve that surface into a sphere and the picture changes: the tension forces around the rim of any patch now have a component pointing toward the centre of the sphere, because the patch is tilted relative to its neighbours. Summed over the whole surface, this net inward pull is exactly balanced by a slightly higher pressure on the concave (inner) side pushing outward. That balance is what fixes the excess pressure.

Liquid drop, or a bubble of air inside a liquid — one curved surface. A spherical drop of radius $r$ and surface tension $S$ has only a single liquid–air interface, so its internal pressure $P_i$ exceeds the external pressure $P_o$ by

\Delta P = P_i - P_o = \frac{2S}{r}

where: $S$ is the surface tension of the liquid forming the curved surface (in N/m), $r$ is the radius of curvature of that surface (in m), and $\Delta P$ is the excess pressure (in Pa) felt on the concave side. The same formula applies to an air bubble trapped inside a liquid, because it too has just one curved liquid–air surface — the liquid is on the outside this time, but the geometry and the force balance are identical, so $S$ in the formula is the surface tension of the surrounding liquid.

Soap bubble in air — two curved surfaces. A soap bubble is a thin spherical film of soap solution with air both inside and outside it. That film therefore has two liquid–air interfaces — an outer one bulging into the outside air, and an inner one bulging into the trapped air — each independently contributing an excess pressure of $2S/r$ . The two add, so the bubble's total excess pressure is doubled:

\Delta P = \frac{4S}{r}

where: the symbols are the same as above, and the film is thin enough that both surfaces share essentially the same radius $r$ . Physically, doubling makes sense once you picture the film as two back-to-back "drop skins," each squeezing the trapped air by its own $2S/r$ .

Absolute pressure — the step examiners actually test. $\Delta P$ is never itself the pressure inside the drop or bubble; it is the extra pressure over whatever pressure already exists outside. For a drop or bubble sitting in open air at atmospheric pressure $P_0$ , the absolute internal pressure is $P_i = P_0 + \Delta P$ . For a bubble formed at depth $h$ inside a liquid of density $\rho$ , the outside pressure itself is the hydrostatic pressure $P_0 + \rho g h$ (from the earlier pressure-with-depth result), so $P_i = P_0 + \rho g h + \Delta P$ — atmospheric pressure, the weight of the liquid column above, and the curvature term all stack together.

Smaller means higher pressure. Because $\Delta P \propto 1/r$ , halving the radius doubles the excess pressure. This single proportionality explains a familiar demonstration: if a small soap bubble and a large one are connected by a short tube, the smaller bubble — being at the higher pressure — pushes air into the larger one and shrinks, sometimes until it vanishes, while the large bubble swells further. The system moves toward the state with the smallest net excess pressure, and a single large bubble stores less surface energy per unit volume than two separate small ones.

Figure

Three circles side by side comparing a liquid drop with one curved surface and excess pressure two S over r, a soap bubble with two curved surfaces shown as concentric circles and excess pressure four S over r, and two connected soap bubbles of unequal size with an arrow showing air flowing from the smaller higher pressure bubble into the larger lower pressure bubble

Derivation AExcess pressure inside a drop from a hemisphere force balance

The goal is to derive $\Delta P = 2S/r$ for a spherical liquid drop of radius $r$ and surface tension $S$ using nothing beyond a force balance on half the drop, and then to extend the same reasoning in one further step to get $\Delta P = 4S/r$ for a soap bubble, where the doubled surface count is the only new ingredient. No calculus is needed: the whole argument rests on two facts — pressure exerts a force equal to pressure times area, and surface tension exerts a force equal to tension times the length of the line along which it acts.

Step 1. Imagine slicing the spherical drop exactly in half with an imaginary flat plane through its centre, giving two identical hemispherical shells, and focus on the forces acting on just one of them. Because the drop is in equilibrium as a whole, each hemisphere must itself be in force balance along the direction perpendicular to the cutting plane — the two halves neither fly apart nor collapse into each other.

Step 2. Identify the outward force trying to separate the two hemispheres. The excess pressure $\Delta P$ acts on the flat circular face of the hemisphere, which has area $\pi r^2$ , so the net outward push from the excess internal pressure is $\Delta P \times \pi r^2$ — the same pressure-times-area idea used for a pressurised tank wall.

Step 3. Identify the inward force holding the two hemispheres together. Surface tension acts along the rim where the cutting plane meets the drop's surface — a circle of circumference $2\pi r$ — pulling the film on each side toward the other with a force per unit length equal to $S$ . The total inward pull along this rim is therefore $S \times 2\pi r$ .

Step 4. Equate the two forces, since the hemisphere is in equilibrium: $\Delta P \times \pi r^2 = S \times 2\pi r$ . Both sides contain a factor of $\pi r$ ; cancelling it leaves $\Delta P \times r = 2S$ , and rearranging gives the drop formula, $\Delta P = \dfrac{2S}{r}$ .

Step 5. Extend to a soap bubble by re-examining what surface tension pulls along the rim. A drop has liquid only on the inside of its one surface, so exactly one ring of film crosses the cutting plane. A soap bubble is a thin film with an outer and an inner surface both crossing the cutting plane at essentially the same radius $r$ , so the tension force at the rim is doubled: $2 \times (S \times 2\pi r)$ instead of $S \times 2\pi r$ . Repeating Step 4 with this doubled tension, $\Delta P \times \pi r^2 = 2S \times 2\pi r$ , gives $\Delta P = \dfrac{4S}{r}$ — exactly twice the drop result, because twice as many surfaces resist the split. Verified: setting $S = 1$ and $r = 1$ in consistent units gives $\Delta P = 2$ for the drop and $\Delta P = 4$ for the bubble, matching the stated ratio of exactly 2, and the algebra reduces cleanly with no leftover factors of $\pi$ on either side.

Worked examples

Board level — excess and absolute pressure inside a soap bubble

A soap bubble of radius $2.0\ \text{mm}$ is blown from a soap solution of surface tension $S = 3.0\times10^{-2}\ \text{N/m}$ . Taking atmospheric pressure $P_0 = 1.013\times10^{5}\ \text{Pa}$ , find (a) the excess pressure inside the bubble, and (b) the absolute pressure of the air inside it.

Step 1. Convert the radius to SI units: $r = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}$ . A soap bubble in open air has two curved surfaces, so the bubble formula $4S/r$ applies, not the drop formula.

Step 2. Apply the bubble formula. $\Delta P = \dfrac{4S}{r} = \dfrac{4 \times 3.0\times10^{-2}}{2.0\times10^{-3}} = \dfrac{0.12}{2.0\times10^{-3}} = 60\ \text{Pa}$ .

Step 3. Add atmospheric pressure to get the absolute pressure inside. $P_{\text{inside}} = P_0 + \Delta P = 1.013\times10^{5} + 60 = 1.0136\times10^{5}\ \text{Pa}$ . Answer: excess pressure $= 60\ \text{Pa}$ ; absolute pressure inside $\approx 1.0136\times10^{5}\ \text{Pa}$ . The excess pressure is tiny next to atmospheric pressure — about $0.06\%$ of it — which is exactly why soap bubbles are so fragile and easily deformed by the faintest breeze.

Competitive level — an air bubble formed underwater

An air bubble of radius $0.5\ \text{mm}$ is formed inside a beaker of water (surface tension of water $S = 7.2\times10^{-2}\ \text{N/m}$ ) at a depth $h = 10\ \text{cm}$ below the free surface. Taking $P_0 = 1.013\times10^{5}\ \text{Pa}$ , $\rho = 1000\ \text{kg/m}^3$ , and $g = 9.8\ \text{m/s}^2$ , find the absolute pressure of the air inside the bubble.

Step 1. Find the pressure of the water immediately outside the bubble, which is the hydrostatic pressure at depth $h$ : $P_{\text{outside}} = P_0 + \rho g h = 1.013\times10^5 + (1000)(9.8)(0.10) = 1.013\times10^5 + 980 = 1.0228\times10^{5}\ \text{Pa}$ .

Step 2. Find the excess pressure due to the bubble's curvature. This is an air bubble surrounded by liquid, so it has a single curved surface just like a drop — the $2S/r$ formula, not $4S/r$ . With $r = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}$ : $\Delta P = \dfrac{2S}{r} = \dfrac{2 \times 7.2\times10^{-2}}{5\times10^{-4}} = \dfrac{0.144}{5\times10^{-4}} = 288\ \text{Pa}$ .

Step 3. Add the excess pressure on top of the surrounding hydrostatic pressure, since the bubble's own curvature is an additional squeeze beyond whatever the water around it is already at. $P_{\text{inside}} = P_{\text{outside}} + \Delta P = 1.0228\times10^{5} + 288 = 1.02568\times10^{5}\ \text{Pa} \approx 1.026\times10^{5}\ \text{Pa}$ . Answer: absolute pressure inside the bubble $\approx 1.026\times10^{5}\ \text{Pa}$ . Timing note: under two minutes once two sub-skills are automatic — reading off $\rho g h$ from the depth-pressure formula, and picking $2S/r$ (not $4S/r$ ) because the surrounding water gives this bubble only one curved surface. A common slip under time pressure is to reach for $4S/r$ out of habit whenever the word "bubble" appears; always check first whether it is a soap film (two surfaces, air both sides) or a gas pocket in a liquid (one surface, liquid outside only).

Common traps

Using $4S/r$ for a plain drop or an air bubble inside a liquid — the factor of 4 applies only to a soap bubble (or any thin film) that has air/gas on both its inner and outer faces, giving two curved surfaces, not one.
Forgetting to add atmospheric (or hydrostatic) pressure to get the absolute pressure inside — $\Delta P = 2S/r$ or $4S/r$ gives only the excess pressure over the surrounding pressure, never the absolute pressure by itself.
Assuming bigger bubbles have higher internal pressure — it is the opposite; excess pressure is inversely proportional to radius, so smaller bubbles and drops always carry the larger excess pressure.
Mixing up which radius to use when a bubble is blown at the end of a tube versus floating freely, or when a bubble sits at some depth — always use the bubble's own current radius of curvature, and remember hydrostatic pressure depends on depth, not on the bubble's size.

Common trapThe word "bubble" does not automatically mean 4S over r

The single biggest mark-loser in this topic is reaching for $\Delta P = 4S/r$ every time a question uses the word "bubble," without checking what kind of bubble it is. A soap bubble in air genuinely has two curved surfaces (outer film face, inner film face) and needs $4S/r$ . But an air bubble formed inside a liquid — water, mercury, any liquid — has only one curved surface, the same as a drop, because there is liquid outside and gas inside, with no second film. That bubble needs $2S/r$ , using the surface tension of the surrounding liquid. Before writing any formula, ask: is there liquid film with gas on both sides (two surfaces, $4S/r$ ), or just one liquid–gas boundary (one surface, $2S/r$ )? Getting this identification wrong is a conceptual error no amount of correct arithmetic afterward can fix.

You should be able to…

Derive, via the hemisphere force-balance argument, the excess pressure formulas for a drop ( $2S/r$ ) and a soap bubble ( $4S/r$ ), and explain in words why the bubble formula carries an extra factor of two.
Correctly identify which formula a given "bubble" needs — a soap film in air (two surfaces, $4S/r$ ) versus an air bubble trapped inside a liquid (one surface, $2S/r$ ).
Combine excess pressure with atmospheric pressure, and with hydrostatic pressure when the drop or bubble is submerged, to compute the absolute pressure inside.
State and apply the inverse relationship between excess pressure and radius, and use it to explain what happens when two unequal bubbles are connected.
Set up and solve full numerical problems that chain radius conversion, the correct excess-pressure formula, and pressure addition into one final absolute-pressure answer.

Ask the tutor: want to see the two-connected-bubbles problem solved numerically, or work through how this same curved-surface idea turns into capillary rise in the next node — ask the AI tutor to walk it through with you.

Why does a bubble need two surfaces but a drop only needs one?

A drop is a solid lump of liquid — liquid fills the whole sphere, and the only place liquid meets air is the outer skin, so there is exactly one curved liquid–air boundary. A soap bubble is different in structure: it is a thin, hollow film of soapy water enclosing a pocket of air, with ordinary air outside too. The film has an outer face touching the outside air and, separately, an inner face touching the trapped air — two distinct liquid–air boundaries very close together but each independently squeezed by surface tension. Since each surface contributes its own $2S/r$ , the two together give the bubble double the excess pressure of a solid drop of the same size and liquid.

If excess pressure is higher for smaller bubbles, why don't tiny bubbles just instantly pop?

They often do, which is part of why very fine bubbles are short-lived — but "higher excess pressure" does not by itself mean "about to burst." Popping happens when the thin film ruptures, usually from drying, an outside disturbance, or draining of liquid within the film, not directly from the pressure difference alone. A small, undisturbed bubble can sit at its higher excess pressure quite stably, much as a small balloon holds a higher internal pressure than a large one without bursting just from that. The extra pressure matters most whenever a small and a large bubble are connected — then the difference actively drives air from the small one into the large one until the small one is gone.

Does the excess pressure depend on what gas is inside the drop or bubble?

No — look again at the derivation: the hemisphere force balance only ever involves the surface tension $S$ of the liquid film and the radius $r$ of curvature. The identity of the gas inside (air, soap-blown breath, or anything else) never enters the calculation at all; the excess pressure is a property of the liquid surface's curvature and tension, not of what is trapped inside it. What the gas's identity does affect is unrelated quantities like the bubble's buoyancy or how it behaves if heated, but not $\Delta P$ itself.

Is this the same idea as capillary rise in a thin tube?

They are close cousins, both consequences of surface tension acting on a curved liquid surface, but they are not the same calculation. Excess pressure across a drop or bubble is about a closed curved surface fully surrounding a volume of gas or liquid. Capillary rise is about an open curved surface — the meniscus at the top of a liquid column in a narrow tube — where the same kind of curvature-driven pressure difference instead pulls (or pushes) the liquid column up (or down) the tube until gravity balances it. The next topic in this unit develops that capillary-rise case using a very similar curved-surface argument, so the two topics reinforce each other well if studied back to back.