MHT-CET Physics · Mechanical Properties of Fluids

Excess Pressure in Drops and Bubbles, and Capillary Rise

A curved liquid surface pushes harder on its concave side: the pressure inside a drop exceeds the outside by 2T/r and inside a soap bubble by 4T/r, and the same curvature, in the meniscus of a narrow tube, lifts liquid to a height h = 2T cos θ/(rρg).

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Why this matters

38 PYQs, none HARD — nearly all one-formula ratios. Seventeen are excess pressure: bubbles with inside pressures of 1.01 and 1.02 atm, the ratio of volumes or masses when one excess pressure is three or four times another, two bubbles merging under isothermal conditions; twenty-one are capillary rise — how the height and the raised mass change with the radius, the liquid, the angle of contact, g on the moon or down a mine, a tube tilted or pushed down. Two cards.

Concept 1 of 2: Excess Pressure Inside a Drop and a Bubble

A drop has one surface, a soap bubble two, so the bubble's excess pressure is twice the drop's: 2T/r for a drop, 4T/r for a bubble. Either way it goes as 1/r — the SMALLER bubble has the higher pressure. So a threefold excess pressure means a third of the radius and a twenty-seventh of the volume or mass. Two bubbles that merge under isothermal conditions keep their total surface: R² = r₁² + r₂².

Definition

Drop (one surface): $\Delta P = \dfrac{2T}{r}$ . Soap bubble (two): $\Delta P = \dfrac{4T}{r}$ . Meniscus in a tube of radius r: $\dfrac{2T}{r}$ .
$r \propto \dfrac{1}{\Delta P}$ , so volume and mass $\propto \dfrac{1}{\Delta P^3}$ : excess pressure × 3 ⇒ volume or mass ÷ 27; × 4 ⇒ ÷ 64.
Given total pressures: subtract the outside first. 1.01 and 1.02 atm in 1 atm: excess 0.01 and 0.02, so radii 2 : 1 and volumes 8 : 1. In general $\dfrac{V_1}{V_2} = \left(\dfrac{P_2 - P_0}{P_1 - P_0}\right)^3$ .
Isothermal merge of two bubbles in vacuum: $R = \sqrt{r_1^2 + r_2^2}$ .
A drop split into 8: each droplet has half the radius and twice the excess pressure, so the big drop's is half a droplet's.

Excess pressure

\Delta P_{\text{drop}} = \frac{2T}{r}, \qquad \Delta P_{\text{bubble}} = \frac{4T}{r}

Worked example

Soap bubble A has inside pressure 1.004 atm and B has 1.008 atm, outside 1 atm. Ratio of their volumes, A to B?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q33Moderate

Example 1 · Mechanical Properties of Fluids · Excess Pressure and Capillary Rise

The pressure inside a soap bubble A is 1.01 atmosphere and that in a soap bubble B is 1.02 atmosphere. The ratio of volume of bubble A to that of B is [Surrounding pressure = 1 atmosphere]

Taking 1.01 : 1.02 as the pressure ratio

Radius depends on the EXCESS pressure. Subtract the outside 1 atm first: 0.01 : 0.02, not 101 : 102.

Using 2T/r for a soap bubble

A bubble has an inner and an outer surface, so 4T/r. A drop, or the meniscus in a tube, has one: 2T/r.

Drill 16 more on excess pressure inside a drop and a bubble

Concept 2 of 2: Capillary Rise

The meniscus pulls up along the tube's circumference with T cos θ per unit length, and the column's weight balances it: h = 2T cos θ/(rρg). So h goes as 1/r — a tube a fifth as wide lifts liquid five times as high — but the MASS lifted, ρπr²h, goes as r: the narrow tube holds a fifth of the mass. h also goes as 1/g, so on the moon (g/6) the rise is six times greater. If the tube is shorter than h, the liquid does not overflow: it reaches the top and flattens its meniscus, so the apparent contact angle grows until h cos θ fits.

Definition

$h = \dfrac{2T\cos\theta}{r\rho g}$ ; upward force $= T \times 2\pi r$ = weight of the column.
$h \propto \dfrac{1}{r}$ (cross-section 4a ⇒ radius × 2 ⇒ h/2; diameter 80% ⇒ h × 1.25); mass raised $\propto r$ (radius r/3 ⇒ m/3).
Same tube, two liquids: $\dfrac{h_1}{h_2} = \dfrac{T_1\rho_2}{T_2\rho_1}$ (T 6 : 5, ρ 4 : 3 ⇒ 9 : 10). T and ρ both doubled ⇒ h unchanged.
$h \propto \dfrac{1}{g}$ : moon (g/6) ⇒ 6h; down a mine, $\dfrac{h_2}{h_1} = \dfrac{R}{R - d}$ .
θ = 90°: no rise or fall. Same rise with same T: $\cos\theta \propto \rho$ , so the densest liquid has the smallest θ.
Tube shorter than h (length h/3 above the water): $\cos\theta' = \tfrac{1}{3}$ . Tube tilted at α to the vertical: the vertical rise stays h, the length of liquid is $h/\cos\alpha$ .
Circumference from force: $2\pi r = \dfrac{F}{T}$ — 105 dyne with T = 70 dyne/cm gives 1.5 cm.

Capillary rise

h = \frac{2T\cos\theta}{r\rho g}

Worked example

Water rises 4 cm in a tube. How high in a tube of half the radius, and how does the mass raised compare?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q16Easy

Example 2 · Mechanical Properties of Fluids · Excess Pressure and Capillary Rise

A liquid rises to a height of 2.4 cm in a glass capillary P. Another glass capillary Q having diameter

80\%

of capillary P is immersed in the same liquid. The rise of liquid in capillary Q is

Thinking the narrow tube holds more water

The narrow tube lifts liquid HIGHER (h ∝ 1/r) but holds LESS of it (m ∝ r). Radius r/5 gives 5h but m/5.

Drill 20 more on capillary rise

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Excess Pressure Inside a Drop and a Bubble
Excess pressure
$\Delta P_{\text{drop}} = \frac{2T}{r}, \qquad \Delta P_{\text{bubble}} = \frac{4T}{r}$
Capillary Rise
Capillary rise
$h = \frac{2T\cos\theta}{r\rho g}$

Watch out for (3)

Taking 1.01 : 1.02 as the pressure ratio
→ Excess Pressure Inside a Drop and a Bubble
Using 2T/r for a soap bubble
→ Excess Pressure Inside a Drop and a Bubble
Thinking the narrow tube holds more water
→ Capillary Rise

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