A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radiu

IV. Exercises · Q3

Q. A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.

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Concept understanding — Excess Pressure in a Drop

Why Does a Drop Have Extra Pressure Inside?

Imagine you're blowing a tiny soap bubble on a wand. You have to push air into it — that push is extra pressure. Now think of a raindrop hanging from a leaf. It's round, not flat. Something is squeezing it into that shape. That something is surface tension.

Surface tension acts like a stretched elastic skin on the liquid surface. It always tries to shrink the surface area. For a spherical drop, the surface is curved outward. The tension pulls tangentially along the surface, and because the surface is curved, that pull has a net inward component — like a rubber band squeezing a ball from all sides. That inward squeeze compresses the liquid inside, raising its pressure above the outside.

So the excess pressure is the extra pressure inside the drop (or bubble) compared to the atmosphere outside. It's what keeps the drop from collapsing.

The Precise Physics

For a spherical liquid drop (like a water droplet in air), there is only one liquid-air interface. Surface tension acts along that single surface. The excess pressure $\Delta P$ inside is given by:

\Delta P = \frac{2T}{R}

where $T$ is the surface tension of the liquid and $R$ is the radius of the drop.

For a soap bubble, there are two liquid-air interfaces — an inner surface and an outer surface, each with its own surface tension. Both surfaces are curved and both contribute to the inward squeeze. So the excess pressure inside a soap bubble is double that of a single-surface drop:

\Delta P = \frac{4T}{R}

\displaystyle \Delta P = \frac{2T}{R}

Why the Factor of 2?

Here's the intuition without heavy math. For a single surface, the net inward force from surface tension on a hemisphere is $T \times (2\pi R)$ (tension times circumference). This force is balanced by the excess pressure acting on the cross-sectional area $\pi R^2$ . Equating:

T \cdot 2\pi R = \Delta P \cdot \pi R^2 \quad \Rightarrow \quad \Delta P = \frac{2T}{R}

For a bubble, the same argument applies twice — once for the outer surface and once for the inner surface. Both pull inward, so the total inward force is doubled, giving $\Delta P = 4T/R$ . …

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Step 1. Bubble B (radius $R_B=4$ cm) has excess pressure over the outside atmosphere $P_0$ of $P_B-P_0=\dfrac{4T}{R_B}$ .

Step 2. Bubble A (radius $R_A=2$ cm), sitting inside bubble B, has its 'outside' at the pressure $P_B$ (the air trapped inside B), so its excess pressure over $P_B$ is $P_A-P_B=\dfrac{4T}{R_A}$ .

Step 3. The total excess pressure of bubble A's air over the OUTSIDE atmosphere $P_0$ is therefore $P_A-P_0=(P_A-P_B)+(P_B-P_0)=\dfrac{4T}{R_A}+\dfrac{4T}{R_B}=4T\left(\dfrac{1}{R_A}+\dfrac{1}{R_B}\right)$ .

Step 4. A single soap bubble of some radius r that has this SAME total excess pressure over the atmosphere satisfies $\dfrac{4T}{r}=4T\left(\dfrac1{R_A}+\dfrac1{R_B}\right)\Rightarrow\dfrac1r=\dfrac1{R_A}+\dfrac1{R_B}\Rightarrow r=\dfrac{R_AR_B}{R_A+R_B}$ .

Step 5. Substituting $R_A=2$ cm, $R_B=4$ cm: $r=\dfrac{2\times4}{2+4}=\dfrac{8}{6}=\dfrac{4}{3}\approx1.33$ cm. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The radii of two soap bubbles are in the ratio of 2 : 1. The ratio of excess pressure inside them is$
(a) 1 : 4
(b) 4 : 1
(c) 2 : 1
(d) 1 : 2
›Reveal solutionSolution

Excess pressure = 4T/r, so pressures are in the inverse ratio of radii: 1:2. Answer (D).

For a soap bubble (two surfaces), the excess pressure inside is P = 4T/r, inversely proportional to the radius.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The excess pressure inside a soap bubble of radius r is (A) 2T/r (B) 4T/r (C) 3T/r (D) 5T/r$
›Reveal solutionSolution

The excess pressure inside a soap bubble of radius r is 4T/r.

For a single liquid drop (one surface), the excess pressure inside is $\Delta P = \dfrac{2T}{r}$ , from the balance between surface tension force and the pressure difference across the curved surface.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Two soap bubbles have radii in the ratio 3 : 2. .............. is the ratio of excess pressure inside them.$
›Reveal solutionSolution

Because excess pressure in a bubble varies as $1/r$ , the ratio of excess pressures is the inverse of the ratio of radii.

For a soap bubble (which has two liquid surfaces, inner and outer), the excess pressure inside is:

$\Delta P = 4T/r$

where $T$ is the surface tension (same for both bubbles, same soap solution) and $r$ is the radius.

Given $r_1 : r_2 = 3 : 2$ :

…

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CBSE 2024Set ANN1 mark
$Q. The excess pressure inside a soap bubble is$
(a) 0
(b) S/R
(c) 2S/R
(d) 4S/R
›Reveal solutionSolution

A soap bubble in air has two surfaces (inside and outside the thin film), so its excess pressure is double the single-surface case: 4S/R.

For a spherical liquid drop, which has only one free surface, surface tension gives an excess pressure inside of

P = 2S/R

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The diameter and surface tension of a water-drop are d and T respectively. The value of excess pressure inside the drop will be equal to (A) 4T/d (B) 2T/d (C) T/d (D) 5T/d$
›Reveal solutionSolution

Excess pressure inside a drop of diameter $d$ is $4T/d$ .

For a spherical liquid drop (one surface), the excess pressure inside due to surface tension is $\Delta P=\dfrac{2T}{r}$ , where $r$ is the drop's ra …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two soap bubbles have radii in the ratio 2 : 1. What is the ratio of excess-pressure inside them ?$
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
›Reveal solutionSolution

The excess pressure inside a soap bubble is P = 4T/r, so it is inversely proportional to radius; a bigger bubble has lower excess pressure.

A soap bubble has two liquid surfaces (inner and outer), so its excess pressure is:

$P = \frac{4T}{r}$

where T is the surface tension of the soap solution and r is the bubble's radius. For two bubbles of radii $r_1$ and $r_2$ :

$\frac{P_1}{P_2} = \frac{r_2}{r_1}$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Pressure inside soapbubble is:$
(a) equal to atmosphere
(b) greater than atmosphere
(c) less than atmosphere
(d) two times
›Reveal solutionSolution

Pressure inside a soap bubble is greater than atmospheric.

A soap bubble has two liquid surfaces (inner and outer). Surface tension curves these surfaces, and the excess pressure inside a soap bubble is

ΔP = 4T/r,

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Air is filled in a bubble. The value of pressure inside it will —$
(a) Decrease
(b) Increase
(c) Not change
(d) None of these
›Reveal solutionSolution

The pressure of the air inside a bubble is greater than outside (it increases) due to surface tension.

A soap bubble has two surfaces, so the excess pressure inside over the outside is $\Delta P = \dfrac{4T}{r}$ (for a liquid drop or air bubble in liquid it is 2T/r).

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The pressure inside two soap bubbles is 1.01 and 1.02 Atmospheres. The ratio of their respective volumes is —$
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 16 : 1
›Reveal solutionSolution
The ratio of the two bubble volumes is 8 : 1.

For a soap bubble the excess pressure over the atmosphere is $\Delta P = \dfrac{4T}{r}$ , so $r = \dfrac{4T}{\Delta P}$ .

Taking atmospheric pressure as 1.00 atm:

Bubble 1: ΔP₁ = 1.01 − 1.00 = 0.01 atm

Bubble 2: ΔP₂ = 1.02 − 1.00 = 0.02 atm …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. There is a small bubble at one end and bigger bubble at other end of a pipe. Which among the following will happen ?$
(a) remains in equilibrium
(b) smaller will grow until they collapse
(c) bigger will grow until they collapse
(d) none of the above
›Reveal solutionSolution

Because excess pressure inside a bubble is P = 4T/r, the smaller bubble (smaller r) has higher internal pressure, so air flows from the smaller bubble into the bigger one -- the bigger bubble grows while the smaller one shrinks and collapses.

For a soap bubble (which has two liquid surfaces, an inner and an outer film) of radius r and surface tension T, the excess pressure inside the bubble over the outside atmospheric pressure is

P_excess = 4T/r

This shows that the excess pressure is INVERSELY proportional to the radius: a smaller bubble has a smaller r, and hence a LARGER excess pressure than a bigger bubble of the same liquid.

…

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Q.A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B.

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