Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is ______ .

Key concepts you need

1. Laplace pressure for a soap bubble
   For a complete soap bubble (two liquid-air surfaces), the extra internal pressure above atmospheric pressure is
   $\Delta P = \frac{4T}{R}$
   where $T$ is surface tension and $R$ is the bubble’s radius.

2. Pressure jump across a single curved liquid–air surface
   Across one surface the jump is
   $\Delta P = \frac{2T}{R}$
   (convex side has higher pressure).

3. Common wall between two bubbles
   The wall has two closely spaced surfaces—one facing each bubble—but their curvatures point opposite ways.
   Result: Pressure difference between the two bubbles becomes
   $P_2 - P_1 = \frac{4T}{R_c}$
   where $R_c$ is the (signed) radius of curvature of the common surface measured from the side of bubble 1.

Logical chain of thought

1. Write individual pressures
   $P_1 = P_0 + \frac{4T}{r_1},\qquad P_2 = P_0 + \frac{4T}{r_2}$
   ($P_0$ is atmospheric pressure.)

2. Relate them through the common wall
   $P_2 - P_1 = \frac{4T}{R_c}$

3. Substitute
   $\frac{4T}{r_2} - \frac{4T}{r_1} = \frac{4T}{R_c}$

4. Cancel $4T$ (common factor)
   $\frac{1}{r_2} - \frac{1}{r_1} = \frac{1}{R_c}$

5. Insert numbers ($r_1 = 2\,\text{cm}$, $r_2 = 4\,\text{cm}$)
   $\frac{1}{4} - \frac{1}{2} = \frac{1}{R_c}\;\Rightarrow\; -\frac{1}{4} = \frac{1}{R_c}$

6. Solve
   $R_c = -4\,\text{cm}$
   The minus sign shows the common wall bulges toward the larger bubble. Magnitude: $4\,\text{cm}$.

What’s going on?

Imagine two balloons made of the thinnest, soapy skin. One balloon is small (2 cm radius) and the other is bigger (4 cm radius). They touch each other, so a new, tiny curved wall of soap forms between them—like a tiny doorway separating the two balloons.

The big idea

1. Inside every soap bubble the air is squeezed a little harder than the air outside.
2. How much harder? Smaller bubbles squeeze more because the skin is more tightly bent.
3. The squeeze (extra pressure) is given by a neat rule called the Laplace pressure law.
4. When two bubbles touch, the extra pressure in each bubble must balance with the pressure jump across that separating doorway.
5. Using the rule, we write a simple little equation and—boom—we find how tightly that doorway is bent (its “radius of curvature”).

What will we end up with?

After the little bit of maths, we discover the tiny doorway is bent just 4 cm in radius (curved toward the bigger bubble).

Step-by-step calculation

Let $T$ be the surface tension of the soap film (same for both bubbles).

1. Pressures inside each bubble

   $P_1 = P_0 + \frac{4T}{r_1},\qquad P_2 = P_0 + \frac{4T}{r_2}$

2. Pressure difference across the common surface

   $P_2 - P_1 = \frac{4T}{R_c}$

3. Equate and simplify

   $\frac{4T}{r_2} - \frac{4T}{r_1} = \frac{4T}{R_c}$ $\Rightarrow \frac{1}{r_2} - \frac{1}{r_1} = \frac{1}{R_c}$

4. Insert given radii ($r_1 = 2\,\text{cm}$, $r_2 = 4\,\text{cm}$)

   $\frac{1}{4} - \frac{1}{2} = \frac{1}{R_c}$ $-\frac{1}{4} = \frac{1}{R_c}$

5. Solve for $R_c$

   $R_c = -4\,\text{cm}$

6. Interpretation
   The negative sign means the common wall is concave toward the larger (4 cm) bubble. Usually, we quote the magnitude:

   $|R_c| = 4\;\text{cm}$

[ \boxed{4,\text{cm}} ]