Question

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 ________.

Answer 1

4 cm (with the common surface concave towards the smaller bubble)

Answer 2

For a soap bubble, the excess pressure is given by

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

where $T$ is the surface tension and $r$ is the radius of the bubble. For two bubbles of radii $r_1$ and $r_2$, their internal pressures are:

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

Since $r_1 = 2$ cm and $r_2 = 4$ cm, $P_1 > P_2$. When they meet, the common interface has an excess pressure difference that satisfies:

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

Substituting the expressions:

$$\frac{4T}{2} - \frac{4T}{4} = \frac{4T}{R}$$ $$2T - T = \frac{4T}{R} \quad \Longrightarrow \quad T = \frac{4T}{R}$$

Cancelling $T$ (assuming $T \neq 0$):

$$R = 4 \text{ cm}$$

The interface is concave towards the smaller bubble, which has the higher internal pressure.