Question

A soap bubble $A$ of radius $0.03m$ and another bubble $B$ of radius $0.04m$ coalesce to form a combined bubble such that the radius of curvature of their common interface is $r$ . Then, the value of $r$ is

• A. $0.24m$
• B. $0.48m$
• C. $0.12m$
• D. none of these

Answer 1

Answer: (C)

Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles $A$ and $B$ be $r$ .

Excess of pressure as compared to the atmosphere inside $A$ is
$p_1=\frac{4T}{r_1}=\frac{4T}{0.03}$
Excess of pressure inside $B$ is
$p_2=\frac{4T}{r_2}=\frac{4T}{0.04}$
In the double bubble, the pressure difference between $A$ and $B$ on either side of the common surface is
$\frac{4T}{0.03}-\frac{4T}{0.04}=\frac{4T}{r}$
$\Rightarrow \frac{1}{0.03}-\frac{1}{0.04}=\frac{1}{r}$
$\Rightarrow r=\frac{0.03\times 0.04}{0.01}=0.12m$