Question

Two soap bubbles coalesce. It is noticed that, whilst joined together, the radii of the two bubbles are a and b where $a>b$. Then the radius of curvature of interface between the two bubbles will be.

• A. $a-b$
• B. $a+b$
• C. $ab/(a-b)$
• D. $ab/(a+b)$

Answer 1

Answer: (C)

$ab/(a-b)$

According to given figure,
Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles be r. Excess of pressure as compared to atmosphere inside A is $\left(4T/a\right)$. In double bubble, the pressure difference between A and B on either side of the common surface.
$=\frac{4T}{b}-\frac{4T}{a}$
This will be equal to $\left(4T/r\right)$
$\therefore \frac{4T}{r}=\frac{4T}{b}-\frac{4T}{a}$
or $\frac{1}{r}=\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{a-b}{ba}$ or $r=\frac{ab}{a-b}$.