Question

When two soap bubbles of radii $a$ and $b$ $(b>a)$ coalesce, the radius of curvature of common surface is :

• A. $\frac{b-a}{ab}$
  
• B. $\frac{ab}{b-a}$
  
• C. $\frac{ab}{a+b}$
  
• D. $\frac{a+b}{ab}$
  

Answer 1

The correct option is B $\frac{ab}{b-a}$



Let the radius of the common surface is $R$ and the atmospheric pressure is $P_0$.

Excess pressure inside the bubble $1$ :
$P_1-P_0=\frac{4S}{b}$ -----$\left(1\right)$

Excess pressure inside the bubble $2$ :
$P_2-P_0=\frac{4S}{a}$ -----$\left(2\right)$

Now, excess pressure at the common surface :
$P_2-P_1=\frac{4S}{R}$ -----$\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$ we have :
$\frac{1}{a}-\frac{1}{b}=\frac{1}{R}$

$\therefore R=\frac{ab}{b-a}$

Therefore, $\left(B\right)$ is the correct answer.