Question

Two separate air bubbles $\text{A}$ and $\text{B}$ of radii $0.004\text{m}$ and $0.002\text{m}$ formed of the same liquid surface tension $0.07\text{N/m}$ come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles.

• A. $0.004\text{m}$
• B. $4\text{m}$
• C. $0.4\text{m}$
• D. $0.0004\text{m}$

Answer 1

The correct option is A $0.004\text{m}$

Let $R$ be the radius of curvature of common surface when bubbles $\text{A}$ and $\text{B}$ of radii $r_1$ ​ and $r_2$ ​ coalesce.

The excess pressure inside the bubbles

$P_1=\frac{4T}{r_1}$

$P_2=\frac{4T}{r_2}$

where, $T$ is surface tension.

$r_2<r_1$

$\therefore p_2>p_1$


i.e. pressure inside the smaller bubble will be more. So, the excess pressure,


$p=p_2-p_1=4T\left(\frac{r_1-r_2}{r_1{r}_2}\right)...\left(1\right)$

This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. For the new bubble, we can also write,

$p=\frac{4T}{R}.....\left(2\right)$

From Eqs. (1) and (2), we get

$R=\frac{r_1{r}_2}{r_1-r_2}=\frac{\left(0.004\right)\left(0.002\right)}{(0.004-0.002)}$

$\therefore R=0.004\text{m}$

Hence, option (c) is correct answer.

Why this question ? This question is designed to emphasize the topic access pressure inside a bubbles and to apply it to calculate common. radius of curvature while two bubbles coalesce.