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.

Answer 1

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

\(P_{1} = \dfrac{4T}{r_1}\)

\(P_{2}= \dfrac{4T}{r_2}\)

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,


\(p=p_{2} - p_{1} = 4T \left ( \dfrac{r_{1}-r_{2}}{r_{1} r_{2}} \right )\,\,\,\,\,\,\,\,\,\,...(1)\)

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,

\(p = \dfrac{4T}{R} . . . . . (2)\)

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

\(R = \dfrac{r_{1}r_{2}}{r_{1} - r_{2}} = \dfrac{\left ( 0.004 \right )\left ( 0.002 \right )}{\left ( 0.004 - 0.002 \right )}\)

\(\therefore R= 0.004~\text{m}\)

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.