Published by:
CGP EDU Academic Team
Published on: September 12, 2026
If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be
Text Solution
Verified by ExpertsThe correct answer is:
C
Let radius of curvature of the common internal film surface of the double bubble formed be $\Gamma$ .
Then, excess of pressure as compared to atmosphere inside A is
$\frac{4T}{r_{1}} \text{ and B is } \frac{4T}{r_{2}}.$
The pressure difference is
$\frac{4T}{r_1} - \frac{4T}{r_2} = \frac{4T}{r'} \\ \Rightarrow r' = \frac{r_1 r_2}{r_2 - r_1} \\ \text{Given, } r_1 = r_2 = r \\ \therefore r' = \frac{r^2}{0} = \infty$
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