Question

When two soap bubbles of radii $r_{1}$ and $r_{2}\left(r_{2}>r_{1}\right)$ coalesce, the radius of curvature of common surface is

• A. $r_{2}-r_{1}$
• B. $\frac{r_{2}-r_{1}}{r_{1} r_{2}}$
• C. $\frac{r_{1} r_{2}}{r_{2}-r_{1}}$
• D. $r_{2}+r_{1}$

Answer 1

Answer: (C)

For soap bubble of radius $r_{1}$.
$P_{1}-P_{0}=\frac{4 \sigma}{r_{1}},$ where $P_{1}$ is the pressure inside it and $P_{0}$
is the pressure of the atmosphere.
Similarly, $P_{2}-P_{0}=\frac{4 \sigma}{r_{2}}$
When bubbles coalesce and form a common surface,
$P_{1}-P_{2}=\frac{4 \sigma}{r_{1}}-\frac{4 \sigma}{r_{2}}=\frac{4 \sigma}{R}$
$\Rightarrow \frac{1}{R}=\frac{1}{r_{1}}-\frac{1}{r_{2}}$ or $R=\frac{r_{1} \cdot r_{2}}{r_{2}-r_{1}}$