Question

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

• A. $0.4\text{m}$
• B. $0.04\text{m}$
• C. $0.004\text{m}$
• D. $0.002\text{m}$

Answer 1

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

Let $R$ be the radius of curvature of common surface when bubbles $A$ and $B$ of radii $r_A$ and $r_B$ coalesce.

The excess pressure in $A$ and $B$ are

$p_A=\frac{4T}{r_A}$

And, $p_B=\frac{4T}{r_B}$

Where, $T$ is the surface tension.


If $R$ is the radius of common interface, then

$p_A-p_B=\frac{4T}{R}$

$\Rightarrow \frac{4T}{r_A}-\frac{4T}{r_B}=\frac{4T}{R}$

$\Rightarrow R=\frac{r_A{r}_B}{r_B-r_A}$

$\Rightarrow R=\frac{0.002\times 0.004}{0.004-0.002}$

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

Hence, option $\left(\text{C}\right)$ is correct.