When two soap bubbles

Question:

When two soap bubbles of radii a and $b$ ( $b>$ a) coalesce, the radius of curvature of common surface is :

  1. $\frac{a b}{b-a}$

  2. $\frac{a+b}{a b}$

  3. $\frac{b-a}{a b}$

  4. $\frac{a b}{a+b}$


Correct Option: 1

Solution:

Excess pressure at common surface is given by

$\mathrm{P}_{\mathrm{ex}}=4 \mathrm{~T}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)=\frac{4 \mathrm{~T}}{\mathrm{r}}$

$\therefore \frac{1}{\mathrm{r}}=\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}$

$\mathrm{r}=\frac{\mathrm{ab}}{\mathrm{b}-\mathrm{a}}$

Leave a comment

None
Free Study Material