question_answer

The radii of two soap bubbles are \[{{r}_{1}}\]and\[{{r}_{2}}\]\[({{r}_{2}}>{{r}_{1}}).\]When they come into contact, the radius of their common interface is:

A) \[{{r}_{2}}-{{r}_{1}}\]

B)  \[\sqrt{r_{1}^{2}-r_{2}^{2}}\]

C)  \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]

D)  \[\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]

Correct Answer: D

Solution :

 Key Idea: Excess pressure inside a bubble of radius R is given by \[P=\frac{4T}{r}.\] Let \[{{\text{P}}_{\text{1}}}\]and \[{{\text{P}}_{2}}\]are pressure differences across the common interface. Let r is radius of curvature of the common surface. \[{{P}_{2}}-{{P}_{1}}=\frac{4T}{r}\] \[\frac{4T}{r}=\frac{4T}{{{r}_{2}}}-\frac{4T}{{{r}_{1}}}\](T = surface tension) \[\frac{1}{r}=\frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}}\] \[r=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1-}}{{r}_{2}}}\]