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 \[{{P}_{1}}\] and \[{{P}_{2}}\] 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)                 \[\frac{1}{r}=\frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}}\]                 \[r=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]