question_answer

A soap bubble A of radius 0.03 and another bubble B of radius 0.04 m are brought together so that the combined bubble has a common interface of radius r, then the value of r is

A)  0.24m        

B)  0.48m

C)  0.12m        

D)  none of these

Correct Answer: C

Solution :

Key Idea: Pressure inside the smaller bubble is greater than that inside the bigger bubble. Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles A and B be r.                 Excess of pressure as compared to atmosphere inside A is                 \[{{P}_{1}}=\frac{4T}{{{r}_{1}}}=\frac{4T}{0.03}\] Excess of pressure inside B is                 \[{{P}_{2}}=\frac{4T}{{{r}_{2}}}=\frac{4T}{0.04}\] In the double bubble the pressure difference between A and B on either side of the common surface is                 \[\frac{4T}{0.03}-\frac{4T}{0.04}=\frac{4T}{r}\] \[\Rightarrow \] \[\frac{1}{0.003}-\frac{1}{0.04}=\frac{1}{r}\] \[\Rightarrow \] \[r=\frac{0.03\times 0.04}{0.01}\]                 = 0.12 m Note: As pressure inside the smaller bubble is greater than that inside the bigger bubble the curvature of the common film will be concave towards the centre of the smaller bubble.