Q. Two soap bubbles $A$ and $B$ are formed at the two open ends of a tube. The bubble $A$ is smaller than bubble$B$. Valve and air can flow freely between the bubbles, then
A) There is no change in the size of the bubbles
B) The two bubbles will become of equal size
C) $A$ will become smaller and $B$ will become larger
D) $B$ will become smaller and $A$ will become larger
Explanation
(c) Excess of pressure inside the bubble, $p = 4S/r$. So smaller is the radius$r$, the larger is the excess of pressure$p$. It means, the pressure of air is more in bubble $A$ than in bubble$B$. So the air will go from bubble $A$ to bubble $B$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.