Two unequal soap bubbles are formed one on each side of a tube closed in the middle by a tap. What happens when the tap is opened to put the two bubbles in communication?

Options

A. No sir passes in any direaction as the pressures are the same on two sides of the tap
B. Larger bubble shrinks and smaller bubble increases in size till they become equal in size
C. Smaller bubble gradually collapses and the bigger one increases in size
D. None of these

Answer

C

Explanation

`P_(A) = P_(0) + (4sigma)/(r), P_(B) = P_(0) + (4sigma)/(r)` {`P_(0) =` atmospheric pressure}.
Clearly `P_(A) gt P_(B),` so air will flow from `A` to `B`
As `r` decreases, pressure will become more and hence more flow of air from `A` to `B`. Ultermately bubble `A` collapses and `B` becomes bigger in size.