Question

The diagram below shows three soups bubbles A, B and C prepared by blowing the capillary tube fitted with stops cocks S. $S_1,S_2$ and $S_3$ When all the three stocks $S_1,S_2$ and $S_3$ are opened and S closed then,

• A. B will start collapsing with the increasing volume of A and C
• B. C will start collapsing with the increasing volume of A and B
• C. Volumes of A, B and C will become equal at equilibrium
• D. C and A will both start collapsing with the increasing volume of B

Answer 1

The correct option is D C and A will both start collapsing with the increasing volume of B

The pressure difference between the inside and outside of a soap bubble is given by the formula, $\Delta P=\frac{4T}{R}$

Thus smaller the radius higher the pressure difference.

Hence the bubble with the smallest radius has the highest pressure.

Hence the pressure inside the soap bubbles is in the order:

$P_C>P_A>P_B$

Thus on opening stop cocks $S_1,S_2,S_3$ , bubble C and A will start collapsing while B increases

Hence option D is correct.