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# The adjoining diagram shows three soap bubbles$A,B$ and $C$ prepared by blowing the capillary tube fitted with stop cocks $S,{S_1},{S_2}$ and ${S_3}$ with stop cock $S$ closed and stop cocks $S,{\text{ }}{S_1}{\text{ }}and{\text{ }}{S_2}$opened :-A) ${\text{B will start collapsing with volumes of A and C increasing }}$B) ${\text{C will start collapsing with volumes of A and B increasing }}$C) ${\text{C and A will both start collapsing with volumes of B increasing }}$D) ${\text{Volumes of A,B,C will becomes equal at equilibrium}}$

Last updated date: 15th Sep 2024
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Hint: To answer such a question, one has to be very much aware of the relation between the pressure and radius of the bubble. We know that the pressure under the soap bubble is always greater than the pressure that exists outside the soap bubble.

From the figure given above we can conclude that the relation between the radius of $\Rightarrow {r_C} < {r_A} < {r_B}$.
As we know that the pressure inside the soap bubble is always greater as compared to the pressure outside the bubble and the relation between the two is given as $\Delta P$.
$\Rightarrow \Delta P = \dfrac{{4s}}{r}$
Where, the symbols have their usual meaning as the $s$ stands for surface tension and the $r$ stands for radius of the bubble.
Therefore, the largest pressure is inside the $C$ as the pressure is lower as the radius is largest. Similarly, the radius of the bubble $A$ is also smaller as compared to that of the $B$ . Therefore, one can understand that the molecules from both the bubbles of smaller radius will move on the bubble of the larger radius and therefore that bubble will expand, and rest will collapse.
Looking at the options one can easily make out that the volume of both $A{\text{ }}and{\text{ }}C$ will collapse and volume of $B$ should increase.