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Question: If the excess pressure in a soap bubble is p, the excess pressure in an air bubble is: (A). \(\dfr...

If the excess pressure in a soap bubble is p, the excess pressure in an air bubble is:
(A). p2\dfrac{p}{2}
(B). P
(C). 2p
(D). 4p

Explanation

Solution

Hint: In order to answer this question, we should be having an idea about the surface tension of the air bubble and also about the soap bubble. To know the difference between the excess amount of pressure present inside the soap bubble and the air bubble we need to know the difference between the surface tension and what happens when excess amount of pressure is present in both the cases. On the comparison between the soap bubble and the air bubble we will find the correct answer.

Complete step by step answer:
We should know that the surface tension of the interface between the liquid and gas always creates a pressure difference. In case of a soap bubble the pressurized bubble of air is contained within a thin, elastic surface of liquid. When the bubble bursts the difference and pressure causes an audible pop.
In case of a soap bubble there will always be a free surface so excess pressure will be doubled then that which is present in the air bubble. So the air bubble is p2\dfrac{p}{2}.
Therefore, the correct answer is Option A.
It is found that the excess of pressure which is present in a liquid drop or air bubble in a liquid is two times the surface tension of the liquid over the radius of the concerned bubble. On the other hand, in the case of a soap bubble that excess of pressure is equal to the four times of surface tension over the radius.

Additional Information:
When a soap bubble is given a negative charge the charged particles present on the surface of the bubble will repel each other, which will cause a stretching force and hence the bubble will enlarge. At the equilibrium if the force caused by the negative charge and the surface tension of the bubble are equal then the bubble will become stable.

Note: We should know that when a soap bubble is charged then it will expand which will lead to the decrease in the surface tension. By decrease we mean that the surface tension will get minimised.