Question

A soap bubble of radius R is surrourded by another soap bubble of radius 2R, as shown. Take surface tension=S. Then, the pressure inside the smaller soap bubble, in excess of the atmosphere pressure will be:

A
4SR
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B
6SR
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C
8SR
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D
3SR
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Solution

The correct option is B 6SRAccording to Laplace equation, excess pressure created with surface tension of the spherical surface of the liquid is equal to △p=2SR. In case of soap bubbles the excess pressure of air inside them is doubled due to the pressure of two interface are inside and the outside △pB=4SR.Excess pressure of air inside the bigger bubble △pS=4S2R=2SRExcess pressure of the air inside the smaller bubble will be △pS=4SRAir pressure difference between the smaller bubble and the atmosphere will be equal to the sum of excess pressure inside the bigger and smaller bubble △p=△pS+△pS⇒2SR+4SR=6SRHence, the answer is 6SR.

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