Pressure inside a soap bubble, formed in vacuum is P. When this bubble in taken in a closed chamber filled with gas at pressure P/2, its radius is reduced to half of its initial value. Now the pressure inside the bubble will be

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3.5 P

2.5 P

1.5 P

4.5 P

answer is B.

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Detailed Solution

Initially, P=4SrFinally, P1−P2=ΔP=4Sr/2⇒P1=P2+2(4Sr)=P2+2P=52P

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