Pressure inside a drop of liquid is equal to two times the atmospheric pressure. If eight such drops of same liquid coalesce, What will be the excess pressure inside it ? Atmospheric pressure is P0.
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a
3P02
b
P02
c
P04
d
3P04
answer is B.
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Detailed Solution
If r be the radius of the drop of liquid, Then pressure inside it, p=p0+2Tr∴p0+2Tr=2p0⇒2Tr=p0For the larger drop, 43πR3=8 x 43πr3⇒R=2r∴p'=p0+2T2r=p0+12p0⇒∆p'=p'-p0=p02