Two spherical soap bubble coalesce. If `V` is the consequent change in volume of the contained air and `S` the change in total surface area, show that
`3PV+4ST=0`
where `T` is the surface tension of soap bubble and `P` is
Atmospheric pressure

Options

A. 4PV+3ST=0
B. 3PV+4ST=0
C. 2PV+3ST=0
D. 3PV+2ST=0

Answer

B

Explanation

When two bubble are combined and from single drop then `P_(1)V_(1)+P_(2)V_(2)=PV_(3)`
`(P+(4T))/(r_(1))4/3 pir_(1)^(3)+(P+(4T)/(r_(2))4/3 pir_(2)^(3)=(P+(4T)/(r)) 4/3 pir^(3)`
`P(4/3 pir_(1)^(3)) +(4T)/(3) (4pi r_(1)^(2))+P(4/3 pir_(2)^(3)) +(4T)/(3) (4pir_(2)^(2))`
`=PV +(4T)/(3) (4pir^(2))`
`PV+(4T)/(3) S+PV +(4T)/(3)S=PV +(4T)/(3)S`
`PV+(4TS)/(3)=0`
`3PV+4TS=0`