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. `3p_(0)V+4ST=0`
B. `4p_(0)V+3ST=0`
C. `p_(0)V+4TS=0`
D. `4p_(0)V+ST=0`

Answer

A

Explanation

Let`R_(1)` and `R_(2)` are the radii of soap, bubbles before and after collapsing. Then given that
`V=2(4/3 piR_(1)^(3))-4/3 piR_(2)^(3)` …(i)
`S=2(8 pi R_(1)^(2))-8piR_(2)^(2)` …(ii)
`P_(1)V_(1)=P_(2)V_(2)`
`:. (P_(0)+(4T)/(R_(1)))(2xx4/3 pi R_(1)^(3))= (p_(0)+(4T)/(R_(2)))(4/3 pi R_(2)^(3))`...(iii)
Solving these three equation we get the desired result.