Assuming the surface tension of rain water to be `72 "dyne"//"cm"`, find the differnce of pressure inside and outside a rain drop of diameter `0.02 cm`. What would this pressure difference amount to , if the drop were to be decreased by evaporation to a diameter of `0.0002 cm` ?

Answer

`1.44 xx 10^(4) dyn//cm^(2)` and `1.44 xx 10^(7) dyn//cm^(2)`

Explanation

`r_(h) = r_(1)- ((r_(1) - r_(2))/(l))L`, `f_(h) = 0.5 xx 10^(-3) - ((0.5 xx 0.25))/(0.1) xx 10^(-3) xx 8 xx 10^(-2)`
`= 0.3 xx 10^(-3) m, (2T_(0))/(r_(h)) = rhogh rArr T_(0) = ((1)/(14) xx 10^(4) xx 9.8 xx 8 xx 10^(-2) xx 0.3 xx 10^(-3))/(2)`
`T_(0) = 0.084 N//m` , For Tube `'B'`
For Temp `0^(@)C (2T_(0))/(r) = rhogh_(1) rArr r = (2T_(0))/(rhogh_(1))(2xx1xx0.084)/((1)/(14) xx 10^(4) xx 9.8 xx 6 xx 10^(-2))`
`r = 0.40 xx 10^(-3)m rArr` For temp `50^(@)C (2T_(50))/(r) = rhogh`
`T_(50) = (rhoghr)/(2) = ((1)/(14) xx 10^(4) 9.8 xx 5.5 xx 10^(-2) xx 0.4 xx 10^(-3))/(2)`
`T_(50) = 0.077 N//m^(2) = (T_(50) - T_(0))/(50 - 0) = (0.077 - 0.084)/(50) = -1.4 xx 10^(-4) N//m^(@)C`
Ans. - `1.4 xx 10^(-4) N//(m - .^(0)C)`