Find the excess pressure inside a drop of mercury of radius `2 mm`, a soap bubble of radius `4 mm` and an air bubble of radius `4 mm` formed inside a tank of water. Surface tension mercury is `0.465 N//m` and soap solution and water are, `0.03 N//m` and `0.076 N//m` repsectively.

Answer

(a) `465 N//m^(2)`, (b) 30 N//m^(2), (c) `38 N//m^(2)`

Explanation

(a)
drop of `r = 2 mm`.
`P_(excess) = (2T)/(R) = (2 xx 0.465)/(2 xx 10^(-3)m) (N)/(m) = 465(N)/(m^(2))`
(b) Soap bubble has `2` films :
so `P_(excess) = (4T)/(R) = (4 xx 0.03(N)/(m))/(4 xx 10^(-3)m) = 30(N)/(m^(2))`
(c) As the air bubble is being formed inside a tank of water, so only one layer is formed.
`P_(excess) = (2T)/(R) = (2 xx 0.076(N)/(m))/(4 xx 10^(-3)m) = 38(N)/(m^(2))`