Question

There is air bubble of radius $1.0\, mm$ in a liquid of surface tension $0.075\, N / m$ and density $1000\, kg / m$. The bubble is at a depth of $10\, cm$ below the free surface. By what amount is the pressure( in pascal) inside the bubble greater than the atmospheric pressure ? Take $g=9.8\, m / s ^{2}$.

Answer 1

Given: $r =1\, mm =10^{-3}, T =0.075\, N / m ,$
$\rho=1000\, kg / m ^{3}, h =10\, cm$
$g =9.8\, m / s ^{2}$
Pressure inside the bubble $=$ liquid pressure $+$ Atm pressure $+\frac{2 T }{ r }$
Therefore pressure inside the bubble is greater than atm. pressure by
$=$ liquid pressure $+\frac{2 T }{ r }= hpg +\frac{2 T }{ r }$
$=0.1 \times 1000 \times 9.8+\frac{2 \times 0.075}{10^{-3}}$
$=980+150=1130\, N / m ^{2}$