Question

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

Answer 1

Answer: 1130.00

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