Question

There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm^-1 and density 1000 kg m^-3. The bubble is at a depth of 10 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure? Take g = 9.8 m s^-2.

• A. 980 Pa
• B. 2260 N/m²
• C. 1130 N/m²
• D. 1260 Pa

Answer 1

The correct option is C

1130 N/m²

Let the atmospheric pressure be $P_0$. The pressure of the liquid just outside the bubble is
$P=P_0+h\rho g$
The pressure inside the bubble is
$P=P^{\prime }+\frac{2s}{r}=P_0+h\rho g+\frac{2s}{r}$
or,$P^{\prime }-P_0$
=$\left(10cm\right)\left(1000kgm{s}^{-3}\right)\left(9.8m{s}^{-2}\right)+\frac{2\times 0.075N{m}^{-1}}{1.0\times {10}^{-3}m}$
=980N$m^{-2}$+150N$m^{-2}$
=1130 pa