Question

The pressure inside an air bubble of radius 2 cm formed 20 cm below an water surface is (surface tension of water = 70 into 10 raise to power minus 3 N/m)

Answer 1

Answer on Question#36969 – Physics - Mechanics

A surface tension of water $\sigma = 70 * 10^{-3} \frac{N}{m}$.

Pressure inside a bubble when it is at depth $h$ below an water surface is

$$P = \rho_w g h + \frac{2\sigma}{R},$$

where $\rho_w$ – density of water.

The pressure inside an air bubble of radius $2\,\mathrm{cm}$ formed $20\,\mathrm{cm}$ below water surface is

$$P = 10^3 \frac{kg}{m^3} * 9.81 \frac{m}{s^2} * 20 * 10^{-2} m + \frac{2 * 70 * 10^{-3} \frac{N}{m}}{2 * 10^{-2} m} = 1969 \frac{N}{m^2}.$$

Answer: $1969 \frac{N}{m^2}$.