An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension $$0.095 J/m^{2}$$ and density $$10^{3}kg/m^{3}$$. The difference between pressure inside the bubble and atmospheric pressure is _____$$N/m^{2}.(\text{ take }g=10m/s^{2})$$

An air bubble of radius 1.0 mm at depth 20 cm in a liquid with surface tension 0.095 J/m² and density 10³ kg/m³.

We first calculate the pressure inside the bubble using

$$P_{inside} = P_{atm} + \rho g h + \frac{2T}{r}$$

Then the difference between this pressure and the atmospheric pressure is given by

$$P_{inside} - P_{atm} = \rho g h + \frac{2T}{r}$$

Substituting the numerical values gives

$$= 10^3 \times 10 \times 0.20 + \frac{2 \times 0.095}{1.0 \times 10^{-3}}$$

$$= 2000 + 190$$

$$= 2190 \text{ N/m}^2$$

The answer is 2190.