An air bubble of radius $$0.1\,cm$$ lies at a depth of $$20\,cm$$ below the free surface of a liquid of density $$1000\,kg/m^3.$$ If the pressure inside the bubble is  $$2100\,N/m^2$$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is  $$(g=10\,m/s^2):$$

An air bubble of radius $$r = 0.1$$ cm = $$0.001$$ m at depth $$h = 20$$ cm = $$0.20$$ m has excess pressure of 2100 N/m² above atmospheric. We need the surface tension.

The pressure inside the bubble exceeds atmospheric pressure by both the hydrostatic pressure and the surface tension pressure:

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

Note: For a bubble in a liquid (not a soap bubble), there is only one surface, so the excess pressure due to surface tension is $$\frac{2T}{r}$$.

$$ 2100 = 1000 \times 10 \times 0.20 + \frac{2T}{0.001} $$

$$ 2100 = 2000 + 2000T $$

$$ 100 = 2000T $$

$$ T = \frac{100}{2000} = 0.05 \text{ N/m} $$

The correct answer is Option (2): 0.05.