There is an air bubble of radius $$1.0$$ mm in a liquid of surface tension $$0.075$$ N m$$^{-1}$$ and density $$1000$$ kg m$$^{-3}$$ at a depth of $$10$$ cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is _____ Pa ($$g = 10$$ m s$$^{-2}$$).

The excess pressure inside an air bubble in a liquid:

$$\Delta P = \frac{2T}{r} + \rho g h$$

where $$T$$ = surface tension, $$r$$ = radius, $$\rho$$ = density, $$h$$ = depth.

Note: An air bubble in liquid has only one surface, so excess pressure due to surface tension is $$\frac{2T}{r}$$ (not $$4T/r$$).

$$\frac{2T}{r} = \frac{2 \times 0.075}{1 \times 10^{-3}} = 150 \text{ Pa}$$

$$\rho g h = 1000 \times 10 \times 0.1 = 1000 \text{ Pa}$$

Total excess pressure above atmospheric: $$150 + 1000 = 1150 \text{ Pa}$$

This matches the answer key value of $$\mathbf{1150}$$.