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

We can use the Young-Laplace equation to find the difference in pressure inside and outside the air bubble due to surface tension: <br/><br/> $\Delta P = 2 \frac{T}{R}$ <br/><br/> where $\Delta P$ is the pressure difference, $T$ is the surface tension, and $R$ is the radius of the bubble. <br/><br/> Plugging in the given values: <br/><br/> $\Delta P = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{1.0 \mathrm{~mm}} = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{10^{-3} \mathrm{~m}} = 150 \mathrm{~Pa}$ <br/><br/> Now, we need to account for the hydrostatic pressure due to the depth of the bubble below the free surface: <br/><br/> $P_{hydrostatic} = \rho g h$ <br/><br/> where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth below the free surface. <br/><br/> Plugging in the given values: <br/><br/> $P_{hydrostatic} = 1000 \mathrm{~kg} \mathrm{~m}^{-3} \cdot 10 \mathrm{~ms}^{-2} \cdot 0.1 \mathrm{~m} = 1000 \mathrm{~Pa}$ <br/><br/> So, the total pressure difference inside the bubble compared to atmospheric pressure is the sum of the pressure difference due to surface tension and hydrostatic pressure: <br/><br/> $\Delta P_{total} = \Delta P + P_{hydrostatic} = 150 \mathrm{~Pa} + 1000 \mathrm{~Pa} = 1150 \mathrm{~Pa}$