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

<p>To find the difference between the pressure inside the air bubble and the atmospheric pressure, we use the fact that at depth $h$ in a liquid of density $\rho$, the external (hydrostatic) pressure exceeds atmospheric by $\rho g h$. Additionally, for a spherical bubble in a liquid, the internal pressure exceeds the external pressure by $\frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble.</p> <p>Hence, the internal pressure of the bubble is</p> <p>$ P_\text{inside} = P_\text{atm} + \rho g h + \frac{2T}{r}. $</p> <p>Therefore, the difference between the bubble's internal pressure and the atmospheric pressure is</p> <p>$ \Delta P = P_\text{inside} - P_\text{atm} = \rho g h + \frac{2T}{r}. $</p> <p>Substituting the given values:</p> <p><p>$\rho = 10^3 \,\text{kg/m}^3$</p></p> <p><p>$g = 10 \,\text{m/s}^2$</p></p> <p><p>$h = 0.20 \,\text{m}$</p></p> <p><p>$T = 0.095 \,\text{J/m}^2$</p></p> <p><p>$r = 1.0 \times 10^{-3} \,\text{m}$</p></p> <p>we compute each term:</p> <p><p><strong>Hydrostatic term</strong>:</p> <p>$ \rho g h = (10^3 \,\text{kg/m}^3) \cdot (10 \,\text{m/s}^2) \cdot (0.20 \,\text{m}) = 2000 \,\text{N/m}^2. $</p></p> <p><p><strong>Surface tension term</strong>:</p> <p>$ \frac{2T}{r} = \frac{2 \times 0.095 \,\text{J/m}^2}{1.0 \times 10^{-3} \,\text{m}} = \frac{2 \times 0.095}{10^{-3}} = 2 \times 95 = 190 \,\text{N/m}^2. $</p></p> <p>Hence,</p> <p>$ \Delta P = 2000 \,\text{N/m}^2 + 190 \,\text{N/m}^2 = 2190 \,\text{N/m}^2. $</p> <p>Thus, the required pressure difference between the inside of the bubble and the atmospheric pressure is</p> <p>$ \boxed{2190 \,\text{N/m}^2.} $</p>