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

<p>$P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r}$</p> <p>Given that the pressure inside the bubble is $P_{\text{atm}} + 2100\,\text{N/m}^2$ and the liquid pressure at a depth of 20 cm is</p> <p>$$ P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2, $$</p> <p>the pressure difference due solely to surface tension is</p> <p>$$ \Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2. $$</p> <p>The Laplace pressure difference for a bubble (which has one interface) is given by</p> <p>$\Delta P = \frac{2S}{r}.$</p> <p>The given bubble radius is 0.1 cm, which in SI units is</p> <p>$r = 0.1\,\text{cm} = 0.001\,\text{m}.$</p> <p>Substituting the known values into the Laplace equation gives</p> <p>$\frac{2S}{0.001} = 100.$</p> <p>Solving for $S$:</p> <p>$2S = 100 \times 0.001 = 0.1,$</p> <p>$S = \frac{0.1}{2} = 0.05\,\text{N/m}.$</p> <p>Thus, the surface tension of the liquid is</p> <p>$\boxed{0.05\,\text{N/m}}.$</p>