QJEEM25Phef6f: JEE Main 2025 Physics - Class 11 JEEM25Phef6f Solution: An air bubble of radius 0.1 cm lies at a depth of 20 cm below…

$P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r}$

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

$P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2,$

the pressure difference due solely to surface tension is

$\Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2.$

The Laplace pressure difference for a bubble (which has one interface) is given by

$\Delta P = \frac{2S}{r}.$

The given bubble radius is 0.1 cm, which in SI units is

$r = 0.1\,\text{cm} = 0.001\,\text{m}.$

Substituting the known values into the Laplace equation gives

$\frac{2S}{0.001} = 100.$

Solving for $S$ :

$2S = 100 \times 0.001 = 0.1,$

$S = \frac{0.1}{2} = 0.05\,\text{N/m}.$

Thus, the surface tension of the liquid is

$\boxed{0.05\,\text{N/m}}.$