QJEEM25Phe7b3: JEE Main 2025 Physics - Class 11 JEEM25Phe7b3 Solution: An air bubble of radius 1.0 mm is observed at a depth 20 cm…

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.

Hence, the internal pressure of the bubble is

$P_\text{inside} = P_\text{atm} + \rho g h + \frac{2T}{r}.$

Therefore, the difference between the bubble's internal pressure and the atmospheric pressure is

$\Delta P = P_\text{inside} - P_\text{atm} = \rho g h + \frac{2T}{r}.$

Substituting the given values:

$\rho = 10^3 \,\text{kg/m}^3$

$g = 10 \,\text{m/s}^2$

$h = 0.20 \,\text{m}$

$T = 0.095 \,\text{J/m}^2$

$r = 1.0 \times 10^{-3} \,\text{m}$

we compute each term:

Hydrostatic term:

$\rho g h = (10^3 \,\text{kg/m}^3) \cdot (10 \,\text{m/s}^2) \cdot (0.20 \,\text{m}) = 2000 \,\text{N/m}^2.$

Surface tension term:

$\frac{2T}{r} = \frac{2 \times 0.095 \,\text{J/m}^2}{1.0 \times 10^{-3} \,\text{m}} = \frac{2 \times 0.095}{10^{-3}} = 2 \times 95 = 190 \,\text{N/m}^2.$

Hence,

$\Delta P = 2000 \,\text{N/m}^2 + 190 \,\text{N/m}^2 = 2190 \,\text{N/m}^2.$

Thus, the required pressure difference between the inside of the bubble and the atmospheric pressure is

$\boxed{2190 \,\text{N/m}^2.}$