QJEEM23Ph2c94: JEE Main 2023 Physics - Class 11 JEEM23Ph2c94 Solution: There is an air bubble of radius in a liquid of surface tension …

We can use the Young-Laplace equation to find the difference in pressure inside and outside the air bubble due to surface tension:

$\Delta P = 2 \frac{T}{R}$

where $\Delta P$ is the pressure difference, $T$ is the surface tension, and $R$ is the radius of the bubble.

Plugging in the given values:

$\Delta P = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{1.0 \mathrm{~mm}} = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{10^{-3} \mathrm{~m}} = 150 \mathrm{~Pa}$

Now, we need to account for the hydrostatic pressure due to the depth of the bubble below the free surface:

$P_{hydrostatic} = \rho g h$

where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth below the free surface.

Plugging in the given values:

$P_{hydrostatic} = 1000 \mathrm{~kg} \mathrm{~m}^{-3} \cdot 10 \mathrm{~ms}^{-2} \cdot 0.1 \mathrm{~m} = 1000 \mathrm{~Pa}$

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:

$\Delta P_{total} = \Delta P + P_{hydrostatic} = 150 \mathrm{~Pa} + 1000 \mathrm{~Pa} = 1150 \mathrm{~Pa}$