JEE Physics 2025 Question with Solution

Given: radius of bubble $r = 1.0 \, \text{mm} = 1 \times 10^{-3} \, \text{m}$, depth $h = 20 \, \text{cm} = 0.2 \, \text{m}$, surface tension $T = 0.095 \, \text{J/m}^2$, density $\rho = 10^3 \, \text{kg/m}^3$, and $g = 10 \, \text{m/s}^2$.

Find: the difference between pressure inside the bubble and atmospheric pressure.

An air bubble inside a liquid has excess pressure due to surface tension, and since it is at depth $h$, hydrostatic pressure also contributes.

First, hydrostatic pressure at depth is

$$P_{\text{hydro}} = \rho g h$$

So,

$$P_{\text{hydro}} = 10^3 \times 10 \times 0.2 = 2000 \, \text{Pa}$$

For an air bubble in a liquid, the excess pressure due to surface tension is

$$P_{\text{surface}} = \frac{2T}{r}$$

Thus,

$$P_{\text{surface}} = \frac{2 \times 0.095}{1 \times 10^{-3}} = 190 \, \text{Pa}$$

Therefore, pressure inside the bubble above atmospheric pressure is

$$P_{\text{diff}} = P_{\text{hydro}} + P_{\text{surface}} = 2000 + 190 = 2190 \, \text{Pa}$$

Therefore, the pressure difference is $2190 \, \text{N/m}^2$ and the correct option is A.

Given: $T = 0.095 \, \text{J/m}^2$, $r = 1.0 \, \text{mm} = 10^{-3} \, \text{m}$, $\rho = 10^3 \, \text{kg/m}^3$, $g = 10 \, \text{m/s}^2$, $h = 0.2 \, \text{m}$.

Find: $P_{\text{inside}} - P_{\text{atm}}$.

At depth $h$, the pressure just outside the bubble is

$$P_{\text{outside}} = P_{\text{atm}} + \rho g h$$

For an air bubble, inside pressure exceeds outside pressure by

$$P_{\text{inside}} - P_{\text{outside}} = \frac{2T}{r}$$

Hence,

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

Substituting the values,

$$P_{\text{inside}} - P_{\text{atm}} = 10^3 \times 10 \times 0.2 + \frac{2 \times 0.095}{10^{-3}}$$ $$P_{\text{inside}} - P_{\text{atm}} = 2000 + 190 = 2190 \, \text{N/m}^2$$

So, the required pressure difference is $2190 \, \text{N/m}^2$.

Note: another approach shown in the solution uses $\frac{4T}{r}$ and still concludes $2190 \, \text{N/m}^2$, but numerically $\frac{4T}{r} = 380 \, \text{Pa}$ would give $2380 \, \text{Pa}$. For an air bubble in a liquid, the correct excess pressure is $\frac{2T}{r}$, which matches option A.