Q 11-09-047JEE MainJEE Main 2025 (23 Jan, Shift 2)Medium
An air bubble of radius $1.0\ \text{mm}$ is observed at a depth of $20\ \text{cm}$ below the free surface of a liquid having surface tension $0.095\ \text{J/m}^2$ and density $10^3\ \text{kg/m}^3$. The difference between the pressure inside the bubble and atmospheric pressure is ______ $\text{N/m}^2$. (Take $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 2190
An air bubble inside a liquid has one surface, so its excess pressure is $2T/r$. The liquid pressure at depth $h$ adds to this:
$$P_{in} - P_0 = \rho gh + \frac{2T}{r} = 10^3\times10\times0.20 + \frac{2\times0.095}{1.0\times10^{-3}}$$
$$= 2000 + 190 = 2190\ \text{N/m}^2$$
Solution by Sreeraj P, M.Sc Physics