Q 11-09-049JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
An air bubble of radius $0.1\ \text{cm}$ lies at a depth of $20\ \text{cm}$ below the free surface of a liquid of density $1000\ \text{kg/m}^3$. If the pressure inside the bubble is $2100\ \text{N/m}^2$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $g = 10\ \text{m/s}^2$)
Answer: (B) $0.05$
An air bubble in a liquid has one surface:
$$P_{in} - P_0 = \rho gh + \frac{2T}{r}$$
$$2100 = 1000\times10\times0.20 + \frac{2T}{0.001} = 2000 + 2000\,T$$
$$T = \frac{100}{2000} = 0.05\ \text{N/m}$$
Solution by Sreeraj P, M.Sc Physics