An air bubble, having a radius of $2 \mathrm{~mm}$, is formed in water having a surface tension of $

Question

An air bubble, having a radius of $2 \mathrm{~mm}$, is formed in water having a surface tension of $8.0 	imes 10^{-2} \mathrm{~N} / \mathrm{m}$. The pressure inside the bubble equals the atmospheric pressure plus the hydrostatic pressure at a point $Z_0$ below the free surface of the water. Taking $g=10 \mathrm{~m} / \mathrm{s}^{2}$ and density of water $=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$, the value of $Z_0$ is

Professor PC Thomas Classes · Answer

(c) : The pressure at a point $Z_0$ below the surface of water is $P_{Z_0} = P_0 + ho g Z_0$. The pressure inside an air bubble in a liquid is $P = P_0 + \frac{2T}{R}$. As per the question, $P_{Z_0} = P$. Therefore, $P_0 + \frac{2T}{R} = P_0 + ho g Z_0$. This simplifies to $Z_0 = \frac{2T}{R ho g}$. Substituting the given values, $Z_0 = \frac{2 	imes 8.0 	imes 10^{-2}}{2 	imes 10^{-3} 	imes 10^3 	imes 10} = \frac{16 	imes 10^{-2}}{20} = 0.8 	imes 10^{-2} \mathrm{~m} = 0.8 \mathrm{~cm}$.