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(A) The pressure inside an air bubble at a depth $h$ is given by $P_{in} = P_0 + \rho gh + \frac{2T}{R}$.Here,$P_0$ is the atmospheric pressure,$\rho$

Vedclass · Accepted Answer

$2190$

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(A) The pressure inside an air bubble at a depth $h$ is given by $P_{in} = P_0 + ho gh + \frac{2T}{R}$.Here,$P_0$ is the atmospheric pressure,$ho$ is the density of the liquid,$g$ is the acceleration due to gravity,$h$ is the depth,$T$ is the surface tension,and $R$ is the radius of the bubble.The difference between the pressure inside the bubble and the atmospheric pressure is $\Delta P = P_{in} - P_0 = ho gh + \frac{2T}{R}$.Given values: $ho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$,$h = 20 \ cm = 0.2 \ m$,$T = 0.095 \ J/m^2$,and $R = 1.0 \ mm = 10^{-3} \ m$.Substituting these values:$\Delta P = (10^3 	imes 10 	imes 0.2) + \frac{2 	imes 0.095}{10^{-3}}$$\Delta P = 2000 + \frac{0.19}{10^{-3}}$$\Delta P = 2000 + 190 = 2190 \ N/m^2$.

An air bubble of radius $1.0 \ mm$ is observed at a depth of $20 \ cm$ below the free surface of a liquid having surface tension $0.095 \ J/m^2$ and density $10^3 \ kg/m^3$. The difference between pressure inside the bubble and atmospheric pressure is . . . . . . $N/m^2$. (Take $g = 10 \ m/s^2$)

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