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(D) Let $T$ be the surface tension of the liquid.The pressure inside an air bubble at a depth $h$ is given by $P_{	ext{in}} = P_0 + ho gh + \frac{2

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$0.05$

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(D) Let $T$ be the surface tension of the liquid.The pressure inside an air bubble at a depth $h$ is given by $P_{	ext{in}} = P_0 + ho gh + \frac{2T}{R}$.Given that the pressure inside the bubble is $2100 \ N/m^2$ greater than the atmospheric pressure $(P_0)$,we have $P_{	ext{in}} - P_0 = 2100 \ N/m^2$.Substituting the given values: $R = 0.1 \ cm = 10^{-3} \ m$,$h = 20 \ cm = 0.2 \ m$,$ho = 1000 \ kg/m^3$,and $g = 10 \ m/s^2$.$2100 = ho gh + \frac{2T}{R}$$2100 = (1000 	imes 10 	imes 0.2) + \frac{2T}{10^{-3}}$$2100 = 2000 + \frac{2T}{10^{-3}}$$100 = \frac{2T}{10^{-3}}$$2T = 100 	imes 10^{-3} = 0.1$$T = 0.05 \ N/m$.

An air bubble of radius $0.1 \ cm$ lies at a depth of $20 \ cm$ below the free surface of a liquid of density $1000 \ kg/m^3$. If the pressure inside the bubble is $2100 \ N/m^2$ greater than the atmospheric pressure,then the surface tension of the liquid in $SI$ unit is (use $g=10 \ m/s^2$)

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