The excess pressure inside a soap bubble of r | vedclass

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(C) Given:Radius of soap bubble, $R = 0.5 \ cm = 0.5 	imes 10^{-2} \ m$Height of oil column, $h = 4 \ mm = 4 	imes 10^{-3} \ m$Density of the oil, $

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$4.5 	imes 10^{-2} \ N \ m^{-1}$

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(C) Given:Radius of soap bubble, $R = 0.5 \ cm = 0.5 	imes 10^{-2} \ m$Height of oil column, $h = 4 \ mm = 4 	imes 10^{-3} \ m$Density of the oil, $ho = 900 \ kg \ m^{-3}$Acceleration due to gravity, $g = 10 \ m \ s^{-2}$Excess pressure inside a soap bubble is given by $P = \frac{4S}{R}$, where $S$ is the surface tension.The pressure exerted by the oil column is $P' = ho g h$.According to the problem, the excess pressure is balanced by the oil column pressure:$\frac{4S}{R} = ho g h$Substituting the values:$\frac{4S}{0.5 	imes 10^{-2}} = 900 	imes 10 	imes 4 	imes 10^{-3}$$\frac{4S}{0.5 	imes 10^{-2}} = 36$$4S = 36 	imes 0.5 	imes 10^{-2}$$4S = 18 	imes 10^{-2}$$S = 4.5 	imes 10^{-2} \ N \ m^{-1}$

The excess pressure inside a soap bubble of radius $0.5 \ cm$ is balanced by the pressure due to an oil column of height $4 \ mm$. If the density of the oil is $900 \ kg \ m^{-3}$, then the surface tension of the soap solution is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

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