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(C) The pressure inside a soap bubble is given by $P = P_{0} + \frac{4T}{R}$,where $P_{0}$ is the atmospheric pressure,$T$ is the surface tension,and

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(C) The pressure inside a soap bubble is given by $P = P_{0} + \frac{4T}{R}$,where $P_{0}$ is the atmospheric pressure,$T$ is the surface tension,and $R$ is the radius of the bubble. The pressure at a depth $Z_{0}$ below the free surface of water is given by $P = P_{0} + ho g Z_{0}$. Equating the two pressures: $P_{0} + ho g Z_{0} = P_{0} + \frac{4T}{R}$. Simplifying,we get $ho g Z_{0} = \frac{4T}{R}$. Solving for $Z_{0}$: $Z_{0} = \frac{4T}{ho g R}$. Substituting the given values: $T = 2.5 	imes 10^{-2}\; N/m$,$R = 1\; mm = 10^{-3}\; m$,$ho = 10^{3}\; kg/m^{3}$,and $g = 10\; m/s^{2}$. $Z_{0} = \frac{4 	imes 2.5 	imes 10^{-2}}{10^{3} 	imes 10 	imes 10^{-3}} = \frac{10^{-1}}{10} = 10^{-2}\; m$. Converting to centimeters: $Z_{0} = 10^{-2} 	imes 100\; cm = 1\; cm$.

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