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(C) The excess pressure inside a soap bubble is given by $P = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble, bec

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$58.8 	imes 10^{-3} \,N/m$

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(C) The excess pressure inside a soap bubble is given by $P = \frac{4T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble, because a soap bubble has two liquid-gas interfaces.The pressure exerted by a water column of height $h$ is given by $P = ho h g$.Given that the excess pressure is equal to the pressure of the water column, we have $\frac{4T}{r} = ho h g$.Rearranging for surface tension $T$, we get $T = \frac{r ho h g}{4}$.Substituting the given values: $r = 3 \,mm = 3 	imes 10^{-3} \,m$, $h = 0.8 \,cm = 8 	imes 10^{-3} \,m$, $ho = 1000 \,kg/m^3$, and $g = 9.8 \,m/s^2$.$T = \frac{(3 	imes 10^{-3} \,m) 	imes (1000 \,kg/m^3) 	imes (8 	imes 10^{-3} \,m) 	imes (9.8 \,m/s^2)}{4}$.$T = \frac{3 	imes 10^{-3} 	imes 10^3 	imes 8 	imes 10^{-3} 	imes 9.8}{4} \,N/m$.$T = \frac{3 	imes 8 	imes 10^{-3} 	imes 9.8}{4} \,N/m = 6 	imes 9.8 	imes 10^{-3} \,N/m = 58.8 	imes 10^{-3} \,N/m$.

If the excess pressure inside a soap bubble of radius $3 \,mm$ is equal to the pressure of a water column of height $0.8 \,cm$, then the surface tension of the soap solution is ( $\rho_{\text{water}} = 1000 \,kg/m^3, g = 9.8 \,m/s^2$ ).

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