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(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$, where $S$ is the surface tension and $R$ is the radius of the bubb

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$7$

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(B) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4S}{R}$, where $S$ is the surface tension and $R$ is the radius of the bubble.This pressure is balanced by the hydrostatic pressure of the liquid column: $\Delta P = ho g h$.Equating the two: $ho g h = \frac{4S}{R}$.Given: $ho = 8 	imes 10^3 \,kg \,m^{-3}$, $g = 10 \,m \,s^{-2}$, $h = 0.04 \,cm = 4 	imes 10^{-4} \,m$, and $S = 0.28 \,N \,m^{-1}$.Substituting the values: $(8 	imes 10^3) 	imes 10 	imes (4 	imes 10^{-4}) = \frac{4 	imes 0.28}{R}$.$32 = \frac{1.12}{R}$.$R = \frac{1.12}{32} \,m = 0.035 \,m = 3.5 \,cm$.The diameter $D = 2R = 2 	imes 3.5 \,cm = 7 \,cm$.

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