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(B) The total pressure required to blow an air bubble at a depth $h$ is the sum of the hydrostatic pressure at that depth and the excess pressure due

Vedclass · Accepted Answer

$0.0066 	imes 10^5$

Vedclass · Answer

(B) The total pressure required to blow an air bubble at a depth $h$ is the sum of the hydrostatic pressure at that depth and the excess pressure due to surface tension at the bubble interface.$1$. Hydrostatic pressure at depth $h = 1 \ cm = 0.01 \ m$ is $P_h = ho g h = 10^3 	imes 10 	imes 0.01 = 100 \ N/m^2$.$2$. The excess pressure due to surface tension for a bubble of radius $r = 0.025 \ cm = 2.5 	imes 10^{-4} \ m$ is $P_s = \frac{2T}{r} = \frac{2 	imes 7 	imes 10^{-2}}{2.5 	imes 10^{-4}} = \frac{14 	imes 10^{-2}}{2.5 	imes 10^{-4}} = 5.6 	imes 10^2 = 560 \ N/m^2$.$3$. Total excess pressure $P = P_h + P_s = 100 + 560 = 660 \ N/m^2$.$4$. Converting to the required scientific notation: $660 \ N/m^2 = 0.0066 	imes 10^5 \ N/m^2$.

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