A drop of water with a volume of 0.05 cm^3 i | vedclass

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Question

(A) The pressure inside the liquid film is less than the atmospheric pressure outside by an amount $\Delta P = T \left( \frac{1}{r_1} + \frac{1}{r_2}

Vedclass · Accepted Answer

$45 \ N$

Vedclass · Answer

(A) The pressure inside the liquid film is less than the atmospheric pressure outside by an amount $\Delta P = T \left( \frac{1}{r_1} + \frac{1}{r_2} ight)$.Given that the film is thin,$r_1 = t/2$ and $r_2 = \infty$,where $t$ is the thickness of the film.Thus,the excess pressure is $\Delta P = \frac{2T}{t}$.The force required to separate the plates is $F = \Delta P 	imes A = \frac{2TA}{t}$.Since the volume $V = A 	imes t$,we have $t = V/A$.Substituting $t$ into the force equation: $F = \frac{2TA^2}{V}$.Given values: $A = 40 \ cm^2 = 40 	imes 10^{-4} \ m^2$,$V = 0.05 \ cm^3 = 0.05 	imes 10^{-6} \ m^3$,and $T = 70 \ dyne/cm = 70 	imes 10^{-3} \ N/m$.$F = \frac{2 	imes (70 	imes 10^{-3} \ N/m) 	imes (40 	imes 10^{-4} \ m^2)^2}{0.05 	imes 10^{-6} \ m^3}$.$F = \frac{140 	imes 10^{-3} 	imes 1600 	imes 10^{-8}}{0.05 	imes 10^{-6}} = \frac{224000 	imes 10^{-11}}{0.05 	imes 10^{-6}} = \frac{2.24 	imes 10^{-6}}{0.05 	imes 10^{-6}} = 44.8 \ N \approx 45 \ N$.

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