A drop of water of volume 0.05, cm^3 is press | vedclass

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(B) The thickness of the water film $d$ is given by $d = \frac{V}{A} = \frac{0.05\, cm^3}{40\, cm^2} = 1.25 	imes 10^{-3}\, cm$.The excess pressure i

Vedclass · Accepted Answer

$44.8$

Vedclass · Answer

(B) The thickness of the water film $d$ is given by $d = \frac{V}{A} = \frac{0.05\, cm^3}{40\, cm^2} = 1.25 	imes 10^{-3}\, cm$.The excess pressure inside the water film is $\Delta P = \frac{2T}{d}$.The force required to separate the plates is $F = \Delta P 	imes A = \frac{2T}{d} 	imes A = \frac{2T 	imes A}{V/A} = \frac{2T 	imes A^2}{V}$.Substituting the values: $F = \frac{2 	imes 70\, dyne/cm 	imes (40\, cm^2)^2}{0.05\, cm^3} = \frac{140 	imes 1600}{0.05} = 4,480,000\, dyne$.Since $1\, Newton = 10^5\, dyne$,$F = \frac{4,480,000}{10^5}\, N = 44.8\, N$.

$A$ drop of water of volume $0.05\, cm^3$ is pressed between two glass plates,as a consequence of which it spreads and occupies an area of $40\, cm^2$. If the surface tension of water is $70\, dyne/cm$,then the normal force required to separate out the two glass plates will be in Newton.

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