A water drop of 0.01 cm^3 is squeezed betwee | vedclass

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(B) Given: Volume of water drop $V = 0.01 \ cm^3$,Area $A = 10 \ cm^2$,Surface tension $T = 70 \ dyne/cm$.When a drop is squeezed between two plates,i

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$14$

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(B) Given: Volume of water drop $V = 0.01 \ cm^3$,Area $A = 10 \ cm^2$,Surface tension $T = 70 \ dyne/cm$.When a drop is squeezed between two plates,it forms a thin film of thickness $t = V/A = 0.01 / 10 = 0.001 \ cm$.The pressure difference between the inside and outside of the film is given by $\Delta P = 2T / t$ (since there are two surfaces of the film in contact with air).$\Delta P = 2 	imes 70 / 0.001 = 140,000 \ dyne/cm^2$.The force required to separate the plates is $F = \Delta P 	imes A$.$F = 140,000 	imes 10 = 1,400,000 \ dyne$.Since $1 \ N = 10^5 \ dyne$,$F = 1,400,000 / 10^5 = 14 \ N$.

$A$ water drop of $0.01 \ cm^3$ is squeezed between two glass plates and spreads into an area of $10 \ cm^2$. If the surface tension of water is $70 \ dyne/cm$,then the normal force required to separate the glass plates from each other will be: (in $N$)

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