A water drop of $0.05\ cm^{3}$ is squeezed between two glass plates and spreads into area of $40\ cm^{2}$. If the surface tension of water is $70\ dyne\ cm^{- 1}$ then the normal force required to separate the glass plates from each other will be

Question

Dr. Abubacker · Accepted Answer

(b) $F\ t = 2\ SA$ or $F = \frac{2SA}{t} = \frac{2\ SA^{2}}{At} = \frac{2\ SA^{2}}{V}$
$F = \frac{2 	imes 70 	imes (40)^{2}}{0.05} = 44.8 	imes 10^{5}dyne$
$= 44.8\ N \approx 45\ N$

Q. A water drop of $0.05\ cm^{3}$ is squeezed between two glass plates and spreads into area of $40\ cm^{2}$. If the surface tension of water is $70\ dyne\ cm^{- 1}$ then the normal force required to separate the glass plates from each other will be

Explanation

Reviewed by Dr. Abubacker

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