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Question

A drop of water of volume 0.05 cm3 is pressed between two glass plates. as a consequence of which it spreads and occupies an area of 40 cm2. If the surface tension of water is 70 dyne/cm, find the normal force required to separate out the two glass plates in newton.

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Solution

Pressure inside the film is less than outside by an amount
p=T[1r1+1r2] /
Where r1 and r2 and the radii of curvature of the meniscus. Here r1=t/2 and r2= , then the force required to separate the two glass plates, between which the liquid film is enclosed, is
F=P×A=2ATt
Where t is the thickness of the film, A = area of film
r=t/2
F=2A2TAt=2A2TV=2×(40×104)2×(70×103)0.05×106=45N

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