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A drop of water in volume 0.05cc is pressed between two glass plates so that is spreads for an area of 40.89cmsquare. If the surface tension of water is 70 dyne/cm, What is the normal force required to separate the two given glass plates? |
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Answer» SOLUTION :PA = F = 2TA/d = `(2 xx 70 xx 40)/ (0.05/40)` ` = 4.5 xx 10^6` dyne |
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