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A spherical water drop of radius R is split into 8 equal droplets. If T is the surface tension of water, what is the work done in the process? |
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Answer» SOLUTION :For a water drop, W = `4piR^2 xx T` `4/3 piR^3 = 4/3 pir^3 xx 8` `therefore` r = r/2` CHANGE in AREA = `4pir^2 xx 8 - 4piR^2` dA = `4pi (8r^2 - R^2) = 4piR^2` `therefore` change in ENERGY = T dA = `4piR^2T` |
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