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The diameter of one drop of water is 0.2 cm. The work done in breaking onedrop into 1000 equal droplets will be:-(surface tension of water = 7 * 10?N/m) |
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Answer» Diameter of one drop of water = 0.2 cmSurface tension = 7 × 10^2 N/mTo find :Work done in breaking onedrop into 1000 EQUAL dropletsSolution :Let R and r be tha radius of big and SMALL droplet respectively• Volume in whole PROCESS would REMAIN sameTherefore,(4/3)×pi×R^3 = 1000×(4/3)×pi×r^3r = R/10 = 0.01 cm• Change in surface area is GIVEN as∆A = 1000×4×pi×r^2 - 4×pi×R^2 = 4×pi×(10R^2 - R^2) = 36×pi×R^2• Therefore, work done would beW = ∆U = T∆AW = 7×10^-2×36×pi×10^-6W = 7.9 × 10^-6 J |
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