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Two small drops of mercury each of radius R coalesce to form a single drop. Find the ratio of the total surface energies before and after the change. |
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Answer» Solution :`2[4/3 pi R^3]d = 4/3 pi r^3d` `THEREFORE r^3 = 2R^3` `therefore r = 2^(1/3)R` `because E = T XX A` `therefore E_1/E_2 = (2(4piR)^2T)/(4pir^2T) = (2R^2)/r^2` = `(2R^2)/(2^(2/3)R^2) = 2^(1/3):1` |
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