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Two small drops of mercury each of radius r coalesce to form a single large drop of radius R. The ratio of the total surface energies before and after the change is …… |
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Answer» `1:2^((1)/(3))` Total volume of N drops = volume of large drop. Each small drop has radius r , then `n(4)/(3)pir^(3)=(4)/(3)piR^(3)` `thereforeR=n^((1)/(3))r` [Here n=2] `thereforeR=2^((1)/(3))r` (The ratio of surface ENERGIES of small drops and large drop) `(E_(i))/E_(f)=(2(4pir^(2)T))/(4piR^(2)T)` `=(2r^(2))/(R^(2))=(2r^(2))/(R^((2)/(3)r^(2)))` `=(2)/(2^((2)/(3)))=(2^((1)/(3)))/(1)` |
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