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Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change isA. `2 : 2^(2//3)`B. `2^(2//3) : 1`C. `2 : 1`D. `1 :2` |
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Answer» Correct Answer - A Suppose R be the radius of bigger drop, Then by equating volumes, we have `2(4//3pir^(3))=(4//3)piR^(3)" or "R=(2)^(1//3)r` Now, surface energy `prop` surface area `therefore" "U_(1)/U_(2)=A_(1)/A_(2)=(2r^(2))/R^(2)=2/2^(2//3)` |
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