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The work done to get 'n' smaller equal size spherical drops from a bigger size spherical drop of water is proportional to : |
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Answer» `1//n^(2//3)`-1 `(4piR^(3))/3=n(4pir^(3))/3rArrR^(3)=NR^(3)rArrr=n^(-1//3)R` Now INCREASE in area=`n4pir^(2)-4piR^(2)=4pi[nr^(2)-R^(2)]` `=4pi[n.n^(-2//3)R^(2)-R^(2)]` `=4piR^(2)[n^(1//3)-1]` `therefore` Work DONE `= T xx` increase in area `W=T.4piR^(2)[n^(1//3)-1]` `rArrWprop(n^(1//3)-1)` Hence the correct choice is (c). |
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