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A drop of radius r is broken into n equal drips. Calulate the work done if surface tension of water is T.A. `4rpiR^(2)nT`B. `4rpiR^(2)T(n^(2//3-1))`C. `4rpiR^(2)T(n^(1//3-1))`D. None of the above |
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Answer» The volume of n smaller drop =volume of bigger drop `:. " " n(4)/(3)pir^(3)=(4)/(3)piR^(2)` `:. " " R=n^(1//3)r` `:. " " r=(R)/(n^(1//3))` Here, R=radius of bigger drop r=radius of smallar drop `:. W=4pi(nr^(2)-R^(2))T` `4pi[n((R)/(n^(1//2)))^(2)-R^(2)]T=4pi{(nR^(2))/(n^(2//3))-R^(2)}T` `=4piR^(2)(n^(1-2//3)-1)T=4piR^(2)T(n^(1//3)-L)` |
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