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The work done to split a liquid drop id radius `R` into `N` identical drops is (taken `sigma` as the surface tension of the liquid)A. `4piR^(2)(N^(-1//3)-1)sigma`B. `4piR^(2)N sigma`C. `4piR^(2)(N^(-1//2)-1)`D. None of these |
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Answer» Correct Answer - A `N((4)/(3) pir^(3))=(4)/(3) piR^(3)` `:. r=R((1)/(N))^(1//3)` `W=sigma (Delta A)` `= sigma (A_(f)-A_(i))` `=sigma [N(4 pi r^(2))- 4 pi R^(2)]` `=4 pi sigma [NR^(3)((1)/(N))^(2//3)-R^(2)]` `=4sigma pi R^(2) (N^(-1//3)-1)`. |
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