n drops of a liquid,each with surface energy | vedclass

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(A) Let $r$ be the radius of each small drop and $R$ be the radius of the large drop.Surface energy of one small drop is $E = T(4 \pi r^2)$.Volume con

Vedclass · Accepted Answer

Some energy will be released in the process.

Vedclass · Answer

(A) Let $r$ be the radius of each small drop and $R$ be the radius of the large drop.Surface energy of one small drop is $E = T(4 \pi r^2)$.Volume conservation: $\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$,which gives $R = n^{1/3} r$.Surface energy of the large drop is $E' = T(4 \pi R^2) = T(4 \pi (n^{1/3} r)^2) = T(4 \pi r^2) n^{2/3} = E n^{2/3}$.Total initial surface energy is $nE$.Total final surface energy is $E' = E n^{2/3}$.Since $n > n^{2/3}$ for $n > 1$,the initial energy is greater than the final energy.Energy released = $nE - E' = nE - E n^{2/3} = E(n - n^{2/3})$.Thus,energy is released in the process.

$n$ drops of a liquid,each with surface energy $E$,join to form a single drop. Which of the following statements is correct?

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