Energy needed in the breaking of a drop of ra | vedclass

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Question

(A) The energy required to break a large drop into smaller drops is equal to the work done against surface tension,which is stored as surface energy.W

Vedclass · Accepted Answer

$(4\pi r^2 n - 4\pi R^2)S$

Vedclass · Answer

(A) The energy required to break a large drop into smaller drops is equal to the work done against surface tension,which is stored as surface energy.Work done $W = S 	imes \Delta A$,where $\Delta A$ is the change in surface area.Initial surface area $A_i = 4\pi R^2$.Final surface area $A_f = n 	imes 4\pi r^2$.Change in area $\Delta A = A_f - A_i = 4\pi r^2 n - 4\pi R^2$.Therefore,the energy needed is $W = S(4\pi r^2 n - 4\pi R^2)$.

Energy needed in the breaking of a drop of radius $R$ into $n$ drops of radius $r$ is given by : (With $S$ surface tension and $P$ atmospheric pressure)

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