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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.S

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.Surface energy $U = S 	imes \Delta A$,where $S$ is the surface tension and $\Delta A$ is the change in surface area.Initial surface area of the drop of radius $R$ is $A_i = 4\pi R^2$.Final surface area of $n$ drops of radius $r$ is $A_f = n 	imes 4\pi r^2$.Change in surface area $\Delta A = A_f - A_i = 4\pi r^2 n - 4\pi R^2$.Therefore,the energy required is $U = (4\pi r^2 n - 4\pi R^2)S$.

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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