The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is
[ $T$ = surface tension of the liquid]

Question

The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is [ T = surface tension of the liquid]

TestHub · Accepted Answer

Volume of the initial drop is equal to the total volume of $n$ droplets.

$\frac{4}{3} π R^{3} = n \times \frac{4}{3} π r^{3}$

$n = \frac{R ^{3}}{r ^{3}}$

Initial surface area of the drop, $A_{1} = 4 π R^{2}$

Final total surface area of $n$ droplets, $A_{2} = n \times 4 π r^{2}$

Substituting the value of $n$ :

$A_{2} = (\frac{R ^{3}}{r ^{3}}) \times 4 π r^{2} = 4 π \frac{R ^{3}}{r}$

Increase in surface area, $Δ A = A_{2} - A_{1} = 4 π \frac{R ^{3}}{r} - 4 π R^{2}$

$Δ A = 4 π R^{2} [\frac{R}{r} - 1]$

Energy required to break the drop, $E = T \times Δ A$

$E = 4 π T R^{2} [\frac{R}{r} - 1]$

Answer: $4 π T R^{2} [\frac{R}{r} - 1]$

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