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A drop of liquid of radius $R = 10^{- 2} m$ having surface tension $S = \frac{0.1}{4 π} {Nm}^{- 1}$ divides itself into K identical drops. In this process the total change in the surface energy $ΔU = 10^{- 3} J . If K = 10^{α}$ the value of $α$ is:

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answer is 6.

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Detailed Solution

By mass conservation,

$ρ \cdot \frac{4}{3} π R^{3} = ρ \cdot K \cdot \frac{4}{3} π r^{3}$

$\Rightarrow R = K^{1 / 3} r$

$∴ ΔU = TΔA = T (K \cdot 4 π r^{2} - 4 π R^{2})$

$= T (K .4 π R^{2} K^{- 2 / 3} - 4 π R^{2})$

$ΔU = 4 π R^{2} T [K^{1 / 3} - 1]$

Putting the values

$\Rightarrow 10^{- 3} = \frac{10^{- 1}}{4 π} \times 4 π \times 10^{- 4} [K^{1 / 3} - 1]$

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