A drop of liquid of radius R = 10 ^ - 2 m having surface tension S...

By mass conservation, ρ . \dfrac{4}{3}π{R}^{3} = ρ . K . \dfrac{4}{3}π{r}^{3}

$\to R = {K}^{\dfrac{1}{3}}r$

∴ ΔU = TΔA = T\left(K . 4π{r}^{2} - 4π{R}^{2}\right)

= T\left(K . 4π{R}^{2}{K}^{- \dfrac{2}{3}} - 4π{R}^{2}\right)

ΔU = 4π{R}^{2}T\left[{K}^{\dfrac{1}{3}} - 1\right]

Putting the value’s \to {10}^{- 3} = \dfrac{{10}^{- 1}}{4π} × 4π × {10}^{- 4}\left[{K}^{\dfrac{1}{3}} - 1\right]

$100 = {K}^{\dfrac{1}{3}} - 1$

\to {K}^{\dfrac{1}{3}} ≅ 100 = {10}^{2}

Given that K = {10}^{α} \to ∴ {10}^{\dfrac{α}{3}} = {10}^{2}

\to \dfrac{α}{3} = 2

\to α = 6