A drop of liquid of radius R = 10 - 2 m having surface tension S = 0.1 4 π N m - 1 divides itself into…

By mass conservation, $\rho .\frac{4}{3} \pi R^3=\rho .K.\frac{4}{3}\pi r^3$

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

$\therefore \mathrm{\Delta }U=T \mathrm{\Delta }A=T \left(K. 4\pi r^2-4\pi R^2\right)$

$=T \left(K. 4\pi R^2{K}^{-\frac{2}{3}}-4\pi R^2\right)$

$\mathrm{\Delta }U=4\pi R^{2}T \left[K^{\frac{1}{3}}-1\right]$

Putting the value’s  $\Rightarrow {10}^{-3}=\frac{{10}^{-1}}{4\pi }\times 4\pi \times {10}^{-4}\left[K^{\frac{1}{3}}-1\right]$

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

$\Rightarrow K^{\frac{1}{3}} \cong 100={10}^2$

Given that  $K={10}^{\alpha } \Rightarrow \therefore {10}^{\frac{\alpha }{3}}={10}^2$

$\Rightarrow \frac{\alpha }{3}=2$

$\Rightarrow \alpha =6$