Question

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

Answer 1

$\frac{4}{3}\pi R^3=K\frac{4}{3}\pi r^3,R={10}^{-2}$

$r=R\times k^{\frac{-1}{3}}$

$dA=(k\times 4\pi r^2)-4\pi R^2$

$dA=4\pi R^2[k^{\frac{1}{3}}-1]$

$\Delta U=T\times dA$

${10}^{-3}=\frac{0.1}{4\pi }\times 4\pi R^2[k^{\frac{1}{3}}-1]$

${10}^2=[k^{\frac{1}{3}}-1]$

$k\approx {10}^6$

$\alpha =6$