Discussion Forum

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

Answer:

Explanation:

From mass conservation

         $\rho\frac{4}{3}\pi R^{3}=\rho K\frac{4}{3}\pi r^{3}$ 

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

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

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

$\triangle U=4\pi R^{2}T[K^{\frac{1}{3}}-1]$

Putting the values, we get

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

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

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

   Given that K= $10^{\alpha}$

$\therefore$    $10^{\frac{\alpha}{3}}=10^{2}$

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

   $\Rightarrow$  $\alpha=6$