Answer:
Option D
Explanation:
Volume of big drop = n(Volume of small drop)
$\Rightarrow$ $\frac{4}{3}\pi R^{3}=n.\frac{4}{3}\pi r^{3}$
where, R= radius of big water drop
and r = radius of small water drop
$\Rightarrow$ $R^{3} =n r^{3}$
$\Rightarrow $ $R= n^{1/3}.r$ ........(i)
surface energy of n drops,
$E_{2}= n \times 4 \pi r^{2}\times T$
Surface energy of big drop ,
$E_{1}= 4 \pi R^{2}T$
$\therefore$ $\frac{E_{2}}{E_{1}}= \frac{nr^{2}}{R^{2}}= \frac{nr^{2}}{(n^{1/3}.r)^{2}}$
$= \frac{nr^{2}}{n^{2/3}.r^{2}}= n^{1/3}$ $ [\because $ from eq.(i)]
$E_{2}:E_{1}=\sqrt[3]{n}:1$
