A big drop is formed by coalescing 1000 small droplets of water. What will be the loss in surface energy if E is surface energy of big drop

$\frac{4}{3}\pi R^3=1000\times \frac{4}{3}\pi r^3\Rightarrow R=10r$ 
(or) $r=\frac{R}{10}$  
Surface energy of 1000 droplets = $1000(T\times 4\pi r^2)$  
$=1000(T\times 4\pi \frac{R^2}{100})=10(T\times 4\pi r^2)$ 
Surface energy of big drop = $T\times 4\pi R^2=E$  
Loss in energy = $10E-E=9E$