Question

$ 64 $ identical water droplets combine to form a large drop. Find the ratio of the total surface energy of $ 64 $ droplets to that of the large drop.

• A. $ 64 : 1 $
• B. $ 4 : 1 $
• C. $ 8 : 1 $
• D. $ 1 : 8 $

Answer 1

Answer: (B)

Let the radius of small water droplet is $r$
Volume of one small water droplet, $V=\frac{4}{3}\pi r^{3}$
Total volume $=nV=64\times \frac{4}{3} \pi\,r^{3}$ $(\because n=64)$
Volume of larger drop $=\frac{4}{3} \pi \,R^{3}$
$\therefore \frac{4}{3}\pi\,R^{3}=64 \times \frac{4}{3}\pi\,r^{3}$
$\Rightarrow R=4r$
Surface energy of one small droplet $=S(4 \pi\, r^{2})$
Surface energy of large droplet $=S(4\,\pi\,R^{2})$
Required ratio $=\frac{64\times S\left(4\pi r^{2}\right)}{S\left(4\pi R^{2}\right)}$
$=\frac{64\,r^{2}}{R^{2}}$
$=\frac{64\,r^{2}}{16\,r^{2}}$
$=4:1$