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 $=\frac{64\times S\left(4\pi r^2\right)}{S\left(4\pi R^2\right)}$
$=\frac{64r^2}{R^2}$
$=\frac{64r^2}{16r^2}$
$=4:1$