Question

If n drops, each of capacitance C, coalesce to form a single big drop, the capacitance of the big drop will be

• A. $\frac{V}{n^{2/3}}$
  
• B. $\frac{V}{n^{1/3}}$
  
• C. $V{n}^{1/3}$
  
• D. $V{n}^{2/3}$

Answer 1

The correct option is D

$V{n}^{2/3}$

If $\rho$ is the density of a small drop and r its radius, then the mass of each small drop is $m=\frac{4\pi }{3}{r}^3\rho .$ If n such drops coalesce to form a big drop of radius R, then the mass of the big drop is n m=$\frac{4\pi }{3}{R}^3\rho .$
Now, by conservation of mass:
n*$\frac{4\pi }{3}{r}^3\rho .$ = $\frac{4\pi }{3}{R}^3\rho .$
Hence, R=$n^{\frac{1}{3}}r.$
Now, the capacitance of a sphere is proportional to its radius.
Hence the capacitance of the big drop will be $C^{\prime }=n^{\frac{1}{3}}C.$
Hence, the correct choice is (d).