Q. 64 small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is
A) 1 : 64
B) 64 : 1
C) 4 : 1
D) 1 : 4
Explanation
(d) $\frac{\sigma_{small}}{\sigma_{Big}} = \frac{q}{Q} \times \frac{R^{2}}{r^{2}} = \frac{q}{(nq)} \times \frac{\left( n^{1/3}r \right)^{2}}{r^{2}} = n^{- 1/3} = (64)^{- 1/3} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.