Thank you for reporting, we will resolve it shortly
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
KCETKCET 2002Electrostatic Potential and Capacitance
Solution:
Here $R^3 = (4r)^3 \, \, i.e. \, R = 4r$
$\frac{\sigma_1}{\sigma_2} = \frac{r^2}{4r^2} = \frac{1}{4}$