Explanation

(c) After combining, the volume remains same $i.e.,\ $
Volume of bigger drop$= N\ \times$ volume of smaller drop
Or
$\frac{4}{3}\pi R^{3} = N \times \frac{4}{3}\pi r^{3}$
Or
$N = \left( \frac{R}{r} \right)^{3}\ldots\ldots\ldots\ldots\ldots(i)$
As charge is conserved, hence
$\mathcal{Q =}Nq\ldots\ldots\ldots\ldots\ldots\ldots.(ii)$
Capacity of bigger drop =$4\pi\varepsilon_{0}R$
Capacity of smaller drop=4$\pi\varepsilon_{0}r$
From Eq. (ii), we have
$\left( 4\pi\varepsilon_{0}R \right)V_{big} = N\left( 4\pi\varepsilon_{0}r \right)V_{small}$
or$\ \left( 4\pi\varepsilon_{0}R \right) \times 40 = N\left( 4\pi\varepsilon_{0}r \right) \times 10$
$or\ 4R = Nr$
$or\ \ \ \frac{R}{r} = \frac{N}{4}\ \ \ \ \ldots\ldots\ldots\ldots.(iii)$
From Eqs. (i) and (iii), we have
$N = \left( \frac{N}{4} \right)^{3}$
or
$N = \frac{N^{3}}{64}$
or
$N^{2} = 64$
OR N=8