Explanation

(b) Let $m$ rows of $n$ series capacitor be taken then minimum number of capacitors required is
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Also effective voltage is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V^{''} = 1000 = n \times 250$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ n\ \ = \frac{1000}{250} = 4$
Also these four capacitor are connected in series then effective capacitance is
$\ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8}$
$\Rightarrow \ \ \ \ \ \ \ C^{''} = 2\mu F$
$\therefore\ \ \ \ \ \ \ \ C^{''''} = \ 16 = 2 \times m$
$\Rightarrow \ \ \ \ \ \ \ \ m = \frac{16}{2} = 8$
Hence $N = m \times n = 8 \times 4 = 32$