b
(b) Suppose \(C = 8\, µF ,\, C' = 16\,µF\)
and \(V = 250\, V,\, V' = 1000\,V\)
Suppose \(m\) rows of given capacitors are connected in parallel and each row contains \(n\) capacitors then potential difference across each capacitor \(V = \frac{{V'}}{n}\) and equivalent capacitance of network \(C' = \frac{{mC}}{n}\) on putting the values we get \(n = 4\) and \(m = 8\)
 Total capacitors \(= n × m = 4 × 8 = 32\)
Short Trick : For such type of problems number of capacitors = \(\frac{{C'}}{C} \times {\left( {\frac{{V'}}{V}} \right)^2} = \frac{{16}}{8}{\left( {\frac{{1000}}{{250}}} \right)^2} = 32\)