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So, $x×50=200$

$⇒x=4$ capacitors

Effective capacitance in a row $=410 $

Now, let there are 'y' such rows,

So, $410 ×y=10$

$⇒y=4$ capacitor

So, the combinations of four rows each having $4$ capacitors each will form a $10μF$ capacitor that can withstand $200V$.

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