$(A)\;\large\frac{1600}{9} \\ (B)\;\large\frac{800}{9} \\ (C)\; 400 \\ (D)\;800 $

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In series,

Equivalent $C=\large\frac{20 \times 40}{20 + 40}$

$\qquad= \large\frac{40}{3}$$ \mu F$

Charge on each capacitor $ =\large\frac{40}{3}$$ \times 400 $

$\qquad= \large\frac{16000}{3}$$\mu c$

In series charge on both capacitor is equal.

=> Total charge $=\large\frac{2 \times 16000}{3} $

$\qquad = \large\frac{32000}{3}$$\mu c$

New capacitors are connected in parallel.

Equivalent $C= 20+40=60 \mu F$

Potential $V'= \large\frac{Q'}{C'}$

$\qquad= \large\frac{32000}{3 \times 60}$

$\qquad= \large\frac{1600}{9}$

Hence A is the correct answer.

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