IIT-JEE 2007 · Paper 1, Physics question 6

Switch open. The capacitors are in series across 9 V: $C_\text{eq}=2\ \mu$F, so each carries 18 μC. The two plates joined at X carry $-18$ and $+18$ μC: the node X is neutral.

Switch closed. X is held at Y's potential. The resistors divide 9 V as 3 V (across 3 Ω) and 6 V (across 6 Ω), so the 3 μF capacitor (in parallel with 3 Ω) has 3 V, charge 9 μC, and the 6 μF capacitor has 6 V, charge 36 μC. The net charge on the plates at X is now $36-9=27$ μC (in magnitude).

Charge transferred. It came through the switch from Y to X: 27 μC.

(C)