Combination of Capacitors question from NEET 1999 (AIPMT)

Answer: (a)

1. Nodes: X = A, Y = D. Arms A–C, A–B, C–D, B–D are all $6\ \mu\text{F}$, with C–B as the bridge.
2. Bridge is balanced ($\frac{6}{6} = \frac{6}{6}$), so the C–B capacitor carries no charge.
3. Each path: two $6\ \mu\text{F}$ in series $= 3\ \mu\text{F}$; two paths in parallel: $6\ \mu\text{F}$.
4. So the answer is (a) $6\ \mu\text{F}$.