In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will: (All resistance given in ohms)

In standard Wheatstone bridge problems where the bridge is unbalanced, the direction of current flow is determined by comparing the potentials at points A and B ($V_A$ and $V_B$). When calculated for the specific resistances given in the AIPMT circuit, the potential at B is found to be greater than the potential at A ($V_B>V_A$). Since conventional current flows from higher potential to lower potential, it will flow from B to A.