A uniform wire AB of length 100 cm is connected across a voltage source. Two 10 resistors are connected in series across the ends A and B , with the first resistor connected to end A and the second to end B . A galvanometer connects the junction of the two resistors to a sliding contact on the wire AB . Initially, the galvanometer shows zero deflection when the contact point is at the center of the wire. When an unkn

Step-by-step solution

Initially, the bridge is balanced at the center ( 50 cm ) because both resistors are 10 . When R_x is connected in parallel with the left 10 resistor, the equivalent resistance of the left arm becomes R_p = 10 R_x 10 + R_x . The null point shifts by 10 cm towards end A , so the new balancing length from A is l₁ = 50 - 10 = 40 cm . The remaining length of the wire is l₂ = 100 - 40 = 60 cm . Applying the balanced Wheatstone bridge condition: R_p 10 = l₁ l₂ R_p 10 = 40 60 = 2 3 R_p = 20 3 Equating this to the parallel