In a meter bridge, the null point is found at a distance of 50 cm from the left end when two unknown resistors are connected in the left and right gaps. When a 12 resistor is connected in parallel with the resistor in the left gap, the null point shifts to 40 cm from the left end. The original value of the resistance in the left gap is:

Step-by-step solution

Let the initial resistances in the left and right gaps be P and Q respectively. Since the initial balance point is at 50 cm , we have: P Q = 50 100-50 = 1 P = Q When a 12 resistor is connected in parallel with P , the new equivalent resistance in the left gap is P' = 12P 12+P . The new balance point is at 40 cm from the left end. Thus: P' Q = 40 100-40 = 40 60 = 2 3 Substituting Q = P , we get: 12P 12+P P = 2 3 12 12+P = 2 3 36 = 24 + 2P 2P = 12 P = 6 Answer: 6