A known resistance R is connected in the left gap and an unknown wire is connected in the right gap of a meter bridge. The balancing length is found to be 20 cm from the left end. The wire is then removed, uniformly stretched to increase its length by 50 % , and reconnected in the right gap. The new balancing length from the left end is:

Step-by-step solution

Let the initial resistance of the wire be X . From the meter bridge balancing condition, R 20 = X 100 - 20 R 20 = X 80 X = 4R When the wire is uniformly stretched to increase its length by 50 % , its new length is L' = 1.5L . Since the volume of the wire remains constant, its cross-sectional area becomes A' = A 1.5 . The new resistance X' is: X' = L' A' = 1.5L A 1.5 = 2.25 ( L A ) = 2.25X Substituting X = 4R : X' = 2.25 4R = 9R Let the new balancing length be l' . Applying the balancing condition again: R l' = X' 1