In a Wheatstone's metre bridge, resistances R and S are connected in left and right gap, null point is obtained as l . After interchanging the resistances in the gaps, null point shifts to the left by 20 cm. If the series resistance of R and S is 2 k , then the value of resistance R is

Step-by-step solution

For a metre bridge, the balancing condition with resistances R and S in the left and right gaps respectively is given by: R S = l 100 - l When the resistances are interchanged, the new balancing length becomes 100 - l . It is given that the null point shifts to the left by 20 cm , so the new balancing length is l - 20 . Equating the two expressions for the new balancing length: 100 - l = l - 20 2l = 120 l = 60 cm Substituting l = 60 cm into the initial balancing condition: R S = 60 100 - 60 = 60 40 = 3 2 S = 2 3 R