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Question:

In a meter bridge, the wire of length 1m has a non-uniform cross-section such that the variation dR/dl of its resistance R with length l is dR/dl ∝ 1/√l. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?

0.2m

0.25m

0.35m

0.3m

Solution:

For the given wire : dR = C dl/√l, where C = constant
Let resistance of part AP is R1 and PB is R2
∴ R1/R2 = R1/R2 or R1 = R2
By balanced Wheatstone bridge concept
Now ∫dR = c∫dl/√l
∴ R1 = c∫₀ˡ √l dl = C . 2 . √l
R2 = c∫l¹ √l dl = C (2√l - 2√l)
Putting R1 = R2
C2√l = C(2√1 - 2√l)
∴ 2√l = 2 - 2√l
4√l = 2
√l = 1/2
l = 1/4 m
→ 0.25m