In a meter bridge, the wire of length 1m has a non-uniform cross-section such that, the variation \frac{dR}{dℓ} of its resistance R with length ℓ \textrm{ }\textrm{ } is \textrm{ }\textrm{ } \frac{dR}{d \text{l}} \textrm{ } \propto \textrm{ } \frac{1}{\sqrt{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?

Engineering

Physics

Instruments

Question

In a meter bridge, the wire of length 1m has a non-uniform cross-section such that, the variation $\frac{dR}{dℓ}$ of its resistance R with length $ℓ is \frac{dR}{d l} \propto \frac{1}{\sqrt{ℓ}}$ . 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?

0.2 m

0.3 m

0.35 m

0.25 m

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$\frac{dR}{dℓ} = \frac{k}{\sqrt{ℓ}} k = constant$

$\int_{0}^{R} dR = k \int_{0}^{1} \frac{dℓ}{\sqrt{ℓ}}$

R = 2k resistance of wire AB.

$Again, \int_{0}^{R / 2} dR = k \int_{0}^{L} \frac{dℓ}{\sqrt{ℓ}} L \to Length AP$

$\frac{R}{2} = k 2 L^{1 / 2}; k = k 2 L^{1 / 2}$

$\Rightarrow L = \frac{1}{4} m = 0 .25 m$

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