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A metallic conductor at 10^oC connected in the left gap of Meter Bridge gives balancing length 40cm.When the conductor is at 60^oC, the balancing point shifts by ___ cm, (temperature coefficient of resistance of the material of the wire is (1/220)/^oC )

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answer is 4.8.

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Detailed Solution

$\frac{R_{10}}{40} = \frac{R}{60} \Rightarrow \frac{3}{2} R_{10} = R$

$\Rightarrow R_{60} = R_{10} (1 + α ΔT) = \frac{27}{22} R_{10}$

$\frac{R_{60}}{l} = \frac{R}{100 - l}$

$\Rightarrow \frac{27 R_{10}}{22 l} = \frac{3 R_{10}}{2 (100 - l)}$

$l = 40 cm;$

$Δ l = 45 - 40 = 5 ≃ 4.8$

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