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Q.

The balance point in meter bridge experiment is obtained at 30cm from the left. If the right gap contains 3.5 $Ω$ ,what is the resistance in the left gap?

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Detailed Solution

$R = S \cdot \frac{l_{1}}{100 - l_{1}}$
Given that $l_{1} = 30 cm, S = 3 .5 Ω$
$∴ R = 3 .5 \times \frac{30}{100 - 30} = 3 .5 \times \frac{30}{70} = 1 .5 Ω$

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