If the balancing point in a meter-bridge from the left is 60cm, compare the resistances in the left and right gaps of meter bridge.

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Detailed Solution

Let R₁, R₂ be the resistances in the left and right gap of the meter bridge, then
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{60}{40} = \frac{l_{1}}{l_{2}} \frac{l_{1}}{l_{2}} = \frac{6}{2} \Rightarrow \frac{3}{2} = 1 .5 Ω$ .

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