Two resistances \(X\) and \(Y\) in the two gaps of a meterbridge give a null point dividing the wire in the ratio 4 : 6. If both resistances are increased by 30 \(\Omega\), the null point divides the wire in the ratio 5 : 6. Calculate the value of X…

Q.23 Two resistances \(X\) and \(Y\) in the two gaps of a meterbridge give a null point dividing the wire in the ratio 4 : 6. If both resistances are increased by 30 \(\Omega\), the null point divides the wire in the ratio 5 : 6. Calculate the value of X and Y.

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: \(X=20\ \Omega\), \(Y=30\ \Omega\)

At balance in a metre bridge, \(\dfrac{X}{Y}=\dfrac{l_X}{l_Y}\).

Case 1: \(\dfrac{X}{Y}=\dfrac{4}{6}\Rightarrow Y=\dfrac{3X}{2}\) ...(1)

Case 2: \(\dfrac{X+30}{Y+30}=\dfrac{5}{6}\Rightarrow 6X+180=5Y+150\) ...(2)

Substituting (1) in (2): \(6X+180=7.5X+150\Rightarrow1.5X=30\Rightarrow X=20\ \Omega\).

Then \(Y=\dfrac{3\times20}{2}=30\ \Omega\).

Check: \(\dfrac{50}{60}=\dfrac{5}{6}\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Two resistances \(X\) and \(Y\) in the two gaps of a meterbridge give a null point dividing the wire in the ratio 4 : 6. If both resistances are increased by 30 \(\Omega\), the null point divides the wire in the ratio 5 : 6. Calculate the value of X…

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