A resistance wire connected in the left gap of a metre bridge balances a $10 \ \Omega$ resistance in the right

Question

A resistance wire connected in the left gap of a metre bridge balances a $10 \ \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5 	ext{ m}$, then the length of $1 \ \Omega$ of the resistance wire is :

Accepted Answer

Initially, $\frac{P}{10} = \frac{l_1}{l_2} = \frac{3}{2} \Rightarrow P = \frac{30}{2} = 15 \ \Omega$. Now Resistance, $R = \frac{ho l}{A}$. $\frac{R_1}{R_2} = \frac{l_1}{l_2} \Rightarrow \frac{15}{1} = \frac{1.5}{l_2} \Rightarrow l_2 = 0.1 	ext{ m} = 1.0 	imes 10^{-1} 	ext{ m}$.

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