# 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 m, then the length of 1$\Omega$  of the resistance wire is: Option: 1 Option: 2 Option: 3 Option: 4

For a meter bridge

$\frac{P}{Q}=\frac{R}{S} =\frac{l}{100-l}$

From question

$\frac{P}{Q}= \frac{3}{2 } \\ and \ S= 10 \Omega\\ \\ So \ \frac{R}{S}=\frac{3}{2}\\ \\ R=\frac{3}{2}*10=15 \Omega$

I.e For 1.5 m long wire its total resistance is 15$\Omega$

So using the unitary method

The length of $1 \Omega$ resistance wire will be = $L=\frac{1.5}{15}=0.1=1\times10^{-1} \ m$

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