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  • A resistance wire connected in the left gap of a metre bridge balances a 10  resistance in the right gap at a point which divides the bridge wire in the ratio 3:2

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 1.0\times 10^{-2}m
Option: 2 1.0\times 10^{-1}m
Option: 3 1.5\times 10^{-1}m
Option: 4 1.5\times 10^{-2}m

Answers (2)

best_answer

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

Posted by

Deependra Verma

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Posted by

Guru G

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