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  • What equal length of an iron wire and a copper-nickel alloy wire,each of 2 mm diameter connected parallel to give an equivalent resistance of ? (Given resistivities

What equal length of an iron wire and a copper-nickel alloy wire,each of 2 mm diameter connected parallel to give an equivalent resistance of 3\Omega ?
(Given resistivities of iron and copper-nickel ally wire are 12\mu \Omega\, cm\, and\, 51\mu \Omega \, cm respectively )
Option: 1 97 m
Option: 2 110 m
Option: 3 90 m
Option: 4 82 m

Answers (1)

best_answer


R_{eq}= \frac{R_{1}R_{2}}{R_{1}+R_{2}}= 3\Omega
R_{1}= \frac{3_{1}l_{1}}{A_{1}}= \frac{12\times 10^{-6}\times 10^{-2}\times l}{A}
R_{2}= \frac{3_{2}l_{2}}{A_{2}}= \frac{51\times 10^{-8}\times l}{A}
A_{1}= A_{2}= \frac{\pi d^{2}}{4}= \frac{\pi \times \left ( 2\times 10^{-8} \right )^{2}}{4}
R_{eq}= \frac{R_{1}R_{2}}{R_{1}+R_{2}}= \frac{12\times 10^{-8}\times 51\times 10^{-8}l}{A\times \left ( 12\times 10^{-8}+51\times 10^{-8} \right )}

3= \frac{ {12}\times 51\times 10^{-8}\times l}{ 63}\times \frac{\pi \times {4}\times 10^{-6}}{ {4}}
\frac{63}{204}\times \pi \times 10^{2}= l
l= \frac{\not{63}^{9}}{204}\times \frac{22}{\not{7}}\times 10^{2}
l=\frac{198}{204}\times 10^{2}\simeq 97\, m
The correct option is (1)

Posted by

vishal kumar

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