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What length of wire of specific resistance $48\times10^{-8}\,\Omega\,\text{m}$ is needed to make a resistance of $4.2\,\Omega$ if the diameter of the wire is $0.4\,\text{mm}$?

1. 4.1 m
2. 3.1 m
3. 2.1 m
4. 1.1 m

Correct answer: 1.1 m

Solution

Using $R=\rho L/A$, the length is $L=RA/\rho$. With diameter $0.4\,\text{mm}$, the radius is $2\times10^{-4}\,\text{m}$, so $A=\pi(2\times10^{-4})^2$. Substituting $R=4.2\,\Omega$ and $\rho=48\times10^{-8}\,\Omega\,\text{m}$ gives $L\approx1.1\,\text{m}$.

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