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$$
\begin{array}{l}{\text { A wire } 6.50 \mathrm{m} \text { long with diameter of } 2.05 \mathrm{mm} \text { has a resist- }} \\ {\text { ance of } 0.0290 \Omega . \text { What material is the wire most likely made of? }}\end{array}
$$

$$
R=\frac{\rho_{L}}{A}
$$

$$
r=\frac{D}{2}
$$

$$
\rightarrow \rho=\frac{A R}{L}=\frac{\pi\left(\frac{D}{2}\right)^{2} * R}{L}
$$

$$
=\frac{\pi\left(\frac{0.00205}{2}\right)^{2} * 0.029}{6.5}=1.47 * 10^{-8} \Omega \cdot \mathrm{m}
$$

$$
\rho=1.47 * 10^{-8} \Omega.m
$$

$$
\text { so, the material is silver. }
$$

$$
\begin{array}{l}{\text { You apply a potential difference of } 4.50 \mathrm{V} \text { between the }} \\ {\text { ends of a wire that is } 2.50 \mathrm{m} \text { in length and } 0.654 \mathrm{mm} \text { in radius. The }} \\ {\text { resulting current through the wire is } 17.6 \mathrm{A} \text { . What is the resistivity }} \\ {\text { of the wire? }}\end{array}
$$

$$
R=\frac{\rho L}{A} \longrightarrow
$$$$
\rho=\frac{R A}{L}
$$

$$
V= I R \rightarrow R=\frac{V}{I}
$$

$$
\longrightarrow \rho=\frac{V A}{I L}
$$

$$
=\frac{4.5 * \pi *\left(6.54 * 10^{-4}\right)^{2}}{176+2.5}=1.37*10^{-7} \Omega \cdot \mathrm{m}
$$

\(∴ \quad \rho=1.37 * 10^{-7} \Omega \cdot \mathrm{m} \)

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