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(D) Given:Length of the wire,$l = 15 \; m$Area of cross-section of the wire,$A = 6.0 	imes 10^{-7} \; m^{2}$Resistance of the wire,$R = 5.0 \; \Omega

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$2 	imes 10^{-7} \; \Omega \, m$

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(D) Given:Length of the wire,$l = 15 \; m$Area of cross-section of the wire,$A = 6.0 	imes 10^{-7} \; m^{2}$Resistance of the wire,$R = 5.0 \; \Omega$The formula for resistance is $R = ho \frac{l}{A}$,where $ho$ is the resistivity.Rearranging for resistivity: $ho = \frac{R 	imes A}{l}$Substituting the values: $ho = \frac{5.0 	imes 6.0 	imes 10^{-7}}{15}$$ho = \frac{30 	imes 10^{-7}}{15} = 2 	imes 10^{-7} \; \Omega \, m$Thus,the resistivity of the material is $2 	imes 10^{-7} \; \Omega \, m$.

$A$ negligibly small current is passed through a wire of length $15 \; m$ and uniform cross-section $6.0 \times 10^{-7} \; m^{2},$ and its resistance is measured to be $5.0 \; \Omega .$ What is the resistivity of the material at the temperature of the experiment?

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