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(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{\ell}{A}$, where $\rho$ is the resistivity, $\ell$ is the length, and $A

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$2$

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(A) The resistance $R$ of a conductor is given by the formula $R = ho \frac{\ell}{A}$, where $ho$ is the resistivity, $\ell$ is the length, and $A$ is the cross-sectional area.For a hollow cylinder, the cross-sectional area $A = \pi(R_{outer}^2 - R_{inner}^2)$.Given: $\ell = 3.14 \, m$, $ho = 2.4 	imes 10^{-8} \, \Omega \, m$, outer diameter $D_{out} = 8 \, mm \implies R_{outer} = 4 	imes 10^{-3} \, m$, and inner diameter $D_{in} = 4 \, mm \implies R_{inner} = 2 	imes 10^{-3} \, m$.Calculating the area $A = \pi((4 	imes 10^{-3})^2 - (2 	imes 10^{-3})^2) = \pi(16 - 4) 	imes 10^{-6} = 12\pi 	imes 10^{-6} \, m^2$.Substituting the values into the resistance formula:$R = \frac{2.4 	imes 10^{-8} 	imes 3.14}{12 	imes 3.14 	imes 10^{-6}}$$R = \frac{2.4}{12} 	imes 10^{-2} = 0.2 	imes 10^{-2} = 2 	imes 10^{-3} \, \Omega$.Comparing this with $n 	imes 10^{-3} \, \Omega$, we get $n = 2$.

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