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

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(D) The resistance $R$ of a wire is given by the formula $R = ho \frac{l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.Given: $ho = 48 	imes 10^{-8} \, \Omega \, m$,$R = 4.2 \, \Omega$,and diameter $d = 0.4 \, mm = 0.4 	imes 10^{-3} \, m$.The radius $r = \frac{d}{2} = 0.2 	imes 10^{-3} \, m$.The area $A = \pi r^2 = 3.14 	imes (0.2 	imes 10^{-3})^2 = 3.14 	imes 0.04 	imes 10^{-6} = 0.1256 	imes 10^{-6} \, m^2$.Rearranging the formula for length: $l = \frac{R \cdot A}{ho}$.Substituting the values: $l = \frac{4.2 	imes 3.14 	imes (0.2 	imes 10^{-3})^2}{48 	imes 10^{-8}}$.$l = \frac{4.2 	imes 3.14 	imes 0.04 	imes 10^{-6}}{48 	imes 10^{-8}} = \frac{0.52752 	imes 10^{-6}}{48 	imes 10^{-8}} = \frac{52.752}{48} = 1.1 \, m$.

What length of the wire of specific resistance $48 \times 10^{-8} \, \Omega \, m$ is needed to make a resistance of $4.2 \, \Omega$ (in $.1$)? (Diameter of wire = $0.4 \, mm$)

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