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(B) The resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area.G

Vedclass · Accepted Answer

$97$

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(B) The resistance $R$ of a wire is given by $R = ho \frac{L}{A}$, where $ho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area.Given diameter $d = 2 \; mm = 2 	imes 10^{-3} \; m$, so radius $r = 1 	imes 10^{-3} \; m$.Area $A = \pi r^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6} \; m^2$.Resistivity of iron $ho_1 = 12 \; \mu\Omega \cdot cm = 12 	imes 10^{-6} 	imes 10^{-2} \; \Omega \cdot m = 1.2 	imes 10^{-7} \; \Omega \cdot m$.Resistivity of alloy $ho_2 = 51 \; \mu\Omega \cdot cm = 51 	imes 10^{-6} 	imes 10^{-2} \; \Omega \cdot m = 5.1 	imes 10^{-7} \; \Omega \cdot m$.For parallel connection, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = 3 \; \Omega$.$R_1 = \frac{ho_1 L}{A}$ and $R_2 = \frac{ho_2 L}{A}$.$R_{eq} = \frac{(ho_1 L / A) \cdot (ho_2 L / A)}{(ho_1 L / A) + (ho_2 L / A)} = \frac{ho_1 ho_2 L}{A(ho_1 + ho_2)} = 3$.$L = \frac{3 A (ho_1 + ho_2)}{ho_1 ho_2} = \frac{3 	imes \pi 	imes 10^{-6} 	imes (1.2 + 5.1) 	imes 10^{-7}}{1.2 	imes 10^{-7} 	imes 5.1 	imes 10^{-7}} = \frac{3 	imes \pi 	imes 6.3 	imes 10^{-13}}{6.12 	imes 10^{-14}} \approx 97 \; m$.

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