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A nichrome wire $50 \mathrm{~cm}$ long and $1 \mathrm{~mm}^2$ cross-section carries a current of $4 \mathrm{~A}$. When connected to 2 volt battery. The resistivity of nichrome wire in $\Omega-\mathrm{m}$ is
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$1 \times 10^{-6}$
$\begin{aligned} l=50 \mathrm{~cm} & =50 \times 10^{-2} \mathrm{~m} \\ A & =1 \mathrm{~mm}^2=1 \times 10^{-4} \mathrm{~m}^2 \\ i & =4 \mathrm{~A}, V=2 \mathrm{volt}\end{aligned}$
$\therefore$ Resistance of wire, $R=\frac{V}{i}=\frac{2}{4}=0.5 \Omega$
$\begin{aligned} \therefore \text { Resistiviry, } \rho & =R \frac{A}{l}=0.5 \times \frac{1 \times 10^{-6}}{50 \times 10^{-2}} \\ & =\frac{0.5 \times 1 \times 10^{-4}}{50} \\ & =1 \times 10^{-4} \Omega-\mathrm{fm}\end{aligned}$
$\therefore$ Resistance of wire, $R=\frac{V}{i}=\frac{2}{4}=0.5 \Omega$
$\begin{aligned} \therefore \text { Resistiviry, } \rho & =R \frac{A}{l}=0.5 \times \frac{1 \times 10^{-6}}{50 \times 10^{-2}} \\ & =\frac{0.5 \times 1 \times 10^{-4}}{50} \\ & =1 \times 10^{-4} \Omega-\mathrm{fm}\end{aligned}$
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