The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is $31.4 \,cm$. The diameter of wire is measured as $2.4 \,cm$. The resistivity of the given wire is measured as $x \times 10^{-3} \,\Omega cm$. The value of $x$ is_______ [Take $\pi=3.14]$
JEE MAIN 2022, Medium
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$1=\rho \frac{\ell}{ A }$$1=\frac{\rho \times 31.4}{\frac{\pi(2.4)^{2}}{4}}$
$\frac{\pi(2.4)^{2}}{4}=\rho \times 314$
$\frac{2.4 \times 2.4}{4}=\rho \times 10$
$\frac{0.6 \times 2.4}{10}=\rho$
$\frac{1.44}{10}=\rho$
$0.144=\rho$
$144 \times 10^{-3}=\rho$
$\frac{\pi(2.4)^{2}}{4}=\rho \times 314$
$\frac{2.4 \times 2.4}{4}=\rho \times 10$
$\frac{0.6 \times 2.4}{10}=\rho$
$\frac{1.44}{10}=\rho$
$0.144=\rho$
$144 \times 10^{-3}=\rho$
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