An aluminium wire is stretched to make its length, $0.4 \,\%$ larger. Then percentage change in resistance is $.....\,\%$
JEE MAIN 2022, Medium
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$R =\frac{\rho \ell}{ A }$
$\frac{\Delta R }{ R }=\frac{\Delta \ell}{\ell}-\frac{\Delta A }{ A }$
$\ell A = k$
$\frac{\Delta \ell}{\ell}+\frac{\Delta A }{ A }=0$
$\frac{\Delta R }{ R }=\frac{2 \Delta \ell}{\ell}$
$\frac{\Delta R }{ R }=2 \times 0.4=0.8 \%$
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