The resistance of a wire is $5 \Omega$. It's new resistance in ohm if stretched to $5$ times of it's original length will be :
JEE MAIN 2023, Medium
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$R _{\text {initial }}=\frac{\rho \ell}{ A }=5 \Omega$
$\because$ Volume of wire is constant in stretching
$V _{ i }= V _{ f }$
$A _{ i } \ell_{ i }= A _{ f } \ell_{ f }$
$A \ell= A ^{\prime}(5 \ell)$
$A ^{\prime}=\frac{ A }{5}$
$R _{ f }=\frac{\rho \ell_{ f }}{ A _{ f }}=\frac{\rho(5 \ell)}{\left(\frac{ A }{5}\right)}$
$=25\left(\frac{\rho \ell}{A}\right)$
$=25 \times 5=125 \Omega$
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