A uniform wire of resistance $100\Omega$ is melted and recast as a wire of length is double that of the original. What would be the resistance of the wire?
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$\ell'=2\ell$
Volume of the wire remains constant.
$\text{A}\ell=\text{A}'\ell'$
$\Rightarrow\text{A}\ell=\text{A}'\times2\ell$
$\Rightarrow\text{A}'=\frac{\text{A}}{2}$
f = Specific resistance
$\text{R}=\frac{\text{f}\ell}{\text{A}};\text{R}'=\frac{\text{f}\ell'}{\text{A}'}$
$100\Omega=\frac{\text{f}2\ell}{\text{A}/2}=\frac{4\text{f}\ell}{\text{A}}=4\text{R}$
$\Rightarrow4\times100\Omega=400\Omega$
Volume of the wire remains constant.
$\text{A}\ell=\text{A}'\ell'$
$\Rightarrow\text{A}\ell=\text{A}'\times2\ell$
$\Rightarrow\text{A}'=\frac{\text{A}}{2}$
f = Specific resistance
$\text{R}=\frac{\text{f}\ell}{\text{A}};\text{R}'=\frac{\text{f}\ell'}{\text{A}'}$
$100\Omega=\frac{\text{f}2\ell}{\text{A}/2}=\frac{4\text{f}\ell}{\text{A}}=4\text{R}$
$\Rightarrow4\times100\Omega=400\Omega$
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