The discussion focuses on the relationship between the dimensions of a cylindrical wire and its resistance. When the length \(\ell\) of the wire is increased by \(x\) percent while maintaining a constant volume and resistivity, the new cross-sectional area \(A_{\textrm{new}}\) is derived as \(A_{\textrm{new}} = \frac{A}{1+\frac{x}{100}}\). The resistance equations show that \(R_{\textrm{new}} \propto \frac{\ell(1+\frac{x}{100})^2}{A}\). By expanding the quadratic term and ignoring the negligible \((\frac{x}{100})^2\) component, the resistance change is approximately \(2x\) percent.
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I said that A\ell=A_{\textrm{new}}\ell\left(1+\frac{x}{100}\right)\implies A_{\textrm{new}}=\frac{A}{1+\frac{x}{100}}. Also, we know that R\propto\frac{\ell}{A}.. Therefore,A cylindrical wire of length \ell and cross-sectional area \textrm{A} has a fixed volume \textrm{V}. If \ell is increased by +x percent and the volume and resistivity stay the same, by what percentage (in terms of x) will the resistance change?