JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 24)
If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.
Answer
44
Explanation
Let $\ell_{0}$ be its initial length and $A_{0}$ be initial area.
Considering volume to be conserved
$$ \begin{aligned} & \text { Vol. }=\ell_{0} A_{0}=\left(1.2 \ell_{0}\right) \mathrm{A} \\\\ & A_{\text {final }}=\frac{A_{0}}{1.2} \\\\ & R_{\text {in }}=\frac{\rho \ell_{0}}{A_{0}} \\\\ & R_{\text {final }}=\frac{\rho 1.2 \ell_{0}}{\frac{A_{0}}{1.2}}=\frac{\rho \ell_{0}}{A_{0}}(1.2)^{2} \end{aligned} $$
$=\mathrm{R}_{\text {in }}(1.44)$
Hence increase $=44 \%$
Considering volume to be conserved
$$ \begin{aligned} & \text { Vol. }=\ell_{0} A_{0}=\left(1.2 \ell_{0}\right) \mathrm{A} \\\\ & A_{\text {final }}=\frac{A_{0}}{1.2} \\\\ & R_{\text {in }}=\frac{\rho \ell_{0}}{A_{0}} \\\\ & R_{\text {final }}=\frac{\rho 1.2 \ell_{0}}{\frac{A_{0}}{1.2}}=\frac{\rho \ell_{0}}{A_{0}}(1.2)^{2} \end{aligned} $$
$=\mathrm{R}_{\text {in }}(1.44)$
Hence increase $=44 \%$
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