Question 61 of 380

If a copper wire is stretched to increase its length by $20$. The percentage increase in resistance of the wire is _____%.

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Answer: 44

Before heating, resistance of wire is ${R}_{0}=\left(\dfrac{\rho {l}_{0}}{{A}_{0}}\right)$
After heating, resistance is $R=\dfrac{\rho l}{A}$
As volume = constant
Therefore,
${A}_{0}{l}_{0}=1.2{l}_{0}A$
$\Rightarrow A=\left(\dfrac{{A}_{0}}{1.2}\right)$
So,
$R=\dfrac{\rho l}{A} =\dfrac{\rho \times \left(1.2{l}_{0}\right)}{\left(\dfrac{{A}_{0}}{1.2}\right)} =\dfrac{\rho {l}_{0}}{{A}_{0}}\times 1.44 =1.44{R}_{0}$
Percentage increase in $R$ $=\dfrac{R-{R}_{0}}{{R}_{0}}\times 100=44$

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