An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :
Solution
<p>$R = {{\rho l} \over A}$</p>
<p>Also volume will remain constant</p>
<p>i.e., Al = constant $\Rightarrow A \propto {1 \over l}$</p>
<p>$\therefore$ $R \propto {l^2}$</p>
<p>${{\Delta R} \over R} = 2{{\Delta l} \over l} = 0.8$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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