A cylindrical metallic wire is stretched to increase its length by 10%. Calculate the percentage increase in its resistance.
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When the same wire is stretched, it's length increases but cross-sectional area decreases. The change in resistance is due to both increase in length and decrease in cross-sectional area.
Valume V = lA = Costans, $\text{A}=\frac{\text{V}}{\text{l}}=$ constant
$\therefore\text{R}=\frac{\rho\text{l}}{\text{A}}=\frac{\rho\text{l}^2}{\text{V}}\propto\text{l}^2$
$\therefore\frac{\text{R}'}{\text{R}}=\Big(\frac{\text{l}'}{\text{l}}\Big)^2$
Given, $\text{l}'=\text{l}+\frac{10}{100}\text{l}=1.1\text{l}$
$\Rightarrow\frac{\text{l}'}{\text{l}}=1.1$
$\frac{\text{R}'}{\text{R}}=(1.1)^2=1.21$
% increase in resistance,
$\frac{\text{R}'-\text{R}}{\text{R}}\times100\%=\Big(\frac{\text{R}'}{\text{R}}-1\Big)\times100\%=(1.21-1\times100\%=21\%$
Valume V = lA = Costans, $\text{A}=\frac{\text{V}}{\text{l}}=$ constant
$\therefore\text{R}=\frac{\rho\text{l}}{\text{A}}=\frac{\rho\text{l}^2}{\text{V}}\propto\text{l}^2$
$\therefore\frac{\text{R}'}{\text{R}}=\Big(\frac{\text{l}'}{\text{l}}\Big)^2$
Given, $\text{l}'=\text{l}+\frac{10}{100}\text{l}=1.1\text{l}$
$\Rightarrow\frac{\text{l}'}{\text{l}}=1.1$
$\frac{\text{R}'}{\text{R}}=(1.1)^2=1.21$
% increase in resistance,
$\frac{\text{R}'-\text{R}}{\text{R}}\times100\%=\Big(\frac{\text{R}'}{\text{R}}-1\Big)\times100\%=(1.21-1\times100\%=21\%$
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