Question

A cylindrical wire is stretched to increase its length by $10\%$. Calculate the percentage increase in resistance.

Answer 1

$\text{Step 1: Finding final length}$

Let initial length be ${\ell }_1$

Percentage increment in length $=10\%$

So, Final length, ${\ell }_2={\ell }_1+\frac{10}{100}\times {\ell }_1=\frac{11{\ell }_1}{10}$

$\text{Step 2: Finding Initial and Final Resistances}$

When the wire is stretched, its volume $V$ remains the same.

We know that, Resistance $R=\frac{\rho \ell }{A}=\frac{\rho {\ell }^2}{A\ell }=\frac{\rho {\ell }^2}{V}$

So, $R_1=\frac{\rho {\ell }_1^2}{V}$ $....\left(1\right)$

and $R_2=\frac{\rho {\ell }_2^2}{V}$ $....\left(2\right)$

$\text{Step 3: Divide equation}\left(1\right)\text{by equation}\left(2\right)$

$\frac{R_1}{R_2}={\left(\frac{{\ell }_1}{{\ell }_2}\right)}^2$

$\Rightarrow R_2={\left(\frac{{\ell }_2}{{\ell }_1}\right)}^2{R}_1={\left(\frac{11}{10}\right)}^2{R}_1$

$\text{Step 4: Percentage change in resistance}$

$\%$ Change $=\frac{R_2-R_1}{R_1}\times 100$

$=\frac{{\left(\frac{11}{10}\right)}^2-1}{1}\times 100$

$=21\%$