Question

A wire is stretched to increase its length by $5\%.$ Calculate percentage change in its resistance.

Answer 1

A wire us stretched increase to length by $5\%$ consider to initial length is $l$ and cross sectional area is $A$

$\therefore$ volume $V=Al$

Now length is increased by $5\%$

$\therefore$ New length $=\frac{105}{100}=\frac{21}{20}l$.

If new area is $l$ considering that the volume of the wire is unchanged

$V=Al=A;\frac{21}{20}l$

$\Rightarrow A=\frac{20}{21}A$

Now if resistanty of the wire is $\rho$ then initial Resistance

$R_1=\rho \frac{l}{A}$

Final Resistance

$R_2=\rho \frac{\frac{21}{20}l}{\frac{20}{21}A}$

$=\frac{\rho l}{A}{\left(\frac{21}{20}\right)}^2$

$\therefore \frac{R_2}{R_1}={\left(\frac{21}{20}\right)}^2=\frac{441}{400}$

$\%$ change in resistance $=\left(\frac{R_2-R_1}{R_1}\right)\times 100\%$

$=10.25\%$