Question

A cylindrical metallic wire is stretched to increase its length by 5%. Calculate the percentage change in resistance.

Answer 1

Step 1: Given that:

Let length and cross-sectional area of the wire is, $landA$ respectively.

Percentage increase in length of cylindrical metallic wire= $5\%$

Step 2: Calculation of percentage change cross-sectional area

We know that volume, $V=Al$

Also on changing the length, there is no change in volume.

Therefore,

$A_1{V}_1=A_2{V}_2$

$A\times l=A_2\times 1.05l$

$A_2=\frac{A}{1.05}$

$A_2=\frac{20}{21}A$

Step 2: Calculation of percentage change in resistance of the metallic wire:

The resistance of a conductor of length l and area of cross-section A is given by

$R=\frac{\text{ρ}l}{A}$

Where ρ is the resistivity of the metallic wire.

Resistivity(ρ) is the property of the material of a conductor.

Now,

When length of wire is increased, then resistance is

$R^{\prime }=\text{ρ}\frac{\frac{21l}{20}}{\frac{20A}{21}}$

$R^{\prime }=\text{ρ}{\left(\frac{21}{20}\right)}^2\frac{l}{A}$

Change in resistance is,

$\frac{R^{\prime }}{R}=\frac{{\left(\frac{21}{20}\right)}^2\text{ρ}\frac{l}{A}}{\text{ρ}\frac{l}{A}}$

$\frac{R^{\prime }}{R}={\left(\frac{21}{20}\right)}^2$

$\frac{R^{\prime }}{R}=\frac{441}{400}$

Percentange change in resistance

$\frac{R^{\prime }-R}{R}\times 100=\frac{441-400}{400}\times 100=10.25\%$

Thus,

The percentage change in resistance will be $10.25\%$.