Question

If a copper wire is stretched to make its cross sectional radius 0.1% thinner,then what is % increase in its resistence?

Answer 1

Step 1, Given data

Radius becomes 0.1% thinner

Step 2, Finding the percentage increases in resistance

We know,

Resistance of the wire i given as

$R=\text{ρ}\frac{L}{A}$ Where, ρ=Resistivity, L= length, A = Cross-sectional area.

We can say that due to the stretching of the wire volume of the wire will remain constant

We can write

$R=\text{ρ}\frac{L}{A}=\text{ρ}\frac{L^2}{AL}=\text{ρ}\frac{L^2}{V}$

In the above equation, ρ and V are constant

So,

$\frac{\Delta R}{R}=2\frac{\Delta L}{L}$

$Or,\%\left(\frac{\Delta R}{R}\right)=2\%\left(\frac{\Delta L}{L}\right)=2\times 0.1=0.2\%$

Hence the percentage increase in resistance is 0.2 %