Question

If a copper wire is stretched to make radius 0.1% thinner, then what is the percentage increase in resistance.

Answer 1

consider initial resistance as $R_1$

Final resistance is$R_2$

Initial length is $L_1$

Final length as $L_2$

Converted % is 0.1 % = $\frac{1}{1000}$

final resistance $R_2=R_1-\frac{R_1}{1000}$

$R_2=0.999R_1$ because volume is constant

$\pi R_1^2{L}_1=\pi R_2^2{L}_2$

since resistance directly proportional to length.

so $R_{1}as{R}_2=\left(0.9960\right)R_2$

change in resistance is $R_2-R_1=0.004$

convert to % change $=0.004\times 100=0.4$%