The resistance of a wire depends on its length, cross-sectional area, and the resistivity of the material. The formula for resistance is:

$R=\rho \frac{L}{A}$

where $R$ is resistance, $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

When the wire is stretched, its volume remains constant. So, if the length increases, the area must decrease to keep volume $V=A\times L$ constant.

Step 1: Let the original length be $L$ and original area be $A$. After stretching, the new length $L^{\prime }=L+0.001L=1.001L$ (since 0.1% increase).

Step 2: Since volume is constant, $AL=A^{\prime }{L}^{\prime }$. So, $A^{\prime }=\frac{AL}{L^{\prime }}=\frac{AL}{1.001L}=\frac{A}{1.001}$.

Step 3: The new resistance $R^{\prime }=\rho \frac{L^{\prime }}{A^{\prime }}=\rho \frac{1.001L}{A/1.001}=\rho \frac{L}{A}\times (1.001{)}^2$.

So, $R^{\prime }=R\times (1.001{)}^2\approx R\times 1.002001$.

Step 4: The percentage increase in resistance is $\frac{R^{\prime }-R}{R}\times 100\%=(1.002001-1)\times 100\%=0.2001\%\approx 0.2\%$.

Therefore, the resistance increases by approximately 0.2%.

Final answer: increase by 0.2%

Related Topics

This problem involves understanding how the resistance of a conductor changes with its dimensions, assuming constant volume. Key related concepts include resistivity, which is an intrinsic property of the material, and the relationship between length, cross-sectional area, and volume.

Formulae

The primary formula used is $R=\rho \frac{L}{A}$. For fractional changes, when volume is constant, the percentage change in resistance is approximately twice the percentage change in length (for small changes). Specifically, if length increases by x%, resistance increases by 2x%.