A wire of length $L$ and cross-sectional area $A$ is shown - Scottish Highers Physics - Question 15 - 2017

When plotting the graph, ensure that both axes are appropriately labeled: the x-axis for Length of wire $L$ (m) and the y-axis for Resistance of wire $R$ ($\times 10^3 \Omega$). Choose suitable scales that cover the range of your data. Plot the points from the table accurately and draw a best fit line to represent the relationship between $R$ and $L$.

Select two points from your line. For example, let’s choose data points (1.5, 5.6) and (3.5, 13.2). The gradient $m$ is calculated as:

$m = \frac{\Delta R}{\Delta L} = \frac{13.2 - 5.6}{3.5 - 1.5} = \frac{7.6}{2} = 3.8 \times 10^{3} \Omega/m$

This gradient indicates the change in resistance per unit length of the wire.

Using the formula for resistivity, we can rearrange the resistance formula: $\rho = m \times A$ Substituting the calculated gradient $m$ and the cross-sectional area: $\rho = 3.8 \times 10^{3} \Omega/m \times 4.52 \times 10^{-6} m^2$ Calculating this gives: $\rho = 1.72 \times 10^{-2} \Omega m$ Thus, the resistivity of the metal wire is approximately $1.72 \times 10^{-2} \Omega m$.