In an experiment to measure the resistivity of the material of a wire, a student measured the length, diameter and the resistance of a sample of nichrome wire - Leaving Cert Physics - Question 4 - 2005

The student used a micrometer to measure the diameter of the wire.

The diameter was measured in three different places to account for any non-uniformity in the wire. Measuring at multiple points allows for an average diameter to be calculated, ensuring that any irregularities do not significantly affect the final measurement.

To calculate the average diameter from the measurements of 0.20 mm, 0.19 mm, and 0.21 mm:

Average diameter = $\frac{0.20 + 0.19 + 0.21}{3} = 0.20 \text{ mm}$.

Using the average diameter calculated:

For $d = 0.20 \text{ mm} = 0.20 \times 10^{-3} \text{ m}$,

The cross-sectional area is calculated as:

$A = \frac{\pi (0.20 \times 10^{-3})^2}{4} = 3.14 \times 10^{-8} \text{ m}^2$

Using the resistance $R = 26.4 \: \Omega$, the area $A = 3.14 \times 10^{-8} \: \text{m}^2$, and the length $L = 685 \: \text{mm} = 0.685 \: \text{m}$:

$\rho = \frac{26.4 \times 3.14 \times 10^{-8}}{0.685} \approx 1.21 \times 10^{-6} \: \Omega \cdot \text{m}$

The student should avoid pulling on the wire while measuring its length, as this can stretch the wire and lead to inaccurate measurements.