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 instrument used to measure the diameter of the wire is a micrometer or a caliper.

The student measured the diameter of the wire in three different places to ensure accuracy and to account for any variations in the wire's thickness. This helps in obtaining an average diameter for more reliable calculations of the wire's cross-sectional area.

The diameters measured were 0.20 mm, 0.19 mm, and 0.21 mm. To calculate the average diameter:

$\text{Average Diameter} = \frac{0.20 + 0.19 + 0.21}{3} = \frac{0.60}{3} = 0.20 \text{ mm}$.

First, we need to convert the diameter to radius by dividing by 2:

$r = \frac{0.20}{2} = 0.10 \text{ mm} = 0.10 \times 10^{-3} \text{ m}$.

Now, calculate the cross-sectional area (A):

$A = \pi r^2 = \pi (0.10 \times 10^{-3})^2 \approx 3.14 \times (1.00 \times 10^{-8}) \approx 3.14 \times 10^{-8} \text{ m}^2$.

Using the resistance R = 26.4 Ω, the cross-sectional area A ≈ 3.14 × 10⁻⁸ m², and the length L = 685 mm = 0.685 m, we can calculate the resistivity:

$\rho = R \cdot \frac{A}{L} = 26.4 \cdot \frac{3.14 \times 10^{-8}}{0.685} \approx 1.21 \times 10^{-6} \text{ Ω m}$.

One precaution taken by the student when measuring the length of the wire was to avoid applying too much tension on the wire, as this could change its length, leading to inaccurate readings.