In an experiment to find the resistivity of the material of a wire, a student took a sample of the wire and measured its length $l$, diameter $d$ and resistance $R$ - Leaving Cert Physics - Question 4 - 2010

The diameter of the wire was measured using a micrometer screw gauge, which allows for precise measurements of small diameters.

To calculate the cross-sectional area $A$ of the wire, we use the formula for the area of a circle:

$A = \pi \left(\frac{d}{2}\right)^2$

First, we need to find the average diameter:

$d_{avg} = \frac{0.21 + 0.26 + 0.22}{3} = 0.23 \text{ mm} = 0.23 \times 10^{-3} \text{ m}$

Now substituting into the area formula:

$A = \pi \left(\frac{0.23 \times 10^{-3}}{2}\right)^2 \approx 4.15 \times 10^{-8} \text{ m}^2$

To find the resistivity $\rho$ of the material, we use the formula:

$\rho = R \cdot \frac{A}{l}$

Using the given values:

• $R = 30 \text{ Ω}$
• $A = 4.15 \times 10^{-8} \text{ m}^2$
• $l = 80 \text{ cm} = 0.8 \text{ m}$

Substituting these values into the formula gives:

$\rho = 30 \cdot \frac{4.15 \times 10^{-8}}{0.8} = 1.56 \times 10^{-7} \text{ Ω m}$

1. Ensure that the wire is straight and properly secured during measurement to prevent inaccuracies in length measurement.
2. Calibrate the measuring instruments (multimeter and micrometer) before use to ensure their readings are accurate.