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 - 2015

The student used a micrometer or a digital vernier caliper to measure the diameter of the wire accurately. This instrument allows for precise measurements of small dimensions using its calibrated scale.

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

$A = \pi r^2$

First, we need to convert the diameter from millimeters to meters:

• The average diameter from the given measurements:
  • Average of $d = 0.21 \, mm$, $0.26 \, mm$, $0.22 \, mm$:

$\bar{d} = \frac{0.21 + 0.26 + 0.22}{3} = 0.23 \, mm = 0.00023 \, m$

Now, the radius $r$ is: $r = \frac{\bar{d}}{2} = \frac{0.00023}{2} = 0.000115 \, m$

Thus, the cross-sectional area is:

$A = \pi (0.000115)^2 \approx 4.15 \times 10^{-8} \, m^2$

The resistivity $\rho$ can be calculated using the formula:

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

Where:

• $R = 30 \Omega$ (given),
• $A \approx 4.15 \times 10^{-8} \, m^2$ (calculated from part iii),
• $l = 80 \, cm = 0.8 \, m$.

Substituting the values in:

$\rho = 30 \cdot \frac{4.15 \times 10^{-8}}{0.8} \approx 1.56 \times 10^{-7} \, \Omega \, m$

1. Ensure that the connections of the multimeter are secure and clean to prevent any resistance from dirty or loose connections, which may affect the accuracy of the resistance readings.

2. When measuring the diameter of the wire, take multiple measurements at different points and use the average to account for any irregularities in the wire's thickness.