Conductive putty can easily be formed into different shapes to investigate the effect of shape on electrical resistance - AQA - A-Level Physics - Question 3 - 2022 - Paper 3

To find the average diameter, we first add the calliper measurements:

$d_{avg} = \frac{34.5 + 34.2 + 32.9 + 33.4 + 34.0}{5} = 33.6 \, mm$

Next, we calculate the uncertainty for a single measurement. The least count of vernier callipers is typically 0.1 mm; hence:

$u = \frac{0.1}{2} = 0.05 \, mm$

Now, we calculate the percentage uncertainty:

\text{Percentage Uncertainty} = \left(\frac{u}{d_{avg}}\right) \times 100 = \left(\frac{0.05}{33.6}\right) \times 100 \approx 0.149 \,\text{%} \approx 0.15 \,\text{%}\n\text{Rounding it gives approximately 2.4%}.

The volume V of a cylinder is given by:

$V = \pi r^2 h$

Where r is the radius. The uncertainty in volume can be found using the formula for propagation of uncertainties:

$\left(\frac{\Delta V}{V}\right) = 2\left(\frac{\Delta r}{r}\right) + \left(\frac{\Delta h}{h}\right)$

Using $d = 33.6 \, mm$, then $r = \frac{d}{2} = 16.8 \, mm$, and the height $h = 71 \, mm$ with uncertainty $\\Delta h= 2 \, mm$.

Now we can find the values:

$V_{cylinder} = \pi (16.8 imes 10^{-3})^2 (71 imes 10^{-3}) \approx 0.0006266 \, m^3$

Calculating the uncertainty:

$\left(\frac{\Delta V}{V}\right) = 2\left(\frac{0.05}{16.8}\right) + \left(\frac{2}{71}\right)$

This yields:

ight)$$ The final answer will need to be converted to mm³.

To determine ρ, we rearrange the formula given:

$R = \frac{\rho L}{A}$ Where A is the cross-sectional area. The graph of R vs. (L^2) can be used to find the resistance for various lengths. From the slope of the graph, we can derive:

$\rho = \text{slope} \times (5.83 \times 10^{-5})$

The appropriate SI unit for ρ is ohm-meter (Ω·m).