Use a long
Materials - OCR A-Level Physics
- Use a long, thin wire (at least 2 m) clamped at one end with masses hung from the other end
- Measure the original length L with a metre rule
- Measure the diameter d with a micrometer screw gaugeA precision instrument for measuring small thicknesses or diameters to ±0.01 mm. at several points and in two perpendicular directions
- Measure the extension for each added mass using a travelling microscopeAn instrument used to measure small displacements by moving a crosshair along a calibrated scale. or a marker and ruler
- Plot stressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). (F/A) against strainThe fractional change in length of a material under stress. It is dimensionless (no units). ($\Delta$L / L). The gradient of the linear region gives the Young modulusThe ratio of stressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). to strainThe fractional change in length of a material under stress. It is dimensionless (no units). for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa).
- Use a reference wire alongside the test wire to compensate for thermal expansion
Examiner Tips and Tricks
- A common practical question: 'Why use a long thin wire?' A long wire produces a larger, more measurable extension for a given strainThe fractional change in length of a material under stress. It is dimensionless (no units)., reducing percentage uncertainty.
- A thin wire reaches higher stressThe force applied per unit cross-sectional area of a material. Measured in pascals (Pa). without needing enormous forces.
Worked Example [4 marks]
A steel wire of original length 2.5 m and diameter 0.50 mm is stretched by a force of 80 N. The extension is 1.2 mm. Calculate the Young modulusThe ratio of stress to strain for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa). of the steel.
Show Solution
1
$Radius = d/2 = 0.25 \text{ mm} = 0.25 \times 10^{-3} \text{ m}. Area = \pi r^{2} = \pi(0.25 \times 10^{-3})^{2} = 1.96 \times 10^{-7} \text{ m}^{2}$
[1]2
$Stress = \frac{F}{A} = \frac{80}{1.96 \times 10^{-7}} = 4.08 \times 10^{8} \text{ Pa}$
[1]3
$Strain = \frac{\Delta L}{L} = \frac{1.2 \times 10^{-3}}{2.5} = 4.8 \times 10^{-4}$
[1]4
$Young modulusThe ratio of stress to strain for a material in the elastic region. A measure of stiffness. Measured in pascals (Pa). = \frac{\sigma}{\varepsilon} = \frac{4.08 \times 10^{8}}{4.8 \times 10^{-4}} = 8.5 \times 10^{11} \text{ Pa} \approx 200 \text{ GPa}$
[1]Answer
\(8.5 \times 10^{11}\) Pa (approximately 200 GPa, consistent with steel)