Practical: Determining the Young Modulus Revision Notes | A Level AQA Physics

1. Why Measuring the Young Modulus Matters

Engineers designing bridges, aircraft, and medical implants need to know exactly how stiff a material is — not roughly, but precisely. The Young modulus gives them that number. In this required practical, you will measure it directly from a stretching wire.

The core idea is straightforward: apply a known force to a wire, measure how much it stretches, and use those measurements to calculate stress and strain. The ratio of stress to strain gives you the Young modulus, $E$. Getting that ratio accurately is where the real challenge lies.

2. The Key Quantities and Equations

Before touching any apparatus, you need to be fluent with three quantities: tensile stress, tensile strain, and the Young modulus itself. Each one builds on the previous, so take them in order.

Notice that stress depends on the area of the wire's cross-section, not its length. A thicker wire of the same material experiences less stress for the same applied force — the load is spread over more area.

Because strain is dimensionless, you will often see very small numbers — a steel wire might strain by only $10^{-3}$ or less before reaching its elastic limit. This is why measuring the extension accurately is so critical.

The combined form $E = FL / (A \Delta L)$ is the version you will use in calculations. Every symbol must carry consistent SI units — metres, newtons, and metres squared — before you substitute.

3. Apparatus and Experimental Setup

The standard AQA method uses a long metal wire (typically copper or steel) clamped at one end and loaded with masses at the other. The wire must be long — at least 1 m, ideally 2 m or more — because a longer wire produces a larger, more measurable extension for the same applied force.

A reference wire is often set up alongside the test wire, both clamped from the same support. The reference wire carries a small fixed load to keep it taut, and a crosspiece with a spirit level bridges the two wires. A micrometer screw or vernier scale on the crosspiece measures the differential extension between the wires. This comparison arrangement cancels out any sag or thermal expansion of the support structure — a clever way to reduce systematic error.

• Long metal wire (copper or steel, ~1–2 m minimum)
• G-clamp or fixed support at the top
• Pulley at the end of the wire to redirect the load vertically
• Slotted masses and hanger (to apply known forces)
• Metre rule or tape measure (to measure original length $L$)
• Micrometer screw gauge (to measure wire diameter)
• Vernier scale or travelling microscope (to measure extension $\Delta L$)
• Reference wire with spirit level (to detect and correct for support movement)

The micrometer is used to measure the diameter of the wire at several positions along its length. You then calculate the cross-sectional area using $A = \pi r^2$, where $r$ is the radius (half the diameter). Multiple measurements along the wire account for any variation in thickness.

4. Reading the Stress-Strain Graph

Once you have your data, plotting stress against strain is the most powerful way to extract the Young modulus. The gradient of the straight-line (elastic) region is $E$ directly — no further rearrangement needed.

The gradient here is $200 \times 10^6 \text{ Pa} / 0.0010 = 2.0 \times 10^{11}$ Pa — consistent with steel. When you draw your best-fit line, use a large triangle spanning as much of the linear region as possible to minimise the percentage uncertainty in the gradient.

The graph also tells you whether you stayed within the elastic limit. If the points curve away from the straight line at higher stresses, the wire was being permanently deformed — those points should not be used to calculate $E$.

5. Worked Calculation: Finding E from Experimental Data

Exam questions on this practical almost always ask you to calculate $E$ from a set of measurements. The method is always the same: calculate stress, calculate strain, divide. Here is a full walkthrough with the kind of data you might collect in the lab.

Worked example: Calculating the Young Modulus from Wire Data

A metal wire of original length 1.80 m and diameter 0.40 mm is loaded with a force of 25 N. The wire extends by 0.90 mm. Calculate the Young modulus of the wire.

1. $L = 1.80$ m\n$d = 0.40 \text{ mm} = 0.40 \times 10^{-3}$ m $= 4.0 \times 10^{-4}$ m\n$\Delta L = 0.90 \text{ mm} = 0.90 \times 10^{-3}$ m $= 9.0 \times 10^{-4}$ m\n$F = 25$ N — Always convert to metres and newtons before substituting. Leaving values in mm is the most common source of errors in this calculation.
2. $r = d/2 = 4.0 \times 10^{-4} / 2 = 2.0 \times 10^{-4}$ m\n$A = \pi r^2 = \pi \times (2.0 \times 10^{-4})^2$\n$= \pi \times (2.0 \times 10^{-4} \times 2.0 \times 10^{-4})$\n$= \pi \times 4.0 \times 10^{-8}$\n$= 1.257 \times 10^{-7}$ m² — Use radius (half the diameter) in the area formula. Note: $(2.0 \times 10^{-4})^2 = 2.0 \times 2.0 \times 10^{-4} \times 10^{-4} = 4.0 \times 10^{-8}$. Carry extra significant figures here to avoid rounding errors in later steps.
3. $\sigma = F/A = 25 / (1.257 \times 10^{-7})$\n$= 1.989 \times 10^{8}$ Pa — Stress = force ÷ area. The result is in pascals. This is approximately $2.0 \times 10^8$ Pa — a reasonable stress for a loaded metal wire.
4. $\varepsilon = \Delta L / L = (9.0 \times 10^{-4}) / 1.80$\n$= 5.0 \times 10^{-4}$ — Strain is dimensionless. Dividing metres by metres gives a pure number. The small value ($5 \times 10^{-4}$) is typical for metals in the elastic region.
5. $E = \sigma / \varepsilon = (1.989 \times 10^{8}) / (5.0 \times 10^{-4})$\n$= 3.98 \times 10^{11}$ Pa — Dividing by $5.0 \times 10^{-4}$ is the same as multiplying by $2.0 \times 10^{3}$. The result is approximately $4.0 \times 10^{11}$ Pa. This value is consistent with a hard metal such as tungsten or a high-stiffness alloy; note that common metals vary widely (e.g. steel ~$2 \times 10^{11}$ Pa, copper ~$1.2 \times 10^{11}$ Pa), so always check your result against the material stated in the question.
6. $E \approx 4.0 \times 10^{11}$ Pa (or N m⁻²) — Always state units. The Young modulus is always in Pa (or equivalently N m⁻²). A value in the range $10^{10}$ to $10^{12}$ Pa is typical for metals.

Answer: $E \approx 4.0 \times 10^{11}$ Pa

6. Uncertainty and Evaluation

AQA exam questions regularly ask you to evaluate this experiment — identifying the largest source of uncertainty and suggesting improvements. Knowing which measurement dominates the uncertainty is the key to a high-mark answer.

The extension $\Delta L$ is typically the most uncertain measurement. For a steel wire under moderate loads, the extension might be only 0.5–1.0 mm. A vernier scale reads to $\pm 0.1$ mm, giving a percentage uncertainty of 10–20% on the extension alone. Compare that to the original length (measured to $\pm 1$ mm over 2 m, giving $< 0.1\%$ uncertainty) and you can see where the effort should go.

The wire diameter is the second critical measurement. Because area scales as $r^2$, any percentage uncertainty in the diameter is doubled in the area. Measuring the diameter at multiple positions and orientations (the wire may not be perfectly circular) and taking a mean reduces this uncertainty significantly.

The reference wire arrangement addresses a different problem: if the support structure bends or the temperature changes during the experiment, both wires are affected equally. The vernier scale measures the difference in position between the two wires, so any common movement cancels out. This is a systematic error reduction, not a random one.

7. Exam Technique and Common Mistakes

This practical appears in AQA papers in several forms: a calculation from given data, a graph interpretation question, or an evaluation of method. Each type has its own traps.

For a 6-mark evaluate question on this practical, structure your answer as: identify the largest uncertainty source and quantify it if possible, explain its effect on $E$, suggest a specific improvement, and state what that improvement achieves. Vague answers such as 'be more careful' do not score marks — the examiner wants a named instrument or technique change with a reason.

Every spec requirement for this required practical connects back to the same core principle: the Young modulus is the gradient of the elastic region of a stress-strain graph, and the quality of your result depends entirely on how accurately you measure force, length, diameter, and extension.