Measuring Young Modulus

Measuring the Young modulus of a material in the laboratory requires careful experimental technique. The challenge is that metals are very stiff — even under substantial loads, the extension of a wire might be only a fraction of a millimetre. Accurate results demand precise measurements and a systematic approach to minimising uncertainty.

The Long Wire Method

The standard A-level experiment uses a long, thin wire under tension. A long wire is used because extension is proportional to original length (ΔL = FL/AE), so a longer wire gives a larger, more measurable extension for the same stress.

Apparatus

A length of wire (typically copper or steel), at least 2–3 metres long
A clamp or fixed support at one end
A pulley at the free end with a weight hanger and slotted masses
A ruler or metre rule to measure the original length
A micrometer screw gauge to measure the wire diameter
A travelling microscope or marker and ruler system to measure extension
Safety goggles (the wire may snap under high loads)

Setup

The wire is clamped firmly at one end (often to a beam or the ceiling) and passed over a pulley. Masses are hung from the free end. A reference marker is attached to the wire near the measured section, and extension is measured using a fixed ruler or a travelling microscope focused on the marker.

Searle’s Apparatus

A more sophisticated version uses Searle’s apparatus, which consists of two identical wires hanging side by side from the same support. One is the test wire (loaded with increasing masses) and the other is the reference wire (kept under a constant small load to keep it taut).

A spirit level bridges the two wires. As the test wire extends, the spirit level tilts. A micrometer on one side of the spirit level is adjusted to bring it back to level, and the micrometer reading gives the extension directly.

Advantages of Searle’s apparatus:

The reference wire compensates for thermal expansion — if the room temperature changes, both wires expand equally, and the relative extension is unaffected.
The micrometer allows extensions to be measured to ±0.01 mm.
The spirit level provides a highly sensitive method for detecting small changes in extension.

Measurement Procedure

Step 1: Measure the original length (L)

Measure the length of wire between the clamped end and the point where extension is measured. Use a metre rule (resolution ±1 mm). Record L in metres.

Step 2: Measure the diameter (d)

Use a micrometer screw gauge (resolution ±0.01 mm). Take at least six readings at different points along the wire and at different orientations (rotating the micrometer by 90°). This accounts for:

The wire not being perfectly circular in cross-section
Variations in diameter along the length
Zero error on the micrometer

Calculate the mean diameter. Then calculate the cross-sectional area: A = π(d/2)².

Step 3: Apply loads and measure extension

Start with a small initial load to straighten the wire and remove any kinks. Then add masses in equal increments (e.g., 100 g or 1.0 N at a time). After each addition, wait for the wire to stop oscillating and reach equilibrium, then record the new position of the marker.

Extension for each load = current reading − initial reading (at the straightening load).

Step 4: Remove loads and check for elastic behaviour

After reaching the maximum load (staying well below the expected yield point), remove masses one at a time and record the extension as the load decreases. If the loading and unloading readings agree, the wire has remained within its elastic limit, confirming the measurements are valid.

Analysing the Results

Method 1: Direct calculation

For each pair of (F, ΔL) values: $E = \frac{FL}{A \Delta L}$

Calculate E for each data point and find the mean. This method is simple but does not make good use of all the data.

Method 2: Graphical method (preferred)

Plot a graph of force (F) against extension (ΔL). The graph should be a straight line through the origin (within the elastic region).

The gradient of this graph = F/ΔL = k (the spring constant of the wire).

Then: E = kL/A = (gradient × L) / A

Alternatively, plot stress against strain. The gradient of this graph directly gives the Young modulus.

The graphical method is preferred because:

It uses all data points, not just one pair
Anomalous results can be identified and excluded
The line of best fit averages out random errors
The linearity of the graph confirms Hooke’s law is obeyed

Worked Example: Full Experimental Calculation

A student performs the Young modulus experiment with a copper wire and records the following:

Mass / kg	Force / N	Extension / mm
0.50	4.91	0.10
1.00	9.81	0.21
1.50	14.72	0.30
2.00	19.62	0.41
2.50	24.53	0.51
3.00	29.43	0.62

Wire length: L = 2.50 m. Mean diameter: d = 0.48 mm.

Step 1: Gradient of F vs ΔL graph. Using first and last points: gradient = (29.43 − 4.91) / (0.62 − 0.10) × 10⁻³ = 24.52 / 5.2 × 10⁻⁴ = 47 150 N m⁻¹

Step 2: Cross-sectional area. A = π(0.24 × 10⁻³)² = 1.810 × 10⁻⁷ m²

Step 3: Young modulus. E = gradient × L / A = 47 150 × 2.50 / 1.810 × 10⁻⁷ = 117 875 / 1.810 × 10⁻⁷ = 6.51 × 10¹¹ Pa ≈ 65 GPa

This is lower than the accepted value of 130 GPa, suggesting systematic error — likely in the diameter (if overestimated, area would be too large and E too small).

Sources of Error and Uncertainty

Systematic errors

Zero error on the micrometer: Always check that the micrometer reads zero when fully closed. If it does not, record the zero error and subtract it from all readings.
The wire may not be truly straight initially: Apply a small pre-load to remove kinks before starting measurements.
Thermal expansion: Use Searle’s apparatus to compensate, or ensure room temperature is stable.

Random errors

Reading the ruler or marker position: Reduce by using a travelling microscope (±0.01 mm resolution vs ±1 mm for a ruler).
Wire diameter measurement: The wire may not be perfectly uniform. Take multiple readings and use the mean.

Detailed Uncertainty Analysis

The percentage uncertainty in each measured quantity:

Quantity	Typical value	Resolution	% uncertainty	Effect on E
Length L	2.50 m	±1 mm	0.04%	Small
Diameter d	0.48 mm	±0.01 mm	2.1%	×2 = 4.2% (via d²)
Extension ΔL	0.50 mm	±0.05 mm	10%	10%
Force F	29.4 N	±0.01 N	0.03%	Negligible

Key insight: The extension has the largest percentage uncertainty, followed by the diameter. This is why:

Using a longer wire increases the extension, reducing its % uncertainty.
Using a travelling microscope reduces the resolution uncertainty in ΔL.
Taking multiple diameter readings and using the mean reduces random error in d.

Combined uncertainty in E: Since E = FL/(AΔL) and A = πd²/4:

% uncertainty in E = % in F + % in L + 2 × (% in d) + % in ΔL ≈ 0.03 + 0.04 + 4.2 + 10 = 14.3%

For E = 130 GPa, this gives ΔE = ±18.6 GPa, so E = 130 ± 19 GPa.

Safety Precautions

Wear safety goggles — wires under tension can snap and recoil.
Place a cushion or sand tray below the weights to prevent damage if they fall.
Do not exceed the elastic limit — increasing the load too far may cause the wire to undergo plastic deformation or break.
Keep feet away from below the hanging masses.
Clamp the support securely to prevent the apparatus from toppling.

Common Exam Mistakes

Not plotting a graph: Single-point calculations waste data and are more susceptible to error. Always use the graphical method.
Ignoring the pre-load: Initial data points where the wire is straightening should be excluded from the gradient calculation.
Forgetting to double the % uncertainty in diameter: Since area = πd²/4, the percentage uncertainty in area is 2 × (% uncertainty in d).
Using the wrong gradient: If the extension axis is in mm, you must convert to m before calculating E, or the answer will be wrong by a factor of 1000.

A-Level Deep Dive: Measuring Young Modulus

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, Core Practical 2 (CP2): determination of the Young modulus of a metal wire requires students to plan and execute the measurement of $E$ from primary measurements of force, original length, diameter and extension, and to evaluate the experiment quantitatively in terms of uncertainty (refer to the official Pearson Edexcel specification document for exact wording). This lesson is the practical counterpart to the Young modulus theory lesson. CP2 is examined directly in 9PH0 Paper 3 — General and Practical Principles in Physics, and the same data-handling skills (gradient extraction, uncertainty propagation, evaluation of dominant error terms) appear in any Paper 1 Mechanics or Materials question that quotes raw data, and across CP1 (viscosity), CP4 (free fall) and CP9 (gas laws). The Edexcel formula booklet provides $E = \sigma/\varepsilon$ but does not spell out $E = FL/(A\Delta L)$ or $A = \pi d^2/4$ — both must be memorised.

Worked example with full mark scheme

Question (8 marks):

A student measures the Young modulus of a copper wire using the Searle's apparatus method. The recorded values are:

Original length $L = 1.850 \pm 0.002$ m
Diameter (mean of six readings, taken with a micrometer): $d = 0.42 \pm 0.01$ mm
Gradient of the force–extension graph (drawn through the origin and within the elastic region): $k = 64.0 \pm 2.0$ N mm⁻¹

(a) Calculate the Young modulus of copper from these data, in Pa, to an appropriate number of significant figures. (4)

(b) Determine the percentage uncertainty in $E$ and quote the final answer with its absolute uncertainty. (4)

Solution with mark scheme:

(a) Step 1 — convert the gradient to SI.

$k = 64.0$ N mm⁻¹ $= 6.40 \times 10^4$ N m⁻¹.

M1 — converting mm to m before substitution. Failing this is the single most common source of "answer out by a factor of 1000".

Step 2 — cross-sectional area.

$d = 0.42 \times 10^{-3}$ m, $r = 0.21 \times 10^{-3}$ m.

$A = \pi r^2 = \pi (0.21 \times 10^{-3})^2 = 1.385 \times 10^{-7}\ \text{m}^2$

M1 — using radius (not diameter) inside the squared term, or equivalently $A = \pi d^2/4$ . The error $A = \pi d^2$ (missing $/4$ ) loses this M1.

Step 3 — apply $E = kL/A$ .

$E = \dfrac{(6.40 \times 10^4)(1.850)}{1.385 \times 10^{-7}} = 8.55 \times 10^{11}\ \text{Pa}$

A1 — substitution carried out correctly with all quantities in SI; answer quoted to 3 sf consistent with the data: $E = 8.6 \times 10^{11}$ Pa. (For copper, the accepted value is $\sim 1.2 \times 10^{11}$ Pa; a candidate obtaining $8.6 \times 10^{11}$ Pa would flag the discrepancy as a likely diameter under-measurement.)

A1 — final value clearly identified, with units.

(b) Step 1 — list the percentage uncertainties.

$\Delta L / L = 0.002/1.850 = 0.108\%$
$\Delta d / d = 0.01/0.42 = 2.38\%$ , doubled because $d$ is squared in $A$ : contribution $= 4.76\%$
$\Delta k / k = 2.0/64.0 = 3.13\%$

M1 — recognising that $\Delta d/d$ doubles in the propagation because $A \propto d^2$ .

Step 2 — combine.

$\dfrac{\Delta E}{E} = \dfrac{\Delta k}{k} + \dfrac{\Delta L}{L} + 2 \cdot \dfrac{\Delta d}{d} = 3.13 + 0.11 + 4.76 = 8.0\%$

M1 — correct addition of fractional uncertainties for a product/quotient.

Step 3 — convert to absolute uncertainty.

$\Delta E = 8.0\% \times 8.55 \times 10^{11} = 6.8 \times 10^{10}$ Pa, rounded to 1 sf: $\pm 0.7 \times 10^{11}$ Pa.

A1 — absolute uncertainty quoted to 1 sf, in the same power of ten as the central value.

Step 4 — final quote.

$E = (8.6 \pm 0.7) \times 10^{11}\ \text{Pa}$

A1 — central value and uncertainty quoted to consistent precision, both in the same units.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (10 marks): A student is given a 2.0 m length of steel wire and the following equipment: two G-clamps, a pulley, slotted masses (100 g each), a metre rule, a micrometer screw gauge, a vernier scale, masking tape and a small marker flag.

(a) Describe a procedure to determine the Young modulus of steel from this equipment. Include a labelled diagram and the measurements you would take. (6)

(b) Identify the dominant source of uncertainty in your procedure and explain quantitatively how it could be reduced. (2)

Mark scheme decomposition by AO:

(a)

B1 (AO1.2) — labelled diagram showing wire clamped at one end, passing horizontally over a pulley, with masses on a hanger; marker flag taped to the wire close to the pulley; metre rule below the wire.
B1 (AO1.2) — measurement plan: original length $L$ from clamp to marker with metre rule; diameter $d$ at three to six points using the micrometer, recording each reading and taking the mean; extension $\Delta L$ from the marker position relative to a fixed reference, read with the vernier scale.
M1 (AO2.1) — strategy of adding masses in equal increments and recording extension after the wire has stopped oscillating; tabulating $F$ against $\Delta L$ .
M1 (AO2.1) — analytical strategy of plotting $F$ against $\Delta L$ , taking the gradient $k$ , then computing $E = kL/A$ with $A = \pi d^2/4$ .
B1 (AO3.5) — checking elastic behaviour by unloading and confirming the wire returns to its original length (loading and unloading data fall on the same straight line through the origin).
B1 (AO3.5) — taking initial measurement at a small pre-load to remove kinks, treating that point as the zero of extension.

(b)

B1 (AO3.4) — identifying $\Delta L$ as dominant: at $\Delta L \sim 1$ mm with vernier resolution $\pm 0.05$ mm the percentage uncertainty is $\sim 5\%$ , larger than the contribution from $d$ for a typical 0.4 mm wire (where $2 \Delta d/d \sim 5\%$ may be comparable, so accept either identification with quantitative justification).
B1 (AO3.4) — reduction: use a longer wire (extension scales with $L$ ) or a travelling microscope with $\pm 0.01$ mm resolution; quantitatively this reduces $\Delta(\Delta L)/\Delta L$ from $\sim 5\%$ to $\sim 1\%$ .