In short

Searle's apparatus measures Young's modulus Y of a long, thin metal wire by loading it and reading the resulting extension with a micrometer spirit level. Two identical wires hang side by side from a common rigid support:

  • the experimental wire, loaded with slotted masses hung from a hook;
  • the reference wire, carrying only a small fixed tension to keep it straight.

A rectangular frame carries a spirit level whose one end pivots on the reference wire and whose other end is raised or lowered by a micrometer screw resting on the experimental wire. When the experimental wire stretches by \Delta L, the screw must be turned by the same distance to re-level the bubble — and the micrometer reads \Delta L to \pm 0.01 mm.

From the article on Hooke's law and Young's modulus, the extension of a wire under load F is

\Delta L \;=\; \frac{F L}{A Y}, \qquad Y \;=\; \frac{F L}{A\,\Delta L} \;=\; \frac{M g L}{\pi r^2 \,\Delta L}.

Plot load F = Mg (vertical) against extension \Delta L (horizontal). The points lie on a straight line through the origin within the elastic limit; the slope s = F/\Delta L equals A Y / L, so

\boxed{\;Y \;=\; \frac{s L}{A} \;=\; \frac{s L}{\pi r^2}.\;}

For a 2 m steel wire of diameter 0.5 mm, each 1 kg load adds about 0.25 mm of extension; the slope is \sim 39\,000 N/m and gives Y \approx 2.0 \times 10^{11} Pa — the steel value.

Why two wires? The reference wire hangs from the same support and is in the same ambient air. Any temperature change, any sag of the ceiling clamp, any drift of the support — all of these happen to both wires equally, so the relative reading on the spirit level is insensitive to them. Only the differential effect of the load — pure elastic extension — shows up. This is the key idea behind the entire design.

The practical is set up on the far wall of the third-year physics laboratory at IIT Kanpur. A steel hook is bolted into the concrete ceiling, six metres above the floor. Two lengths of steel wire — 0.5 mm diameter, 2 metres long, cut from the same roll so they are genuinely identical — hang from the hook. On each wire's lower end sits a small brass frame. Between the two frames is clamped a bubble level with a micrometer screw at one end. An aluminium weight-hanger dangles from the experimental wire; a box of slotted masses — 100 g, 200 g, 500 g, 1000 g — sits on the table below. The reference wire carries only a small 500-gram dead load, just enough to keep it taut and straight.

A student attaches the first slotted mass. The experimental wire stretches by 0.25 mm — too small to see by eye, but the bubble on the spirit level between the two wires tilts visibly to one side. The student turns the micrometer screw until the bubble is centred again; the micrometer reads 0.25 mm. She adds another mass. Another 0.25 mm appears on the micrometer. Another. Another. Six loads in, she has five data points on a straight line, the graph is pencil-drawn in the practical file, the slope is computed, and the value of Young's modulus for the wire comes out at 2.01 \times 10^{11} Pa. The known value for mild steel is 2.00 \times 10^{11} Pa. The experiment — designed by G. F. C. Searle at Cambridge a hundred years ago and in continuous use ever since — has just measured one of the most fundamental material properties of steel to within half a per cent, in less than an afternoon, with equipment that cost less than a new smartphone.

This article is the companion to the theoretical article Hooke's Law and Young's Modulus, which derives Y = FL/(A\,\Delta L) from the stress-strain proportionality. This one is the NCERT / CBSE / JEE-syllabus practical: the apparatus, the procedure, the derivation of why two wires are used, and the graphical analysis you will actually hand in with your practical file. If you have a lab session in Roorkee or Pune or Chennai tomorrow where you will perform Searle's experiment, read this first.

Why Young's modulus is hard to measure

Before the apparatus, pause on the numbers. A 2-metre steel wire of 0.5 mm diameter loaded with 2 kg stretches by about 0.5 mm — one part in four thousand. A wooden ruler held next to the wire sees nothing at all; you cannot perceive 0.5 mm over 2 m with your eyes. Rough experimental approaches — eyeballing a pointer against a millimetre scale, or weighing a spring-like extension against a spring balance — give Y to maybe 20%, which is uselessly bad for a quantity that distinguishes steel from brass from aluminium by factors well within a factor of two.

To get 1% precision on Y, you need to measure \Delta L to about 1% of 0.5 mm, which is \pm 0.005 mm. That is one-fifth the diameter of a human hair. You need a micrometer-scale length readout. Searle's apparatus is the low-cost, mechanical way to achieve this. Optical methods (lever-and-mirror, Michelson interferometer) also work and are even more precise, but they are not the CBSE practical; Searle's is.

The apparatus

Here is what you will find on the bench of a typical CBSE / ISC / JEE-practical laboratory.

Searle's apparatus for Young's modulusTwo identical vertical wires hang from a rigid ceiling clamp at the top. The left wire is the reference wire, carrying only a small fixed tension from a hanger at the bottom labelled W (reference). The right wire is the experimental wire, carrying a much larger variable load from a hanger labelled W + n m g. Between the bottoms of the two wires sits a horizontal rectangular frame carrying a spirit level; the left end of the frame is hinged on the reference wire and the right end rests on a vertical micrometer screw attached to the experimental wire's frame. When the experimental wire stretches, the frame tilts until the screw is turned to restore it. The original length L of the wires is labelled between the clamp and the frames. The bubble of the spirit level is shown centred when the frame is horizontal.rigid ceiling clampreference wireexperimental wireLspirit level (bubble)hingemicrometer screwW₀reference loadW₀+ Mref + slotted masses
Searle's apparatus. Two identical wires of length $L$ hang from a common rigid support. The reference wire (left) carries a small fixed tension $W_0$ to keep it straight; the experimental wire (right) carries the same $W_0$ plus variable slotted masses $M$. A rectangular frame spans the two wires near their lower ends. Its left end pivots freely on the reference wire; its right end is supported by a vertical micrometer screw attached to the experimental wire's frame. The spirit level sits horizontally on top of this frame. When the experimental wire stretches by $\Delta L$, the right side of the frame drops by $\Delta L$, tilting the bubble. Turning the micrometer screw raises that side back — the amount you turn equals $\Delta L$.

The components are:

Before any loading, the micrometer is brought to a "zero reading" at which the bubble is centred. This zero reading is recorded. Every subsequent reading is a difference from this zero, which is the extension \Delta L of the experimental wire.

The two-wire trick — why it's the key idea

The first instinct is: why use two wires? Can't you just hang one wire and measure its elongation against a ruler glued to the wall?

You cannot. For three reasons.

First, temperature. Steel has a linear thermal expansion coefficient \alpha \approx 12 \times 10^{-6} per °C. A 2-metre wire in a lab whose temperature rises by just 1 °C during the experiment expands by L\alpha \Delta T = 2 \times 12 \times 10^{-6} \times 1 = 2.4 \times 10^{-5} m = 0.024 mm — which is ten per cent of the total elastic extension you are trying to measure. Even a class-11 lab has ambient temperature drift of 1–2 °C during an hour-long practical, especially on a hot Delhi afternoon.

Second, support sag. A ceiling clamp is not infinitely rigid. Add 5 kg of load, and the clamp and the building itself sag microscopically — maybe 0.01 mm. You cannot tell this apart from the wire stretching.

Third, wire straightening. A wire that has been on a shelf is not perfectly straight; under increasing load, the kinks and coils pull out before the material itself starts stretching elastically. If you measured against a fixed ruler you would record this straightening as extension and badly over-estimate \Delta L.

Putting a reference wire alongside the experimental wire, hung from the same clamp, and measuring the relative displacement of their lower ends, kills all three. A 1 °C temperature rise stretches both wires by 0.024 mm — the spirit level frame descends by 0.024 mm at both ends, the bubble does not tilt, the micrometer does not move. A clamp sag of 0.01 mm drops both wires by 0.01 mm together — again, no tilt. Residual kinks in the experimental wire are straightened out during an initial loading cycle (see procedure) before data taking begins. Only the differential extension — the part of the experimental wire's elongation that the reference wire does not share — shows up on the micrometer, and that is pure elastic \Delta L.

Searle's 1907 design for this apparatus is a textbook case of differential measurement: when you cannot beat the noise, arrange to measure the signal relative to a twin sample that carries the same noise. The same trick appears in Wheatstone's bridge (measure resistance ratios, not absolute resistances), in modern atomic clocks (compare against a second identical clock), and in gravitational-wave detectors (the two LIGO arms). Searle's apparatus is the mechanical version.

Deriving Y from the load–extension graph

From Hooke's law and Young's modulus, within the elastic limit:

F \;=\; Y A \frac{\Delta L}{L} \qquad \Longleftrightarrow \qquad \Delta L \;=\; \frac{F L}{A Y} \tag{1}

For a wire of length L and radius r, A = \pi r^2.

Step 1. The load on the experimental wire is F = Mg, where M is the slotted mass added beyond the initial reference load.

Why: the initial reference load W_0 is present on both wires and produces an equal small extension on each. Its contribution is already baked into the zero reading of the micrometer; only the extra mass M produces measurable differential extension.

Step 2. Substituting into (1):

\Delta L \;=\; \frac{M g L}{\pi r^2 Y}. \tag{2}

Step 3. This is linear in M. So if you plot \Delta L on the x-axis and F = Mg on the y-axis, you get a straight line through the origin with slope

s \;=\; \frac{F}{\Delta L} \;=\; \frac{\pi r^2 Y}{L}. \tag{3}

Why: rearranging (2) gives F/\Delta L = \pi r^2 Y / L. Both sides are ratios of independently-measured quantities; s is the slope of the load-versus-extension line you plot from your data.

Step 4. Solve for Y:

\boxed{\;Y \;=\; \frac{s L}{\pi r^2}\;} \tag{4}

Why: the geometric factors L and \pi r^2 are measured separately (metre scale for L, screw gauge for the diameter 2r). The slope s is read from the graph. Multiplying gives Y in SI units (Pa).

Equation (4) is the working formula. Notice three things:

  1. Small r is good. If the wire is thinner, the extension for a given load is larger (equation (2)), so the relative error in reading \Delta L is smaller. But a too-thin wire yields under small loads or snaps.

  2. Large L is good, for the same reason. Longer wire gives larger \Delta L per newton. Lab ceiling height limits this to about 2 m in practice.

  3. No dependence on the mass of the wire itself. The wire's own weight produces a tiny elongation (a tension that grows from 0 at the bottom to mg_{\text{wire}} at the top), but this too appears in the zero reading before data taking and does not affect the slope.

The procedure

The canonical CBSE / JEE-syllabus procedure is:

Step 1: measure wire dimensions.

Why: Y \propto L/r^2. A 1% error in r is a 2% error in Y (the square). Diameter measurement is the single biggest error source in a typical undergraduate run, so measure it carefully.

Step 2: set the reference load. Add the reference load W_0 (usually the 0.5 kg hanger alone) to both hangers — both the reference wire and the experimental wire. This straightens both wires and brings them to a common initial stress state.

Step 3: initial cycling. Before starting real data, add and remove the full anticipated load once or twice. This "works" the wire and pulls out any residual kinks or residual plastic sets. If the micrometer reading does not return to its initial value after the load is removed, the wire is still being straightened; cycle again. Record data only after the return is repeatable.

Step 4: take the zero reading. With only the reference load W_0 on the experimental wire (no slotted mass), turn the micrometer screw until the spirit level's bubble is centred. Record the micrometer reading d_0.

Step 5: load and read — incremental. Gently add the first slotted mass M_1 = 0.5 kg to the experimental wire's hanger. The bubble tilts. Turn the micrometer screw (slowly — fast turning risks jarring the frame) until the bubble is centred again. Record the new micrometer reading d_1. The extension is \Delta L_1 = d_1 - d_0.

Wait 30 seconds after adding each mass — the wire takes a moment to reach equilibrium because internal friction damps the small post-load oscillations. Reading too fast gives \Delta L values systematically smaller than the true static extension.

Step 6: continue up the loading curve. Add each successive mass and read: M = 1.0 kg, M = 1.5 kg, M = 2.0 kg, M = 2.5 kg, M = 3.0 kg. For a 0.5 mm steel wire, 3 kg is well within the elastic limit (steel yields at stress \sim 2.5 \times 10^8 Pa, giving a yield load of \sim 40 kg for a 0.5 mm wire). Record the six or seven load-extension pairs.

Step 7: unload. Remove the masses one at a time, reading \Delta L at each step again. The unloading points should fall on the same line as the loading points — this is your check that the wire is in its elastic regime and no plastic deformation occurred. If the unloading points consistently sit below the loading points, the wire yielded somewhere and the data from that point onward is unreliable.

Step 8: plot and compute. Plot F = Mg (newtons) on the y-axis against \Delta L (metres) on the x-axis. Draw the best-fit straight line through the points and through the origin. Measure the slope s in N/m. Then apply equation (4):

Y \;=\; \frac{s L}{\pi r^2} \qquad (\text{units: Pa}).

For steel, expect Y \approx 2 \times 10^{11} Pa. For brass, Y \approx 1 \times 10^{11} Pa. Aluminium gives Y \approx 7 \times 10^{10} Pa.

Interactive: explore the load–extension line

The figure below simulates a Searle's experiment on a 2 m steel wire of diameter 0.5 mm with Y = 2.00 \times 10^{11} Pa. Drag the red dot to change the load (from 0 to 4 kg); the extension \Delta L updates live and the point slides along the line. The slope s = F/\Delta L is displayed, as is the inferred Y = sL/(\pi r^2) — which should read a constant 2.00 × 10¹¹ Pa as long as you stay within the elastic limit.

Interactive load–extension line for a steel wire A straight line from the origin showing the linear relationship between load F in newtons and extension delta L in metres for a 2 m steel wire of diameter 0.5 mm. A draggable dot lets the reader pick a load; the extension follows. The slope equals pi r squared Y over L and is used to extract Young's modulus. extension ΔL (mm) load F (N) 0 0.2 0.4 0.6 0.8 1.0 1.2 0 10 20 30 40 slope s = F / ΔL = π r² Y / L drag load up or down
Drag the red dot along the vertical axis to change the load $F$. The horizontal position of the dot is the corresponding extension $\Delta L$. The line $F = (A Y / L)\,\Delta L$ is fixed. Its slope $s \approx 39\,270$ N/m combines with $L$ and $\pi r^2$ to return $Y = 2.00 \times 10^{11}$ Pa — the steel value.

Worked examples

Example 1: Steel wire in an NCERT lab

A JEE student in Roorkee performs Searle's experiment on a mild steel wire. She measures L = 1.983 m with a metre scale; the mean diameter from five screw-gauge readings is d = 0.506 mm. After initial cycling, her loading data is:

M (kg) \Delta L (mm)
0.5 0.25
1.0 0.49
1.5 0.74
2.0 0.98
2.5 1.24
3.0 1.48

Find Young's modulus for her wire.

Load-versus-extension data for a steel wireA graph with load F on the vertical axis in newtons from 0 to 35 and extension delta L on the horizontal axis in mm from 0 to 1.6. Six data points are plotted, falling on a straight line through the origin. A best-fit line is drawn. The slope of the line is indicated as approximately 19.87 kN per m.extension ΔL (mm)load F (N)00.40.81.21.60102030slope s ≈ 19.87 kN/m= π r² Y / L
Six data points from Example 1, falling cleanly on a line through the origin within the elastic regime. The slope $s = F/\Delta L$ is the number used to extract $Y$.

Step 1. Convert units. \Delta L values in mm → multiply by 10^{-3} for metres. Load values: multiply M by g = 9.80 m/s² to get F in newtons. So the data becomes: (F, \Delta L) pairs in SI units:

  • (4.90, 2.5\times 10^{-4}), (9.80, 4.9\times 10^{-4}), (14.70, 7.4\times 10^{-4}), (19.60, 9.8\times 10^{-4}), (24.50, 1.24\times 10^{-3}), (29.40, 1.48\times 10^{-3}).

Why: the derivation of Y demands SI units throughout so the final answer comes out in Pa without hidden factors.

Step 2. Compute the slope using the mean of F/\Delta L for each point (since all go through the origin):

  • 4.90/(2.5\times 10^{-4}) = 19\,600
  • 9.80/(4.9\times 10^{-4}) = 20\,000
  • 14.70/(7.4\times 10^{-4}) = 19\,865
  • 19.60/(9.8\times 10^{-4}) = 20\,000
  • 24.50/(1.24\times 10^{-3}) = 19\,758
  • 29.40/(1.48\times 10^{-3}) = 19\,865

Mean: s = (19\,600 + 20\,000 + 19\,865 + 20\,000 + 19\,758 + 19\,865)/6 = 19\,848 N/m.

Why: for a line through the origin, a good estimator of the slope is the average of individual F/\Delta L ratios. A least-squares fit through the origin gives essentially the same number (it weights larger-load points a bit more, which is fine because they have smaller relative error).

Step 3. Compute A = \pi r^2. With d = 0.506 mm, r = 0.253 \times 10^{-3} m:

A \;=\; \pi (0.253 \times 10^{-3})^2 \;=\; \pi \times 6.4 \times 10^{-8} \;=\; 2.011 \times 10^{-7}\ \text{m}^2.

Why: the wire's cross-section is the area on which the load is distributed. The formula needs A not d.

Step 4. Compute Y using equation (4):

Y \;=\; \frac{s L}{A} \;=\; \frac{19\,848 \times 1.983}{2.011 \times 10^{-7}} \;=\; \frac{39\,358}{2.011 \times 10^{-7}} \;=\; 1.957 \times 10^{11}\ \text{Pa}.

Why: s in N/m times L in m gives N. Dividing by A in m² gives N/m² = Pa, which is the SI unit of Young's modulus.

Step 5. Compare to textbook value for mild steel, Y = 2.00 \times 10^{11} Pa. The student's answer is 1.96 \times 10^{11} Pa — within 2% of the accepted value, which is excellent agreement for a first-year undergraduate practical.

Result: Y = 1.96 \times 10^{11} Pa. Agreement with the textbook steel value to within 2%.

What this shows: A metre scale, a screw gauge, a micrometer, and a handful of kilogram weights measure Young's modulus of steel to two significant figures in one afternoon — a number that would have been unimaginable to anyone before the twentieth century, now routinely measured by every first-year engineering student at IITs, RECs, and every state engineering college in India.

Example 2: Brass wire, and an error analysis

A second student uses the same apparatus to test a brass wire of similar dimensions: L = 1.985 m, d = 0.508 mm. Her slope from the load-vs-extension plot is s = 9\,920 \pm 70 N/m. (The uncertainty is from the scatter of her data about the best-fit line.) She also records the uncertainties in the geometric measurements: \Delta L_{\text{wire}} = \pm 1 mm on a 1985 mm length, \Delta d = \pm 0.01 mm on a 0.508 mm diameter. Find (a) her value of Y, (b) the fractional uncertainty in Y, and (c) the absolute uncertainty.

Error propagation chart for Young's modulusA block diagram showing the formula Y equals s L over pi r squared. Below each input variable is a box with the variable's fractional uncertainty: slope s has 0.7 per cent, length L has 0.05 per cent, radius r has 1.97 per cent. Arrows combine these to produce the total fractional uncertainty in Y of approximately 4.1 per cent, shown in a final box.slope sΔs/s = 0.7%length LΔL/L = 0.05%radius rΔr/r = 1.97%radius²2 × 1.97% = 3.94%fractional uncertainty in Y√(0.7² + 0.05² + 3.94²) ≈ 4.0%
Error propagation in Searle's experiment. The dominant contribution comes from $r$ — because $Y \propto 1/r^2$, a 2% uncertainty in $r$ becomes a 4% uncertainty in $Y$ through the squaring.

Step 1. Compute the central value. A = \pi (0.254 \times 10^{-3})^2 = 2.027 \times 10^{-7} m².

Y \;=\; \frac{s L}{A} \;=\; \frac{9\,920 \times 1.985}{2.027 \times 10^{-7}} \;=\; \frac{19\,691}{2.027 \times 10^{-7}} \;=\; 9.71 \times 10^{10}\ \text{Pa}.

Why: same equation (4) as before; only the numbers change for brass.

Step 2. Fractional uncertainty propagation. Young's modulus from (4) is Y = sL/(\pi r^2). Taking logs and differentials:

\frac{\Delta Y}{Y} \;=\; \sqrt{\left(\frac{\Delta s}{s}\right)^2 + \left(\frac{\Delta L}{L}\right)^2 + \left(2\,\frac{\Delta r}{r}\right)^2}.

Why: for a multiplicative formula Y = k s^{a} L^{b} r^{c} (here a=1, b=1, c=-2), the fractional uncertainty in Y is the RSS (root-sum-square) of the fractional uncertainties in each factor, each weighted by its exponent's absolute value. The exponent 2 on r means radius errors are doubled in their impact.

Step 3. Numbers.

  • \Delta s / s = 70/9\,920 = 0.00706 = 0.71\%
  • \Delta L / L = 0.001 / 1.985 = 0.000504 = 0.050\%
  • \Delta r / r = 0.01/0.508 = 0.0197 = 1.97\% (since \Delta d/d passes directly through when r = d/2)
  • 2 \Delta r / r = 3.94\%
\frac{\Delta Y}{Y} \;=\; \sqrt{0.71^2 + 0.05^2 + 3.94^2}\ \% \;=\; \sqrt{0.50 + 0.0025 + 15.52}\ \% \;=\; \sqrt{16.03}\ \% \;=\; 4.00\%.

Step 4. Absolute uncertainty. \Delta Y = 0.04 \times 9.71 \times 10^{10} = 0.39 \times 10^{10} Pa.

Result: Y = (9.7 \pm 0.4) \times 10^{10} Pa — consistent with the textbook brass value of 9.9 \times 10^{10} Pa.

What this shows: The dominant error in Searle's experiment is the diameter measurement — not the micrometer reading, and not the metre scale. Improve the screw gauge measurement (more readings, a more precise gauge) and you gain more than any other change to the procedure. This is a broadly useful lesson: in a formula where one variable appears squared, a small error in that variable dominates everything else.

Sources of error and typical magnitudes

A summary of error sources in order of typical size for a Searle's experiment performed in a school / first-year engineering lab:

Adding the dominant errors in quadrature, a careful CBSE experiment returns Y to about \pm 3\%, and a careful JEE / first-year-engineering experiment with care in diameter measurement reaches \pm 2\%.

Common confusions

If you came here to do the experiment, read the formula, and turn in the practical file, you already have what you need. What follows is for readers who want the formal error propagation, the weight-of-wire correction, the alternative optical-lever method, and the connection to the full stress–strain curve.

The weight-of-wire correction

Equation (2) treated the wire's own weight as negligible, but for a precise experiment there is a subtle correction. The tension at the top of the wire is the full load plus the wire's own weight; the tension at the bottom is just the load. A 2 m long, 0.5 mm diameter steel wire has mass \rho A L = 7850 \times 2.01 \times 10^{-7} \times 2 = 3.2 \times 10^{-3} kg — about 3 grams. Negligible against a 1 kg load at, say, 1% of the load, making the correction tiny. But for a thicker wire (say 2 mm diameter, 80 g mass) and a small test load (say 500 g), the self-weight becomes a 16% contributor to the top-of-wire tension.

The corrected formula treats the wire as a column where tension varies linearly with position. The extension integrates to

\Delta L \;=\; \frac{L}{A Y}\left(F_{\text{external}} + \frac{1}{2}W_{\text{wire}}\right),

where W_{\text{wire}} = \rho A L g is the wire's weight. The factor of \frac{1}{2} appears because the average tension along the wire is F_{\text{ext}} + W_{\text{wire}}/2. For CBSE wires (a few grams under kg loads), this correction is negligible and ignored.

The optical-lever (Michelson-interferometer) variant

Searle's apparatus reaches 0.01 mm resolution — fine for a school lab, but not for a research lab. A more precise version uses an optical lever: a small mirror is attached to the experimental wire's frame, and a laser reflecting off the mirror travels to a distant screen. A tilt of the mirror by \theta moves the reflected spot on the screen by 2D\theta, where D is the distance to the screen. With D = 3 m and a tilt of 0.01\,\text{mm}/5\,\text{cm} = 2 \times 10^{-4} rad (where 5 cm is the lever arm), the spot moves by 1.2 mm — forty times amplified.

A Michelson interferometer can do even better: count fringes as the wire extends. Each fringe corresponds to a half-wavelength (about 0.3\ \mum = 0.0003 mm) of extension. Research measurements of Y use this method and reach 0.1% precision, dominated then by the diameter measurement (which hasn't improved since the screw gauge — modern methods use a non-contact laser micrometer).

Least-squares fit and uncertainty on the slope

Instead of averaging F/\Delta L ratios, you can perform a formal least-squares fit to the line through the origin. The best-fit slope for n points (x_i, y_i) on y = sx (no intercept) is

\hat{s} \;=\; \frac{\sum_i x_i y_i}{\sum_i x_i^2},

and its standard uncertainty is

\sigma_{\hat{s}} \;=\; \sqrt{\frac{\sum_i (y_i - \hat{s} x_i)^2}{(n-1) \sum_i x_i^2}}.

For Example 1's data, \sum x_i y_i = 9.22 \times 10^{-2}, \sum x_i^2 = 4.65 \times 10^{-6}, giving \hat{s} = 19\,828 N/m with \sigma_{\hat{s}} \approx 180 N/m (about 1% statistical uncertainty) — very close to the simple average of ratios. For a small dataset of 6 points on a clean line, the simple-average method and least-squares disagree only at the level of the third significant figure.

The full stress–strain curve — beyond Hooke's law

For large loads, the wire leaves the Hookean regime and enters the non-linear part of its stress–strain curve — plastic flow, ultimate tensile stress, and eventual fracture. For mild steel, the proportional limit is at about 200 MPa (stress), i.e. a strain of 10^{-3}, i.e. a 2 mm extension of a 2 m wire. Beyond this, Y is no longer constant — the stress rises more slowly than the strain. A Searle's apparatus cannot access this regime because the extensions become too large for the micrometer screw's range and because the wire no longer returns to its original length on unloading (the load–extension line hysteresises).

For the full stress–strain curve, see the article on Elastic Potential Energy and Stress–Strain Curves. Engineers use tensile testing machines — essentially a vastly scaled-up Searle's rig with a computerised load cell and a position encoder — to map the full curve and find yield stress, ultimate strength, and fracture strain. The Indian standards lab (NPL, Delhi) runs such equipment for certification of structural steels used in buildings and bridges across the country.

Historical note

Searle designed this apparatus at the Cavendish Laboratory in Cambridge in 1907, publishing it in his Experimental Elasticity textbook. It was adopted by Indian physics labs almost immediately — Cambridge textbooks travelled with the colonial education system — and has been standard in undergraduate Indian physics syllabi since the 1920s. A century later, every IIT, every NIT, every state engineering college, and every serious BSc physics programme in India runs this experiment essentially unchanged from Searle's original design. This is unusual for a physics experiment — most have been superseded by electronic or optical replacements. Searle's apparatus survives because it makes the physics directly visible: you load a wire, it stretches, and the micrometer reads it. No black box. The concept of Young's modulus is demonstrated in the act of measurement.

Where this leads next