In short

Hooke's law says that within the elastic limit, stress is directly proportional to strain: \sigma = Y\varepsilon for a wire stretched along its length. The proportionality constant Y is Young's modulus — a measure of the material's stiffness, measured in pascals (Pa = N/m²). For a rod of length L and cross-sectional area A pulled by a force F and extending by \Delta L, the formula becomes Y = \dfrac{F L}{A\,\Delta L}. Steel has Y \approx 2 \times 10^{11} Pa; rubber has Y \approx 10^6 Pa — steel is two hundred thousand times stiffer. Beyond the elastic limit, the material yields and Hooke's law fails. Searle's apparatus measures Y by hanging a known load on a wire and reading the tiny extension with a micrometer.

Hang a 1 kg weight from a 2-metre-long steel wire thinner than a matchstick. The wire stretches — not by much. By about 0.01 millimetres. Add another kilogram. It stretches by another 0.01 mm. Add a third. Another 0.01 mm. The relationship is brutally simple: doubling the load doubles the stretch; tripling the load triples it. This is not an approximation; within a wide range of loads, it is exact to the precision of the best instruments we have.

That exactness is Hooke's law — and it holds for almost every solid material in almost every situation that does not involve hammers, explosives, or permanent damage. When you sit on a chair, the chair's legs compress by a few micrometres, exactly proportional to your weight. When you drive over the Bandra–Worli Sea Link, the massive cables that hold up the deck stretch by a few millimetres under the load, exactly proportional to the traffic. Hooke's law is the foundation on which every building, bridge, bone, and bicycle frame is designed.

This article derives it, attaches the correct proportionality constant (Young's modulus), compares that constant across materials in the Indian context (railway rails, bamboo scaffolding, a cricket bat), shows you what happens when Hooke's law fails, and walks through how you measure Y in a JEE lab with Searle's apparatus.

Hooke's law — the proportionality

Take a uniform wire of original length L, cross-sectional area A, attached to the ceiling. Hang a weight F from the bottom. The wire stretches by a small amount \Delta L.

Assumptions. The wire has uniform cross-section and material. The load is applied gradually (no jerk). The extension \Delta L is small compared to L. The wire is well below its elastic limit — that is, releasing the load returns the wire exactly to its original length. Temperature is constant.

Experiments done repeatedly — by Robert Hooke in 1676, and by every undergraduate physics student since — show the same result: for any given wire, as long as the load is small enough,

\Delta L \propto F.

Double the load, you get double the extension. Not the square of the load, not the square root — just the load. Linear.

But this is only the first piece. Try a longer wire (same material, same thickness): the extension doubles if you double L.

\Delta L \propto L.

And try a thicker wire (same material, same length): the extension halves if you double A.

\Delta L \propto \frac{1}{A}.

Put all three together:

\Delta L \propto \frac{FL}{A}.

Why these three proportionalities

Each of them is intuitive once you picture the material.

Factor out the quantities that depend on the specimen's shape (namely L and A) to leave only a material-dependent constant. Define the stress as force per unit area and the strain as fractional extension (see Stress and Strain for the full setup):

\sigma = \frac{F}{A}, \qquad \varepsilon = \frac{\Delta L}{L}.

Then the combined proportionality \Delta L \propto FL/A rearranges into

\frac{\Delta L}{L} \propto \frac{F}{A} \quad\Longleftrightarrow\quad \varepsilon \propto \sigma.

Why: dividing both sides of \Delta L \propto FL/A by L puts the strain on the left and the stress on the right, with a dimensionless proportionality constant pulling the two sides together. The geometry factors L and A have vanished — what remains is a statement about the material alone.

This is Hooke's law in its most useful form:

\boxed{\; \sigma = Y \varepsilon \;}

where Y is the Young's modulus — a property of the material that captures how stiff it is.

Young's modulus — the material constant

Rearranging Hooke's law to solve for Y:

Y = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta L / L} = \frac{F L}{A\, \Delta L}.

This is the working formula every JEE student should have permanently available. It converts four measurable things — load F, original length L, cross-sectional area A, and extension \Delta L — into a material property.

Units and a physical meaning

The units of Y are pascals:

[Y] = \frac{\text{N/m}^2}{\text{dimensionless}} = \text{N/m}^2 = \text{Pa}.

Since strain is dimensionless, Y has the same units as stress. But for typical materials, Y is enormous — typically 10^9 to 10^{11} Pa — so you usually see it quoted in gigapascals (GPa = 10^9 Pa).

There is a clean way to interpret Y physically: if you could apply enough stress to double the length of a material (strain \varepsilon = 1) while staying within its Hookean range, the required stress would equal Y. No real material actually survives strain = 1 — steel breaks at strain around 0.002 — but that is the intuition the number captures. Y is the stress you would need to apply to achieve 100% strain if the material obeyed Hooke's law forever.

The material landscape

Here is a table of Young's modulus for materials you will encounter in Indian engineering and biology:

Material Young's modulus Y (GPa) Context
Diamond 1,050 Hardest natural material
Tungsten 411 Light bulb filament
Steel (mild) 200 Railway rails, bridge cables
Iron (pure) 170 Cast iron pipes
Brass 100 Locks, bell metal
Copper 120 Household wiring
Aluminium 70 Metro coaches, kitchen foil
Glass 70 Window panes
Concrete 30 High-rise construction
Bone (compact) 15 Human femur
Wood (sal, along grain) 12 Indian hardwood furniture
Bamboo (along grain) 18 Scaffolding across India
Ice 9 Glaciers
Nylon 3 Fishing line, ropes
Rubber (natural) 0.01 – 0.1 Tyres, rubber bands

Rubber's Y is smaller than steel's by a factor of two million. That is why you can stretch a rubber band with your fingers (you are applying only a few newtons and getting centimetres of extension) but you cannot stretch a steel wire of similar size at all. Bamboo — at 18 GPa along its fibre direction — is stiffer than oak. This is why scaffolders in Mumbai still prefer bamboo over steel for low-rise construction: it is strong, stiff, light, cheap, and bends without snapping.

Young's modulus bar chart for common materials on a logarithmic scale A horizontal bar chart showing Young's modulus in gigapascals for nine materials on a logarithmic scale: rubber at 0.05, ice at 9, bone at 15, bamboo at 18, concrete at 30, aluminium at 70, copper at 120, steel at 200, and diamond at 1050. Y (GPa, log scale) 0.1 1 10 100 1000 rubber ice bone bamboo concrete aluminium copper steel diamond
Young's modulus spans five orders of magnitude across everyday materials. Note the log-scale axis: each tick is a factor of ten. Rubber is near the left end; steel and diamond are near the right.

Stress–strain curve — where Hooke's law lives

Hooke's law is not a law of nature. It is an empirical observation that works within a restricted range. Push a material hard enough and the proportionality breaks. The stress–strain curve is a plot of stress versus strain as you steadily increase the load on a material until it breaks. It tells the full story.

For a typical ductile metal like mild steel, the curve looks like this:

Stress-strain curve for a ductile metal showing elastic, yield, plastic, and fracture regions A stress-strain curve rises linearly from the origin to the proportional limit, continues almost linearly to the elastic limit, dips slightly at the upper yield point, then flattens at the lower yield point before rising through the plastic region to a maximum at the ultimate tensile strength, then falling to a final fracture point. strain ε stress σ proportional limit P elastic limit E yield point Y ultimate U fracture F elastic (Hookean) plastic (irreversible) ε_E ε_U ε_F
The stress-strain curve for a typical ductile metal (like mild steel). The Hookean region is the straight line from origin to the proportional limit $P$. Beyond the elastic limit $E$, the wire does not return to its original length when unloaded. At the yield point $Y$, the material begins to flow plastically. It continues to carry increasing load up to the ultimate tensile strength $U$, then necks down and fails at the fracture point $F$.

Five important features are marked on the curve.

Proportional limit (P)

Up to stress \sigma_P, the curve is a straight line — stress is proportional to strain. Here Hooke's law holds exactly. The slope of this line is Young's modulus:

Y = \frac{d\sigma}{d\varepsilon} \quad\text{in the Hookean region}.

Below P, the material is behaving like a linear spring. Every JEE problem involving wires assumes you are operating in this region.

Elastic limit (E)

Slightly beyond P lies the elastic limit. Between P and E, the curve is still nearly linear, but Hooke's law is no longer exactly true — the curve bends slightly. What still holds, up to E, is elasticity: if you remove the load, the wire returns to its original length.

For many materials, P and E are almost indistinguishable; the separation between them is small enough that it is often ignored, and students use "elastic limit" to mean both.

Yield point (Y)

Just past the elastic limit, the material begins to yield — it deforms permanently. If you remove the load after this point, the wire stays slightly longer than it was. A small increase in stress past Y produces a big increase in strain. For mild steel, there is often an upper yield point where the stress momentarily dips (because dislocations start propagating through the crystal), followed by a lower yield point where flow begins in earnest.

The region between E and Y is formally the end of the elastic regime; beyond Y, you are in plastic deformation.

Ultimate tensile strength (U)

Past the yield point, the material continues to hold more load — just not linearly. As strain grows, the wire hardens (a process called work hardening), and stress rises until reaching the ultimate tensile strength \sigma_U — the highest stress the material can support. For mild steel, \sigma_U \approx 400 MPa; for a high-strength steel cable, it can reach 2 GPa.

Fracture (F)

Beyond the ultimate, the material begins to narrow — a region called necking — where the cross-section shrinks rapidly at one location. The engineering stress (calculated using the original cross-section) appears to drop, but the true stress at the neck keeps rising. Eventually the neck becomes so thin that the material tears. This is the fracture point.

What "elastic limit" means operationally

For the purposes of JEE and most Indian physics textbooks, "the elastic limit" refers to the boundary between reversible and irreversible deformation. Within it: Hooke's law (approximately), full recovery upon unloading. Outside it: permanent deformation, failure of Hooke's law. The proportional limit is sometimes called the limit of proportionality and is stricter (demands exact linearity); the elastic limit is slightly more forgiving.

Measuring Young's modulus — Searle's apparatus

How do you measure Young's modulus in a laboratory? The naive approach — hang a weight, measure how much it stretches — fails because the extension is incredibly small. A 1-m steel wire of 1-mm² area under a 1 kg load stretches by

\Delta L = \frac{FL}{AY} = \frac{9.8 \times 1}{10^{-6} \times 2 \times 10^{11}} = 4.9 \times 10^{-5} \text{ m} = 0.049 \text{ mm}.

Fifty micrometres. You need a precision instrument to see that, and you need to eliminate thermal drift, support sag, and the natural flexing of the support as you add weights. Searle's apparatus solves all of these at once.

The apparatus

Two identical wires, P and Q, hang side by side from a common rigid support. They are both made of the material under test (say steel), of the same length, and of nearly identical cross-section. A heavy fixed load on wire P keeps it permanently taut — it is the reference wire, and it stays at a fixed length throughout the experiment. Wire Q, the experimental wire, carries the load you are varying.

A horizontal frame connects the bottoms of the two wires. The frame is hinged to wire P (so it pivots as Q stretches) and carries a spirit level resting on a micrometer screw mounted on wire Q. Adjusting the micrometer screw raises or lowers one end of the frame until the spirit level reads horizontal. The reading on the micrometer tells you exactly how much Q has stretched relative to P.

Searle's apparatus for measuring Young's modulus Two wires hang side by side from a rigid ceiling bracket. The left wire P carries a fixed heavy weight and serves as a reference. The right wire Q has a micrometer screw at its bottom and carries a variable weight on a hanger. A horizontal frame with a spirit level connects the bottoms of the two wires. The spirit level indicates when the frame is horizontal, and the micrometer reads the relative extension of Q with respect to P. rigid ceiling support P (reference) Q (experiment) spirit level micrometer screw M (fixed) W (variable)
Searle's apparatus. Wire $P$ carries a fixed load and stays taut and at constant length. Wire $Q$ carries the variable load $W$. The micrometer screw at the bottom of $Q$ is adjusted until the spirit level reads horizontal — the reading then gives the extension of $Q$ relative to $P$ directly, eliminating thermal drift and support sag (which affect both wires equally).

The trick is the reference wire P. It experiences the same temperature as Q (they are right next to each other), so any thermal expansion of Q is cancelled by the identical thermal expansion of P. Any sag in the support likewise affects both wires equally and is cancelled. What the micrometer reads is the differential extension of Q — a much cleaner measurement than an absolute one.

The measurement procedure

  1. Initialise. Add just enough weight to Q to take out any slack and set the micrometer to zero at the horizontal-level position.
  2. Load in steps. Add weights to Q in equal increments (say 0.5 kg, 1.0 kg, 1.5 kg, up to about 4 kg). At each step, adjust the micrometer until the spirit level reads horizontal, and record the micrometer reading as the extension.
  3. Unload symmetrically. Remove the weights in reverse order, reading the micrometer each time. If all readings during unloading match the loading values, you are in the elastic regime.
  4. Plot extension \Delta L (vertical axis) against load F (horizontal axis).
  5. Fit a straight line. The slope is \Delta L / F. Compute Y from
Y = \frac{FL}{A\, \Delta L} = \frac{L}{A \cdot (\text{slope of } \Delta L \text{ vs } F)}.
  1. Measure the wire's radius r with a screw gauge (to 0.01 mm) and the wire's length L with a metre rule. Then A = \pi r^2.

Error analysis

The formula Y = FL / (A \Delta L) relates Y to four measured quantities. A fractional-error propagation gives

\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\cdot\frac{\Delta r}{r} + \frac{\Delta(\Delta L)}{\Delta L},

Why: for products and quotients of measured quantities, fractional errors add. The factor of 2 in front of \Delta r/r comes from A = \pi r^2 — a fractional error in r doubles when you square it.

The dominant source of error is usually the extension \Delta L itself (because it is small) and the radius r (because of its quadratic sensitivity). Using a screw gauge with 0.01 mm least count for r typically keeps \Delta r/r below 1%. The micrometer screw for \Delta L can reach 0.001 mm, keeping that term manageable. See Errors in Measurement for the full propagation framework.

Worked examples

Example 1: Stretching a steel wire with a hanging load

A steel wire of length L = 3.0 m and radius r = 0.5 mm is hung from a ceiling. A mass of m = 5.0 kg is attached to its lower end. If Y_{\text{steel}} = 2.0 \times 10^{11} Pa, find the extension of the wire. Assume the mass is attached gently (no impact).

Steel wire of length 3 m and radius 0.5 mm supporting a 5 kg mass A vertical wire attached to a ceiling support at the top, with a 5 kg mass hanging at the bottom. Arrows show the weight W acting downward at the hanging mass and the tension T acting upward at the top of the wire. ceiling L = 3 m 5 kg W = 49 N r = 0.5 mm A = πr² = 7.85 × 10⁻⁷ m² ΔL = FL/(AY) = 0.94 mm
The wire holds a 5 kg mass at its end. Gravity pulls the mass down with 49 N, and the wire supports it; the same 49 N tension acts along the wire, stretching it by 0.94 mm.

Step 1. Identify the knowns and compute the force.

L = 3.0 m, r = 0.5 mm = 5.0 \times 10^{-4} m, m = 5.0 kg, Y = 2.0 \times 10^{11} Pa. The tension in the wire (tensile force) equals the weight of the hanging mass:

F = mg = 5.0 \times 9.8 = 49 \text{ N}.

Why: in equilibrium, the wire tension must support the hanging mass. The mass is stationary, so net force on it is zero — tension equals weight.

Step 2. Compute the cross-sectional area.

A = \pi r^2 = \pi \times (5.0 \times 10^{-4})^2 = \pi \times 2.5 \times 10^{-7} = 7.85 \times 10^{-7} \text{ m}^2.

Why: for a circular wire, A = \pi r^2. Work entirely in SI — metres squared, not mm².

Step 3. Apply Hooke's law in the form Y = FL/(A\Delta L), solved for \Delta L.

\Delta L = \frac{FL}{AY} = \frac{49 \times 3.0}{7.85 \times 10^{-7} \times 2.0 \times 10^{11}}.

Numerator: 147 N·m.

Denominator: 7.85 \times 2.0 \times 10^{-7+11} = 15.7 \times 10^{4} = 1.57 \times 10^{5}.

\Delta L = \frac{147}{1.57 \times 10^{5}} = 9.4 \times 10^{-4} \text{ m} = 0.94 \text{ mm}.

Why: carry the powers of ten carefully. The numerator is of order 10², the denominator of order 10⁵, so the answer is of order 10⁻³ — about a millimetre, which matches the order-of-magnitude for a steel wire under small loads.

Step 4. Check dimensionality and order of magnitude.

Dimensions: \dfrac{\text{N} \cdot \text{m}}{\text{m}^2 \cdot \text{N/m}^2} = \dfrac{\text{N}\cdot\text{m}}{\text{N}} = \text{m}. Good.

Order of magnitude: less than 1 mm extension for a 3-m wire under a 5-kg load — consistent with the marble-scale intuition that steel is very stiff.

Result: The wire stretches by \Delta L = 0.94 mm.

What this shows: Even a relatively thin (0.5-mm radius) steel wire under a 5-kg load stretches by less than a millimetre over 3 metres. This is why structural steel is so reliable — it is stiff enough that bridge cables and suspension wires change length imperceptibly under load.

Example 2: Comparing steel and copper wires loaded equally

Two wires — one steel, one copper — have the same original length (L = 2.0 m) and the same cross-sectional area (A = 1.0 mm² = 1.0 \times 10^{-6} m²). Both are loaded with the same force F = 100 N. Given Y_{\text{steel}} = 2.0 \times 10^{11} Pa and Y_{\text{copper}} = 1.2 \times 10^{11} Pa, find the extension of each wire and their ratio.

Steel and copper wires side by side, same dimensions, same load Two vertical wires of the same initial length hang from a common support. Both carry the same weight at their lower ends. The copper wire stretches more than the steel wire because its Young's modulus is smaller. steel copper 100 N 100 N Y = 2×10¹¹ Y = 1.2×10¹¹ ΔL = 1 mm ΔL = 1.67 mm
Same length, area, and load — but the copper wire stretches 1.67× more than the steel wire, because its Young's modulus is smaller by the same factor.

Step 1. Write out \Delta L for each wire.

\Delta L_{\text{steel}} = \frac{FL}{AY_{\text{steel}}}, \qquad \Delta L_{\text{copper}} = \frac{FL}{AY_{\text{copper}}}.

Why: Hooke's law in extension form. Same F, L, A — the only difference is Y.

Step 2. Compute each extension.

\Delta L_{\text{steel}} = \frac{100 \times 2.0}{10^{-6} \times 2.0 \times 10^{11}} = \frac{200}{2.0 \times 10^{5}} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}.
\Delta L_{\text{copper}} = \frac{100 \times 2.0}{10^{-6} \times 1.2 \times 10^{11}} = \frac{200}{1.2 \times 10^{5}} = 1.67 \times 10^{-3} \text{ m} = 1.67 \text{ mm}.

Step 3. Compute the ratio.

\frac{\Delta L_{\text{copper}}}{\Delta L_{\text{steel}}} = \frac{Y_{\text{steel}}}{Y_{\text{copper}}} = \frac{2.0}{1.2} = \frac{5}{3} \approx 1.67.

Why: \Delta L \propto 1/Y when everything else is fixed, so the ratio of extensions is the inverse ratio of Young's moduli. This relationship is the most useful form of Hooke's law for comparison problems.

Result: The steel wire stretches by 1.0 mm; the copper wire stretches by 1.67 mm — a factor of 5/3 more.

What this shows: When you apply the same force to two wires of the same shape but different materials, the extensions are inversely proportional to Young's modulus. This is why steel is used where low stretch matters (elevator cables, concrete reinforcement), while softer metals like copper are preferred where small flexibility is desirable (house wiring, musical instrument strings).

Example 3: Designing a crane cable

A crane cable is made of steel (Y = 2.0 \times 10^{11} Pa). It is 20 m long and must lift a maximum load of 2000 kg. To keep the extension under the maximum load below 2.0 mm, what is the minimum cross-sectional area the cable must have?

Step 1. Identify the knowns.

L = 20 m, m = 2000 kg, \Delta L_{\max} = 2.0 mm = 2.0 \times 10^{-3} m, Y = 2.0 \times 10^{11} Pa.

Force required to lift: F = mg = 2000 \times 9.8 = 1.96 \times 10^4 N.

Why: the cable must support the load against gravity. At the moment of just starting to lift (no acceleration), tension equals weight.

Step 2. Rearrange Hooke's law for A.

Y = \frac{FL}{A\,\Delta L} \quad\Longrightarrow\quad A = \frac{FL}{Y\,\Delta L}.

Step 3. Plug in.

A = \frac{(1.96 \times 10^4) \times 20}{(2.0 \times 10^{11}) \times (2.0 \times 10^{-3})} = \frac{3.92 \times 10^5}{4.0 \times 10^{8}} = 9.8 \times 10^{-4} \text{ m}^2 = 9.8 \text{ cm}^2.

Why: carry units carefully. N·m in the numerator divided by Pa·m = N/m² · m = N/m in the denominator gives m² in the result. Converting m² to cm² involves multiplying by 10⁴.

Step 4. Translate to a radius.

For a circular cable, A = \pi r^2, so

r = \sqrt{A/\pi} = \sqrt{9.8/\pi} \text{ cm} = \sqrt{3.12} \text{ cm} \approx 1.77 \text{ cm}.

The cable must have a minimum radius of about 1.77 cm (or diameter 3.5 cm), corresponding to a thumb-width steel rope.

Result: The minimum cross-sectional area is \approx 9.8 cm² (radius \approx 1.77 cm).

What this shows: Engineering design routinely uses Young's modulus to fix material and dimensions. In practice, safety margins push the required area higher — a 2x factor of safety would double the area, giving a cable of radius 2.5 cm. The underlying calculation is pure Hooke's law.

Common confusions

If you came here to use Hooke's law, compute extensions, and understand Young's modulus, you have what you need. What follows is for readers who want to see where Hooke's law comes from microscopically, why all three elastic moduli are related, and how Searle's measurement is designed for minimum error.

Hooke's law from atomic potentials

Every solid is a lattice of atoms held together by interatomic forces. Between any two neighbouring atoms, there is a potential energy U(r) that has a minimum at some equilibrium separation r_0 — where the attractive and repulsive forces balance. Near the minimum, Taylor-expand:

U(r) = U(r_0) + \frac{1}{2} U''(r_0)(r - r_0)^2 + O((r-r_0)^3).

Why: any smooth function near a minimum looks like a parabola to leading order. The first derivative vanishes at the minimum (by definition), so the leading non-constant term is quadratic.

The force between the atoms is

F(r) = -\frac{dU}{dr} = -U''(r_0)(r - r_0) + O((r-r_0)^2).

The linear term says that for small displacements from equilibrium, the interatomic force is proportional to the displacement — a spring. U''(r_0) is the spring constant per bond. Multiply by the number of bonds in parallel (proportional to A) and divide by the distance of bonds in series (proportional to L), and you recover the macroscopic Hooke's law. The material constant that falls out is proportional to U''(r_0) divided by the atomic spacing — this is Young's modulus, microscopically.

Why Hooke's law fails beyond the elastic limit is now clear: the Taylor expansion breaks down. Cubic and higher terms become important at large strains, and the elegant linear behaviour disappears.

Relations between Y, B, and G

A material has three independent elastic constants — Young's modulus Y, the bulk modulus B, and the shear modulus G — but for an isotropic material (one whose properties are the same in every direction), only two are independent. The third is determined by the others through Poisson's ratio \nu:

Y = 2G(1 + \nu) = 3B(1 - 2\nu).

For a typical metal, \nu \approx 0.3, giving Y \approx 2.6 G and Y \approx 1.2 B — Young's modulus is the intermediate-sized one of the three. Rubber, with \nu \approx 0.5 (nearly incompressible), has Y approaching 3G and B \gg Y.

Poisson's ratio

When you stretch a wire, it doesn't just get longer — it also gets thinner. The ratio of lateral contraction (fractional reduction in diameter) to longitudinal extension (fractional increase in length) is Poisson's ratio:

\nu = -\frac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{longitudinal}}}.

For most materials, \nu is between 0.2 and 0.35 — you see a visible narrowing of a stretched rubber band, but not of a stretched steel wire (the fractional change is tiny). A Poisson's ratio of 0.5 would mean perfectly incompressible — volume preserved during stretching. Rubber comes close.

A few exotic materials — called auxetic — have negative Poisson's ratio. When stretched, they get thicker. These are engineered foams used in body armour and specialised fabrics; they are not found in nature.

Error minimisation in Searle's apparatus — why the method works

For Y = FL / (A\Delta L) with A = \pi r^2, the fractional error is

\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta L)}{\Delta L}.

The dominant term is typically 2\Delta r / r, because r is small (making fractional errors large) and it is squared (doubling the impact). A vernier or screw gauge with 0.01 mm least count and r \sim 0.5 mm gives \Delta r / r \sim 2\%, contributing 4% to \Delta Y / Y. This is the reason Indian lab setups go to lengths to measure wire radius at multiple points and average.

The reference-wire idea in Searle's apparatus is a form of differential measurement — a strategy that appears across physics (Wheatstone bridge for resistance, Mach-Zehnder interferometer for phase, Cassegrain pointing corrections in telescopes). The principle is: common-mode effects (both wires warming together, the whole support sagging together) subtract out of the difference, leaving only the differential effect you want to measure.

Hooke's law in more general stress states

Everything above deals with uniaxial tension — pulling a wire. For a solid subjected to a general stress state (compression in one direction, tension in another, shear), Hooke's law generalises to a tensor form:

\sigma_{ij} = C_{ijkl}\, \varepsilon_{kl},

where \sigma_{ij} is the stress tensor, \varepsilon_{kl} is the strain tensor, and C_{ijkl} is the stiffness tensor — in general, 81 numbers for an anisotropic material (like a single crystal of quartz). For an isotropic material, symmetry collapses this to just Y and \nu (or equivalently B and G) — the pair that appears in every introductory physics course. This tensor form is what engineers use when designing bridges, aircraft wings, and the ducts in a Mangalyaan satellite.

For the curious student, the key takeaway is: Hooke's law is the first term in a local Taylor expansion of the elastic response. Beyond that first term, the material's full nonlinear behaviour takes over — and that is where metallurgy, materials science, and engineering design live.

Where this leads next