Physics · Mechanical Properties of Solids

Hooke's Law

Hooke's law is the single relationship that turns elasticity into a quantitative subject. NCERT §8.3 states it in one line: for small deformations, stress is proportional to strain, and the constant of proportionality is the modulus of elasticity. Every elastic modulus you will meet — Young's, shear, bulk — is just this law specialised to one kind of stress. This deep-dive states the law precisely, derives stress = E × strain, connects it to the spring law $F=-kx$, fixes the limit beyond which it fails, and works through the NEET PYQs that test it.

Statement of Hooke's law

When a deforming force acts on a solid, the body develops an internal restoring force; the restoring force per unit area is the stress, and the fractional change in dimension is the strain. Robert Hooke established by experiment in 1678 that, for small deformations, these two quantities rise in step. The law is stated as follows.

Hooke's law. Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it.

In symbols, the law and its constant form are

$$\text{stress} \propto \text{strain} \qquad\Longrightarrow\qquad \text{stress} = E \times \text{strain},$$

where $E$ is the constant of proportionality, called the modulus of elasticity (NCERT writes it as $k$ in Eq. 8.6; the symbol does not matter). The proportionality holds only along the initial portion of the deformation; the precise cut-off is the proportional limit, treated below.

The modulus of elasticity

The modulus of elasticity is the heart of the law. It is defined by rearranging the constant form:

$$E = \frac{\text{stress}}{\text{strain}}.$$

Three facts about $E$ are tested directly by NEET, and each follows from this definition.

Property of EStatementReason
Unit$\text{N m}^{-2}$ or pascal (Pa)Strain is a pure ratio, so $E$ inherits the unit of stress, $\text{F}/\text{A}$.
Dimensions$[\text{M L}^{-1}\text{T}^{-2}]$Same as stress, for the same reason — strain is dimensionless.
What it depends onOnly the nature of the materialIts value is independent of the magnitude of the applied stress and strain.

Because strain carries no units, the modulus is numerically equal to the stress that would, by linear extrapolation, produce unit strain (a 100% change in dimension). Real materials fracture long before that, so the modulus is an idealised slope, not an achievable stress. A large modulus signals a stiff material that yields little strain for a given stress — steel, for instance, has a much larger Young's modulus than copper, brass or aluminium, which is precisely why steel is judged the more elastic of the metals.

Reading the law off the graph

The cleanest picture of Hooke's law is the first stretch of the stress–strain curve. When a wire is loaded in steps and the stress is plotted against the strain, the early data fall on a straight line through the origin. The slope of that line is the modulus of elasticity.

Strain ε Stress σ P — proportional limit Δε Δσ slope = E O Hooke's law obeyed law fails
Within the proportional region O→P the graph is a straight line; its slope is the modulus of elasticity. Beyond the proportional limit P the curve bends and Hooke's law no longer holds.

The spring analogue F = −kx

The same linearity that governs a stretched wire governs a helical spring, and the spring is where most students first meet it. For a wire of natural length $L$, cross-sectional area $A$ and Young's modulus $Y$, Hooke's law $\sigma = Y\varepsilon$ reads $F/A = Y(\Delta L/L)$. Solving for the force gives

$$F = \left(\frac{YA}{L}\right)\Delta L.$$

The bracket is a constant for a given wire, so the applied force is proportional to the extension — exactly the spring relation. Writing the extension as $x$ and the bracketed stiffness as $k=YA/L$, and noting that the spring's restoring force opposes the displacement, gives the familiar form

$$F = -k x.$$

QuantityStretched wire (general elastic law)Helical spring (special case)
Law$\sigma = E\,\varepsilon$$F = -kx$
Linear instrain $\varepsilon = \Delta L / L$extension $x$
Constantmodulus of elasticity $E$spring constant $k = YA/L$
Unit of constant$\text{N m}^{-2}$$\text{N m}^{-1}$
Validityup to the proportional limitup to the spring's elastic limit

The minus sign in $F=-kx$ records direction: the restoring force always points back toward the natural length, opposing the deformation. The general elastic law $\sigma = E\varepsilon$ is usually quoted in magnitudes, so the sign is dropped, but the physics is identical — the body resists, in proportion to how far it has been pushed from equilibrium.

F L ΔL σ = Y · (ΔL/L) F x F = −k x
Left — a wire elongates by $\Delta L$, with stress proportional to strain. Right — a spring extends by $x$, with force proportional to extension. The two are the same linear elastic response; the spring constant $k=YA/L$ packages the wire's geometry and material into one number.

Foundation of every elastic modulus

Stress and strain take three different forms — longitudinal, shearing and volumetric — and Hooke's law applies to each. The proportionality constant simply takes a different name in each case. This is why Hooke's law is the foundation on which all of the elastic moduli are built: each modulus is the constant of one Hooke-type relation.

Type of stressHooke's law formModulus (constant)
Longitudinal (tensile / compressive)$\dfrac{F}{A} = Y\,\dfrac{\Delta L}{L}$Young's modulus $Y$
Shearing$\dfrac{F}{A} = G\,\theta$Shear modulus (rigidity) $G$
Hydraulic$p = B\,\dfrac{\Delta V}{V}$Bulk modulus $B$

Read the table as three sentences of one grammar: stress equals modulus times strain. Each row is Hooke's law; the modulus on the right is the slope of the corresponding straight-line region. The full shapes of these curves — yield point, plastic region, fracture — belong to the stress–strain curve; here the point is only that the straight initial portion is Hooke's law in every case.

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Before this page

Hooke's law links two quantities you must define first — see stress and strain for the precise definitions and units that this law connects.

Validity and limitations

Hooke's law is powerful but narrow. It is an empirical law — NCERT states that it "is found to be valid for most materials" but is not derived from a deeper principle — and it carries clear limits that NEET tests repeatedly.

  • Valid only up to the proportional limit. The straight-line region ends at the proportional limit; beyond it stress and strain are no longer proportional, even though the material may stay elastic a little further up to the yield point.
  • Small deformations only. The proportionality is a small-strain approximation; large strains break it.
  • Not obeyed by elastomers. Rubber and biological tissue such as the aorta stretch to several times their length and still recover, yet their stress–strain curves are not linear — they do not obey Hooke's law over most of their elastic range.
  • One constant is not always enough. A deforming force in one direction generally produces strains in other directions too (described by Poisson's ratio); a single elastic constant cannot capture every such case.

Worked examples

NCERT Example 8.1 (adapted)

A structural steel rod of radius $10~\text{mm}$ and length $1.0~\text{m}$ is stretched along its length by a force of $100~\text{kN}$. Young's modulus of structural steel is $2.0\times10^{11}~\text{N m}^{-2}$. Find the stress, the elongation, and the strain.

Stress. $A = \pi r^2 = 3.14\times(10^{-2})^2 = 3.14\times10^{-4}~\text{m}^2$, so $\sigma = F/A = (1.0\times10^{5})/(3.14\times10^{-4}) = 3.18\times10^{8}~\text{N m}^{-2}.$

Strain (Hooke's law). Since $\sigma = Y\varepsilon$, $\ \varepsilon = \sigma/Y = (3.18\times10^{8})/(2.0\times10^{11}) = 1.59\times10^{-3}\ (\approx 0.16\%).$

Elongation. $\Delta L = \varepsilon L = (1.59\times10^{-3})(1.0) = 1.59\times10^{-3}~\text{m} = 1.59~\text{mm}.$ The whole solution is a single application of stress $= E\times$ strain.

NEET 2024

The elastic limit of steel is $8\times10^{8}~\text{N m}^{-2}$ and its Young's modulus is $2\times10^{11}~\text{N m}^{-2}$. Find the maximum elongation of a steel wire of length $1~\text{m}$.

Maximum strain. The largest stress allowed is the elastic limit. Within the linear region, $\varepsilon_{\max} = \sigma_{\max}/Y = (8\times10^{8})/(2\times10^{11}) = 4\times10^{-3}.$

Maximum elongation. $\Delta L_{\max} = \varepsilon_{\max}\,L = (4\times10^{-3})(1) = 4\times10^{-3}~\text{m} = 4~\text{mm}.$ Answer: 4 mm. The PYQ is solved entirely by stress $=Y\times$ strain at the limiting stress.

Significance for NEET

Hooke's law is the workhorse of the whole chapter. Once you recognise that a problem stays within the proportional region, the relation $\sigma = E\varepsilon$ converts any one of force, area, length, elongation or modulus into any other. Crane ropes, beams and pillars are designed by demanding that the working stress stay safely below the proportional and elastic limits so that the deformation is recoverable; that engineering judgement is explored in applications of elastic behaviour. For the exam itself, the high-yield skills are recognising stress $=F/A$, writing $\varepsilon = \sigma/E$, and remembering that the modulus is a slope fixed by the material alone.

Quick recap

Hooke's law in one breath

  • Within the elastic limit, stress $\propto$ strain; in constant form, stress $= E \times$ strain.
  • The constant $E$ is the modulus of elasticity — the slope of the straight region, unit $\text{N m}^{-2}$, set only by the material.
  • The modulus is the SLOPE of $\sigma$–$\varepsilon$, not the area; the area is the elastic energy density $\tfrac12\sigma\varepsilon$.
  • Spring analogue: $F = -kx$ with $k = YA/L$ — Hooke's law specialised to a spring; the minus sign is the restoring direction.
  • Young's, shear and bulk moduli are each Hooke's constant for one type of stress — the law is their common foundation.
  • Valid only up to the proportional limit (which precedes the elastic limit); empirical; not obeyed by elastomers.

NEET PYQ Snapshot — Hooke's Law

Three PYQs that turn on the single relation stress = E × strain and the spring–shear connection.

NEET 2024

The maximum elongation of a steel wire of $1~\text{m}$ length, if the elastic limit of steel and its Young's modulus are $8\times10^{8}~\text{N m}^{-2}$ and $2\times10^{11}~\text{N m}^{-2}$ respectively, is:

  1. 4 mm
  2. 0.4 mm
  3. 40 mm
  4. 8 mm
Answer: (1) 4 mm

Hooke's law. Maximum strain $=\sigma_{\max}/Y = (8\times10^{8})/(2\times10^{11}) = 4\times10^{-3}$. Then $\Delta L_{\max} = \varepsilon_{\max} L = 4\times10^{-3}\times 1 = 4~\text{mm}$.

NEET 2023

A wire is suspended from the ceiling and stretched by a weight $W$ attached at its free end. The longitudinal stress at any point of cross-sectional area $A$ of the wire is:

  1. Zero
  2. $2W/A$
  3. $W/A$
  4. $W/2A$
Answer: (3) W/A

Stress before strain. Hooke's law needs the stress, and the tension at any cross-section equals the weight $W$ (not $2W$, despite the equal pull from the ceiling). Hence $\sigma = W/A$ — the input to $\varepsilon = \sigma/Y$.

NEET 2022

Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of the same dimensions.

  1. Both (A) and (R) true; (R) is the correct explanation of (A)
  2. Both (A) and (R) true; (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
Answer: (3) A true, R false

Spring ↔ Hooke's law. Stretching a coil twists its wire — the deformation is shear, so the spring's stiffness is governed by the shear modulus: (A) is true. But steel has greater tensile strength than copper, so (R) is false.

FAQs — Hooke's Law

Short answers to the Hooke's-law questions NEET aspirants get wrong most often.

Does Hooke's law apply right up to the breaking point of a wire?
No. Hooke's law — stress proportional to strain — holds only along the initial straight portion of the stress–strain curve, up to the proportional limit. Beyond that point the curve bends, and even though the material may still be elastic for a little further (up to the yield point), stress and strain are no longer proportional. The law fails well before fracture.
What exactly is the modulus of elasticity in Hooke's law?
The modulus of elasticity is the constant of proportionality in stress = E × strain. Geometrically it is the slope of the straight-line region of the stress–strain graph — it is NOT the area under the graph. It depends only on the nature of the material, not on the magnitude of the stress or strain applied. Its SI unit is the same as that of stress, N m⁻², because strain is dimensionless.
Is Hooke's law a fundamental law of physics or an empirical one?
It is empirical. NCERT states it plainly: "Hooke's law is an empirical law and is found to be valid for most materials." It is a good approximation for small deformations of most solids but is not derived from a deeper principle, and some materials — notably elastomers such as rubber and the tissue of the aorta — do not obey it even within their large elastic ranges.
How is the spring law F = −kx related to Hooke's law?
F = −kx is Hooke's law specialised to a helical spring. For a stretched wire, force is proportional to extension because stress (F/A) is proportional to strain (ΔL/L); rearranging gives F = (YA/L)·ΔL, a force proportional to extension with an effective stiffness YA/L. A coil spring obeys the same linear force–extension relation with spring constant k. Both express the single idea that within the elastic limit the restoring response is linear in the deformation.
Does every modulus — Young's, shear and bulk — come from Hooke's law?
Yes. Each modulus is simply the proportionality constant of Hooke's law for one particular kind of stress and strain: Young's modulus for longitudinal stress and longitudinal strain, shear modulus for shearing stress and shearing strain, and bulk modulus for hydraulic stress and volume strain. They are three faces of the same statement, stress = modulus × strain.
Why does a material that stretches less count as more elastic?
Elasticity measures resistance to deformation, not the amount of stretch. A material with a larger modulus needs a larger stress to produce a given strain, so it stretches less for the same load. Steel has a far larger Young's modulus than rubber, so steel is more elastic than rubber even though rubber visibly stretches more. NCERT flags the everyday intuition that "more stretchy means more elastic" as a misnomer.