What is the difference between stress and strain in the context of deformation of solids?

In the context of deformation of solids, stress is defined as the force applied per unit area within materials. It is measured in pascals (Pa). Strain, on the other hand, is the measure of deformation representing the displacement between particles in the material body. It is a dimensionless quantity as it is the ratio of change in dimension to the original dimension. Understanding these concepts is crucial for Cambridge International A Levels Physics as they form the basis for analyzing material behavior under force.

How are stress and strain related in Hooke's Law?

Hooke's Law states that, within the elastic limit of a material, the stress applied to it is directly proportional to the strain it produces. This relationship is expressed as σ = Eε, where σ is the stress, E is the Young's modulus (a constant for a given material), and ε is the strain. This fundamental principle is essential for Cambridge International A Levels students to understand how materials respond to forces and return to their original shape when the force is removed.

What are the types of stress and how do they affect materials?

There are three primary types of stress: tensile stress, compressive stress, and shear stress. Tensile stress occurs when forces act to stretch a material, compressive stress occurs when forces act to compress or shorten a material, and shear stress occurs when forces act parallel to the surface. Each type of stress affects materials differently, influencing their deformation and potential failure. Understanding these types is crucial for analyzing material behavior in Cambridge International A Levels Physics.

Why is Young's modulus important in the study of stress and strain?

Young's modulus, also known as the modulus of elasticity, is a measure of the stiffness of a solid material. It quantifies the relationship between tensile stress and tensile strain in the linear elasticity regime of a uniaxial deformation. A higher Young's modulus indicates a stiffer material. For Cambridge International A Levels students, understanding Young's modulus is vital for predicting how materials will deform under various forces, which is essential for engineering and material science applications.

What is the significance of the elastic limit in the deformation of solids?

The elastic limit is the maximum stress that a material can withstand without undergoing permanent deformation. Beyond this point, the material will not return to its original shape when the stress is removed. For Cambridge International A Levels Physics, understanding the elastic limit is crucial as it determines the safe working limits of materials and helps in designing structures that can withstand applied forces without permanent damage.

Why is "Quoting stress in N/cm²" a common mistake in Stress and strain?

Why it happens: Convenient. How to avoid it: Stress in SI is Pa = N/m². Convert area to m² before computing. Source: 9702 Examiner Reports 2022-2024

Why is "Multiplying strain by 100 to give percentage when not asked" a common mistake in Stress and strain?

Why it happens: Reading mistake. How to avoid it: Strain is DIMENSIONLESS. Quote as decimal or scientific (e.g. 7.5 × 10^-4) unless 'percentage' explicitly requested. Source: 9702 Examiner Reports 2022-2024

Deformation of SolidsCambridge International A Levels Physics (9702)

Stress and strain

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Stress and Strain Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Hooke's law. Stress, strain, Young modulus. Elastic strain energy.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

6.1.1 — Apply Hooke's law.
6.1.2 — Compute stress, strain, Young modulus.
6.1.3 — Calculate elastic PE.

Hooke's law

$F = kx$ within limit.

Hooke's law. Extension proportional to force, up to LIMIT OF PROPORTIONALITY. $F = kx$ where $k$ is the spring constant.

Force-extension graph. Straight line through origin within limit.

Elastic limit. Beyond this, body doesn't return to original length when force removed.

Plastic deformation. Permanent change of shape; occurs beyond elastic limit.

Cambridge tip. State 'within elastic limit' or 'within limit of proportionality' when applying $F = kx$ .

$F = kx$ .
Linear up to limit of proportionality.
Beyond elastic limit: permanent deformation.

See the full worked example for stress and strain →

Stress, strain, Young modulus

Material-independent ratios.

Stress. $\sigma = F/A$ . Force per cross-sectional area. Pa.

Strain. $\varepsilon = \Delta L/L$ . Fractional extension. Dimensionless.

Young modulus. $E = \sigma/\varepsilon = FL/(A\Delta L)$ . Material constant. Pa.

Why useful. Hooke's $k$ depends on dimensions (length, area). Young modulus depends only on MATERIAL.

Typical values.

Steel: $E \approx 2 \times 10^{11}$ Pa.
Copper: $\sim 1.3 \times 10^{11}$ Pa.
Rubber: $\sim 10^7$ Pa.

Example. 2 m wire, $A = 1.2 \times 10^{-7}$ m², 5 kg mass, extends 1.5 mm.

$F = 49.05$ N → $\sigma \approx 4.09 \times 10^8$ Pa.
$\varepsilon = 1.5 \times 10^{-3}/2 = 7.5 \times 10^{-4}$ .
$E \approx 5.5 \times 10^{11}$ Pa.

Cambridge tip. Mark scheme rewards SI units throughout.

$\sigma = F/A$ , $\varepsilon = \Delta L/L$ , $E = \sigma/\varepsilon$ .
$E$ is material constant.
Steel ~ $2 \times 10^{11}$ Pa.

See the full worked example for stress and strain →

Elastic strain energy

$\frac{1}{2}kx^2$ in spring.

Elastic PE in Hookean spring. Area under $F$ - $x$ graph (triangle). $E_p = \tfrac{1}{2}kx^2 = \tfrac{1}{2}Fx$

General. Even for non-linear: $E_p = \int F \, dx$ = area under $F$ - $x$ graph.

Example. Spring $k = 200$ N/m stretched 0.10 m. $E = \frac{1}{2}(200)(0.01) = 1.0$ J.

Energy storage. Springs, rubber bands, archery bows — used to store and release energy.

Cambridge tip. Always specify Hookean before $\frac{1}{2}kx^2$ .

$E_p = \frac{1}{2}kx^2$ (Hookean).
Area under $F$ - $x$ graph.
Used for energy storage.

See the full worked example for stress and strain →

How it’s examined

Stress/strain in every Paper 1 or 2. Most-tested: Young modulus calc (7-8 marks), Hooke + elastic PE (5-7 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Stress and strain

Step-by-step solutions to past-paper-style questions on stress and strain, written exactly the way a tutor would explain them at the board.

1Young modulus from data (7 marks)
Extended• Adapted from 9702/22 May/Jun 2024• Young modulus
Question
A wire of length $2.00$ m and cross-sectional area $1.20 \times 10^{-7}$ m² extends by $1.5$ mm when a 5.0 kg mass is hung on it. Take $g = 9.81$ . Find (a) stress, (b) strain, (c) Young modulus. (7 marks)
Step-by-step solution
1. Step 1
  Force on wire. $F = mg = 5.0 \times 9.81 = 49.05$ N.
2. Step 2
  (a) Stress = $F/A$ .
  $\sigma = \dfrac{49.05}{1.20 \times 10^{-7}} \approx 4.09 \times 10^8 \text{ Pa}$
3. Step 3
  (b) Strain = $\Delta L / L$ .
  $\varepsilon = \dfrac{1.5 \times 10^{-3}}{2.00} = 7.5 \times 10^{-4}$
4. Step 4
  (c) Young modulus = stress/strain.
  $E = \dfrac{4.09 \times 10^8}{7.5 \times 10^{-4}} \approx 5.45 \times 10^{11} \text{ Pa}$
Answer
(a) $\sigma \approx 4.09 \times 10^8$ Pa. (b) $\varepsilon = 7.5 \times 10^{-4}$ . (c) $E \approx 5.5 \times 10^{11}$ Pa.
2Hooke's law (5 marks)
Extended• Hooke's law
Question
A spring extends 4.0 cm under a 2.0 N load. Find (a) spring constant $k$ , (b) extension under 3.5 N. (5 marks)
Step-by-step solution
1. Step 1
  $F = kx$ .
  $k = F/x = 2.0/0.040 = 50 \text{ N/m}$
2. Step 2
  Extension under 3.5 N.
  $x = F/k = 3.5/50 = 0.070 \text{ m} = 7.0 \text{ cm}$
Answer
$k = 50$ N/m; $x = 7.0$ cm.
3Elastic strain energy (5 marks)
Extended• elastic energy
Question
A spring of stiffness $k = 200$ N/m is stretched 0.10 m. Find the elastic PE stored. (5 marks)
Step-by-step solution
1. Step 1
  Elastic PE = $\frac{1}{2}kx^2$ (Hookean spring).
  $E = \tfrac{1}{2} \times 200 \times (0.10)^2 = 1.0 \text{ J}$
Answer
$E = 1.0$ J.

Key Formulae — Stress and strain

The formulae you need to memorise for stress and strain on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Hooke's law
$F = kx$
$k$
spring constant (N/m)
$x$
extension
When to use
Within elastic region. Force proportional to extension.
Stress
$\sigma = \dfrac{F}{A}$
When to use
Force per unit cross-sectional area. Units: Pa.
Strain
$\varepsilon = \dfrac{\Delta L}{L}$
When to use
Fractional extension. Dimensionless.
Young modulus
$E = \dfrac{\sigma}{\varepsilon} = \dfrac{FL}{A \Delta L}$
When to use
Material property; constant within elastic region. Units: Pa.
Elastic PE (Hookean)
$E_p = \tfrac{1}{2}kx^2 = \tfrac{1}{2}Fx$
When to use
Energy stored in a spring obeying Hooke's law. Area under $F$ - $x$ graph.

Key Definitions and Keywords — Stress and strain

Definitions to memorise and the exact keywords mark schemes credit for stress and strain answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Stress
Examiner keyword
Force per unit cross-sectional area. $\sigma = F/A$ . Units: Pa.
Strain
Examiner keyword
Fractional extension: $\varepsilon = \Delta L / L$ . Dimensionless.
Young modulus
Examiner keyword
Ratio of stress to strain within elastic region. Material constant. Units: Pa.
Hooke's law
Examiner keyword
Extension is proportional to applied force, up to the limit of proportionality.

Common Mistakes and Misconceptions — Stress and strain

The traps other students keep falling into on stress and strain questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Quoting stress in N/cm²
9702 Examiner Reports 2022-2024
Why it happens
Convenient.
How to avoid it
Stress in SI is Pa = N/m². Convert area to m² before computing.
✕Multiplying strain by 100 to give percentage when not asked
9702 Examiner Reports 2022-2024
Why it happens
Reading mistake.
How to avoid it
Strain is DIMENSIONLESS. Quote as decimal or scientific (e.g. $7.5 \times 10^{-4}$ ) unless 'percentage' explicitly requested.

Practice questions

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Past paper style quiz

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Stress and strain — frequently asked questions

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Stress and strain

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Young modulus from data (7 marks)

2Hooke's law (5 marks)

3Elastic strain energy (5 marks)

Hooke's law

Stress

Strain

Young modulus

Elastic PE (Hookean)

Stress

Strain

Young modulus

Hooke's law

✕Quoting stress in N/cm²

✕Multiplying strain by 100 to give percentage when not asked

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Stress and strain — frequently asked questions