PhysicsCBSE Class 11 · Mechanical Properties of Solids

Stress, Strain and Young's Modulus Class 11

Updated 17 October 2026 · Fact-checked

Stress is force per unit area (σ = F/A) and longitudinal strain is the fractional change in length (ΔL/L). Within the proportional limit, stress is proportional to strain (Hooke's law) and Young's modulus is Y = σ/ε = FL/(AΔL), a property of the material. The extension of a wire is ΔL = FL/(AY).

The straight start of the curve has slope Y (Hooke's law); past the elastic limit the wire is permanently stretched until it fractures.

Illustration drawn for QuestionBuddy.

Understand Stress, Strain and Young's Modulus

Elasticity. A body that regains its original shape when the deforming force is removed is *elastic*; one that stays deformed is *plastic*. The restoring force per unit area inside a deformed body is the stress, $σ = F / A$ (unit N/m² = Pa, dimension $[M L^{- 1} T^{- 2}]$ ). The relative change in dimension is the strain, a pure number. Longitudinal strain is $ε = Δ L / L$ .

Hooke's law and Young's modulus. For small strains, stress is proportional to strain. The ratio is Young's modulus,
$ $Y = \frac{σ}{ε} = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}, Δ L = \frac{F L}{A Y} .$ $
$Y$ depends on the material only, not on the dimensions of the wire. Typical order: steel $\sim 2 \times 1 0^{11}$ Pa, copper about $1 \times 1 0^{11}$ Pa, rubber a very small value, so steel is the more elastic (it needs more stress for the same strain).

Stress-strain curve (ductile metal). Starting from the origin: a straight line up to the *proportional limit*; slightly beyond it, the *elastic limit* (the body still recovers on unloading); then the *yield point* where strain grows rapidly with little extra stress (permanent deformation); the curve reaches the *ultimate tensile strength* and the wire *fractures*. Materials with a long plastic region are *ductile* (can be drawn into wires); those that break soon after the elastic limit are *brittle*. For rubber the curve is not linear and shows a large strain for small stress.

Wire as a spring. For a load $F$ , $F = \frac{A Y}{L} Δ L = k Δ L$ with $k = A Y / L$ . Two wires in parallel (same extension) share the load in the ratio $A_{1} Y_{1} / L_{1} : A_{2} Y_{2} / L_{2}$ ; wires in series carry the same force and their extensions add.

Breaking stress and design. The breaking stress of a material is the stress at which it fractures; the factor of safety is the ratio of breaking stress to the working stress. A wire hanging under its own weight has the greatest stress at the top, $σ = ρ g L$ .

Stress, Strain and Young's Modulus: formulas and key facts

Stress: $σ = \frac{F}{A}$; Pa.
Longitudinal strain: $ε = \frac{Δ L}{L}$; No unit.
Young's modulus: $Y = \frac{σ}{ε} = \frac{F L}{A Δ L}$; Material constant.
Extension: $Δ L = \frac{F L}{A Y}$; Hooke's law range.
Wire as spring: $k = \frac{A Y}{L}$; $F = k Δ L$ .
Stress at the top of a hanging wire: $σ = ρ g L$; Own weight.

How to solve Stress, Strain and Young's Modulus questions

Find the area from the diameter, form the stress and strain, and use Y as the ratio, remembering that Y is fixed for the material.

1Convert the diameter or radius to area, A = πr², in m².
2Compute stress F/A and strain ΔL/L (a ratio, no units).
3Use Y = stress/strain, or ΔL = FL/(AY) to find the unknown.
4For comparisons, write ΔL ∝ FL/(r²Y) and cancel the common quantities.
5For load sharing, equal extensions give F ∝ AY/L.
6Check that the stress is below the elastic limit before using Hooke's law.

Stress, Strain and Young's Modulus in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A hanging rope swing

Suppose a steel cable of length 4.0 m and cross-section 2.0 × 10⁻⁴ m² holds a load of 4000 N (Y = 2.0 × 10¹¹ Pa).

Find the stress and the extension.

Try it first, then show the solutionSolution

σ = 4000/2.0 × 10⁻⁴ = 2.0 × 10⁷ Pa.
ΔL = σL/Y = 2.0 × 10⁷ × 4/2.0 × 10¹¹ = 4.0 × 10⁻⁴ m.

Answer: 2.0 × 10⁷ Pa; 0.4 mm.

Memory hook: Stress divided by Y gives the strain.

Choosing a wire

Suppose a worker needs a wire to stretch by no more than 1 mm under a 300 N load over a length of 2 m. The material has Y = 1.0 × 10¹¹ Pa.

Find the minimum cross-sectional area.

Try it first, then show the solutionSolution

A = FL/(Y ΔL) = 300 × 2/(1.0 × 10¹¹ × 10⁻³).
A = 6.0 × 10⁻⁶ m² = 6 mm².

Answer: 6 mm² (or more).

Memory hook: Bigger area, smaller extension.

Tricks to remember Stress, Strain and Young's Modulus

Short memory aids. Say each one aloud once, then use it on a question the same day.

Y is the slope
Y is the slope of the linear part of the stress-strain graph.
Use it when: Reading graphs.
L over r squared
ΔL ∝ L/r² for the same load and material.
Use it when: Comparison questions.
Parallel wires share by AY/L
Equal extension means force proportional to AY/L.
Use it when: Load sharing.
Stress at the top
The top of a hanging wire carries the whole weight below it.
Use it when: Breaking length.

Stress, Strain and Young's Modulus: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Stress, strain and Y · 2 marks · Very short answerBoard level

A load of 80 N is hung from a wire 4 m long with cross-section 2 mm², and the wire stretches by 0.8 mm. Find the stress, the strain and Young's modulus of the wire.

Try it first, then show the solutionSolution

Stress = 80/2 × 10⁻⁶ = 4 × 10⁷ Pa; strain = 0.8 × 10⁻³/4 = 2 × 10⁻⁴. (1)
Y = 4 × 10⁷/2 × 10⁻⁴ = 2 × 10¹¹ Pa. (1)

Answer: 4 × 10⁷ Pa; 2 × 10⁻⁴; 2 × 10¹¹ Pa.

Example 2: Extension of a copper wire · 3 marks · Short answerBoard level

A copper wire (Y = 1.1 × 10¹¹ Pa) 3.0 m long and 1.5 mm in diameter carries a 6.0 kg mass (g = 10 m/s²). Find its extension.

Try it first, then show the solutionSolution

F = 60 N; A = π(0.75 × 10⁻³)² = 1.77 × 10⁻⁶ m². (1)
ΔL = FL/(AY) = 60 × 3/(1.77 × 10⁻⁶ × 1.1 × 10¹¹). (1)
= 180/1.94 × 10⁵ = 9.3 × 10⁻⁴ m ≈ 0.93 mm. (1)

Answer: About 0.93 mm.

Example 3: Two wires of the same material · 3 marks · Short answerHarder board level

Wire A has length L and radius r. Wire B has length 2L and radius 2r, and is of the same material. Both carry the same load. Compare (a) their extensions, (b) their stresses and (c) the elastic energy stored.

Try it first, then show the solutionSolution

ΔL ∝ L/r²: A: L/r²; B: 2L/4r² = L/2r². (a) ΔL_A : ΔL_B = 2 : 1. (1)
(b) Stress ∝ 1/r²: σ_A : σ_B = 4 : 1. (1)
(c) U = ½FΔL with equal F: U_A : U_B = 2 : 1. (1)

Answer: (a) 2 : 1; (b) 4 : 1; (c) 2 : 1.

Example 4: Load shared by two wires · 4 marks · Long answerHarder board level

A 600 N load hangs from a light rigid bar supported by two vertical wires of equal length 1 m. Wire P has Y = 2.0 × 10¹¹ Pa and area 1.0 mm²; wire Q has Y = 1.0 × 10¹¹ Pa and area 1.0 mm². The bar stays horizontal. Find the load carried by each wire, the stresses, and the extension.

Try it first, then show the solutionSolution

Equal extensions: F ∝ AY/L, so F_P : F_Q = 2 : 1. (1)
F_P = 400 N, F_Q = 200 N. (1)
Stresses: 4 × 10⁸ Pa and 2 × 10⁸ Pa. (1)
ΔL = 400/(10⁻⁶ × 2 × 10¹¹) = 2 mm for P; 200/(10⁻⁶ × 10¹¹) = 2 mm for Q (equal, as required). (1)

Answer: 400 N and 200 N; 4 × 10⁸ and 2 × 10⁸ Pa; 2 mm each.

Example 5: Analysing load-extension data · 4 marks · Long answerHOTS / competency

A 2.0 m wire of cross-section 1.0 mm² gives these readings of load (N) and extension (mm): (100, 1.0), (200, 2.0), (300, 3.0), (400, 4.0), (500, 6.0). The wire recovers fully after 400 N but not after 500 N. (a) Find Young's modulus from the linear part. (b) Find the stress at the elastic limit. (c) Estimate the extension under 250 N. (d) Find the energy stored at 400 N and say why the 500 N reading is not used.

Try it first, then show the solutionSolution

(a) Y = FL/(AΔL) = 100 × 2/(10⁻⁶ × 1.0 × 10⁻³) = 2 × 10¹¹ Pa. (1)
(b) Stress = 400/10⁻⁶ = 4 × 10⁸ Pa. (1)
(c) Linear: ΔL = 250 × 2/(10⁻⁶ × 2 × 10¹¹) = 2.5 mm. (1)
(d) U = ½ × 400 × 4 × 10⁻³ = 0.8 J. The 500 N point lies beyond the elastic limit, so the extension is not proportional to the load. (1)

Answer: (a) 2 × 10¹¹ Pa; (b) 4 × 10⁸ Pa; (c) 2.5 mm; (d) 0.8 J.

Where students lose marks in Stress, Strain and Young's Modulus

The slips examiners see most often, and what to write instead.

Mistake: Using the diameter as the radius in A = πr².
Missing the factor of 2.
Do this: r = d/2; convert mm to m before squaring.
Usually costs: The answer mark.
Mistake: Leaving the area in mm² with the force in N and expecting Pa.
Unit mismatch.
Do this: 1 mm² = 10⁻⁶ m².
Usually costs: The answer mark.
Mistake: Saying Y depends on the length of the wire.
Confusing a constant with a dimension.
Do this: Y is a material property; k = AY/L depends on dimensions.
Usually costs: Concept mark.
Mistake: Using Hooke's law beyond the elastic limit.
Ignoring the graph.
Do this: Check that the load is within the linear region.
Usually costs: 1 mark.

Exam tips

Write the formula ΔL = FL/(AY) and the area conversion in the first line.
For ratio questions, cancel everything common and keep only L/r².
State the elastic limit when you use readings from a graph.
Give units for stress (Pa) and note that strain has none.
Describe regions of the curve in order: proportional, elastic limit, yield, ultimate, fracture.

Practice questions on Stress, Strain and Young's Modulus

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
Two wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio 1 : 2. Under the same load, the ratio of their extensions is
(a) 1 : 1 (b) 2 : 1 (c) 1 : 2 (d) 1 : 4
Hint
ΔL ∝ L/r².
Answer
(b) 2 : 1.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Young's modulus of a wire doubles if the wire is stretched to twice its length.
Reason (R): Young's modulus is the ratio of stress to strain.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Y is a material property.
Answer
(d). A is false: Y depends on the material only; R is true.
Q3 · 2 marks · Very short answerBoard level
A steel rod 1 m long with cross-section 1 cm² (Y = 2 × 10¹¹ Pa) carries a tensile load of 10⁴ N. Find its extension.
Hint
ΔL = FL/AY.
Answer
ΔL = 10⁴ × 1/(10⁻⁴ × 2 × 10¹¹) = 5 × 10⁻⁴ m = 0.5 mm.
Q4 · 3 marks · Short answerHarder board level
The load-extension graph of a wire of length 1.5 m and cross-section 0.5 mm² is a straight line of slope 3.0 × 10⁴ N/m. Find Young's modulus of the wire material.
Hint
Slope = F/ΔL = AY/L.
Answer
Y = slope × L/A = 3.0 × 10⁴ × 1.5/(0.5 × 10⁻⁶) = 9 × 10¹⁰ Pa.
Q5 · 4 marks · Long answerHOTS / competency
A steel wire of breaking stress 2.4 × 10⁸ Pa and density 8000 kg/m³ hangs vertically. (a) Find the greatest length it can have before it breaks under its own weight (g = 10 m/s²). (b) What length is safe with a factor of safety of 3? (c) Explain where the wire is most likely to break.
Hint
σ at the top = ρgL.
Answer
(a) L = σ/(ρg) = 2.4 × 10⁸/(8000 × 10) = 3000 m. (b) 3000/3 = 1000 m. (c) At the top, which supports the weight of the entire wire below.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Young's modulus?Show answer
Stress/strain = FL/(AΔL).
2.Unit of strain?Show answer
None.
3.Extension of a wire?Show answer
FL/(AY).
4.Wire spring constant?Show answer
AY/L.
5.Ductile material?Show answer
One with a large plastic region.
6.Factor of safety?Show answer
Breaking stress / working stress.

Stress, Strain and Young's Modulus: frequently asked questions

Is steel more elastic than rubber?

Yes. For the same stress steel shows a much smaller strain, so it has a much larger Young's modulus.

What is the elastic limit?

The maximum stress up to which the body recovers fully when the load is removed.

Why are thick wires used for heavy loads?

Larger area reduces stress, so the wire stays below its breaking stress.

How do the elastic limit and the proportional limit differ?

Up to the proportional limit stress is proportional to strain; the elastic limit is a little beyond it where recovery is still complete.

What is the SI unit of Young's modulus?

Pascal (N/m²).