Mechanical Properties of Solids ISC Class 11: Elasticity Notes

What Mechanical Properties of Solids is about

Weightage: Mechanical Properties of Solids belongs to unit 7 of the ISC Class XI Physics syllabus. CISCE weights units 7-9 together (Properties of Bulk Matter; Heat and Thermodynamics; Behaviour of Perfect Gases and Kinetic Theory of Gases) at 20 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Elasticity, plasticity and the elastic limit

A deforming force changes the shape or size of a body; inside the body, restoring forces between molecules oppose the change. Elasticity is the property by which a body regains its original shape and size when the deforming force is removed; plasticity is the opposite. The elastic limit is the greatest stress up to which a body remains elastic. A perfectly rigid body (no deformation at all) is an idealisation. 'More elastic' in physics means a larger modulus: steel is more elastic than rubber because the same stress produces a much smaller strain in steel and it recovers fully. Quartz is nearly perfectly elastic; putty is nearly perfectly plastic.

Worked example

A rubber band and a steel wire of the same length and cross-section carry the same load. Which stretches more, and which is more elastic? Explain.

Try it first, then show the solutionSolution

1. Same stress in both; the rubber band shows a much larger strain.
2. Modulus = stress/strain, so steel has the far larger modulus (about $2\times 10^{11}$ Pa against about $10^6$ Pa for rubber).

Answer: Rubber stretches more, but steel is more elastic (larger Y)

Stress and strain

Stress = restoring force per unit area = $\frac{F}{A}$ (N m⁻², Pa). Strain = change in dimension ÷ original dimension (dimensionless). Three types: (1) Longitudinal: force along the length; tensile stress stretches, compressive stress shortens; longitudinal strain $=\frac{\Delta L}{L}$. (2) Volumetric (hydraulic): equal normal pressure on all faces; volume strain $=\frac{\Delta V}{V}$. (3) Shear (tangential): force parallel to a face; shear strain = angle of shear θ ≈ $\frac{\Delta x}{L}$ (lateral displacement of the face ÷ its distance from the fixed face). Stress has the same unit as pressure but acts as an internal restoring force, not an external push.

Worked example

A wire 2 m long with a cross-section of 1 mm² carries a load of 100 N and extends by 1 mm. Find the stress and strain.

Try it first, then show the solutionSolution

1. Stress = F/A = 100/(1 × 10⁻⁶) = 1 × 10⁸ N m⁻²
2. Strain = ΔL/L = 1 × 10⁻³/2 = 5 × 10⁻⁴

Answer: $10^8$ Pa; $5\times 10^{-4}$ (so Y = $2\times 10^{11}$ Pa)

Hooke's law and the stress-strain curve

Hooke's law: within the elastic limit, stress ∝ strain; stress/strain = modulus of elasticity. Stress-strain curve for a metal wire under increasing load: (1) OA, a straight line up to the proportional limit A: Hooke's law holds; (2) up to B, the elastic limit: the wire still recovers if unloaded, though the graph may curve slightly; (3) beyond B, at the yield point C, strain increases rapidly with little extra stress (plastic flow); unloading leaves a permanent set; (4) the curve rises to D, the ultimate tensile strength (maximum stress); (5) the wire thins (necks) and breaks at E, the fracture point. Ductile materials (copper, aluminium) have a large gap between D and E and can be drawn into wires; brittle ones (glass, cast iron) break just after the elastic limit. Elastomers such as rubber can be stretched hugely and still recover, but do not obey Hooke's law.

Worked example

On a stress-strain graph, the straight part for metal X has a greater slope than for metal Y, and X breaks soon after its elastic limit while Y stretches a long way before breaking. Compare the two.

Try it first, then show the solutionSolution

1. Slope of the linear part = Young's modulus, so X has the larger Y: it is stiffer.
2. X breaks soon after the elastic limit: brittle. Y has a long plastic region: ductile.

Answer: X: stiffer and brittle; Y: less stiff and ductile (better for drawing into wires)

Young's modulus

Young's modulus is the ratio of longitudinal stress to longitudinal strain within the elastic limit: $Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$ (Pa). For a wire of radius r carrying a mass M: $Y=\frac{MgL}{\pi r^2\Delta L}$. Typical values: steel $2\times 10^{11}$ Pa, copper $1.1\times 10^{11}$ Pa, aluminium $0.7\times 10^{11}$ Pa. Y depends only on the material (and temperature), not on the length or thickness of the wire; but the extension $\Delta L=\frac{FL}{AY}$ grows with length and falls with area. Y is defined for solids only, since liquids and gases have no fixed length.

Worked example

A steel wire 2 m long and 1 mm in diameter carries a 10 kg load. Find its extension. ($Y=2\times 10^{11}$ Pa, g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. A = π(0.5 × 10⁻³)² = 7.85 × 10⁻⁷ m²
2. $\Delta L=\frac{FL}{AY}=\frac{98\times 2}{7.85\times 10^{-7}\times 2\times 10^{11}}=1.25\times 10^{-3}m$

Answer: 1.25 mm

Determination of Young's modulus (Searle's method)

Two long, identical wires hang from the same rigid support. The reference wire carries a fixed load to keep it taut and carries the main scale; the experimental wire carries a vernier (or a spirit level with a micrometer screw) and a hanger. Using a reference wire cancels errors from the support yielding and from temperature changes, which affect both wires equally. Measure the length L with a metre rule and the diameter with a screw gauge at several places. Add loads in equal steps and read the extension each time; repeat while unloading and average. Plot load (mass M) against extension: a straight line through the origin within the elastic limit. Then $Y=\frac{gL}{\pi r^2}\times \frac{M}{\Delta L}$, using the slope. The wire is long and thin so that the extension is large enough to measure accurately.

Worked example

In Searle's experiment, a wire 3 m long and 1 mm in diameter extends by 0.2 mm for each kilogram added. Find Young's modulus. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. A = 7.85 × 10⁻⁷ m²; extension per newton = 0.2 × 10⁻³/9.8 m
2. $Y=\frac{9.8\times 3}{7.85\times 10^{-7}\times 0.2\times 10^{-3}}=1.87\times 10^{11}Pa$

Answer: About $1.9\times 10^{11}$ Pa (close to steel)

Bulk modulus and modulus of rigidity

Bulk modulus: when a uniform pressure change p acts on all faces, $B=-\frac{p}{\Delta V/V}$ (the minus sign makes B positive, as the volume decreases). Its reciprocal, $k=\frac{1}{B}$, is the compressibility. Gases are far more compressible than liquids, and liquids more than solids (water: B ≈ $2.2\times 10^9$ Pa). B is defined for solids, liquids and gases. Modulus of rigidity (shear modulus): $G=\frac{F/A}{\theta }$, where F acts tangentially on a face of area A and θ is the shear strain. Only solids resist shear, so G is defined for solids alone. For most materials G is roughly one-third of Y.

Worked example

(a) By what fraction does the volume of water decrease under an extra pressure of $10^7$ Pa? (B = $2.2\times 10^9$ Pa) (b) A rubber block 5 cm high has a top face of area 0.01 m². A tangential force of 100 N shifts the top face by 2 mm. Find the modulus of rigidity.

Try it first, then show the solutionSolution

1. (a) $\frac{\Delta V}{V}=\frac{p}{B}=\frac{10^7}{2.2\times 10^9}=4.5\times 10^{-3}$, about 0.45%.
2. (b) Shear stress = 100/0.01 = 10⁴ Pa; strain = 2/50 = 0.04
3. G = 10⁴/0.04 = 2.5 × 10⁵ Pa

Answer: (a) 0.45% (b) $2.5\times 10^5$ Pa

Poisson's ratio

When a wire is stretched it becomes longer and thinner. Longitudinal strain = $\frac{\Delta L}{L}$; lateral strain = $-\frac{\Delta d}{d}$ (the diameter decreases). Poisson's ratio $\sigma =\frac{\text{lateral strain}}{\text{longitudinal strain}}=\frac{\Delta d/d}{\Delta L/L}$ (magnitudes), a pure number depending only on the material. Theory allows −1 to 0.5; real materials lie between about 0 and 0.5 (steel about 0.28, rubber close to 0.5). The fractional change of volume of a stretched wire is $\frac{\Delta V}{V}=\frac{\Delta L}{L}(1-2\sigma )$, so σ = 0.5 means the volume does not change.

Worked example

A wire 2 m long and 2 mm in diameter is stretched by 1 mm, and its diameter decreases by 0.0003 mm. Find Poisson's ratio.

Try it first, then show the solutionSolution

1. Lateral strain = 0.0003/2 = 1.5 × 10⁻⁴; longitudinal strain = 1/2000 = 5 × 10⁻⁴
2. σ = 1.5 × 10⁻⁴/5 × 10⁻⁴ = 0.3

Answer: 0.3

Work done in stretching a wire: elastic energy

As a wire is stretched slowly, the tension rises from 0 to F in proportion to the extension, so the average force is F/2 and the work done is $W=\frac{1}{2}F\Delta L$ = ½ × load × extension. This work is stored in the wire as elastic (strain) potential energy. Per unit volume: $u=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{1}{2}Y(\text{strain}{)}^2$. The energy is released when the wire returns to its natural length (a catapult or bow). If the wire is taken beyond its elastic limit, part of the work becomes heat and is not recovered. (The syllabus treats elastic energy qualitatively.)

Worked example

The steel wire above extends 1.25 mm under a load of 98 N. How much elastic energy does it store?

Try it first, then show the solutionSolution

1. $U=\frac{1}{2}F\Delta L=\frac{1}{2}\times 98\times 1.25\times 10^{-3}$
2. U = 0.061 J

Answer: About 0.06 J

Physics in real life: Mechanical Properties of Solids

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.

Topics in this chapter

• Elastic Moduli, Poisson's Ratio and Elastic Energy
• Stress, Strain and Young's Modulus

Mechanical Properties of Solids formula sheet

Stress and strain

$\text{stress}=\frac{F}{A},\text{strain}=\frac{\Delta L}{L},\frac{\Delta V}{V},\theta$

Stress in Pa; strain has no unit.

Hooke's law

$\frac{\text{stress}}{\text{strain}}=\text{modulus of elasticity}$

Within the proportional limit.

Young's modulus

$Y=\frac{FL}{A\Delta L}=\frac{MgL}{\pi r^2\Delta L}$

Steel ≈ $2\times 10^{11}$ Pa.

Bulk modulus and compressibility

$B=-\frac{p}{\Delta V/V},k=\frac{1}{B}$

Defined for solids, liquids, gases.

Modulus of rigidity

$G=\frac{F/A}{\theta },\theta \approx \frac{\Delta x}{L}$

Solids only.

Poisson's ratio

$\sigma =\frac{\Delta d/d}{\Delta L/L}$

Between 0 and 0.5 in practice.

Extension of a wire

$\Delta L=\frac{FL}{AY}$

∝ L, ∝ 1/r².

Elastic energy

$U=\frac{1}{2}F\Delta L,u=\frac{1}{2}\times \text{stress}\times \text{strain}$

u is energy per unit volume.

Mechanical Properties of Solids: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Mechanical Properties of Solids

The question patterns to expect, and the method that works for each.

Tricks to remember Mechanical Properties of Solids

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

• Y, B, G for length, bulk, glide

  Y stretches length; B squeezes bulk volume; G (rigidity) makes layers glide.

  Use it when: Matching each modulus with its strain.

• The curve's alphabet

  Proportional, Elastic, Yield, Ultimate, Fracture: 'Please Enjoy Your Usual Food'.

  Use it when: Labelling the stress-strain curve in order.

• Extension rule

  ΔL ∝ F L / (r² Y): longer stretches more, thicker stretches less.

  Use it when: Comparing two wires.

• Half load times extension

  Energy in a wire is the triangle under the load-extension line: ½ F ΔL.

  Use it when: Strain-energy questions.

• Poisson stops at a half

  σ = 0.5 means no volume change; real materials stay below it.

  Use it when: Poisson's ratio limits.

Mechanical Properties of Solids: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 2 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Mechanical Properties of Solids

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

• ✗Mistake: Using the diameter as the radius

  Wire data are usually given as diameter, and students put it straight into πr².

  Wrong working

  d = 1 mm: A = π(1 × 10⁻³)² = 3.14 × 10⁻⁶ m².

  Right working

  r = 0.5 mm: A = π(0.5 × 10⁻³)² = 7.85 × 10⁻⁷ m² (four times smaller).

  ✓Do this: Halve the diameter before squaring.

  Usually costs: 1 mark (the answer is out by a factor of 4).

• ✗Mistake: Thinking 'more elastic' means 'stretches more'

  In everyday language rubber is 'elastic', but in physics elasticity is measured by the modulus.

  Wrong working

  Rubber is more elastic than steel because it stretches more.

  Right working

  Steel is more elastic: it has a much larger Y and recovers fully.

  ✓Do this: Compare moduli, not extensions.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting units of area

  Areas in mm² or cm² are used without converting to m².

  Wrong working

  Stress = 100 N/1 mm² = 100 Pa.

  Right working

  1 mm² = 10⁻⁶ m², so stress = 10⁸ Pa.

  ✓Do this: Convert all lengths to metres before squaring.

  Usually costs: 1 mark.

• ✗Mistake: Mixing up the elastic limit and the breaking point

  Students say a wire 'breaks at its elastic limit'.

  Wrong working

  Beyond the elastic limit the wire breaks.

  Right working

  Beyond the elastic limit the wire is permanently deformed (plastic); it breaks only at the fracture point, after the ultimate tensile strength.

  ✓Do this: Learn the order: proportional limit, elastic limit, yield point, ultimate strength, fracture.

  Usually costs: 1 mark.

• ✗Mistake: Using the full load instead of the average in strain energy

  Students write W = F × ΔL for stretching a wire.

  Wrong working

  Energy = 98 × 1.25 × 10⁻³ = 0.12 J.

  Right working

  The force rises from 0 to F, so W = ½FΔL = 0.061 J.

  ✓Do this: Remember the triangle under the load-extension graph.

  Usually costs: 1 mark.

Practice questions on Mechanical Properties of Solids

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

1. Q1 · 1 mark · MCQExam level

   The SI unit of stress is:
   (a) N m
   (b) N m⁻¹
   (c) N m⁻²
   (d) no unit

   Hint

   Force per area.

   Answer

   (c) N m⁻² (Pa)

2. Q2 · 1 mark · MCQExam level

   For real materials, Poisson's ratio lies between:
   (a) 0 and 0.5
   (b) 0.5 and 1
   (c) 1 and 2
   (d) −2 and −1

   Hint

   σ = 0.5 means no change of volume.

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   Define elastic limit.

   Hint

   Recovery.

   Answer

   The greatest stress up to which a body regains its original shape and size when the deforming force is removed.

4. Q4 · 1 mark · Very short answerExam level

   Why are springs made of steel rather than copper?

   Hint

   Compare Y and elastic limit.

   Answer

   Steel has a larger Young's modulus and a higher elastic limit, so it gives a larger restoring force and recovers fully under larger loads.

5. Q5 · 2 marks · Short answerExam level

   Distinguish between ductile and brittle materials, giving one example of each.

   Hint

   Size of the plastic region.

   Answer

   Ductile materials have a large plastic region between the elastic limit and fracture and can be drawn into wires (copper). Brittle materials break soon after the elastic limit (glass, cast iron).

6. Q6 · 2 marks · Short answerHarder exam level

   A wire stretches by 2 mm under a certain load. Another wire of the same material, twice as long and of twice the radius, carries the same load. Find its extension.

   Hint

   ΔL ∝ L/r².

   Answer

   ΔL ∝ 2/4 = ½ of the first: 1 mm.

7. Q7 · 2 marks · Short answerHarder exam level

   A steel cube of side 10 cm has its lower face fixed. A tangential force of 5000 N acts on its upper face. Find the shear strain and the displacement of the upper face. (G = $4\times 10^{10}$ Pa)

   Hint

   Area 0.01 m².

   Answer

   Stress = 5 × 10⁵ Pa; strain = 1.25 × 10⁻⁵; displacement = 1.25 × 10⁻⁵ × 0.1 = 1.25 × 10⁻⁶ m.

8. Q8 · 2 marks · Short answerHarder exam level

   A wire 1 m long with a cross-section of 1 mm² (Y = $2\times 10^{11}$ Pa) is stretched by 1 mm. Find the force needed and the energy stored.

   Hint

   F = YAΔL/L.

   Answer

   F = 2 × 10¹¹ × 10⁻⁶ × 10⁻³/1 = 200 N; U = ½ × 200 × 10⁻³ = 0.1 J.

9. Q9 · 2 marks · Short answerHarder exam level

   An aluminium block is lowered to a depth of 1000 m in fresh water. Find the fractional decrease in its volume. (B = $7.5\times 10^{10}$ Pa, g = 9.8 m s⁻²)

   Hint

   p = ρgh.

   Answer

   p = 1000 × 9.8 × 1000 = 9.8 × 10⁶ Pa; ΔV/V = 9.8 × 10⁶/7.5 × 10¹⁰ = 1.3 × 10⁻⁴.

10. Q10 · 3 marks · Short answerHarder exam level

    The length of a wire increases by 0.5% when stretched. If Poisson's ratio is 0.25, find the percentage decrease in its diameter and the percentage change in its volume.

    Hint

    ΔV/V = (ΔL/L)(1 − 2σ).

    Answer

    Δd/d = 0.25 × 0.5% = 0.125% decrease; ΔV/V = 0.5% × (1 − 0.5) = 0.25% increase.

11. Q11 · 3 marks · Short answerHarder exam level

    A 4 m steel wire of cross-section 2 mm² hangs with a 20 kg load. Find the stress, strain, extension and the elastic energy stored. (Y = $2\times 10^{11}$ Pa, g = 9.8 m s⁻²)

    Hint

    Then U = ½FΔL.

    Answer

    Stress = 196/2 × 10⁻⁶ = 9.8 × 10⁷ Pa; strain = 4.9 × 10⁻⁴; ΔL = 1.96 mm; U = ½ × 196 × 1.96 × 10⁻³ = 0.19 J.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Define bulk modulus and compressibility. [2]
    (b) Explain why gases have a much smaller bulk modulus than solids. [1]
    (c) Find the bulk modulus of a liquid whose volume falls from 100 cm³ to 99.5 cm³ when the pressure rises by $1.2\times 10^7$ Pa. [2]

    Hint

    ΔV/V = 0.5/100.

    Answer

    (a) B = −p/(ΔV/V), the ratio of volumetric stress to volumetric strain; compressibility = 1/B, the fractional decrease in volume per unit increase in pressure. (b) Gas molecules are far apart with weak forces, so a small pressure change produces a large change in volume. (c) B = 1.2 × 10⁷/0.005 = 2.4 × 10⁹ Pa.

13. Q13 · 3 marks · Short answerHOTS / competency

    A light horizontal rod 1 m long hangs from a steel wire at one end and a brass wire at the other; both wires have the same length and cross-section. Where must a load be hung on the rod so that it stays horizontal? ($Y_{steel}=2Y_{brass}$)

    Hint

    Equal extensions need $T_s/Y_s=T_b/Y_b$.

    Answer

    Equal extensions: $T_s=2T_b$. With the load W at x from the steel end, moments give $T_s=W(1-x)$ and $T_b=Wx$, so 1 − x = 2x and x = 1/3 m from the steel wire.

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. 1.State Hooke's law.Show answer

   Within the elastic limit, stress is proportional to strain.

2. 2.Formula for Young's modulus of a wire?Show answer

   Y = FL/(AΔL).

3. 3.What is compressibility?Show answer

   The reciprocal of the bulk modulus.

4. 4.Which modulus exists only for solids?Show answer

   Modulus of rigidity (and Young's modulus).

5. 5.Define Poisson's ratio.Show answer

   Lateral strain ÷ longitudinal strain.

6. 6.Energy stored in a stretched wire?Show answer

   ½ × load × extension.

7. 7.Why is a reference wire used in Searle's method?Show answer

   To cancel errors from the support yielding and from temperature changes.

8. 8.What happens beyond the yield point?Show answer

   Plastic flow: strain rises rapidly and a permanent set remains.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define stress, strain and the elastic limit?
• ☐Can you draw and label the stress-strain curve?
• ☐Can you find the extension of a wire from Y?
• ☐Can you compare the extensions of two wires by ratios?
• ☐Can you describe Searle's method and say why a reference wire is used?
• ☐Can you use the bulk modulus and modulus of rigidity in numericals?
• ☐Can you find Poisson's ratio from changes in length and diameter?
• ☐Can you find the energy stored in a stretched wire?

Mechanical Properties of Solids: frequently asked questions