Mechanical Properties of Solids Class 11 Notes | CBSE Physics Chapter 8

Chapter summary

Mechanical Properties of Solids explains how solids respond to deforming forces through stress and strain, Hooke’s law, and the stress-strain curve that distinguishes elastic, plastic, ductile and brittle behaviour. It defines the three elastic moduli (Young’s, bulk and shear) along with Poisson’s ratio, compressibility and the elastic potential energy stored in a stretched body. For NEET it is a high-yield, formula-driven chapter whose moduli and elastic-energy numericals appear regularly and also support fluids and other mechanics topics.

Chapter notes

Table of Contents


Key Concepts

1. Elastic Behaviour of Solids

A solid is made of atoms held in a regular lattice by inter-atomic bonds that act like tiny springs. When you apply a force, these “springs” stretch or compress, and the body deforms. Remove the force, and the bonds pull the atoms back - the body regains its shape.

Elasticity is the property of a body by which it regains its original shape and size after the deforming force is removed. A steel ball is highly elastic; putty or wet clay is not.

Plasticity is the opposite property - the body retains the new shape after the force is removed. A body with no tendency to return is called a perfectly plastic body.


2. Stress

Stress is the internal restoring force set up per unit area of cross-section when a body is deformed. In magnitude it equals the applied force per unit area at equilibrium.

Stress = F/A

  • SI unit: N/m² or pascal (Pa)
  • Dimensional formula: [ML⁻¹T⁻²]
  • Stress is a scalar (more precisely a tensor), not a vector.

Types of Stress

  • Tensile / Compressive (longitudinal) stress: force perpendicular to area, causing change in length.
  • Shearing (tangential) stress: force parallel to the surface, causing change in shape.
  • Hydraulic (volume) stress: force applied uniformly from all sides by a fluid, causing change in volume.

3. Strain

Strain is the fractional change in configuration (length, shape, or volume) of a body under stress. It is a ratio of two like quantities, so it has no units and no dimensions.

Types of Strain

  • Longitudinal strain = change in length / original length = ΔL/L
  • Shearing strain = relative displacement / distance between layers = Δx/L = tan θ ≈ θ (for small angles)
  • Volume strain = change in volume / original volume = ΔV/V

4. Hooke’s Law

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

Stress ∝ Strain, so Stress = E × Strain

  • The constant of proportionality E is called the modulus of elasticity.
  • Its value depends only on the material, not on the dimensions of the body.
  • SI unit of modulus: N/m² or pascal (same as stress, since strain is unitless).

Key idea: Hooke’s law holds only up to the elastic limit. Beyond it, stress and strain are no longer proportional.


5. Stress-Strain Curve

If a metal wire is loaded gradually and the stress is plotted against strain, we get a characteristic curve with distinct regions.

[DIAGRAM: Stress (y-axis) vs strain (x-axis) curve - straight line O→A (proportional limit), A→B elastic limit (yield point), B→D plastic region, D the ultimate tensile strength, E the fracture point.]

  • O to A - Proportional limit: stress ∝ strain; Hooke’s law obeyed; the graph is a straight line.
  • A to B - Elastic limit (yield point): the wire still returns to its original length on removing the load, but the line curves.
  • B to D - Plastic region: beyond the yield point, strain increases with little extra stress and a permanent (plastic) set remains.
  • D - Ultimate tensile strength: the maximum stress the material can bear.
  • E - Fracture point: the wire finally breaks.

Note: A large gap between the yield point and fracture point means the material is ductile (can be drawn into wires); a small gap means it is brittle (breaks soon after the elastic limit, like glass).


6. Young’s Modulus (Y)

Within the elastic limit, the ratio of longitudinal stress to longitudinal strain is a constant called Young’s modulus.

Y = Longitudinal stress / Longitudinal strain = (F/A) / (ΔL/L) = FL / (A·ΔL)

  • SI unit: N/m² or pascal; dimensions: [ML⁻¹T⁻²]
  • Steel has a higher Young’s modulus than rubber, so steel is more elastic and harder to stretch.
  • A high Y means a small strain for a given stress - the material is stiff.

7. Shear Modulus / Modulus of Rigidity (G)

The ratio of shearing (tangential) stress to shearing strain within the elastic limit is the shear modulus or modulus of rigidity.

G = Shearing stress / Shearing strain = (F/A) / θ = F / (A·θ)

  • SI unit: N/m² or pascal
  • G is generally about one-third of Young’s modulus (G ≈ Y/3) for most metals.
  • Only solids have rigidity; liquids and gases cannot sustain a shear stress.

8. Bulk Modulus (K)

The ratio of hydraulic (volume) stress to the volume strain within the elastic limit is the bulk modulus.

K = −p / (ΔV/V) = −pV / ΔV

  • The negative sign shows that an increase in pressure decreases the volume.
  • SI unit: N/m² or pascal
  • The reciprocal of bulk modulus is the compressibility, k = 1/K.
  • Solids have the highest K, then liquids, then gases - solids are least compressible.

9. Poisson’s Ratio (σ)

When a wire is stretched, it becomes longer but thinner - its length increases while its diameter decreases. The ratio of lateral strain to longitudinal strain is called Poisson’s ratio (σ).

σ = Lateral strain / Longitudinal strain = −(Δd/d) / (ΔL/L)

  • It is a pure ratio, so it has no units and no dimensions.
  • Its theoretical range is −1 to 0.5; for most materials it lies between 0.2 and 0.4.

10. Elastic Potential Energy in a Stretched Wire

Work done in stretching a wire is stored in it as elastic potential energy. Since the force builds up gradually from 0 to F, the average force is F/2.

U = ½ × Stress × Strain × Volume = ½ × F × ΔL

The energy stored per unit volume (elastic energy density) is:

u = ½ × Stress × Strain = ½ × Y × (Strain)²

Key idea: A catapult, a bow, and a spring all store elastic potential energy this way before releasing it as kinetic energy.


11. Applications of Elastic Behaviour

The choice of materials and shapes in engineering comes straight from this chapter.

  • Beams and bridges: a girder is given an I-shape (I-section) so it resists bending with less material, because the depression of a loaded beam δ ∝ Wl³ / (Ybd³).
  • Cranes and ropes: the rope of a crane is made thick (large area) so the stress stays safely below the elastic limit for the maximum load.
  • Pillars: pillars with distributed (cross-section) ends carry more load than those with rounded ends, so building columns are designed accordingly.
  • Mountains: the maximum height of a mountain on Earth (~10 km) is limited by the elastic strength of rock at its base.

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Weightage in Board & Entrance Exams

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 11)3–4 marksStress-strain definitions, Young’s modulus numericals, stress-strain curve
JEE Main / Advanced1–2 questionsYoung’s modulus, elastic energy, composite wires, bulk modulus
NEET1–2 questionsElastic moduli definitions, Poisson’s ratio, stress-strain graph reading

[TABLE: Question-type split - VSA (1 mark): definitions of stress, strain, elastic limit, units; SA (2–3 marks): Young’s modulus derivation, stress-strain curve, Poisson’s ratio; LA (5 marks): elastic energy derivation, comparison of three moduli, applications of elasticity.]


Important Definitions

TermDefinition
ElasticityProperty by which a body regains its original shape and size after the deforming force is removed
StressInternal restoring force per unit area: Stress = F/A; unit Pa
StrainFractional change in configuration (length, shape, or volume); dimensionless
Hooke’s lawWithin the elastic limit, stress ∝ strain
Elastic limitMaximum stress up to which a body returns to its original size on removing the load
Young’s modulus (Y)Ratio of longitudinal stress to longitudinal strain: Y = FL/(A·ΔL)
Shear modulus (G)Ratio of shearing stress to shearing strain: G = F/(A·θ)
Bulk modulus (K)Ratio of volume stress to volume strain: K = −pV/ΔV
Poisson’s ratio (σ)Ratio of lateral strain to longitudinal strain; dimensionless
Elastic potential energyEnergy stored in a stretched body: U = ½ × F × ΔL

Solved Examples

Example 1

A steel wire of length 2 m and cross-sectional area 1 mm² is stretched by a force of 100 N. Find the stress in the wire.

Answer: Stress = F/A = 100 / (1 × 10⁻⁶) = 1 × 10⁸ N/m² (10⁸ Pa).

Example 2

A wire of original length 4 m stretches by 2 mm under a load. Find the longitudinal strain.

Answer: Strain = ΔL/L = (2 × 10⁻³) / 4 = 5 × 10⁻⁴ (no units).

Example 3

A force of 200 N stretches a wire of length 3 m and area 2 × 10⁻⁶ m² by 1.5 mm. Find Young’s modulus.

Answer: Y = FL / (A·ΔL) = (200 × 3) / (2 × 10⁻⁶ × 1.5 × 10⁻³) = 600 / (3 × 10⁻⁹) = 2 × 10¹¹ N/m².

Example 4

The bulk modulus of water is 2.2 × 10⁹ N/m². Find the pressure required to reduce a given volume of water by 0.1%.

Answer: ΔV/V = 0.1% = 10⁻³. p = K × (ΔV/V) = 2.2 × 10⁹ × 10⁻³ = 2.2 × 10⁶ N/m².

Example 5

A wire is stretched by 1 mm under a force of 50 N. Find the elastic potential energy stored in it.

Answer: U = ½ × F × ΔL = ½ × 50 × (1 × 10⁻³) = 0.025 J (2.5 × 10⁻² J).

Example 6

When a wire is stretched, its length increases by 0.2% and its diameter decreases by 0.05%. Find Poisson’s ratio.

Answer: σ = lateral strain / longitudinal strain = 0.05% / 0.2% = 0.0005 / 0.002 = 0.25.


Important Questions for Board Exams

1-Mark Questions (VSA)

  1. Define stress and give its SI unit.
  2. Why has strain no units or dimensions?
  3. Which is more elastic - steel or rubber? Justify.
  4. What does the negative sign in the formula for bulk modulus indicate?
  5. Define Poisson’s ratio and state its dimensions.

2–3-Mark Questions (SA)

  1. State Hooke’s law and define modulus of elasticity. Give its SI unit.
  2. Draw and explain the stress-strain curve for a metal wire, marking the elastic limit, yield point, and fracture point.
  3. Distinguish between Young’s modulus, shear modulus, and bulk modulus.
  4. Why are girders given an I-shape? Explain using the depression formula.

5-Mark Questions (LA)

  1. Define Young’s modulus and derive its expression Y = FL/(A·ΔL). Explain why solids are more elastic than liquids and gases.
  2. Derive an expression for the elastic potential energy stored per unit volume in a stretched wire, and show that u = ½ × stress × strain.
  3. Explain three applications of the elastic behaviour of materials in everyday engineering (bridges, cranes, pillars).

Quick Revision Points

  • Elasticity = tendency to regain shape; plasticity = tendency to keep the new shape
  • Stress = F/A (unit Pa, dimensions [ML⁻¹T⁻²]); strain = fractional change (dimensionless)
  • Three stresses: tensile/compressive, shearing, hydraulic - giving length, shape, volume strain
  • Hooke’s law: stress ∝ strain within the elastic limit; stress = E × strain
  • Stress-strain curve: proportional limit → elastic limit (yield) → plastic region → ultimate strength → fracture
  • Young’s modulus Y = FL/(A·ΔL); higher Y means stiffer material
  • Shear modulus G = F/(A·θ) ≈ Y/3; only solids have rigidity
  • Bulk modulus K = −pV/ΔV; compressibility = 1/K; gases most compressible
  • Poisson’s ratio σ = lateral strain / longitudinal strain; range −1 to 0.5
  • Elastic energy U = ½ × F × ΔL; energy density u = ½ × stress × strain = ½ Y(strain)²
  • Applications: I-shaped girders, thick crane ropes, designed pillars, mountain-height limit

Next Chapter: Chapter 9 - Mechanical Properties of Fluids

🃏 Flash Cards: Mechanical Properties of Solids

Class 11 Physics · Chapter 9 – swipe through all 8 cards to understand the whole chapter.

🔩Start here1/8

Stress and Strain

A deforming force makes a solid develop an internal restoring force; we measure it per unit area as stress, and the deformation as strain.

stress = F / A · strain = ΔL / L

Stress unit: N·m⁻2 (Pa), dimensions [M L⁻1 T⁻2] — same as pressure. Strain is a pure ratio (no units).

  • Three stresses: tensile/compressive (length), shearing (shape), hydraulic (volume)
  • Strain types: longitudinal ΔL/L, shearing Δx/L ≈ θ, volume ΔV/V
  • Stress is the cause (force), strain is the effect (deformation)
📏Core law2/8

Hooke’s Law

For small deformations, stress is directly proportional to strain.

stress = E · strain

E (modulus of elasticity) depends only on material and temperature — not on the body’s size or shape.

  • Valid only within the elastic limit, where the body fully recovers
  • Elasticity = regains shape; plasticity (putty, mud) = keeps deformed shape
  • Larger E means more elastic — steel is more elastic than rubber
📈Key graph3/8

Stress-Strain Curve

Plotting stress vs strain as load grows reveals a material’s whole mechanical behaviour.

Proportional limit → Yield point → Plastic region → Ultimate strength → Fracture

Area under the curve = energy absorbed per unit volume (not force).

  • Ductile (copper, gold): large plastic region — can be drawn into wires
  • Brittle (cast iron, glass): fracture soon after elastic limit
  • Elastomers (rubber): no straight line; loading ≠ unloading path (hysteresis)
🧱Core formula4/8

Young’s Modulus

The stiffness score for stretching or compressing — tensile stress over longitudinal strain.

Y = (F/A) / (ΔL/L) = F L / (A ΔL) ⇒ ΔL = F L / (A Y)

Steel Y ≈ 2 × 1011 Pa. Larger Y ⇒ stiffer ⇒ stretches less. 1 mm2 = 10⁻6 m2.

  • Extension ∝ L / r2 for a given material and force
  • Y is a property of the material, independent of the wire’s dimensions
  • Same load: a thicker wire stretches less (double radius ⇒ ¼ stretch)
🌊Core formula5/8

Bulk Modulus

The stiffness score for all-round (hydraulic) squeezing — pressure over volume strain.

B = −p / (ΔV / V) · compressibility k = 1 / B

Minus sign: volume shrinks as pressure rises, keeping B positive.

  • Solids: very high B (nearly incompressible); gases: very low B
  • Compressibility is the reciprocal of bulk modulus
  • Liquids and gases have only bulk modulus (no shear/Young’s)
✂️Core formula6/8

Shear (Rigidity) Modulus

The stiffness score for shape-changing tangential stress — shear stress over shear angle.

G = (F/A) / θ · G ≈ Y / 3

All moduli have units of pressure (Pa) since strain is dimensionless.

  • θ is the small shear angle (in radians)
  • Only solids have a shear modulus — fluids flow instead of resisting shear
  • Solids have all three moduli (Y, B, G); fluids have only B
🔻Key ratio7/8

Poisson’s Ratio

Stretch a wire lengthwise and it grows thinner sideways; this ratio measures by how much.

σ = −(Δd/d) / (ΔL/L) · ΔV/V = (1 − 2σ) ΔL/L

Dimensionless. Range −1 ≤ σ ≤ 0.5; most metals 0.2–0.4 (≈ 0.3).

  • It is a ratio of two strains, not a modulus
  • σ = 0.5 means perfectly incompressible (ΔV = 0 on stretching)
  • Keep the factor of 2 in (1 − 2σ) for the volume-change shortcut
🏹Energy + uses8/8

Elastic Energy & Applications

Work done in stretching a wire is stored as elastic potential energy, and these ideas drive real engineering.

U = ½ F ΔL = ½ × stress × strain × volume · u = ½ (stress)2 / Y

Energy ∝ (extension)2 — double the stretch ⇒ 4× the energy. The ½ is the triangle area.

  • Energy density u = ½ × stress × strain = area under stress-strain graph
  • I-beams: sag δ ∝ W l3 / (b d3 Y), so depth d (cubed) beats width
  • Mountain height is capped where base stress exceeds rock’s elastic limit
Swipe Click a card to focus 8 cards
📝 Practice Mechanical Properties of Solids — 7 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2020
A wire of length L and area of cross-section A hangs from a fixed support. When a mass M is suspended from its free end, its length becomes L₁. The expression for Young’s modulus of the material is:
Correct answer: C. The extension is Δ L = L₁ – L and the stretching force is Mg. Young’s modulus Y = (stress)/(strain) = (Mg/A)/((L₁-L)/L) = (MgL)/(A(L₁ – L)).
🔎 See the full step-by-step solution in the app →
Q2NEET 2019
The stress-strain curves are drawn for two materials X and Y. For material X the ultimate strength point and the fracture point are close together, while for material Y they are far apart. Materials X and Y are likely to be (respectively):
Correct answer: B. When the ultimate-strength and fracture points are close, the material breaks soon after reaching maximum stress with little plastic deformation — that is brittle (X). When they are far apart, there is a large plastic region between them — that is ductile (Y). Hence X is brittle and Y is ductile.
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Q3NEET 2019
A block of mass M is suspended by a long wire of length L, so that the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
Correct answer: B. The elastic potential energy equals the work done against the restoring force, which builds up linearly from 0 to Mg. So U = (1)/(2) × (force) × (extension) = (1)/(2) (Mg)(l) = (1)/(2) Mgl.
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Q4NEET 2018
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second has cross-sectional area 3A. If the length of the first wire is increased by Δ l on applying a force F, how much force is needed to stretch the second wire by the same amount Δ l?
Correct answer: C. Same volume: A L₁ = 3A L₂ ⇒ L₂ = L₁/3. Extension Δ l = (FL)/(AY). For wire 1: Δ l = (F L₁)/(AY). For wire 2: Δ l = (F₂ L₂)/(3A Y) = (F₂ (L₁/3))/(3AY) = (F₂ L₁)/(9AY). Equating the two gives F₂ = 9F.
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Q5NEET 2017
The bulk modulus of a spherical object is B. If it is subjected to uniform pressure p, the fractional decrease in its radius (Δ R)/(R) is:
Correct answer: D. Bulk modulus B = (p)/(Δ V/V), so (Δ V)/(V) = (p)/(B). For a sphere V ∝ R³, hence (Δ V)/(V) = 3(Δ R)/(R). Therefore 3(Δ R)/(R) = (p)/(B), giving (Δ R)/(R) = (p)/(3B).
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Q6NEET 2014
Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δ l. Which of the following is the correct proportionality?
Correct answer: B. Volume fixed: V = A l ⇒ A = V/l. From Y = (Fl)/(A Δ l), Δ l = (Fl)/(AY) = (Fl)/((V/l)Y) = (F l²)/(VY). With F, V, Y constant, Δ l ∝ l².
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Q7NEET 2013
Four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
Correct answer: A. Δ L = (FL)/(AY) = (FL)/((π d²/4)Y) ∝ (L)/(d²). Compute L/d²: (A) 50/0.5² = 200; (B) 100/1² = 100; (C) 200/2² = 50; (D) 300/3² = 33.3. The largest is (A).
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Frequently Asked Questions

What is the difference between stress and strain?

Stress is the internal restoring force developed per unit area of a deformed solid, measured in pascals (N per square metre), so it is the cause. Strain is the fractional deformation, such as change in length divided by original length, and is a pure ratio with no units, so it is the effect.

What is Young’s modulus and what is its formula?

Young’s modulus is the ratio of tensile (longitudinal) stress to longitudinal strain and measures how much a material resists stretching or compression. Its formula is Y equals (F by A) divided by (delta L by L), which equals F L by (A delta L); a larger Y means a stiffer material that stretches less, with steel around 2 times 10 to the power 11 pascals.

What is the difference between Young’s modulus, bulk modulus and shear modulus?

Young’s modulus governs stretching or compression along a length, bulk modulus governs all-round volume change under pressure, and shear (rigidity) modulus governs change of shape under tangential stress. Solids have all three, whereas liquids and gases have only a bulk modulus because they flow instead of resisting shear.

What is Poisson’s ratio and what is its range?

Poisson’s ratio is the ratio of lateral strain to longitudinal strain, that is minus (delta d by d) divided by (delta L by L), when a body is stretched. It is dimensionless, theoretically lies between minus 1 and 0.5, and for most metals is about 0.2 to 0.4.

Is Mechanical Properties of Solids important for NEET and how do I score in it?

Yes, it is part of the NEET physics syllabus and usually contributes about one question, but it is high yield because the concepts are clear and formula based. Focus on the three moduli, Hooke’s law within the elastic limit, the stress-strain curve, and elastic potential energy U equals half times stress times strain times volume, then practise numericals.

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