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Class 11 Physics · NCERT Chapter 8

Mechanical Properties of Solids Class 11 Physics Notes

Complete, exam-ready notes on the mechanical properties of solids: elasticity, stress and strain, Hooke's law and the stress–strain curve, Young's modulus, shear and bulk moduli, Poisson's ratio, elastic potential energy and the practical applications of elastic behaviour — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

ChapterMechanical Properties of SolidsClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

State Hooke's law and define Young's modulus.

Within the elastic limit, stress is directly proportional to strain: stress ∝ strain. Young's modulus is the ratio of longitudinal stress (F/A) to longitudinal strain (ΔL/L): Y = (F/A)/(ΔL/L) = FL/(AΔL). It measures a material's stiffness under stretching.

Elasticity, Stress and Strain

Elasticity

Elasticity is the property of a body by which it regains its original shape and size when the deforming force is removed. A body that completely regains its original configuration is perfectly elastic; one that retains the deformation is plastic.

Stress

Stress is the internal restoring force acting per unit area of a cross-section, developed in a deformed body. Its SI unit is N m⁻² (pascal). Longitudinal stress = F/A, shearing stress = tangential force/area, and volumetric stress = F/A when the force is applied uniformly to compress the volume.

Strain

Strain is the fractional change in dimensions of a body due to stress. It has no unit. Longitudinal strain = ΔL/L, shearing strain = angular deformation θ (in radians), and volumetric strain = ΔV/V.

  • Perfect elasticity is an ideal; real materials retain some permanent deformation (plasticity).
  • A wire under tension stretches longitudinally and thins transversely.
  • Rubber can be stretched immensely but is not very strong; steel is stiff and strong.

Units to remember

Stress is measured in N m⁻² (pascal). Strain is dimensionless. Both are fundamental to comparing the behaviour of wires, rods and columns.

MCQ Questions

1The property by which a body regains its original shape on removal of the deforming force is called —

2The SI unit of stress is —

3Longitudinal strain is defined as —

4Which of the following is a dimensionless quantity? —

1 Mark Questions

  1. Define elasticity.
  2. Define stress and state its SI unit.
  3. Define longitudinal strain.

2 Mark Questions

  1. Distinguish between stress and strain.
  2. Define longitudinal, shearing and volumetric strain with their expressions.
  3. A wire of length 2 m stretches by 2 mm under load. Find the longitudinal strain.

3 Mark Questions

  1. Define elasticity, stress and strain, and explain how a stretched wire regains its length when the load is removed.
  2. A steel wire of diameter 1 mm and length 2 m is stretched by a force of 50 N. If Young's modulus of steel is 2 × 10¹¹ N m⁻², find the increase in its length.

Hooke's Law and the Stress–Strain Curve

Hooke's law

Within the elastic limit, the stress produced in a body is directly proportional to the strain: stress ∝ strain, or stress = constant × strain. The constant of proportionality is the modulus of elasticity of the material.

Stress=E⋅strain,E=stressstrain\text{Stress} = E \cdot \text{strain}, \qquad E = \frac{\text{stress}}{\text{strain}}
Hooke's law and the modulus of elasticity

The stress–strain curve for a ductile material shows a linear elastic region up to the proportional limit, followed by yielding, plastic flow and finally fracture. The slope of the straight-line portion is Young's modulus. The yield point marks where permanent (plastic) deformation begins.

  • Proportional limit: where stress and strain stop being proportional.
  • Elastic limit: the greatest stress recoverable without permanent strain.
  • Ductile materials (copper, steel) deform greatly before breaking; brittle materials (glass, cast iron) fracture with little strain.

Elastic limit vs breaking point

Do not confuse the elastic limit (recoverable) with the breaking/fracture point (where the material fails). A wire stays elastic only below its elastic limit; beyond it deforms plastically.

MCQ Questions

1Hooke's law states that, within the elastic limit, —

2The ratio of stress to strain is called the —

3The material that fractures with very little plastic deformation is called —

4The slope of the linear portion of the stress–strain curve gives —

1 Mark Questions

  1. State Hooke's law.
  2. Define the modulus of elasticity.
  3. Distinguish between a ductile and a brittle material.

2 Mark Questions

  1. Explain the stress–strain curve for a ductile material and identify the elastic limit and yield point.
  2. A material obeys Hooke's law up to a stress of 2 × 10⁸ N m⁻². Which region of the curve corresponds to this?
  3. Give one example each of a ductile and a brittle material.

3 Mark Questions

  1. Draw and explain the stress–strain curve for a metallic wire, marking the proportional limit, elastic limit, yield point and breaking point.
  2. A wire of length 4 m and area 10⁻⁶ m² stretches by 1 mm under a load of 20 kg. Assuming Hooke's law applies, find the stress and strain, and the Young's modulus of the wire.

The Three Moduli of Elasticity

Young's modulus

Young's modulus Y quantifies a material's resistance to stretching (longitudinal stress divided by longitudinal strain): Y = (F/A)/(ΔL/L) = FL/(AΔL). It is measured in N m⁻². Steel has a very high Young's modulus, meaning it is stiff.

Y=F/AΔL/L=FLAΔLY = \frac{F/A}{\Delta L / L} = \frac{FL}{A\Delta L}
Young's modulus

Shear (rigidity) modulus

Shear modulus G measures resistance to shearing (tangential stress divided by shearing strain): G = (F/A)/θ, where θ is the angular deformation in radians. It applies when a force acts parallel to a surface, deforming the shape but not the volume.

Bulk modulus

Bulk modulus B measures resistance to volume change under uniform pressure: B = −ΔP/(ΔV/V). The negative sign indicates that an increase in pressure decreases volume. The reciprocal of the bulk modulus is the compressibility.

G=F/Aθ,B=−ΔPΔV/VG = \frac{F/A}{\theta}, \qquad B = -\frac{\Delta P}{\Delta V / V}
Shear and bulk moduli
  • Bulk modulus of solids > liquids > gases (gases are most compressible).
  • Compressibility = 1/B.
  • For an increase in pressure the volume always decreases, so ΔV is negative and B is positive.

Solids are nearly incompressible

A solid has a very large bulk modulus — applying pressure scarcely changes its volume. This is why solids are described as practically incompressible compared with gases.

MCQ Questions

1Young's modulus is the ratio of —

2The bulk modulus of a material is the ratio of —

3The compressibility of a material is the reciprocal of its —

4Among solids, liquids and gases, the highest bulk modulus is possessed by —

1 Mark Questions

  1. Define Young's modulus.
  2. Define the bulk modulus of a material.
  3. What is compressibility?

2 Mark Questions

  1. A wire of length 2 m and area 10⁻⁶ m² is stretched by 1 mm under a load of 5 kg. Find its Young's modulus. (g = 10 m/s².)
  2. Distinguish between rigidity modulus and bulk modulus.
  3. Why do gases have a much smaller bulk modulus than solids?

3 Mark Questions

  1. Define the three elastic moduli — Young's, shearing and bulk — and give an expression and an application for each.
  2. A uniform pressure of 2 × 10⁶ N m⁻² is applied to a solid of bulk modulus 10¹⁰ N m⁻². Find the fractional change in its volume.

Poisson's Ratio

Poisson's ratio

When a wire is stretched longitudinally it also contracts transversely. Poisson's ratio σ is the ratio of the lateral (transverse) strain to the longitudinal strain: σ = (Δd/d)/(ΔL/L). For most materials it lies between 0 and 0.5.

σ=−Δd/dΔL/L\sigma = \frac{-\Delta d / d}{\Delta L / L}
Poisson's ratio

The lateral strain is opposite in sign to the longitudinal strain. Values are generally small (about 0.3 for steel, 0.5 for rubber which is nearly incompressible). Poisson's ratio relates the three moduli, so it connects compression behaviour with shear and stretching response.

  • A wire stretched becomes thinner — this is the physical effect of lateral contraction.
  • σ = 0 means no lateral contraction; σ = 0.5 means the volume is conserved.
  • Poisson's ratio has no unit.

Rubber is nearly incompressible

Rubber has a Poisson's ratio close to 0.5, so its volume stays almost constant when stretched — it thins a lot. Steel at ~0.3 changes volume more noticeably.

MCQ Questions

1Poisson's ratio is the ratio of —

2For a material that is nearly incompressible, Poisson's ratio is close to —

3Poisson's ratio has —

4When a wire is stretched, its diameter —

1 Mark Questions

  1. Define Poisson's ratio.
  2. State the typical range of Poisson's ratio for most materials.
  3. What is the physical effect of lateral contraction in a stretched wire?

2 Mark Questions

  1. A wire of diameter 1 mm is stretched so that its longitudinal strain is 2 × 10⁻³. If Poisson's ratio is 0.3, find the change in its diameter.
  2. Explain why rubber has a Poisson's ratio near 0.5 while steel is near 0.3.
  3. Define lateral and longitudinal strain, and write the expression for Poisson's ratio.

3 Mark Questions

  1. Define Poisson's ratio and show that it is dimensionless. State the significance of the two limiting values 0 and 0.5.
  2. A wire of length 3 m and diameter 1 mm is stretched by 3 mm. If Poisson's ratio is 0.25, find the decrease in its diameter.

Elastic Potential Energy and Applications

Elastic potential energy

When a body is deformed elastically, the work done in deforming it is stored as elastic potential energy. The energy stored per unit volume of a stretched wire is U = ½ × stress × strain = ½Y(strain)².

U=12×stress×strain=12Y(ΔLL)2U = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2}Y\left(\frac{\Delta L}{L}\right)^2
Elastic potential energy per unit volume

Elastic behaviour has countless practical applications. Beams and girders are designed so their maximum stress remains safely below the breaking point. Cranes lift loads using steel cables chosen for high Young's modulus and high breaking strength. The flexural rigidity of a beam is used to support roofs and bridges without excessive bending.

  • A spring stores energy ½kx² — a special case of elastic energy.
  • Ropes, belts and tendons work on elastic tension.
  • Structural designers use a safety factor (breaking stress / working stress) typically 5–10.

Design for safety

Engineers choose a working stress well below the breaking stress, divided by a factor of safety. This protects against overloads, fatigue and material defects.

MCQ Questions

1The elastic potential energy stored per unit volume in a stretched wire is —

2The elastic energy stored in a stretched spring of spring constant k and extension x is —

3The ratio of breaking stress to working stress is called the —

4Ropes, belts and crane cables primarily rely on the material's —

1 Mark Questions

  1. Write the expression for the elastic potential energy per unit volume of a stretched wire.
  2. Define the factor of safety in structural design.
  3. Give one practical application of elastic behaviour.

2 Mark Questions

  1. A steel wire of Young's modulus 2 × 10¹¹ N m⁻² is stretched to a strain of 10⁻³. Find the energy stored per unit volume.
  2. Explain why crane cables are made of steel with a high Young's modulus rather than a low one.
  3. A spring of spring constant 200 N m⁻¹ is stretched by 5 cm. Find the elastic energy stored in it.

3 Mark Questions

  1. Derive an expression for the elastic potential energy stored per unit volume in a stretched wire in terms of Young's modulus and strain.
  2. Discuss the practical applications of elastic behaviour in the design of beams, columns and crane cables, and explain the role of the factor of safety.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Stress

stress=FA\text{stress} = \frac{F}{A}

Strain

strain=ΔLL\text{strain} = \frac{\Delta L}{L}

Young's modulus

Y=FLAΔLY = \frac{FL}{A\Delta L}

Shear modulus

G=F/AθG = \frac{F/A}{\theta}

Bulk modulus

B=−ΔPΔV/VB = -\frac{\Delta P}{\Delta V/V}

Poisson's ratio

σ=Δd/dΔL/L\sigma = \frac{\Delta d/d}{\Delta L/L}

Elastic energy/volume

U=12×stress×strainU = \frac{1}{2}\times\text{stress}\times\text{strain}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Within the elastic limit stress ∝ strain (Hooke's law).
  • Y = FL/(AΔL); G = (F/A)/θ; B = −ΔP/(ΔV/V).
  • Bulk modulus: solids > liquids > gases.
  • Compressibility = 1/B.
  • Poisson's ratio σ = lateral strain / longitudinal strain; typical 0–0.5.
  • Elastic energy per unit volume = ½ × stress × strain = ½Y(strain)².
  • Steel is preferred in cables for its high Y and breaking strength.

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Young's modulus of a wire

A wire of length 2 m and area of cross-section 2 × 10⁻⁶ m² is stretched by 1 mm by a load of 10 kg. Find Young's modulus of the wire. (g = 10 m/s².)

  1. F = mg = 10 × 10 = 100 N; ΔL = 10⁻³ m; L = 2 m; A = 2 × 10⁻⁶ m².
  2. Y = FL/(AΔL) = 100 × 2 ÷ (2 × 10⁻⁶ × 10⁻³).
  3. Y = 200 ÷ (2 × 10⁻⁹) = 10¹¹ N m⁻².

Answer

1 × 10¹¹ N m⁻²

2

Elastic energy stored

A steel wire of Young's modulus 2 × 10¹¹ N m⁻² is stretched to a strain of 2 × 10⁻³. Find the elastic energy stored per unit volume.

  1. U = ½Y(strain)² = ½ × 2 × 10¹¹ × (2 × 10⁻³)².
  2. U = ½ × 2 × 10¹¹ × 4 × 10⁻⁶ = 4 × 10⁵ J m⁻³.

Answer

4 × 10⁵ J m⁻³

3

Effect of pressure on a solid

A uniform pressure of 5 × 10⁶ N m⁻² is applied to a solid of bulk modulus 10¹⁰ N m⁻². Find the fractional change in volume.

  1. B = −ΔP/(ΔV/V).
  2. ΔV/V = −ΔP/B = −5 × 10⁶ / 10¹⁰ = −5 × 10⁻⁴.
  3. The magnitude of the fractional change is 5 × 10⁻⁴.

Answer

5 × 10⁻⁴ (volume decreases)

FAQ

Common questions

What is the difference between stress and strain?

Stress is the internal restoring force per unit area (F/A, in N m⁻²) developed in a deformed body. Strain is the fractional change in dimensions (ΔL/L, ΔV/V) caused by the stress. Stress is the cause, strain is the effect, and they are proportional within the elastic limit.

Why is steel preferred to rubber for supporting heavy loads?

Steel has a very high Young's modulus (≈ 2 × 10¹¹ N m⁻²) and high breaking strength, so it stretches little under load and can support large forces. Rubber has a low modulus and stretches greatly, so it is not used for structural support.

What is the significance of the three elastic moduli?

Young's modulus measures resistance to stretching, the shear (rigidity) modulus measures resistance to change of shape, and the bulk modulus measures resistance to change of volume. Together they fully describe a material's elastic response.

Why are solids almost incompressible?

Solids have a very large bulk modulus, meaning a huge pressure produces only a tiny fractional change in volume. The strong intermolecular forces resist any change in the spacing of the constituent atoms.

How is elastic energy stored and released?

When a body is deformed elastically, work done is stored as elastic potential energy (per unit volume U = ½ × stress × strain). On removal of the load the body returns to its original shape, releasing this energy — the basis of springs and trampolines.

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