Class 11 Physics · NCERT Chapter 8

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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.

Chapter:Mechanical Properties of SolidsClass:Class 11Subject:PhysicsBoard:CBSEExams:CBSE · JEE · NEET

7 Key Formulas3 Solved Examples40 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

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

Definition

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.

Definition

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.

Definition

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

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

The SI unit of stress is —

Longitudinal strain is defined as —

Which 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

Definition

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.

Hooke's law and the modulus of elasticity

Stress = E \cdot strain, E = \frac{stress}{strain}

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

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

The ratio of stress to strain is called the —

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

The 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

Definition

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.

Young's modulus

Y = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}

Definition

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.

Definition

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.

Shear and bulk moduli

G = \frac{F / A}{θ}, B = - \frac{Δ P}{Δ V / V}

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

Young's modulus is the ratio of —

The bulk modulus of a material is the ratio of —

The compressibility of a material is the reciprocal of its —

Among 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

Definition

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.

Poisson's ratio

σ = \frac{- Δ d / d}{Δ L / L}

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

Poisson's ratio is the ratio of —

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

Poisson's ratio has —

When 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

Definition

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.

Elastic potential energy per unit volume

U = \frac{1}{2} \times stress \times strain = \frac{1}{2} Y (\frac{Δ L}{L})^{2}

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

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

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

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

Ropes, 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.

NCERT Exercise Questions

From the textbook

All NCERT questions for this chapter — NCERT Physics Part I, Chapter 8 (Mechanical Properties of Solids). Questions keep their NCERT numbering so you can match them to the book.

Exercises (NCERT Q8.1–Q8.16)

8.1) A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young's modulus of steel to that of copper?
8.2) Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
8.3) The stress-strain graphs for materials A and B are shown in Fig. 8.10. The graphs are drawn to the same scale. (a) Which of the materials has the greater Young's modulus? (b) Which of the two is the stronger material?
8.4) Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young's modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.
8.5) Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
8.6) The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
8.7) Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
8.8) A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
8.9) A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N m⁻², what is the maximum load the cable can support?
8.10) A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
8.11) A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
8.12) Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
8.13) What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m⁻³?
8.14) Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
8.15) Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.
8.16) How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answers at a glance

1
Q8.1 → Ratio Y_steel : Y_copper = 1.79 : 1.
2
Q8.2 → (a) Y = 7.5 × 10¹⁰ N m⁻² (b) approximate yield strength = 3 × 10⁸ N m⁻².
3
Q8.3 → (a) Material A (b) material A.
4
Q8.4 → (a) False (b) true.
5
Q8.5 → Steel wire elongates 1.49 × 10⁻⁴ m; brass wire 1.30 × 10⁻⁴ m.
6
Q8.6 → Vertical deflection = 3.92 × 10⁻⁷ m.
7
Q8.7 → Compressional strain of each column = 7.22 × 10⁻⁷.
8
Q8.8 → Strain ≈ 3.65 × 10⁻³.
9
Q8.9 → Maximum load ≈ 7.07 × 10⁴ N.
10
Q8.10 → d_copper : d_iron = 1.31 : 1.
11
Q8.11 → Elongation ≈ 1.54 × 10⁻⁴ m (using the textbook's ω = 2 rev/s convention).
12
Q8.12 → B_water = 2.026 × 10⁹ Pa; B_water/B_air ≈ 2.0 × 10⁴ — air is far more compressible.
13
Q8.13 → Density at 80 atm ≈ 1.034 × 10³ kg m⁻³.
14
Q8.14 → Fractional change in volume = 2.73 × 10⁻⁵.
15
Q8.15 → Volume contraction = 5 × 10⁻⁸ m³ (0.05 cm³).
16
Q8.16 → Pressure change required = 2.2 × 10⁶ N m⁻².

CBSE Pattern Questions

How CBSE asks this chapter

Physics board paper structure: Section A (MCQ/VSA 1M), Section B (2M), Section C (3M), Section D (5M). Mechanical Properties of Solids typically carries ~6 marks: 1-mark MCQs on stress and strain definitions, a 2-mark numerical on Young's modulus, a 3-mark question on the stress–strain curve or Poisson's ratio, and a 5-mark question on elastic potential energy or a practical application of elasticity.

MCQ Questions

Hooke's law states that, within the elastic limit, stress is

The modulus of rigidity relates shearing stress to

In Young's modulus Y = stress/strain, the strain for a wire is computed as

The elastic potential energy stored per unit volume of a stretched wire is

1 Mark Questions

1
Define stress and give its SI unit.
2
Define strain. Why is strain dimensionless?
3
State Hooke's law.
4
Define Poisson's ratio.

2 Mark Questions

1
A steel wire of length 2 m and cross-section 1 mm² is stretched by a force of 200 N. If Young's modulus of steel is 2 × 10¹¹ N/m², find the extension. (Ans: Δl = Fl/AY = 2 × 10⁻³ m)
2
A wire of radius 0.5 mm under a load of 5 kg shows an extension of 1 mm over 2 m. Find Young's modulus (g = 10 m/s²). (Ans: Y = 6.4 × 10¹¹ N/m²)
3
A cube of side 10 cm is subjected to a shear of 0.01 mm by a tangential force of 10⁴ N. Find the modulus of rigidity. (Ans: η = 10¹⁰ N/m²)

3 Mark Questions

1
Draw and explain the stress–strain curve for a ductile material, labelling proportionality limit, elastic limit, yield point and fracture point.
2
Derive the expression for the elastic potential energy stored per unit volume of a stretched wire. (Ans: u = ½ × stress × strain)
3
Describe an experiment to determine Young's modulus of a wire using Searle's method.

5 Mark Questions

1
Derive the expression for the elastic potential energy of a stretched wire and explain how it depends on stress and strain. Hence find the energy stored in a steel wire of given dimensions.
2
Explain Poisson's ratio, and derive the relation between Young's modulus, bulk modulus and modulus of rigidity for an isotropic material.

PDF: Assignments & Tests

Printable study material

Free downloadable PDFs for Mechanical Properties of Solids — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

Free PDF Downloads & Assignments

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Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.

Mechanical Properties of Solids Assignment 1 — Topic-wise Questions

Covers all five topics: elasticity, stress and strain; Hooke's law and the stress–strain curve; the three moduli of elasticity; Poisson's ratio; and elastic potential energy and applications. Arranged topic by topic.

mechanical-properties-of-solids-class-11-notes-assignment-1.pdf

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Mechanical Properties of Solids Assignment 2 — Challenge + Integrated Problems

Higher-difficulty set with multi-topic integration problems (stress and strain, Young's modulus, bulk and rigidity modulus, Poisson's ratio, stored energy) and full worked answers. Includes an answer key.

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Mechanical Properties of Solids Test 1 — CBSE Pattern (90 min, 40 marks)

Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.

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Mechanical Properties of Solids Test 2 — CBSE Pattern (90 min, 40 marks)

Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.

mechanical-properties-of-solids-class-11-notes-test-2-cbse.pdf

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Stress

stress = \frac{F}{A}

Strain

strain = \frac{Δ L}{L}

Young's modulus

Y = \frac{F L}{A Δ L}

Shear modulus

G = \frac{F / A}{θ}

Bulk modulus

B = - \frac{Δ P}{Δ V / V}

Poisson's ratio

σ = \frac{Δ d / d}{Δ L / L}

Elastic energy/volume

U = \frac{1}{2} \times stress \times strain

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

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.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Young's modulus of a wire

Problem StatementA 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.)

Step-by-Step Solution

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

Final Answer

1 × 10¹¹ N m⁻²

Elastic energy stored

Problem StatementA 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.

Step-by-Step Solution

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

Final Answer

4 × 10⁵ J m⁻³

Effect of pressure on a solid

Problem StatementA 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.

Step-by-Step Solution

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

Final Answer

5 × 10⁻⁴ (volume decreases)

FAQ

Frequently asked 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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