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CLASS 11 · PHYSICS · CHAPTER NOTES + WORKSHEET

Mechanical Properties of Solids

Understand elasticity, stress, strain and elastic moduli, then practise exam-style numericals. These are original StudyNest learning notes; check your school syllabus and teacher's conventions.

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1. Core concepts

Elasticity and plasticity

Elasticity is the ability of a material to regain its original shape and size after the deforming force is removed, provided the elastic limit is not exceeded. Plastic deformation remains after the force is removed.

Stress

Stress is restoring force per unit area. Its SI unit is pascal (Pa), equivalent to N m⁻².

Stress = Force / Area   |   σ = F/A

Strain

Strain is the ratio of change in dimension to the original dimension. It is dimensionless.

Longitudinal strain = ΔL/L

Hooke's law

Within the proportional limit, stress is directly proportional to strain.

Stress ∝ Strain

2. Elastic moduli

Young's modulus (Y)

For a wire or rod under longitudinal loading, Young's modulus is longitudinal stress divided by longitudinal strain.

Y = (F/A)/(ΔL/L) = FL/(AΔL)

Bulk modulus (B)

Bulk modulus measures resistance to uniform volume change under pressure. The minus sign indicates that volume decreases when pressure increases.

B = −ΔP / (ΔV/V)

Shear modulus (G)

Shear modulus is shear stress divided by shear strain.

G = (F/A) / (x/L)

For an isotropic material, the elastic moduli are connected by material-dependent relations. Use the relation supplied in your syllabus when a question requires it.

3. Worked examples

Example 1 — Stress

A force of 120 N acts normally on an area of 3 × 10⁻⁴ m². Find the stress.

Stress = F/A = 120/(3 × 10⁻⁴) = 4 × 10⁵ Pa.

Example 2 — Extension

A wire has length 2 m, area 1 × 10⁻⁶ m² and Young's modulus 2 × 10¹¹ Pa. A force of 50 N is applied. Find extension.

ΔL = FL/(AY) = (50 × 2)/[(1 × 10⁻⁶)(2 × 10¹¹)] = 5 × 10⁻⁴ m = 0.5 mm.

4. Numerical practice worksheet

Total: 30 marks · Show formula, substitution, units and final answer. Take π only if needed; assume ideal uniform materials.

Q1 (2 marks). A force of 80 N acts on an area of 4 × 10⁻⁴ m². Calculate the stress.
Q2 (2 marks). A wire of original length 2.5 m extends by 1.0 mm. Find the longitudinal strain.
Q3 (2 marks). A 150 N force acts on a rod of cross-sectional area 5 × 10⁻⁵ m². Find the longitudinal stress.
Q4 (3 marks). A wire 2 m long and of area 1.5 × 10⁻⁶ m² extends by 1 mm under a force of 60 N. Calculate Young's modulus.
Q5 (3 marks). A wire has Young's modulus 2 × 10¹¹ Pa, length 1.5 m and area 2 × 10⁻⁶ m². Find its extension under a 40 N load.
Q6 (3 marks). Pressure on a solid increases by 5 × 10⁶ Pa and its volume decreases by 2 × 10⁻⁴ of its original volume. Calculate the bulk modulus.
Q7 (3 marks). A wire 4 m long and area 2 × 10⁻⁶ m² is pulled by 100 N. If its extension is 1 mm, find Young's modulus.
Q8 (4 marks). A metal wire of length 3 m and diameter 2 mm is stretched by 0.6 mm under a force of 50 N. Calculate its stress, strain and Young's modulus. Use π = 3.14.
Q9 (4 marks). A cube of volume 1 × 10⁻³ m³ experiences a pressure increase of 2 × 10⁷ Pa. If its bulk modulus is 1 × 10¹⁰ Pa, calculate the change in volume and state whether volume increases or decreases.
Q10 (4 marks). A wire of length 2 m and area 1 × 10⁻⁶ m² is stretched by a 25 N force. If Y = 1 × 10¹¹ Pa, calculate extension and strain.

5. Answer check

Try every question before opening this section.

Show concise answers
  1. 2 × 10⁵ Pa
  2. 4 × 10⁻⁴
  3. 3 × 10⁶ Pa
  4. 8 × 10¹⁰ Pa
  5. 1.5 × 10⁻⁴ m (0.15 mm)
  6. 2.5 × 10¹⁰ Pa
  7. 2 × 10¹¹ Pa
  8. Area = 3.14 × 10⁻⁶ m²; stress ≈ 1.59 × 10⁷ Pa; strain = 2 × 10⁻⁴; Y ≈ 7.96 × 10¹⁰ Pa
  9. ΔV = −2 × 10⁻⁶ m³; volume decreases
  10. Extension = 5 × 10⁻⁴ m; strain = 2.5 × 10⁻⁴

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