Chapter 9: Mechanical Properties of Solids

Introduction

Key Points

• Solid bodies have definite shape and size but are not perfectly rigid.
• Elasticity: Property by which a body regains original shape and size after removal of force.
• Elastic substances: Materials that regain original shape after deformation.
• Plasticity: Inability to regain original shape; plastic substances like plastic, bakelite.

Solved Examples

Ex 1: Calculate max height of mountain if ρ = 3 × 10³ kg/m³, g = 10 m/s², elastic limit = 3 × 10⁸ N/m².

• Use h < elastic limit/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A. 1 × 10⁸ N/m²
  B. 3 × 10⁸ N/m²
  C. 5 × 10⁸ N/m²
  D. 7 × 10⁸ N/m²
• Q2: The pressure at the base of a mountain is given by?
  A. hρg
  B. ρg / h
  C. h / ρg
  D. ρ / h g
• Q3: If the density of rock is 4 × 10³ kg/m³ and elastic limit is 4 × 10⁸ N/m², what is max height of mountain?
  A. 5 km
  B. 10 km
  C. 8 km
  D. 12 km

Answer Key

• Q1: B - Elastic limit of typical rock is 3 × 10⁸ N/m².
• Q2: A - Pressure at base is hρg (height × density × gravity).
• Q3: A - Max height = 4 × 10⁸/(4 × 10³ × 10) = 10⁴ / 4 = 2.5 × 10³ m = 2.5 km (closest option 5 km).

Memory Aids

• Max height h = Elastic limit/(ρg).
• Stress = pressure at base = hρg.
• Elastic limit prevents rock flow.

Elastic Behaviour of Solids

Key Points

• Particles in solids are closely packed with stable interatomic distances.
• Deforming force changes these distances; removal restores original shape due to interatomic forces.
• Spring-ball model: balls = atoms, springs = interatomic forces.

Solved Examples

Ex 1: Explain why a stretched solid returns to original shape.
Solution: Interatomic forces act like springs restoring original distances.

Practice Set

• Q1: What represents interatomic forces in the spring-ball model?
  A. Balls
  B. Springs
  C. Both
  D. None
• Q2: What happens to interatomic distances when a deforming force is applied?
  A. Remain same
  B. Change
  C. Increase only
  D. Decrease only
• Q3: Why do solids regain shape after deformation?
  A. Plasticity
  B. Interatomic restoring forces
  C. External force
  D. None

Answer Key

• Q1: B - Springs represent interatomic forces.
• Q2: B - Interatomic distances change under force.
• Q3: B - Restoring interatomic forces bring back original shape.

Memory Aids

• Spring-ball model: springs = forces, balls = atoms.
• Deformation changes atomic distances; restoring forces reverse it.

Stress and Strain

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: σ = F/A
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: ε_l = ΔL/L
  • Tangential (shearing) strain: ε_T = Δx/L = tan θ
  • Volume strain: ε_v = ΔV/V

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: ε_l = 0.01/2 = 0.005

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. ΔV/V
  B. ΔL/L
  C. Δx/L
  D. Force per unit area
• Q3: Shearing strain is:
  A. ΔL/L
  B. ΔV/V
  C. Δx/L = tan θ
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (σ = F/A)
• Longitudinal strain = ΔL / L
• Shearing strain = Δx / L = tan θ
• Volume strain = ΔV / V

Stress

• Stress = Restoring force per unit area when deforming force applied.
• Stress is a tensor quantity.
• Formula: \( \sigma = \frac{F}{A} \)
• SI unit: N/m² (Pascal), Dimensions: [ML^-1T^-2]
• Types of stress:
  • Longitudinal stress: along length
  • Tensile stress: stretching force normal to cross-section
  • Compressive stress: compressing force normal to cross-section
  • Tangential (shearing) stress: force parallel to cross-section

Strain

• Strain = Relative change in configuration due to deforming force.
• Types of strain:
  • Longitudinal strain: \( \varepsilon_l = \frac{\Delta L}{L} \)
  • Tangential (shearing) strain: \( \varepsilon_T = \frac{\Delta x}{L} = \tan \theta \)
  • Volume strain: \( \varepsilon_v = \frac{\Delta V}{V} \)

Solved Examples

Ex 1: Calculate longitudinal strain if a rod of length 2 m elongates by 0.01 m.
Solution: \( \varepsilon_l = \frac{0.01}{2} = 0.005 \)

Practice Set

• Q1: Stress is defined as:
  A. Force per unit volume
  B. Force per unit area
  C. Change in length per unit length
  D. Change in volume per unit volume
• Q2: Longitudinal strain is:
  A. \( \frac{\Delta V}{V} \)
  B. \( \frac{\Delta L}{L} \)
  C. \( \frac{\Delta x}{L} \)
  D. Force per unit area
• Q3: Shearing strain is:
  A. \( \frac{\Delta L}{L} \)
  B. \( \frac{\Delta V}{V} \)
  C. \( \frac{\Delta x}{L} = \tan \theta \)
  D. Force per unit area

Answer Key

• Q1: B - Stress = force per unit area.
• Q2: B - Longitudinal strain = change in length/original length.
• Q3: C - Shearing strain = relative displacement/separation = tan θ.

Memory Aids

• Stress = Force/Area (\( \sigma = F/A \))
• Longitudinal strain = \( \Delta L / L \)
• Shearing strain = \( \Delta x / L = \tan \theta \)
• Volume strain = \( \Delta V / V \)

Hooke’s Law

Concept and Formula

• Discovered by Robert Hooke in 1676.
• For small deformations, stress is directly proportional to strain.
• Mathematically: Stress ∝ Strain or Stress = K × Strain.
• Modulus of elasticity K = Stress/Strain is a constant for a material.
• SI unit of K is Pascal (N/m²), dimension [ML^-1T^-2].

Solved Examples

• Ex 1: A wire experiences a stress of 200 N/m² and strain of 0.001. Find modulus of elasticity.
  Solution: K = Stress/Strain = 200/0.001 = 2 × 10⁵ N/m².

Practice Set

• Q1: Hooke’s Law states that stress is proportional to:
  A) Force
  B) Strain
  C) Displacement
  D) Energy
• Q2: If the strain in a material is doubled, the stress will:
  A) Remain same
  B) Double
  C) Halve
  D) Become zero
• Q3: A wire has a modulus of elasticity of 2 × 10¹¹ N/m². If the strain is 5 × 10^-4, what is the stress?
  A) 1 × 10⁸ N/m²
  B) 1 × 10⁷ N/m²
  C) 1 × 10⁶ N/m²
  D) 1 × 10⁵ N/m²

Answer Key

• Q1: B - Stress is directly proportional to strain as per Hooke’s Law.
• Q2: B - Stress doubles if strain doubles due to direct proportionality.
• Q3: A - Stress = Modulus × Strain = 2 × 10¹¹ × 5 × 10^-4 = 1 × 10⁸ N/m².

Memory Aids

• Stress ∝ Strain (Hooke’s Law)
• Modulus of Elasticity = Stress / Strain
• SI unit of modulus = Pascal (N/m²)

Concept and Formula

• Discovered by Robert Hooke in 1676.
• For small deformations, stress is directly proportional to strain.
• Mathematically: Stress ∝ Strain or Stress = K × Strain.
• Modulus of elasticity K = Stress/Strain is a constant for a material.
• SI unit of K is Pascal (N/m²), dimension [ML^-1T^-2].

Solved Examples

• Ex 1: A wire experiences a stress of 200 N/m² and strain of 0.001. Find modulus of elasticity.
  Solution: K = Stress/Strain = 200/0.001 = 2 × 10⁵ N/m².

Practice Set

• Q1: Hooke’s Law states that stress is proportional to:
  A) Force
  B) Strain
  C) Displacement
  D) Energy
• Q2: If the strain in a material is doubled, the stress will:
  A) Remain same
  B) Double
  C) Halve
  D) Become zero
• Q3: A wire has a modulus of elasticity of 2 × 10¹¹ N/m². If the strain is 5 × 10^-4, what is the stress?
  A) 1 × 10⁸ N/m²
  B) 1 × 10⁷ N/m²
  C) 1 × 10⁶ N/m²
  D) 1 × 10⁵ N/m²

Answer Key

• Q1: B - Stress is directly proportional to strain as per Hooke’s Law.
• Q2: B - Stress doubles if strain doubles due to direct proportionality.
• Q3: A - Stress = Modulus × Strain = 2 × 10¹¹ × 5 × 10^-4 = 1 × 10⁸ N/m².

Memory Aids

• Stress ∝ Strain (Hooke’s Law)
• Modulus of Elasticity = Stress / Strain
• SI unit of modulus = Pascal (N/m²)

Concept and Formula

• Discovered by Robert Hooke in 1676.
• For small deformations, stress is directly proportional to strain.
• Mathematically: Stress ∝ Strain or Stress = K × Strain.
• Modulus of elasticity K = Stress/Strain is a constant for a material.
• SI unit of K is Pascal (N/m²), dimension [ML^-1T^-2].

Solved Examples

• Ex 1: A wire experiences a stress of 200 N/m² and strain of 0.001. Find modulus of elasticity.
  Solution: K = Stress/Strain = 200/0.001 = 2 × 10⁵ N/m².

Practice Set

• Q1: Hooke’s Law states that stress is proportional to:
  A) Force
  B) Strain
  C) Displacement
  D) Energy
• Q2: If the strain in a material is doubled, the stress will:
  A) Remain same
  B) Double
  C) Halve
  D) Become zero
• Q3: A wire has a modulus of elasticity of 2 × 10¹¹ N/m². If the strain is 5 × 10^-4, what is the stress?
  A) 1 × 10⁸ N/m²
  B) 1 × 10⁷ N/m²
  C) 1 × 10⁶ N/m²
  D) 1 × 10⁵ N/m²

Answer Key

• Q1: B - Stress is directly proportional to strain as per Hooke’s Law.
• Q2: B - Stress doubles if strain doubles due to direct proportionality.
• Q3: A - Stress = Modulus × Strain = 2 × 10¹¹ × 5 × 10^-4 = 1 × 10⁸ N/m².

Memory Aids

• Stress ∝ Strain (Hooke’s Law)
• Modulus of Elasticity = Stress / Strain
• SI unit of modulus = Pascal (N/m²)

Concept and Formula

• Discovered by Robert Hooke in 1676.
• For small deformations, stress is directly proportional to strain.
• Mathematically: Stress \( \propto \) Strain or Stress = K × Strain.
• Modulus of elasticity \( K = \frac{Stress}{Strain} \) is a constant for a material.
• SI unit of \( K \) is Pascal (N/m²), dimension \( [ML^{-1}T^{-2}] \).

Solved Examples

• Ex 1: A wire experiences a stress of 200 N/m² and strain of 0.001. Find modulus of elasticity.
  Solution: \( K = \frac{Stress}{Strain} = \frac{200}{0.001} = 2 \times 10^{5} \) N/m².

Practice Set

• Q1: Hooke’s Law states that stress is proportional to:
  A) Force
  B) Strain
  C) Displacement
  D) Energy
• Q2: If the strain in a material is doubled, the stress will:
  A) Remain same
  B) Double
  C) Halve
  D) Become zero
• Q3: A wire has a modulus of elasticity of \( 2 \times 10^{11} \) N/m². If the strain is \( 5 \times 10^{-4} \), what is the stress?
  A) \( 1 \times 10^{8} \) N/m²
  B) \( 1 \times 10^{7} \) N/m²
  C) \( 1 \times 10^{6} \) N/m²
  D) \( 1 \times 10^{5} \) N/m²

Answer Key

• Q1: B - Stress is directly proportional to strain as per Hooke’s Law.
• Q2: B - Stress doubles if strain doubles due to direct proportionality.
• Q3: A - Stress = Modulus × Strain = \( 2 \times 10^{11} \times 5 \times 10^{-4} = 1 \times 10^{8} \) N/m².

Memory Aids

• Stress ∝ Strain (Hooke’s Law)
• Modulus of Elasticity = Stress / Strain
• SI unit of modulus = Pascal (N/m²)

Stress-Strain Curve

Explanation

• Stress-strain curve shows material behavior under tension.
• From O to A: Linear region, obeys Hooke's Law, material returns to original shape.
• From A to B: Non-linear but elastic, material still regains shape.
• Point B: Yield point (elastic limit), beyond which permanent deformation occurs.
• From B to D: Plastic region, strain increases rapidly; D is ultimate tensile strength.
• Point E: Fracture point, material breaks; stress here is breaking stress.
• Larger D-E gap means more ductile; smaller means more brittle.

Solved Examples

• Ex 1: Calculate stress at yield point if force = 1000 N and cross-sectional area = 0.5 mm².
  • Stress = Force / Area = 1000/(0.5 × 10^-6) = 2 × 10⁹ Pa

Practice Set

• Q1: What does point B represent on the stress-strain curve?
  • A) Proportional limit
  • B) Yield point
  • C) Ultimate tensile strength
  • D) Fracture point
• Q2: What happens beyond point B?
  • A) Material returns to original shape
  • B) Permanent deformation occurs
  • C) Stress and strain are proportional
  • D) Material breaks immediately
• Q3: If the distance between D and E is large, the material is:
  • A) Brittle
  • B) Ductile
  • C) Elastic
  • D) Plastic

Answer Key

• Q1: B - Yield point is where permanent deformation starts.
• Q2: B - Beyond B, material deforms permanently.
• Q3: B - Large D-E gap indicates ductility.

Memory Aids

• O-A: Hooke's Law region (linear)
• B: Yield point = start of plastic deformation
• D: Ultimate tensile strength (max stress)
• E: Fracture point (breaking stress)

Explanation

• Stress-strain curve shows material behavior under tension.
• From O to A: Linear region, obeys Hooke's Law, material returns to original shape.
• From A to B: Non-linear but elastic, material still regains shape.
• Point B: Yield point (elastic limit), beyond which permanent deformation occurs.
• From B to D: Plastic region, strain increases rapidly; D is ultimate tensile strength.
• Point E: Fracture point, material breaks; stress here is breaking stress.
• Larger D-E gap means more ductile; smaller means more brittle.

Solved Examples

• Ex 1: Calculate stress at yield point if force = 1000 N and cross-sectional area = 0.5 mm².
  • Stress = Force / Area = 1000/(0.5 × 10^-6) = 2 × 10⁹ Pa

Practice Set

• Q1: What does point B represent on the stress-strain curve?
  • A) Proportional limit
  • B) Yield point
  • C) Ultimate tensile strength
  • D) Fracture point
• Q2: What happens beyond point B?
  • A) Material returns to original shape
  • B) Permanent deformation occurs
  • C) Stress and strain are proportional
  • D) Material breaks immediately
• Q3: If the distance between D and E is large, the material is:
  • A) Brittle
  • B) Ductile
  • C) Elastic
  • D) Plastic

Answer Key

• Q1: B - Yield point is where permanent deformation starts.
• Q2: B - Beyond B, material deforms permanently.
• Q3: B - Large D-E gap indicates ductility.

Memory Aids

• O-A: Hooke's Law region (linear)
• B: Yield point = start of plastic deformation
• D: Ultimate tensile strength (max stress)
• E: Fracture point (breaking stress)

Explanation

• Stress-strain curve shows material behavior under tension.
• From O to A: Linear region, obeys Hooke's Law, material returns to original shape.
• From A to B: Non-linear but elastic, material still regains shape.
• Point B: Yield point (elastic limit), beyond which permanent deformation occurs.
• From B to D: Plastic region, strain increases rapidly; D is ultimate tensile strength.
• Point E: Fracture point, material breaks; stress here is breaking stress.
• Larger D-E gap means more ductile; smaller means more brittle.

Solved Examples

• Ex 1: Calculate stress at yield point if force = 1000 N and cross-sectional area = 0.5 mm².
  • Stress = Force / Area = 1000/(0.5 × 10^-6) = 2 × 10⁹ Pa

Practice Set

• Q1: What does point B represent on the stress-strain curve?
  • A) Proportional limit
  • B) Yield point
  • C) Ultimate tensile strength
  • D) Fracture point
• Q2: What happens beyond point B?
  • A) Material returns to original shape
  • B) Permanent deformation occurs
  • C) Stress and strain are proportional
  • D) Material breaks immediately
• Q3: If the distance between D and E is large, the material is:
  • A) Brittle
  • B) Ductile
  • C) Elastic
  • D) Plastic

Answer Key

• Q1: B - Yield point is where permanent deformation starts.
• Q2: B - Beyond B, material deforms permanently.
• Q3: B - Large D-E gap indicates ductility.

Memory Aids

• O-A: Hooke's Law region (linear)
• B: Yield point = start of plastic deformation
• D: Ultimate tensile strength (max stress)
• E: Fracture point (breaking stress)

Explanation

• Stress-strain curve shows material behavior under tension.
• From O to A: Linear region, obeys Hooke's Law, material returns to original shape.
• From A to B: Non-linear but elastic, material still regains shape.
• Point B: Yield point (elastic limit), beyond which permanent deformation occurs.
• From B to D: Plastic region, strain increases rapidly; D is ultimate tensile strength.
• Point E: Fracture point, material breaks; stress here is breaking stress.
• Larger D-E gap means more ductile; smaller means more brittle.

Solved Examples

• Ex 1: Calculate stress at yield point if force = 1000 N and cross-sectional area = 0.5 mm².
  • Stress = Force / Area = \( \frac{1000}{0.5 \times 10^{-6}} = 2 \times 10^{9} \) Pa

Practice Set

• Q1: What does point B represent on the stress-strain curve?
  • A) Proportional limit
  • B) Yield point
  • C) Ultimate tensile strength
  • D) Fracture point
• Q2: What happens beyond point B?
  • A) Material returns to original shape
  • B) Permanent deformation occurs
  • C) Stress and strain are proportional
  • D) Material breaks immediately
• Q3: If the distance between D and E is large, the material is:
  • A) Brittle
  • B) Ductile
  • C) Elastic
  • D) Plastic

Answer Key

• Q1: B - Yield point is where permanent deformation starts.
• Q2: B - Beyond B, material deforms permanently.
• Q3: B - Large D-E gap indicates ductility.

Memory Aids

• O-A: Hooke's Law region (linear)
• B: Yield point = start of plastic deformation
• D: Ultimate tensile strength (max stress)
• E: Fracture point (breaking stress)

Elastic After Effect, Elastic Fatigue, Elastomer

Explanation

• Elastic after effect: Delay in regaining original shape after removing force.
• Elastic fatigue: Reduced elasticity due to repeated deformation cycles.
• Elastomer: Materials like rubber or aorta tissue that can stretch to large strains.

Solved Examples

• Ex 1: Identify an elastomer from options: rubber, steel, glass.
  • Answer: Rubber (large strain capability)

Practice Set

• Q1: What is elastic after effect?
  • A) Permanent deformation
  • B) Delay in regaining shape
  • C) Immediate recovery
  • D) Material breaking
• Q2: Elastic fatigue occurs due to:
  • A) Single stretch
  • B) Repeated deformation
  • C) Heating only
  • D) Cooling only
• Q3: Which is an elastomer?
  • A) Steel
  • B) Rubber
  • C) Glass
  • D) Copper

Answer Key

• Q1: B - Elastic after effect is delay in shape recovery.
• Q2: B - Elastic fatigue is due to repeated deformation.
• Q3: B - Rubber is an elastomer.

Memory Aids

• Elastic after effect = Delay in recovery
• Elastic fatigue = Loss of elasticity by repeated stress
• Elastomer = Large strain materials (rubber, aorta tissue)

Explanation

• Elastic after effect: Delay in regaining original shape after removing force.
• Elastic fatigue: Reduced elasticity due to repeated deformation cycles.
• Elastomer: Materials like rubber or aorta tissue that can stretch to large strains.

Solved Examples

• Ex 1: Identify an elastomer from options: rubber, steel, glass.
  • Answer: Rubber (large strain capability)

Practice Set

• Q1: What is elastic after effect?
  • A) Permanent deformation
  • B) Delay in regaining shape
  • C) Immediate recovery
  • D) Material breaking
• Q2: Elastic fatigue occurs due to:
  • A) Single stretch
  • B) Repeated deformation
  • C) Heating only
  • D) Cooling only
• Q3: Which is an elastomer?
  • A) Steel
  • B) Rubber
  • C) Glass
  • D) Copper

Answer Key

• Q1: B - Elastic after effect is delay in shape recovery.
• Q2: B - Elastic fatigue is due to repeated deformation.
• Q3: B - Rubber is an elastomer.

Memory Aids

• Elastic after effect = Delay in recovery
• Elastic fatigue = Loss of elasticity by repeated stress
• Elastomer = Large strain materials (rubber, aorta tissue)

Explanation

• Elastic after effect: Delay in regaining original shape after removing force.
• Elastic fatigue: Reduced elasticity due to repeated deformation cycles.
• Elastomer: Materials like rubber or aorta tissue that can stretch to large strains.

Solved Examples

• Ex 1: Identify an elastomer from options: rubber, steel, glass.
  • Answer: Rubber (large strain capability)

Practice Set

• Q1: What is elastic after effect?
  • A) Permanent deformation
  • B) Delay in regaining shape
  • C) Immediate recovery
  • D) Material breaking
• Q2: Elastic fatigue occurs due to:
  • A) Single stretch
  • B) Repeated deformation
  • C) Heating only
  • D) Cooling only
• Q3: Which is an elastomer?
  • A) Steel
  • B) Rubber
  • C) Glass
  • D) Copper

Answer Key

• Q1: B - Elastic after effect is delay in shape recovery.
• Q2: B - Elastic fatigue is due to repeated deformation.
• Q3: B - Rubber is an elastomer.

Memory Aids

• Elastic after effect = Delay in recovery
• Elastic fatigue = Loss of elasticity by repeated stress
• Elastomer = Large strain materials (rubber, aorta tissue)

Explanation

• Elastic after effect: Delay in regaining original shape after removing force.
• Elastic fatigue: Reduced elasticity due to repeated deformation cycles.
• Elastomer: Materials like rubber or aorta tissue that can stretch to large strains.

Solved Examples

• Ex 1: Identify an elastomer from options: rubber, steel, glass.
  • Answer: Rubber (large strain capability)

Practice Set

• Q1: What is elastic after effect?
  • A) Permanent deformation
  • B) Delay in regaining shape
  • C) Immediate recovery
  • D) Material breaking
• Q2: Elastic fatigue occurs due to:
  • A) Single stretch
  • B) Repeated deformation
  • C) Heating only
  • D) Cooling only
• Q3: Which is an elastomer?
  • A) Steel
  • B) Rubber
  • C) Glass
  • D) Copper

Answer Key

• Q1: B - Elastic after effect is delay in shape recovery.
• Q2: B - Elastic fatigue is due to repeated deformation.
• Q3: B - Rubber is an elastomer.

Memory Aids

• Elastic after effect = Delay in recovery
• Elastic fatigue = Loss of elasticity by repeated stress
• Elastomer = Large strain materials (rubber, aorta tissue)

Variation in Elasticity

Explanation

• Effect of temperature:
  • Elasticity decreases as temperature increases.
  • Annealing (heating-cooling cycles) reduces elasticity.
• Effect of impurities:
  • Elastic impurity increases elasticity.
  • Plastic impurity decreases elasticity.
• Elastic relaxation time:
  • Time delay in regaining original shape after force removal.
  • Negligible for phosphor, bronze, silver, gold.
  • Minimum for quartz, high for glass.

Solved Examples

• Ex 1: Which metal has negligible elastic relaxation time?
  • Answer: Gold

Practice Set

• Q1: How does temperature affect elasticity?
  • A) Increases elasticity
  • B) Decreases elasticity
  • C) No effect
  • D) First increases then decreases
• Q2: What happens if impurity is more elastic?
  • A) Elasticity decreases
  • B) Elasticity increases
  • C) No change
  • D) Material breaks
• Q3: Which material has high elastic relaxation time?
  • A) Quartz
  • B) Gold
  • C) Glass
  • D) Silver

Answer Key

• Q1: B - Elasticity decreases with temperature rise.
• Q2: B - Elastic impurity increases elasticity.
• Q3: C - Glass has high elastic relaxation time.

Memory Aids

• Elasticity ↓ with temperature ↑
• Elastic impurity ↑ elasticity
• Relaxation time: Gold ≈ 0, Glass high

Explanation

• Effect of temperature:
  • Elasticity decreases as temperature increases.
  • Annealing (heating-cooling cycles) reduces elasticity.
• Effect of impurities:
  • Elastic impurity increases elasticity.
  • Plastic impurity decreases elasticity.
• Elastic relaxation time:
  • Time delay in regaining original shape after force removal.
  • Negligible for phosphor, bronze, silver, gold.
  • Minimum for quartz, high for glass.

Solved Examples

• Ex 1: Which metal has negligible elastic relaxation time?
  • Answer: Gold

Practice Set

• Q1: How does temperature affect elasticity?
  • A) Increases elasticity
  • B) Decreases elasticity
  • C) No effect
  • D) First increases then decreases
• Q2: What happens if impurity is more elastic?
  • A) Elasticity decreases
  • B) Elasticity increases
  • C) No change
  • D) Material breaks
• Q3: Which material has high elastic relaxation time?
  • A) Quartz
  • B) Gold
  • C) Glass
  • D) Silver

Answer Key

• Q1: B - Elasticity decreases with temperature rise.
• Q2: B - Elastic impurity increases elasticity.
• Q3: C - Glass has high elastic relaxation time.

Memory Aids

• Elasticity ↓ with temperature ↑
• Elastic impurity ↑ elasticity
• Relaxation time: Gold ≈ 0, Glass high

Explanation

• Effect of temperature:
  • Elasticity decreases as temperature increases.
  • Annealing (heating-cooling cycles) reduces elasticity.
• Effect of impurities:
  • Elastic impurity increases elasticity.
  • Plastic impurity decreases elasticity.
• Elastic relaxation time:
  • Time delay in regaining original shape after force removal.
  • Negligible for phosphor, bronze, silver, gold.
  • Minimum for quartz, high for glass.

Solved Examples

• Ex 1: Which metal has negligible elastic relaxation time?
  • Answer: Gold

Practice Set

• Q1: How does temperature affect elasticity?
  • A) Increases elasticity
  • B) Decreases elasticity
  • C) No effect
  • D) First increases then decreases
• Q2: What happens if impurity is more elastic?
  • A) Elasticity decreases
  • B) Elasticity increases
  • C) No change
  • D) Material breaks
• Q3: Which material has high elastic relaxation time?
  • A) Quartz
  • B) Gold
  • C) Glass
  • D) Silver

Answer Key

• Q1: B - Elasticity decreases with temperature rise.
• Q2: B - Elastic impurity increases elasticity.
• Q3: C - Glass has high elastic relaxation time.

Memory Aids

• Elasticity ↓ with temperature ↑
• Elastic impurity ↑ elasticity
• Relaxation time: Gold ≈ 0, Glass high

Explanation

• Effect of temperature:
  • Elasticity decreases as temperature increases.
  • Annealing (heating-cooling cycles) reduces elasticity.
• Effect of impurities:
  • Elastic impurity increases elasticity.
  • Plastic impurity decreases elasticity.
• Elastic relaxation time:
  • Time delay in regaining original shape after force removal.
  • Negligible for phosphor, bronze, silver, gold.
  • Minimum for quartz, high for glass.

Solved Examples

• Ex 1: Which metal has negligible elastic relaxation time?
  • Answer: Gold

Practice Set

• Q1: How does temperature affect elasticity?
  • A) Increases elasticity
  • B) Decreases elasticity
  • C) No effect
  • D) First increases then decreases
• Q2: What happens if impurity is more elastic?
  • A) Elasticity decreases
  • B) Elasticity increases
  • C) No change
  • D) Material breaks
• Q3: Which material has high elastic relaxation time?
  • A) Quartz
  • B) Gold
  • C) Glass
  • D) Silver

Answer Key

• Q1: B - Elasticity decreases with temperature rise.
• Q2: B - Elastic impurity increases elasticity.
• Q3: C - Glass has high elastic relaxation time.

Memory Aids

• Elasticity ↓ with temperature ↑
• Elastic impurity ↑ elasticity
• Relaxation time: Gold ≈ 0, Glass high

Applications of Elastic Behaviour of Materials

Key Applications

• Materials in bridges rely on elastic strength; bridges become unsafe after prolonged use due to loss of elasticity.
• Maximum height of mountains estimated using elastic limit of rocks.
• Pressure at mountain base: stress = hρg.
• Elastic limit of typical rock: 3 × 10⁸ N/m².
• Stress must be below elastic limit to prevent rock flow.
• Maximum height formula: h < 3 × 10⁸/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.
• Height of Mount Everest is approximately 9 km, consistent with calculation.

Solved Examples

Ex 1: Calculate max height of mountain if ρ = 3 × 10³ kg/m³, g = 10 m/s², elastic limit = 3 × 10⁸ N/m².

• Use h < elastic limit/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A. 1 × 10⁸ N/m²
  B. 3 × 10⁸ N/m²
  C. 5 × 10⁸ N/m²
  D. 7 × 10⁸ N/m²
• Q2: The pressure at the base of a mountain is given by?
  A. hρg
  B. ρg / h
  C. h / ρg
  D. ρ / h g
• Q3: If the density of rock is 4 × 10³ kg/m³ and elastic limit is 4 × 10⁸ N/m², what is max height of mountain?
  A. 5 km
  B. 10 km
  C. 8 km
  D. 12 km

Answer Key

• Q1: B - Elastic limit of typical rock is 3 × 10⁸ N/m².
• Q2: A - Pressure at base is hρg (height × density × gravity).
• Q3: A - Max height = 4 × 10⁸/(4 × 10³ × 10) = 10⁴ / 4 = 2.5 × 10³ m = 2.5 km (closest option 5 km).

Memory Aids

• Max height h = Elastic limit/(ρg).
• Stress = pressure at base = hρg.
• Elastic limit prevents rock flow.

Key Applications

• Materials in bridges rely on elastic strength; bridges become unsafe after prolonged use due to loss of elasticity.
• Maximum height of mountains estimated using elastic limit of rocks.
• Pressure at mountain base: stress = hρg.
• Elastic limit of typical rock: 3 × 10⁸ N/m².
• Stress must be below elastic limit to prevent rock flow.
• Maximum height formula: h < 3 × 10⁸/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.
• Height of Mount Everest is approximately 9 km, consistent with calculation.

Solved Examples

Ex 1: Calculate max height of mountain if ρ = 3 × 10³ kg/m³, g = 10 m/s², elastic limit = 3 × 10⁸ N/m².

• Use h < elastic limit/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A. 1 × 10⁸ N/m²
  B. 3 × 10⁸ N/m²
  C. 5 × 10⁸ N/m²
  D. 7 × 10⁸ N/m²
• Q2: The pressure at the base of a mountain is given by?
  A. hρg
  B. ρg / h
  C. h / ρg
  D. ρ / h g
• Q3: If the density of rock is 4 × 10³ kg/m³ and elastic limit is 4 × 10⁸ N/m², what is max height of mountain?
  A. 5 km
  B. 10 km
  C. 8 km
  D. 12 km

Answer Key

• Q1: B - Elastic limit of typical rock is 3 × 10⁸ N/m².
• Q2: A - Pressure at base is hρg (height × density × gravity).
• Q3: A - Max height = 4 × 10⁸/(4 × 10³ × 10) = 10⁴ / 4 = 2.5 × 10³ m = 2.5 km (closest option 5 km).

Memory Aids

• Max height h = Elastic limit/(ρg).
• Stress = pressure at base = hρg.
• Elastic limit prevents rock flow.

Key Applications

• Materials in bridges rely on elastic strength; bridges become unsafe after prolonged use due to loss of elasticity.
• Maximum height of mountains estimated using elastic limit of rocks.
• Pressure at mountain base: stress = hρg.
• Elastic limit of typical rock: 3 × 10⁸ N/m².
• Stress must be below elastic limit to prevent rock flow.
• Maximum height formula: h < 3 × 10⁸/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.
• Height of Mount Everest is approximately 9 km, consistent with calculation.

Solved Examples

Ex 1: Calculate max height of mountain if ρ = 3 × 10³ kg/m³, g = 10 m/s², elastic limit = 3 × 10⁸ N/m².

• Use h < elastic limit/(ρg) = 3 × 10⁸/(3 × 10³ × 10) = 10⁴ m = 10 km.

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A. 1 × 10⁸ N/m²
  B. 3 × 10⁸ N/m²
  C. 5 × 10⁸ N/m²
  D. 7 × 10⁸ N/m²
• Q2: The pressure at the base of a mountain is given by?
  A. hρg
  B. ρg / h
  C. h / ρg
  D. ρ / h g
• Q3: If the density of rock is 4 × 10³ kg/m³ and elastic limit is 4 × 10⁸ N/m², what is max height of mountain?
  A. 5 km
  B. 10 km
  C. 8 km
  D. 12 km

Answer Key

• Q1: B - Elastic limit of typical rock is 3 × 10⁸ N/m².
• Q2: A - Pressure at base is hρg (height × density × gravity).
• Q3: A - Max height = 4 × 10⁸/(4 × 10³ × 10) = 10⁴ / 4 = 2.5 × 10³ m = 2.5 km (closest option 5 km).

Memory Aids

• Max height h = Elastic limit/(ρg).
• Stress = pressure at base = hρg.
• Elastic limit prevents rock flow.

Key Applications

• Materials in bridges rely on elastic strength; bridges become unsafe after prolonged use due to loss of elasticity.
• Maximum height of mountains estimated using elastic limit of rocks.
• Pressure at mountain base: \( \text{stress} = h \rho g \).
• Elastic limit of typical rock: \(3 \times 10^{8} \text{ N/m}^2\).
• Stress must be below elastic limit to prevent rock flow.
• Maximum height formula: \( h < \frac{3 \times 10^{8}}{\rho g} = \frac{3 \times 10^{8}}{3 \times 10^{3} \times 10} = 10^{4} \text{ m} = 10 \text{ km} \).
• Height of Mount Everest is approximately 9 km, consistent with calculation.

Solved Examples

Ex 1: Calculate max height of mountain if \( \rho = 3 \times 10^{3} \text{ kg/m}^3 \), \( g = 10 \text{ m/s}^2 \), elastic limit = \(3 \times 10^{8} \text{ N/m}^2\).

• Use \( h < \frac{\text{elastic limit}}{\rho g} = \frac{3 \times 10^{8}}{3 \times 10^{3} \times 10} = 10^{4} \text{ m} = 10 \text{ km} \).

Practice Set

• Q1: What is the elastic limit of a typical rock?
  A. \(1 \times 10^{8} \text{ N/m}^2\)
  B. \(3 \times 10^{8} \text{ N/m}^2\)
  C. \(5 \times 10^{8} \text{ N/m}^2\)
  D. \(7 \times 10^{8} \text{ N/m}^2\)
• Q2: The pressure at the base of a mountain is given by?
  A. \(h \rho g\)
  B. \(\rho g / h\)
  C. \(h / \rho g\)
  D. \(\rho / h g\)
• Q3: If the density of rock is \(4 \times 10^{3} \text{ kg/m}^3\) and elastic limit is \(4 \times 10^{8} \text{ N/m}^2\), what is max height of mountain?
  A. 5 km
  B. 10 km
  C. 8 km
  D. 12 km

Answer Key

• Q1: B - Elastic limit of typical rock is \(3 \times 10^{8} \text{ N/m}^2\).
• Q2: A - Pressure at base is \(h \rho g\) (height × density × gravity).
• Q3: A - Max height \(= \frac{4 \times 10^{8}}{4 \times 10^{3} \times 10} = 10^{4} / 4 = 2.5 \times 10^{3} \text{ m} = 2.5 \text{ km}\) (closest option 5 km).

Memory Aids

• Max height \(h = \frac{\text{Elastic limit}}{\rho g}\).
• Stress = pressure at base = \(h \rho g\).
• Elastic limit prevents rock flow.

Beam Deflection

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight W.
• Length of beam = l, breadth = b, depth = d.
• Deflection δ due to load: δ = Wl³/(4bd³Y), where Y is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection δ for W=100 N, l=2 m, b=0.05 m, d=0.01 m, Y=2 × 10¹¹ Pa.

• Use formula: δ = 100 × 2³/(4 × 0.05 × 0.01³ × 2 × 10¹¹).
• Calculate and simplify for δ.

Practice Set

• Q1: Deflection δ is directly proportional to?
  A. Wl³
  B. bd³
  C. Y
  D. bl³
• Q2: If depth d doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with l=1 m, b=0.1 m, d=0.02 m, Y=2 × 10¹¹ Pa, and load W=50 N, find deflection δ.
  A. 1.56 × 10^-5 m
  B. 3.12 × 10^-5 m
  C. 6.25 × 10^-6 m
  D. 1.25 × 10^-4 m

Answer Key

• Q1: A - Deflection δ is proportional to load W and cube of length l³.
• Q2: C - Doubling depth d decreases deflection by 2³ = 8 times.
• Q3: A - Calculate δ = 50 × 1³/(4 × 0.1 × 0.02³ × 2 × 10¹¹) = 1.56 × 10^-5 m.

Memory Aids

• Deflection δ = Wl³/(4bd³Y).
• Deflection increases with W and l³, decreases with b, d³, and Y.

Key Points

• Beam supported at ends, loaded at center with weight \(W\).
• Length of beam = \(l\), breadth = \(b\), depth = \(d\).
• Deflection \(\delta\) due to load: \( \delta = \frac{W l^{3}}{4 b d^{3} Y} \), where \(Y\) is Young's modulus.
• Deflection increases with load and cube of length; decreases with breadth, cube of depth, and Young's modulus.

Solved Examples

Ex 1: Calculate deflection \(\delta\) for \(W=100 \text{ N}\), \(l=2 \text{ m}\), \(b=0.05 \text{ m}\), \(d=0.01 \text{ m}\), \(Y=2 \times 10^{11} \text{ Pa}\).

• Use formula: \( \delta = \frac{100 \times 2^{3}}{4 \times 0.05 \times 0.01^{3} \times 2 \times 10^{11}} \).
• Calculate and simplify for \(\delta\).

Practice Set

• Q1: Deflection \(\delta\) is directly proportional to?
  A. \(W l^{3}\)
  B. \(b d^{3}\)
  C. \(Y\)
  D. \(b l^{3}\)
• Q2: If depth \(d\) doubles, deflection changes by?
  A. Doubles
  B. Halves
  C. Decreases by 8 times
  D. Increases by 8 times
• Q3: For a beam with \(l=1 \text{ m}\), \(b=0.1 \text{ m}\), \(d=0.02 \text{ m}\), \(Y=2 \times 10^{11} \text{ Pa}\), and load \(W=50 \text{ N}\), find deflection \(\delta\).
  A. \(1.56 \times 10^{-5} \text{ m}\)
  B. \(3.12 \times 10^{-5} \text{ m}\)
  C. \(6.25 \times 10^{-6} \text{ m}\)
  D. \(1.25 \times 10^{-4} \text{ m}\)

Answer Key

• Q1: A - Deflection \(\delta\) is proportional to load \(W\) and cube of length \(l^{3}\).
• Q2: C - Doubling depth \(d\) decreases deflection by \(2^{3} = 8\) times.
• Q3: A - Calculate \(\delta = \frac{50 \times 1^{3}}{4 \times 0.1 \times 0.02^{3} \times 2 \times 10^{11}} = 1.56 \times 10^{-5} \text{ m}\).

Memory Aids

• Deflection \(\delta = \frac{W l^{3}}{4 b d^{3} Y}\).
• Deflection increases with \(W\) and \(l^{3}\), decreases with \(b\), \(d^{3}\), and \(Y\).

Hollow vs Solid Shaft Strength

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: τ_solid = πηr⁴/(2L).
• Torque for hollow shaft: τ_hollow = πη(r₂⁴ - r₁⁴)/(2L).
• Ratio: τ_solid/τ_hollow = r⁴/(r₂⁴ - r₁⁴).
• Or τ_hollow/τ_solid = ((r₂² + r₁²)(r₂² - r₁²))/r⁴ = (r₂² + r₁²)/r² > 1 (since r² = r₂² - r₁²).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with r=0.05 m, r₂=0.06 m, r₁=0.03 m, same η and L.

• Calculate ratio τ_hollow/τ_solid = (0.06² + 0.03²)/0.05² = (0.0036 + 0.0009)/0.0025 = 1.8 > 1.
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. r²
  B. r³
  C. r⁴
  D. r
• Q2: Torque ratio τ_hollow/τ_solid is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with r₂=0.07 m, r₁=0.04 m, and solid shaft radius r=0.06 m, find τ_hollow/τ_solid.
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft ∝ r⁴.
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - τ_hollow/τ_solid = (0.07² + 0.04²)/0.06² = (0.0049 + 0.0016)/0.0036 = 1.5.

Memory Aids

• Torque solid shaft ∝ r⁴.
• Hollow shaft stronger: τ_hollow/τ_solid = (r₂² + r₁²)/r² > 1.

Key Points

• Hollow shaft is stronger than solid shaft of same mass, length, and material.
• Torque for unit twist in solid shaft: \( \tau_{solid} = \frac{\pi \eta r^{4}}{2 L} \).
• Torque for hollow shaft: \( \tau_{hollow} = \frac{\pi \eta (r_{2}^{4} - r_{1}^{4})}{2 L} \).
• Ratio: \( \frac{\tau_{solid}}{\tau_{hollow}} = \frac{r^{4}}{r_{2}^{4} - r_{1}^{4}} \).
• Or \( \frac{\tau_{hollow}}{\tau_{solid}} = \frac{(r_{2}^{2} + r_{1}^{2})(r_{2}^{2} - r_{1}^{2})}{r^{4}} = \frac{r_{2}^{2} + r_{1}^{2}}{r^{2}} > 1 \) (since \( r^{2} = r_{2}^{2} - r_{1}^{2} \)).

Solved Examples

Ex 1: Compare torque for solid and hollow shafts with \(r=0.05 \text{ m}\), \(r_{2}=0.06 \text{ m}\), \(r_{1}=0.03 \text{ m}\), same \(\eta\) and \(L\).

• Calculate ratio \( \frac{\tau_{hollow}}{\tau_{solid}} = \frac{0.06^{2} + 0.03^{2}}{0.05^{2}} = \frac{0.0036 + 0.0009}{0.0025} = 1.8 > 1 \).
• Hollow shaft requires more torque, hence stronger.

Practice Set

• Q1: Torque for solid shaft is proportional to?
  A. \(r^{2}\)
  B. \(r^{3}\)
  C. \(r^{4}\)
  D. \(r\)
• Q2: Torque ratio \( \frac{\tau_{hollow}}{\tau_{solid}} \) is always?
  A. Less than 1
  B. Equal to 1
  C. Greater than 1
  D. Zero
• Q3: For hollow shaft with \(r_{2}=0.07 \text{ m}\), \(r_{1}=0.04 \text{ m}\), and solid shaft radius \(r=0.06 \text{ m}\), find \( \frac{\tau_{hollow}}{\tau_{solid}} \).
  A. 1.2
  B. 1.5
  C. 1.8
  D. 2.0

Answer Key

• Q1: C - Torque for solid shaft \(\propto r^{4}\).
• Q2: C - Hollow shaft torque ratio is always greater than 1.
• Q3: B - \( \frac{\tau_{hollow}}{\tau_{solid}} = \frac{0.07^{2} + 0.04^{2}}{0.06^{2}} = \frac{0.0049 + 0.0016}{0.0036} = 1.5 \).

Memory Aids

• Torque solid shaft \(\propto r^{4}\).
• Hollow shaft stronger: \( \frac{\tau_{hollow}}{\tau_{solid}} = \frac{r_{2}^{2} + r_{1}^{2}}{r^{2}} > 1 \).