Chapter 9: Mechanical Properties of Solids (Class 11)

1. Define Elasticity.

Property of a body to regain its original shape/size after removal of deforming force.

2. Define Plasticity.

Property of a body by which it does NOT regain its original shape/size after removal of deforming force.

3. Define Stress.

Internal restoring force per unit area of cross-section.

4. What is the SI unit of Stress?

N/m² or Pascal (Pa).

5. Define Strain.

Ratio of change in configuration to the original configuration.

6. What is the unit of Strain?

It has no unit (dimensionless ratio).

7. State Hooke’s Law.

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

8. What is the constant E in Hooke’s law (Stress = E × Strain) called?

Modulus of Elasticity.

9. What does a larger value of Young’s modulus (Y) indicate?

The material is more elastic or stiffer (requires more stress for same longitudinal strain).

10. Define Longitudinal Strain.

Change in length / Original length.

11. Define Volumetric Strain.

Change in volume / Original volume.

12. What type of stress causes change in volume?

Volumetric Stress (Pressure).

13. What is the dimensional formula for Modulus of Elasticity?

[ML⁻¹T⁻²] (Same as Stress).

14. Define Poisson’s ratio.

Ratio of lateral strain to longitudinal strain within the elastic limit.

15. What is Elastic Potential Energy?

Energy stored in a body due to its deformation (work done during deformation).

16. Give an example of a nearly perfectly elastic material.

Quartz fibre.

17. Give an example of a plastic material.

Putty, Mud.

18. What is the reciprocal of Bulk Modulus called?

Compressibility.

1. Differentiate between Elastic and Plastic bodies.

• Elastic bodies regain original shape/size after removing deforming force (e.g., steel spring).
• Plastic bodies do NOT regain original shape/size after removing deforming force (e.g., putty).

2. Define the three types of stress (Longitudinal, Volumetric, Shear).

• Longitudinal: Force perpendicular to area, changes length (Tensile/Compressive).
• Volumetric: Uniform force perpendicular to entire surface, changes volume (like Pressure).
• Shear: Force parallel/tangential to surface, changes shape.

3. Define the three types of strain (Longitudinal, Volumetric, Shear).

• Longitudinal: Change in length / Original length (ΔL/L).
• Volumetric: Change in volume / Original volume (ΔV/V).
• Shear: Angle of deformation (θ) or Δx/L.

4. What is Elastic Limit? What happens beyond it?

• Elastic Limit: Max stress upto which body returns to original configuration completely after force removal.
• Beyond Elastic Limit: Body gets permanently deformed (plasticity) or fractures (breaks).

5. Define Young’s Modulus (Y). What does it measure?

• Y = Longitudinal Stress / Longitudinal Strain = (F/A)/(ΔL/L).
• It measures the material’s resistance to change in length (stiffness against stretching/compression).
• SI Unit: N/m² or Pa.

6. Define Bulk Modulus (B). What does its reciprocal measure?

• B = Volumetric Stress / Volumetric Strain = P / (-ΔV/V).
• It measures the material’s resistance to change in volume under pressure.
• Its reciprocal (1/B) is called Compressibility (k), measuring how easily volume changes.

7. Define Shear Modulus (G). Which state of matter does it apply to primarily?

• G = Shear Stress / Shear Strain = (F_tangential/A) / θ.
• It measures the material’s resistance to change in shape (rigidity).
• It applies primarily to Solids, as fluids cannot sustain shear stress.

8. Why is steel considered more elastic than rubber?

• Elasticity means resistance to deformation.
• For the same applied stress (force/area), rubber stretches much more (larger strain) than steel (smaller strain).
• Since Modulus E = Stress/Strain, a smaller strain for the same stress means steel has a much higher Young’s modulus (Y) than rubber.
• Therefore, steel resists deformation more strongly and is considered more elastic.

9. What is Poisson’s Ratio? Is it dimensionless?

• Ratio of Lateral Strain (change in diameter/original diameter) to Longitudinal Strain (change in length/original length).
• Symbol: σ (sigma) or ν (nu).
• σ = Lateral Strain / Longitudinal Strain.
• Yes, it is dimensionless because it’s a ratio of two strains (which are themselves dimensionless).

10. Write the expression for Elastic Potential Energy stored per unit volume.

• Energy Density = ½ × Stress × Strain.
• OR Energy Density = ½ × Y × (Strain)².
• OR Energy Density = ½ × (Stress)² / Y.
• Unit: Joules per cubic metre (J/m³).

11. Draw a qualitative Stress-Strain curve for a ductile material like copper.

(Description of the graph)

• Initial straight line (O to A): Proportional region (Hooke’s Law).
• Curve up to B: Elastic region (returns to original shape). Point B = Elastic Limit / Yield Point.
• Curve continues, large plastic deformation region (C to D): Strain increases significantly for small stress increase.
• Point D: Ultimate Tensile Strength (max stress).
• Curve drops slightly then Point E: Fracture Point (breaks).
• Region between B and E represents plastic behavior.

12. Which is generally greater for metals: Young’s Modulus (Y) or Bulk Modulus (B)?

• For most common metals, the Bulk Modulus (B) is generally slightly smaller than or comparable to Young’s Modulus (Y).
• For example, Steel: Y ≈ 200 GPa, B ≈ 160 GPa. Copper: Y ≈ 110 GPa, B ≈ 140 GPa. Aluminum: Y ≈ 70 GPa, B ≈ 70 GPa.
• Shear Modulus (G) is usually the smallest of the three (roughly Y/3).

13. What is the difference between Tensile stress and Compressive stress?

• Both are types of Longitudinal Stress (Force applied perpendicular to area).
• Tensile Stress: Occurs when forces pull on the object, tending to increase its length (Stretching).
• Compressive Stress: Occurs when forces push on the object, tending to decrease its length (Compressing).

14. What are elastomers?

• Materials which can be stretched to cause large strains and still regain their original form.
• They do NOT obey Hooke’s Law (stress is not proportional to strain over large range).
• Example: Rubber used in aorta of heart, rubber bands.

15. The breaking stress for a metal is 7.8 × 10⁹ N/m². Find the maximum length of a wire of this metal which can hang vertically without breaking under its own weight (Density of metal = 7800 kg/m³; g ≈ 10 m/s²).

• Let max length = L, area = A, density = ρ.
• Weight of wire W = Mass × g = (Volume × Density) × g = (A × L × ρ) × g.
• This weight acts at the top support, causing max stress there.
• Max Stress = Weight / Area = (A L ρ g) / A = L ρ g.
• Given Max Stress = 7.8 × 10⁹ N/m².
• L ρ g = 7.8 × 10⁹
• L × 7800 × 10 = 7.8 × 10⁹
• L × 7.8 × 10⁴ = 7.8 × 10⁹
• L = (7.8 × 10⁹) / (7.8 × 10⁴) = 10⁹⁻⁴ = 10⁵ m.
• Answer: 10⁵ m (or 100 km).

1. State and explain Hooke’s Law. Discuss the Stress-Strain curve for a metallic wire.

Hooke’s Law: Within the elastic limit, stress is directly proportional to strain (Stress = E × Strain), where E is Modulus of Elasticity.

Stress-Strain Curve (for a ductile metal):

1. O to A (Proportional Limit): Straight line. Stress ∝ Strain. Hooke’s law obeyed perfectly.
2. A to B (Elastic Limit/Yield Point): Slightly curved. Stress is not strictly proportional, but the material still returns to original shape when unloaded. Point B is the yield point/elastic limit.
3. B to D (Plastic Region): If loaded beyond B, material enters plastic region. Strain increases rapidly for small stress increase. Permanent deformation occurs even after unloading.
4. D (Ultimate Tensile Strength): Max stress the material can withstand. Beyond this, ‘necking’ starts (wire becomes thinner at one point).
5. D to E (Fracture): Stress decreases slightly but strain increases rapidly until fracture occurs at Point E.

Brittle materials fracture shortly after the elastic limit.

2. Define Young’s Modulus, Bulk Modulus, and Shear Modulus. Write their formulas and SI units.

1. Young’s Modulus (Y):

• Definition: Ratio of longitudinal stress to longitudinal strain (within elastic limit).
• Formula: Y = (F/A) / (ΔL/L) = FL / AΔL.
• Measures: Resistance to change in length (Stiffness).
• SI Unit: N/m² or Pa.

2. Bulk Modulus (B):

• Definition: Ratio of volumetric stress (Pressure P) to volumetric strain (within elastic limit).
• Formula: B = P / (-ΔV/V) = -PV / ΔV.
• Measures: Resistance to change in volume (Incompressibility).
• SI Unit: N/m² or Pa.

3. Shear Modulus (G or η):

• Definition: Ratio of shear stress to shear strain (within elastic limit).
• Formula: G = (F_tangential/A) / θ = (F_tangentialL) / (AΔx).
• Measures: Resistance to change in shape (Rigidity).
• SI Unit: N/m² or Pa.

3. Explain the concept of Elastic Potential Energy. Derive the expression for energy stored per unit volume in a stretched wire.

Elastic Potential Energy: Work done in deforming an elastic body (against internal restoring forces) gets stored as potential energy within it. When the force is removed, this energy helps restore the body’s shape.

Derivation for Stretched Wire:

• Consider a wire of length L, area A, Young’s Modulus Y, stretched by ΔL by force F.
• Stress = F/A, Strain = ΔL/L. Assume Hooke’s law holds: F/A = Y(ΔL/L) => F = (YA/L)ΔL.
• Force is not constant; it increases from 0 to F as extension increases from 0 to ΔL.
• Work done (dW) for small extension (dl) = Force(l) × dl. Where Force(l) = (YA/L)l.
• dW = (YA/L) l dl.
• Total Work Done (W) for extension ΔL = ∫dW = ∫₀^ΔL (YA/L) l dl.
• W = (YA/L) [l²/2]₀^ΔL = (YA/L) (ΔL²/2).
• This Work Done is stored as Elastic Potential Energy (U). U = ½ (YA/L) ΔL².

Energy per unit Volume (Energy Density):

• Volume = A × L.
• Energy Density (u) = U / Volume = [½ (YA/L) ΔL²] / (AL)
• u = ½ × (Y) × (ΔL/L)² = ½ × Y × (Strain)².
• Since Y = Stress/Strain, we can write: u = ½ × (Stress/Strain) × Strain² = ½ × Stress × Strain.
• Also, Strain = Stress/Y, so: u = ½ × Stress × (Stress/Y) = ½ × (Stress)² / Y.

4. Define Poisson’s ratio. What does it signify? Can it be negative?

Poisson’s Ratio (σ or ν): Within the elastic limit, it is the ratio of lateral strain (strain perpendicular to the applied force) to the longitudinal strain (strain parallel to the applied force).

• Formula: σ = – (Lateral Strain) / (Longitudinal Strain) = – (ΔD/D) / (ΔL/L).
• The negative sign is often included because longitudinal extension (ΔL>0) usually causes lateral contraction (ΔD<0), making σ positive.
• Significance: It measures the tendency of a material to contract/expand in directions perpendicular to the direction of stretching/compression. It links changes in dimensions along different axes.
• It is a dimensionless quantity.
• Values: Theoretically, σ can range from -1 to +0.5. For most isotropic materials, experimental values lie between 0 and 0.5 (e.g., steel ≈ 0.3, rubber ≈ 0.5).
• Negative Poisson’s Ratio?: Yes, some materials called **auxetic materials** contract laterally when compressed longitudinally (or expand laterally when stretched), exhibiting a negative Poisson’s ratio. These are special engineered or natural structures.

5. A steel wire of length 1m and radius 1mm is stretched by a force of 100π N. If Young’s modulus Y=2×10¹¹ N/m², find the stress, strain, and elongation.

• Given: L = 1m, Radius r = 1mm = 1×10⁻³m, Force F = 100π N, Y = 2×10¹¹ N/m².
• Area (A): A = πr² = π(10⁻³)² = π × 10⁻⁶ m².
• Stress (σ): σ = F/A = (100π N) / (π × 10⁻⁶ m²) = 100 / 10⁻⁶ = 100 × 10⁶ = 1 × 10⁸ N/m² (or Pa).
• Strain (ε): From Hooke’s Law, Strain = Stress / Y.
  ε = (1 × 10⁸ N/m²) / (2 × 10¹¹ N/m²) = 0.5 × 10⁻³ = 5 × 10⁻⁴.
• Elongation (ΔL): Strain = ΔL / L => ΔL = Strain × L.
  ΔL = (5 × 10⁻⁴) × (1 m) = 5 × 10⁻⁴ m = 0.5 × 10⁻³ m = 0.5 mm.

Answer: Stress=10⁸ Pa, Strain=5×10⁻⁴, Elongation=0.5 mm.

6. What do you understand by elastic fatigue?

Elastic Fatigue: It is the phenomenon where an elastic material loses its strength, or its elastic properties temporarily deteriorate, due to repeated alternating deforming forces applied to it.

• When a material is subjected to continuous, rapid cycles of stress and strain, its internal structure gets disturbed.
• It may not return exactly to its original state immediately after the force is removed, or it might require less stress to reach its elastic limit, or its breaking point might lower.
• This ‘tiredness’ of the elastic material is called elastic fatigue.
• Example: This is why old bridges are sometimes declared unsafe after prolonged use, as the materials undergo fatigue due to repeated stress cycles from traffic and temperature changes. Bridges need regular inspection for signs of fatigue.
• The elastic properties can often be regained after a period of rest, but repeated fatigue can lead to eventual failure.

7. Explain ductile and brittle materials based on the stress-strain curve.

The shape of the stress-strain curve beyond the elastic limit differentiates ductile and brittle materials:

• Ductile Materials:
  • These materials exhibit a significant **plastic region** after the yield point (elastic limit).
  • They undergo considerable deformation (elongation/strain) before fracturing.
  • Stress-strain curve shows a large difference between the yield point and the fracture point.
  • Examples: Copper, Aluminum, Mild Steel. They can be drawn into wires (ductility).
• Brittle Materials:
  • These materials show very little or **no plastic deformation**.
  • They fracture (break) suddenly soon after the elastic limit is crossed, with very little elongation.
  • The yield point, ultimate strength, and fracture point are very close on the stress-strain curve.
  • Examples: Glass, Cast Iron, Ceramics, Concrete. They tend to shatter when overloaded.

8. Two wires of same length and material but radii r₁ and r₂ are stretched by the same force. Find the ratio of stresses and strains produced in them.

Let Force = F, Length = L, Material has Young’s Modulus Y.

• Wire 1: Radius r₁, Area A₁ = πr₁².
• Wire 2: Radius r₂, Area A₂ = πr₂².

Ratio of Stresses:

• Stress σ = Force / Area.
• σ₁ = F / A₁ = F / (πr₁²)
• σ₂ = F / A₂ = F / (πr₂²)
• Ratio: σ₁σ₂ = (F / A₁)(F / A₂) = A₂A₁ = πr₂²πr₁² = r₂²r₁².

Ratio of Strains:

• Young’s Modulus Y = Stress / Strain => Strain ε = Stress / Y.
• Since material is same, Y is same for both (Y₁=Y₂=Y).
• ε₁ = σ₁ / Y
• ε₂ = σ₂ / Y
• Ratio: ε₁ε₂ = (σ₁ / Y)(σ₂ / Y) = σ₁σ₂.
• From stress ratio, ε₁ε₂ = r₂²r₁².

Answer: Ratio of Stresses (σ₁/σ₂) = r₂²/r₁²; Ratio of Strains (ε₁/ε₂) = r₂²/r₁².

9. What is the elastic potential energy stored in a wire of length L, area A, stretched by an amount x, if its Young’s Modulus is Y?

• Force required to produce extension x: F = (YA/L)x (from Y = (F/A)/(x/L)).
• This force varies from 0 to F as extension goes from 0 to x. Average force = (0+F)/2 = F/2.
• Work Done (Stored Energy U) = Average Force × Extension.
• U = (F/2) × x.
• Substitute F = (YA/L)x:
• U = (1/2) × [(YA/L)x] × x
• U = ½ (YA/L) x²

(Alternatively, using Energy Density = ½ × Y × (Strain)²: Strain = x/L. Volume = AL. U = (½ × Y × (x/L)²) × (AL) = ½ Y (x²/L²) AL = ½ (YA/L) x²).

Answer: U = ½ (YA/L) x²

10. Differentiate between Pressure and Stress.

• Stress: Internal restoring force per unit area developed *within* a deformed body. Can be tensile, compressive, or shear. A tensor quantity (though often treated as scalar/vector in simple cases). Depends on internal forces. (Vastu ke andar develop hua internal force per unit area).
• Pressure: External force acting perpendicularly per unit area *on* a surface (usually of a fluid). Always compressive. A scalar quantity. Depends on external forces. (Kisi satah par perpendicularly lag raha bahari force per unit area).
• While both have same dimension [ML⁻¹T⁻²] and SI unit (Pa or N/m²), they represent different physical concepts related to internal restoring forces vs external applied forces. Volumetric stress is numerically equal to pressure.