Mechanical Properties of Solids: CBSE Class 11 Physics Notes & Practice

Elastic and Plastic Behaviour of Solids

In a solid, atoms or molecules are arranged in a stable, structured lattice. Under normal conditions, they are held in equilibrium by inter-atomic forces. When an external deforming force is applied, these atoms are displaced from their equilibrium positions, causing a change in the shape or size of the solid.

Once the external force is removed, internal inter-atomic forces drive the atoms back to their original configuration. The property of a body by virtue of which it tends to regain its original shape and size after the removal of the deforming force is called elasticity. If a body does not regain its original shape and behaves as if it has permanent deformation, it is called plasticity (e.g., mud or putty). No solid is perfectly elastic or perfectly plastic; all real materials lie somewhere in between.

Fundamental Terms of Elasticity

Stress

The internal restoring force per unit cross-sectional area of a deformed body. Expressed as Stress = Force / Area. Its SI unit is N/m² or Pascal (Pa).

Strain

The ratio of the change in dimension to the original dimension of the body. It is a pure ratio and has no units or dimensions.

Hooke's Law

Within the elastic limit, the stress developed in a body is directly proportional to the strain produced in it. Stress ∝ Strain, or Stress / Strain = Modulus of Elasticity.

Young's Modulus (Y)

The ratio of longitudinal stress to longitudinal strain. It characterizes a material's resistance to change in length.

Understanding the Stress-Strain Curve

1. Proportional Limit (Region OA) — In this initial region, stress is strictly proportional to strain. Hooke's law is fully obeyed here. Point A is called the proportional limit.
2. Elastic Limit (Point B) — From A to B, stress and strain are not proportional, but the material remains elastic. If the load is removed, the body returns to its original dimensions. Point B is the yield point, and the corresponding stress is called the yield strength.
3. Plastic Deformation & Permanent Set (Region BC to D) — Beyond point B, if the stress increases, the strain increases rapidly. Even if the load is removed at point C, the material does not regain its original length and retains a 'permanent set'. This is plastic deformation.
4. Ultimate Tensile Strength & Fracture (Points D & E) — Point D is the ultimate tensile strength of the material. Beyond D, necking occurs, and even with reduced applied force, the material continues to stretch until it breaks at fracture point E. If D and E are close, the material is brittle; if they are far apart, it is ductile.

Common Exam Traps and Board Tips

• Trap 1: Stress vs Pressure: Although both have the unit N/m², pressure is an external force acting perpendicularly per unit area, whereas stress is the internal restoring force developed per unit area of a deformed body.
• Trap 2: Dimension Conversion: CBSE numericals often give wire diameters in mm and length changes in mm. Always convert these parameters to standard SI units (meters) before plugging them into elastic moduli equations. Remember: $1\text{ mm}^2 = 10^{-6}\text{ m}^2$.
• Trap 3: Rubber is not more elastic than steel! In physics, a material that resists deformation more and requires a larger restoring force is 'more elastic'. Steel has a much higher Young's modulus than rubber, making steel more elastic than rubber.

Worked Numerical Examples

• Example 1: A structural steel rod of length 2.0 m and cross-sectional area 2.0 × 10^-4 m² is stretched by a force of 40 kN. Calculate the stress and elongation. (Take Young's Modulus of steel Y = 2.0 × 10^11 N/m²). Step 1: Write down the given values. Length (L) = 2.0 m Area (A) = 2.0 × 10^-4 m² Force (F) = 40 kN = 40 × 10^3 N = 4 × 10^4 N Young's Modulus (Y) = 2.0 × 10^11 N/m² Step 2: Calculate the stress. Stress = Force / Area = (4 × 10^4 N) / (2.0 × 10^-4 m²) = 2.0 × 10^8 N/m². Step 3: Calculate the elongation (ΔL) using the formula Y = (Force × L) / (Area × ΔL). Rearranging for ΔL: ΔL = (Force × L) / (Area × Y) ΔL = ((4 × 10^4 N) × 2.0 m) / ((2.0 × 10^-4 m²) × (2.0 × 10^11 N/m²)) ΔL = (8 × 10^4) / (4.0 × 10^7) = 2.0 × 10^-3 m = 2.0 mm. Final Answer: Stress is 2.0 × 10^8 N/m² and the elongation is 2.0 mm.
• Example 2: Determine the volume contraction of a solid copper cube of side 10 cm when subjected to a hydraulic pressure of 7.0 × 10^6 Pa. (Take Bulk Modulus of copper B = 140 × 10^9 Pa). Step 1: Identify given variables. Side of copper cube = 10 cm = 0.1 m Original Volume (V) = (0.1 m)³ = 1.0 × 10^-3 m³ Hydraulic pressure (ΔP) = 7.0 × 10^6 Pa Bulk Modulus (B) = 140 × 10^9 Pa Step 2: Use the Bulk Modulus formula: B = -ΔP / (ΔV / V) We are looking for the magnitude of volume contraction, ΔV = (ΔP × V) / B. Step 3: Substitute and solve. ΔV = ((7.0 × 10^6 Pa) × (1.0 × 10^-3 m³)) / (140 × 10^9 Pa) ΔV = (7.0 × 10^3) / (140 × 10^9) = 0.05 × 10^-6 m³ = 5.0 × 10^-8 m³. Final Answer: The volume contraction of the solid copper cube is 5.0 × 10^-8 m³.

Practice Questions with Solutions

• Q: A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Find the load applied. (Y_copper = 1.1 × 10^11 Pa, Y_steel = 2.0 × 10^11 Pa). A: Step 1: Write down parameters. Diameter (d) = 3.0 mm = 3.0 × 10^-3 m Area (A) = πr² = π(d/2)² = 3.14 × (1.5 × 10^-3)² = 7.065 × 10^-6 m² L_c = 2.2 m, L_s = 1.6 m Total elongation (ΔL_c + ΔL_s) = 0.70 mm = 7.0 × 10^-4 m Let the common stretching tension (load) be F. Step 2: Express elongation for each wire. ΔL_c = (F × L_c) / (A × Y_c) and ΔL_s = (F × L_s) / (A × Y_s) Step 3: Substitute the sum equation. ΔL_c + ΔL_s = (F / A) [(L_c / Y_c) + (L_s / Y_s)] 7.0 × 10^-4 = (F / (7.065 × 10^-6)) [(2.2 / (1.1 × 10^11)) + (1.6 / (2.0 × 10^11))] 7.0 × 10^-4 = (F / (7.065 × 10^-6)) [2.0 × 10^-11 + 0.8 × 10^-11] 7.0 × 10^-4 = (F / (7.065 × 10^-6)) [2.8 × 10^-11] Step 4: Solve for F. F = (7.0 × 10^-4 × 7.065 × 10^-6) / (2.8 × 10^-11) F = (4.9455 × 10^-9) / (2.8 × 10^-11) = 176.6 N Final answer: The applied load is approximately 177 N.
• Q: Explain why a spring is made of steel and not copper. A: Step 1: Analyze the physical mechanism of a spring. A spring is designed to undergo elastic deformation and recover perfectly. The material of the spring should develop a strong restoring force upon deformation. Step 2: Compare Young's Modulus values. Young's modulus of steel is about 2.0 × 10^11 Pa, whereas for copper it is 1.1 × 10^11 Pa. Steel is more elastic than copper. Step 3: Draw final conclusion. For a given amount of deformation, steel generates a much greater restoring force than copper and exhibits quicker recovery without permanent deformation. Final answer: Steel is used because it has a higher modulus of elasticity than copper, allowing the spring to withstand large deforming forces and recover its original shape effectively.
• Q: What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10^3 kg/m³? (Bulk Modulus of water = 2.2 × 10^9 Pa). A: Step 1: Convert pressure to SI units. Pressure change ΔP = 80.0 atm = 80 × 1.013 × 10^5 Pa = 8.1 × 10^6 Pa. Step 2: Relate density change to volume change. Bulk modulus B = -V (ΔP / ΔV) => -ΔV / V = ΔP / B. Since mass (M) is constant, density ρ = M/V. Taking differentials, dρ/ρ = -dV/V. Thus, fractional change in density Δρ / ρ = ΔP / B. Step 3: Solve for density change Δρ. Δρ = ρ_surface (ΔP / B) = (1.03 × 10^3 kg/m³) (8.1 × 10^6 Pa / 2.2 × 10^9 Pa) Δρ = 1.03 × 10^3 3.68 × 10^-3 = 3.79 kg/m³. Step 4: Calculate final density. ρ_deep = ρ_surface + Δρ = 1030 kg/m³ + 3.79 kg/m³ = 1033.8 kg/m³. Final answer: The density of water at that depth is approximately 1.034 × 10³ kg/m³.
• Q: Under what conditions is Poisson's ratio defined? State its theoretical and practical limits. A: Step 1: Define Poisson's ratio. When a wire is subjected to a longitudinal stretching force, it elongates in length but contracts in diameter (laterally). Poisson's ratio (σ) is defined as the ratio of lateral strain to longitudinal strain. σ = - (Δd / d) / (ΔL / L) Step 2: State theoretical limits. From the conditions of positive bulk modulus and shear modulus in elasticity theory, the theoretical limits of Poisson's ratio are between -1 and +0.5. Step 3: State practical limits. For real physical materials, stretching a material almost always leads to lateral contraction (not expansion). Hence, the practical values of Poisson's ratio are always positive, falling between 0 and 0.5 (e.g., around 0.3 for metals). Final answer: Poisson's ratio is defined within the elastic limit. Its theoretical limits are -1 to +0.5, and practical limits are 0 to 0.5.

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