● CBSE Class 11 Physics — Chapter 9: Mechanical Properties of Solids
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Mechanical Properties of Solids describes deformation under external forces. Elasticity is the property to regain original shape; Plasticity is permanent deformation. Hooke's Law: Stress σ = E * Strain ε within elastic limit. Stress-Strain Curve points: Proportional Limit A (σ ∝ ε), Yield Point B / Elastic Limit (permanent set), Tensile Strength D (max stress), Breaking Point E (fracture). Young's Modulus Y = (F * L) / (A * ΔL) [Steel Y = 2 × 10^11 N/m² > Rubber Y = 10^7 N/m², so steel is more elastic!]. Shear Modulus G = (F/A) / θ. Bulk Modulus B = -ΔP / (ΔV/V); Compressibility K = 1/B. Poisson's Ratio ν = lateral strain / longitudinal strain (theoretical limits -1 ≤ ν ≤ 0.5, practical 0 ≤ ν ≤ 0.5). Elastic Potential Energy in stretched wire: U = ½ * F * ΔL = ½ * Y * A * (ΔL)² / L. Energy Density u = ½ * σ * ε = ½ Y ε². Maximum mountain height on Earth h_max ≈ 10 km (due to rock breaking stress).
Why is steel considered more elastic than rubber in physics? What causes bridges to sag over decades of heavy traffic, and why are structural beams manufactured in an "I" shape rather than solid rectangular blocks? How high can a mountain theoretically grow on Earth before collapsing under its own weight? Mechanical Properties of Solids forms Chapter 9 of the CBSE Class 11 Physics syllabus and is a scoring chapter for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers Elasticity vs Plasticity, Hooke's Law, complete analysis of the Stress-Strain Curve for metallic wires, Young's Modulus, Shear Modulus, Bulk Modulus, Poisson's Ratio limits, Elastic Potential Energy derivations, engineering applications (I-girders & maximum mountain height), and five step-by-step solved entrance exam numericals.
Core Pillars of Solid Elasticity & Engineering Physics
Stress-Strain Curve
Proportional, Yield & Fracture Points
Young's Modulus
Y = (F L)/(A ΔL) & Steel > Rubber
Elastic Moduli
Bulk (B), Shear (G) & Poisson (ν)
Elastic Energy
U = ½ F ΔL & u = ½ σε Density
Table of Contents
- 1. Elasticity vs Plasticity, Stress & Strain Types
- 2. Hooke's Law & Complete Stress-Strain Curve Analysis
- 3. Moduli of Elasticity: Young's, Shear & Bulk Modulus
- 4. Poisson's Ratio (ν) & Theoretical/Practical Limits
- 5. Elastic Potential Energy in Stretched Wire & Energy Density
- 6. Engineering Applications: I-Beams & Max Mountain Height
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Elasticity vs Plasticity, Stress & Strain Types
When a deforming force acts on a body, internal restoring forces arise to oppose the deformation:
✨ Elasticity vs Plasticity
• Elasticity: Property by virtue of which a body regains its original shape and size after removal of deforming forces (e.g. Steel spring, Rubber band). Perfectly elastic body: Quartz fiber, Phosphor bronze.
• Plasticity: Property by virtue of which a body undergoes permanent deformation and does not regain its original shape (e.g. Putty, Clay, Paraffin wax).
• Plasticity: Property by virtue of which a body undergoes permanent deformation and does not regain its original shape (e.g. Putty, Clay, Paraffin wax).
📏 Stress & Strain Definitions
• Stress (σ): Internal restoring force per unit cross-sectional area: $\sigma = \frac{F}{A}$ (SI Unit: N/m² or Pa).
Types: Longitudinal (Tensile/Compressive), Volumetric ($\Delta P$), Shearing ($\tau = F_t / A$).
• Strain (ε): Fractional change in dimensions: $\epsilon = \frac{\text{Change in Dimension}}{\text{Original Dimension}}$ (Dimensionless!).
Types: Longitudinal ($\Delta L / L$), Volumetric ($\Delta V / V$), Shearing ($\theta = \Delta x / L$).
Types: Longitudinal (Tensile/Compressive), Volumetric ($\Delta P$), Shearing ($\tau = F_t / A$).
• Strain (ε): Fractional change in dimensions: $\epsilon = \frac{\text{Change in Dimension}}{\text{Original Dimension}}$ (Dimensionless!).
Types: Longitudinal ($\Delta L / L$), Volumetric ($\Delta V / V$), Shearing ($\theta = \Delta x / L$).
2. Hooke's Law & Complete Stress-Strain Curve Analysis
Hooke's Law & Modulus of Elasticity
Robert Hooke (1676) stated that for small deformations, stress is directly proportional to strain:
\text{Stress} \propto \text{Strain} \implies \text{Stress} = E \times \text{Strain} \implies E = \frac{\text{Stress}}{\text{Strain}}
Where $E$ is the Modulus of Elasticity of the material.
\text{Stress} \propto \text{Strain} \implies \text{Stress} = E \times \text{Strain} \implies E = \frac{\text{Stress}}{\text{Strain}}
Where $E$ is the Modulus of Elasticity of the material.
Stress-Strain Curve for a Stretched Metallic Wire
| Region / Point | Curve Feature | Physical Material Behavior |
|---|---|---|
| Region OA (Proportional Limit) | Linear straight line | Hooke's law is strictly obeyed ($\sigma \propto \epsilon$). Wire returns to original length completely. Slope $= Y$. |
| Point B (Yield Point / Elastic Limit) | Upper yield boundary | Maximum stress up to which material regains original shape when load is removed. Stress at B is Yield Strength ($\sigma_y$). |
| Region BC (Plastic Region) | Non-linear curve | Strain increases faster than stress. If load is removed at C, wire retains permanent strain $OO'$ called Permanent Set. |
| Point D (Ultimate Tensile Strength) | Peak maximum stress | Maximum stress material can withstand before necking/thinning begins. Stress at D is Tensile Strength ($\sigma_u$). |
| Point E (Fracture / Breaking Point) | End of curve | Wire breaks/snaps at point E. If distance DE is large ⇒ Ductile (Copper, Steel); If D & E are close ⇒ Brittle (Glass, Cast Iron). |
✨ Why Steel is More Elastic than Rubber in Physics!
In physics, elasticity is measured by the magnitude of restoring force developed for a given strain ($Y = \frac{\text{Stress}}{\text{Strain}}$).
Consider equal length $L$ and area $A$ of steel and rubber wires subjected to equal stretching force $F$:
$\Delta L_{\text{rubber}} \gg \Delta L_{\text{steel}} \implies Y_{\text{steel}} = \frac{F L}{A \Delta L_{\text{steel}}} \gg Y_{\text{rubber}} = \frac{F L}{A \Delta L_{\text{rubber}}}$.
Since $Y_{\text{steel}} \approx 2 \times 10^{11}\text{ N/m}^2$ while $Y_{\text{rubber}} \approx 10^7\text{ N/m}^2$, Steel is 20,000 times more elastic than rubber!
Consider equal length $L$ and area $A$ of steel and rubber wires subjected to equal stretching force $F$:
$\Delta L_{\text{rubber}} \gg \Delta L_{\text{steel}} \implies Y_{\text{steel}} = \frac{F L}{A \Delta L_{\text{steel}}} \gg Y_{\text{rubber}} = \frac{F L}{A \Delta L_{\text{rubber}}}$.
Since $Y_{\text{steel}} \approx 2 \times 10^{11}\text{ N/m}^2$ while $Y_{\text{rubber}} \approx 10^7\text{ N/m}^2$, Steel is 20,000 times more elastic than rubber!
3. Moduli of Elasticity: Young's, Shear & Bulk Modulus
| Modulus of Elasticity | Definition & Formula | Units & Typical Values |
|---|---|---|
| Young's Modulus ($Y$) | $Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L} = \frac{M g L}{\pi r^2 \Delta L}$ | N/m² or Pa. Steel: $200\text{ GPa}$, Copper: $110\text{ GPa}$, Aluminium: $70\text{ GPa}$. |
| Shear Modulus ($G$ or $\eta$) | $G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F / A}{\theta} = \frac{F L}{A \Delta x}$ | N/m² or Pa. Typically $G \approx Y/3$ for metals. (Only solids have shear modulus!). |
| Bulk Modulus ($B$) | $B = \frac{\text{Hydraulic Stress}}{\text{Volumetric Strain}} = \frac{-\Delta P}{\Delta V / V} = -V \frac{\Delta P}{\Delta V}$ | N/m² or Pa. Water: $2.2\text{ GPa}$, Air: $1.0 \times 10^{-4}\text{ GPa}$. (Solids > Liquids > Gases). |
| Compressibility ($K$) | $K = \frac{1}{B} = -\frac{1}{V} \frac{\Delta V}{\Delta P}$ | N¯¹·m² or Pa¯¹. Reciprocal of Bulk Modulus. Gases are highly compressible. |
4. Poisson's Ratio (ν) & Theoretical/Practical Limits
When a wire is stretched longitudinally, its length increases while its diameter contracts laterally:
Poisson's Ratio (ν) Formula & Limits
\nu = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = -\frac{\Delta d / d}{\Delta L / L} = -\frac{L \Delta d}{d \Delta L}
• Theoretical Limits: $-1 \le \nu \le 0.5$.
• Practical Limits for Real Solids: $0 \le \nu \le 0.5$.
• Cork: $\nu \approx 0$ (no lateral expansion when compressed).
• Paraffin Rubber / Liquids: $\nu \approx 0.5$ (Incompressible, constant volume!).
• Steel: $\nu \approx 0.28\text{ to }0.30$.
• Theoretical Limits: $-1 \le \nu \le 0.5$.
• Practical Limits for Real Solids: $0 \le \nu \le 0.5$.
• Cork: $\nu \approx 0$ (no lateral expansion when compressed).
• Paraffin Rubber / Liquids: $\nu \approx 0.5$ (Incompressible, constant volume!).
• Steel: $\nu \approx 0.28\text{ to }0.30$.
5. Elastic Potential Energy in Stretched Wire & Energy Density
Work done against internal interatomic restoring forces during wire stretching is stored as Elastic Potential Energy ($U$):
Energy Formulas in Stretched Wire
1. Total Elastic Potential Energy ($U$):
$U = \frac{1}{2} \times \text{Stretching Force} \times \text{Elongation} = \frac{1}{2} F \Delta L$.
Since $F = \frac{Y A \Delta L}{L} \implies$ U = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}.
2. Energy Density ($u = U / \text{Volume}$):
$u = \frac{\frac{1}{2} F \Delta L}{A L} = \frac{1}{2} \left( \frac{F}{A} \right) \left( \frac{\Delta L}{L} \right) = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.
Since $\text{Stress} = Y \times \text{Strain} \implies$ u = \frac{1}{2} Y \epsilon^2 = \frac{\sigma^2}{2Y}.
$U = \frac{1}{2} \times \text{Stretching Force} \times \text{Elongation} = \frac{1}{2} F \Delta L$.
Since $F = \frac{Y A \Delta L}{L} \implies$ U = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}.
2. Energy Density ($u = U / \text{Volume}$):
$u = \frac{\frac{1}{2} F \Delta L}{A L} = \frac{1}{2} \left( \frac{F}{A} \right) \left( \frac{\Delta L}{L} \right) = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.
Since $\text{Stress} = Y \times \text{Strain} \implies$ u = \frac{1}{2} Y \epsilon^2 = \frac{\sigma^2}{2Y}.
6. Engineering Applications: I-Beams & Max Mountain Height
🏗️ Bending of Beams & I-Shaped Girders
A rectangular beam of length $L$, breadth $b$, and depth $d$ loaded with weight $W$ at center sags by depression:
δ = (W L³) / (4 Y b d³)
To minimize sag $\delta$:
• Increase depth $d$ (since $\delta \propto 1/d^3$).
• Use I-shaped Girders for bridges & cranes: Provides large depth $d$ and high bending resistance while eliminating excess material from central neutral axis, reducing structural weight and preventing buckling!
δ = (W L³) / (4 Y b d³)
To minimize sag $\delta$:
• Increase depth $d$ (since $\delta \propto 1/d^3$).
• Use I-shaped Girders for bridges & cranes: Provides large depth $d$ and high bending resistance while eliminating excess material from central neutral axis, reducing structural weight and preventing buckling!
🏔️ Maximum Height of Mountain on Earth
At mountain base, rock pressure $P = h \rho g$ must not exceed ultimate shearing strength of mountain rock ($\sigma_s \approx 3 \times 10^8\text{ N/m}^2$):
$h_{\text{max}} \rho g = \sigma_s \implies h_{\text{max}} = \frac{\sigma_s}{\rho g}$.
With rock density $\rho \approx 3000\text{ kg/m}^3$ and $g = 9.8\text{ m/s}^2$:
$h_{\text{max}} = \frac{3 \times 10^8}{3000 \times 9.8} \approx \mathbf{10\text{ km}}$.
This explains why Mount Everest (8.85 km) is near the maximum possible mountain height on Earth!
$h_{\text{max}} \rho g = \sigma_s \implies h_{\text{max}} = \frac{\sigma_s}{\rho g}$.
With rock density $\rho \approx 3000\text{ kg/m}^3$ and $g = 9.8\text{ m/s}^2$:
$h_{\text{max}} = \frac{3 \times 10^8}{3000 \times 9.8} \approx \mathbf{10\text{ km}}$.
This explains why Mount Everest (8.85 km) is near the maximum possible mountain height on Earth!
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
Q1. [JEE Main] A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate: (a) Stress, (b) Elongation, (c) Strain. (Given: Y for steel = 2.0 x 10^11 N/m^2).
Given: $r = 10\text{ mm} = 10^{-2}\text{ m}, L = 1.0\text{ m}, F = 100\text{ kN} = 10^5\text{ N}, Y = 2 \times 10^{11}\text{ N/m}^2$.
Area $A = \pi r^2 = 3.1416 \times (10^{-2})^2 = 3.1416 \times 10^{-4}\text{ m}^2$.
(a) Stress $\sigma = \frac{F}{A} = \frac{10^5}{3.1416 \times 10^{-4}} = \mathbf{3.18 \times 10^8\text{ N/m}^2}$.
(b) Elongation $\Delta L = \frac{F L}{A Y} = \frac{10^5 \times 1.0}{3.1416 \times 10^{-4} \times 2 \times 10^{11}} = \frac{10^5}{6.283 \times 10^7} = \mathbf{1.59 \times 10^{-3}\text{ m}} = \mathbf{1.59\text{ mm}}$.
(c) Strain $\epsilon = \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3}}{1.0} = \mathbf{1.59 \times 10^{-3}}$.
Area $A = \pi r^2 = 3.1416 \times (10^{-2})^2 = 3.1416 \times 10^{-4}\text{ m}^2$.
(a) Stress $\sigma = \frac{F}{A} = \frac{10^5}{3.1416 \times 10^{-4}} = \mathbf{3.18 \times 10^8\text{ N/m}^2}$.
(b) Elongation $\Delta L = \frac{F L}{A Y} = \frac{10^5 \times 1.0}{3.1416 \times 10^{-4} \times 2 \times 10^{11}} = \frac{10^5}{6.283 \times 10^7} = \mathbf{1.59 \times 10^{-3}\text{ m}} = \mathbf{1.59\text{ mm}}$.
(c) Strain $\epsilon = \frac{\Delta L}{L} = \frac{1.59 \times 10^{-3}}{1.0} = \mathbf{1.59 \times 10^{-3}}$.
(a) Stress = 3.18 x 10^8 N/m^2 | (b) Elongation = 1.59 mm | (c) Strain = 1.59 x 10^-3.
Q2. [NEET] 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. Calculate the load applied. (Y_steel = 2.0 x 10^11 N/m^2, Y_copper = 1.1 x 10^11 N/m^2).
Both wires experience same tension $F$ and have same cross-sectional area $A = \pi (1.5 \times 10^{-3})^2 = 7.07 \times 10^{-6}\text{ m}^2$.
Total elongation $\Delta L_{\text{total}} = \Delta L_{\text{copper}} + \Delta L_{\text{steel}} = \frac{F L_c}{A Y_c} + \frac{F L_s}{A Y_s} = \frac{F}{A} \left( \frac{L_c}{Y_c} + \frac{L_s}{Y_s} \right)$.
$\frac{L_c}{Y_c} = \frac{2.2}{1.1 \times 10^{11}} = 2.0 \times 10^{-11}$; $\frac{L_s}{Y_s} = \frac{1.6}{2.0 \times 10^{11}} = 0.8 \times 10^{-11}$.
$\Delta L_{\text{total}} = \frac{F}{7.07 \times 10^{-6}} \times (2.8 \times 10^{-11}) = F \times 3.96 \times 10^{-6}\text{ m}$.
$0.70 \times 10^{-3} = F \times 3.96 \times 10^{-6} \implies F = \frac{0.70 \times 10^{-3}}{3.96 \times 10^{-6}} = \mathbf{176.8\text{ N}}$.
Total elongation $\Delta L_{\text{total}} = \Delta L_{\text{copper}} + \Delta L_{\text{steel}} = \frac{F L_c}{A Y_c} + \frac{F L_s}{A Y_s} = \frac{F}{A} \left( \frac{L_c}{Y_c} + \frac{L_s}{Y_s} \right)$.
$\frac{L_c}{Y_c} = \frac{2.2}{1.1 \times 10^{11}} = 2.0 \times 10^{-11}$; $\frac{L_s}{Y_s} = \frac{1.6}{2.0 \times 10^{11}} = 0.8 \times 10^{-11}$.
$\Delta L_{\text{total}} = \frac{F}{7.07 \times 10^{-6}} \times (2.8 \times 10^{-11}) = F \times 3.96 \times 10^{-6}\text{ m}$.
$0.70 \times 10^{-3} = F \times 3.96 \times 10^{-6} \implies F = \frac{0.70 \times 10^{-3}}{3.96 \times 10^{-6}} = \mathbf{176.8\text{ N}}$.
Stretching Load F = 176.8 N.
Q3. [CBSE Board] Calculate the work done in stretching a steel wire of length 2 m and cross-sectional area 1 mm^2 by 1 mm. (Y = 2 x 10^11 N/m^2).
Given: $L = 2\text{ m}, A = 1\text{ mm}^2 = 10^{-6}\text{ m}^2, \Delta L = 1\text{ mm} = 10^{-3}\text{ m}, Y = 2 \times 10^{11}\text{ N/m}^2$.
Work done formula: $W = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$.
$W = \frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6} \times (10^{-3})^2}{2} = \frac{2 \times 10^5 \times 10^{-6}}{4} = \frac{0.2}{4} = \mathbf{0.05\text{ Joules}}$.
Work done formula: $W = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$.
$W = \frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6} \times (10^{-3})^2}{2} = \frac{2 \times 10^5 \times 10^{-6}}{4} = \frac{0.2}{4} = \mathbf{0.05\text{ Joules}}$.
Work Done W = 0.05 J.
Q4. [JEE Main] The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression ΔV/V of water at the bottom of the ocean. (Given: Bulk modulus of water B = 2.2 x 10^9 N/m^2, density of water ρ = 1000 kg/m^3, g = 10 m/s^2).
Hydraulic pressure at depth $h = 3000\text{ m}$: $\Delta P = \rho g h = 1000 \times 10 \times 3000 = 3.0 \times 10^7\text{ N/m}^2$.
Bulk modulus formula: $B = \frac{\Delta P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{\Delta P}{B}$.
$\frac{\Delta V}{V} = \frac{3.0 \times 10^7}{2.2 \times 10^9} = \mathbf{1.36 \times 10^{-2}} = \mathbf{1.36\%}$.
Bulk modulus formula: $B = \frac{\Delta P}{\Delta V / V} \implies \frac{\Delta V}{V} = \frac{\Delta P}{B}$.
$\frac{\Delta V}{V} = \frac{3.0 \times 10^7}{2.2 \times 10^9} = \mathbf{1.36 \times 10^{-2}} = \mathbf{1.36\%}$.
Fractional Volume Compression ΔV/V = 1.36%.
Q5. A steel wire of length 4 m and diameter 5 mm is stretched by 2 mm. Calculate the lateral contraction in diameter if Poisson's ratio for steel is 0.3.
Given: $L = 4\text{ m}, d = 5\text{ mm}, \Delta L = 2\text{ mm} = 2 \times 10^{-3}\text{ m}, \nu = 0.3$.
Poisson's ratio formula: $\nu = \frac{\Delta d / d}{\Delta L / L} \implies \frac{\Delta d}{d} = \nu \times \frac{\Delta L}{L}$.
$\Delta d = \nu \times d \times \frac{\Delta L}{L} = 0.3 \times 5\text{ mm} \times \frac{2 \times 10^{-3}}{4} = 1.5 \times 0.5 \times 10^{-3}\text{ mm} = \mathbf{7.5 \times 10^{-4}\text{ mm}} = \mathbf{0.75\text{ \mu m}}$.
Poisson's ratio formula: $\nu = \frac{\Delta d / d}{\Delta L / L} \implies \frac{\Delta d}{d} = \nu \times \frac{\Delta L}{L}$.
$\Delta d = \nu \times d \times \frac{\Delta L}{L} = 0.3 \times 5\text{ mm} \times \frac{2 \times 10^{-3}}{4} = 1.5 \times 0.5 \times 10^{-3}\text{ mm} = \mathbf{7.5 \times 10^{-4}\text{ mm}} = \mathbf{0.75\text{ \mu m}}$.
Lateral Contraction in Diameter Δd = 0.75 μm.
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8. Frequently Asked Questions (FAQ)
Why is steel more elastic than rubber in physics?
Because Young's Modulus measures resistance to deformation ($Y = \text{Stress}/\text{Strain}$). Steel requires a vastly larger stress for the same strain than rubber ($Y_{\text{steel}} = 2 \times 10^{11}\text{ N/m}^2$ vs $Y_{\text{rubber}} = 10^7\text{ N/m}^2$), making steel 20,000 times more elastic.
What are the theoretical and practical limits of Poisson's Ratio (ν)?
The theoretical limits for Poisson's ratio are $-1 \le \nu \le 0.5$. However, for all practical isotropic solid materials, Poisson's ratio ranges between $0 \le \nu \le 0.5$.
Why are structural beams manufactured with an I-shaped cross-section?
Bending depression $\delta = (WL^3)/(4Ybd^3)$ depends inversely on depth cubed ($1/d^3$). An I-shaped girder provides large depth $d$ and high resistance to bending while eliminating material from the central neutral axis, reducing structural weight and preventing side-buckling.
What is the formula for elastic potential energy in a stretched wire?
The total elastic potential energy is $U = \frac{1}{2} F \Delta L = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$, and energy per unit volume (energy density) is $u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \epsilon^2$.
What limits the maximum height of a mountain on Earth to ~10 km?
The mountain's base rock experiences pressure $P = h \rho g$. If $P$ exceeds the ultimate shearing strength of rock ($\sigma_s \approx 3 \times 10^8\text{ N/m}^2$), the base rock yields and flows plastically, limiting $h_{\text{max}} = \sigma_s / (\rho g) \approx 10\text{ km}$.
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