Class 11 Physics: Mechanical Properties of Solids | Physics Q&A

CBSE Class 11 › Unit VII: Properties of Bulk Matter

Mechanical Properties of Solids

NCERT Chapter 8 • Stress, Strain, Hooke’s Law & Elastic Moduli

NCERT 2025–26 Unit VII • ~6 Marks JEE Main • 1 Question

Suspension bridge illustrating elasticity and stress distribution in solids. — Elasticity: The property that allows structures to withstand forces and return to shape.

1. Elasticity and Plasticity

Elasticity is the property of a body to regain its original size and shape when deforming forces are removed. Example: Steel, Rubber.

Plasticity is the inability of a body to regain its original shape, resulting in permanent deformation. Example: Putty, Mud.

2. Stress and Strain

When a body is deformed, internal restoring forces develop.

2.1 Stress ( $\sigma$ )

The restoring force per unit area. Unit: $\text{N/m}^2$ or Pascal (Pa).

\sigma = \dfrac{F}{A}

2.2 Strain ( $\varepsilon$ )

The fractional change in dimension. It is dimensionless.

\varepsilon = \dfrac{\Delta L}{L} \quad (\text{Longitudinal})

Visual comparison of Tensile, Compressive, and Shearing stress and strain. — Stress is the restoring force per unit area; Strain is the deformation produced.

Types of Stress & Strain

Type	Deformation	Formulae
Longitudinal	Change in Length	Tensile (stretch) or Compressive (squeeze)
Shearing (Tangential)	Change in Shape	$\sigma_s = F/A$ , $\theta = \Delta x / L$
Hydraulic (Volume)	Change in Volume	$\sigma_v = P$ (Pressure), $\varepsilon_v = -\Delta V / V$

3. Hooke’s Law

For small deformations, stress is directly proportional to strain.

\text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = k \times \text{Strain}

The constant of proportionality $k$ is the Modulus of Elasticity.

4. Stress-Strain Curve

The behavior of a wire under increasing load is described by the stress-strain curve.

Typical Stress-Strain curve for mild steel showing elastic and plastic regions. — This curve defines a material’s strength, ductility, and elasticity.

Key Regions:

Proportional Limit (A): Hooke’s law is valid. Linear region.
Elastic Limit / Yield Point (B): Max stress for elastic behavior. Beyond B, deformation is permanent.
Plastic Region (B to D): Small stress change causes large strain. Material flows.
Fracture Point (E): The wire breaks.

Ductile materials (like copper) have a large plastic region. Brittle materials (like glass) break soon after the elastic limit.

Elastomers:

Substances like rubber, elastin in arteries, and synthetic polymers can be stretched to several times their original length and still return to their original shape. They do not obey Hooke’s law over most of their range and have no well-defined plastic region.

5. Elastic Moduli

5.1 Young’s Modulus ( $Y$ )

Ratio of Longitudinal Stress to Longitudinal Strain. Measure of stiffness.

Y = \dfrac{\sigma_l}{\varepsilon_l} = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\Delta L}

Misconception Alert:

In everyday language, we say rubber is “more elastic” because it stretches more. But in physics, elasticity means resistance to deformation. Steel has a much higher Young’s modulus than rubber, so Steel is more elastic than Rubber in the scientific sense.

Problem 1: A steel wire of length 2 m and cross-section $10^{-5} \text{ m}^2$ is stretched by a load of 5 kg. Calculate change in length. ( $Y_{steel} = 2 \times 10^{11}$ Pa, $g=10$ m/s²)

Show Answer

$F = mg = 5 \times 10 = 50$ N.
$\Delta L = \dfrac{FL}{AY} = \dfrac{50 \times 2}{10^{-5} \times 2 \times 10^{11}}$
$= \dfrac{100}{2 \times 10^6} = 5 \times 10^{-5}$ m = 0.05 mm.

5.2 Shear Modulus ( $G$ or $\eta$ )

Ratio of Shearing Stress to Shearing Strain.

G = \dfrac{\text{Shear Stress}}{\text{Shear Strain}} = \dfrac{F/A}{\theta} \approx \dfrac{F/A}{\Delta x/L}

Generally, $G \approx Y/3$ .

5.3 Bulk Modulus ( $B$ )

Ratio of Hydraulic Stress to Volume Strain.

B = \dfrac{-P}{\Delta V/V}

Compressibility ( $k$ ): The reciprocal of Bulk Modulus ( $k = 1/B$ ).

Compressibility Across States:

Solids are least compressible ( $B \sim 10^{10}-10^{11}$ Pa), liquids more ( $B \sim 10^9$ Pa), and gases most compressible ( $B \sim 10^5$ Pa for air). Gases are about a million times more compressible than solids!

Problem 2: Find the density of water at a depth where pressure is 80 atm. Given density at surface $\rho_0 = 1000$ kg/m³ and $B_{water} = 2 \times 10^9$ Pa. (1 atm $\approx 10^5$ Pa)

Show Answer

$P = 80 \times 10^5 = 8 \times 10^6$ Pa.
Volume Strain $\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{8 \times 10^6}{2 \times 10^9} = 4 \times 10^{-3}$ .
Density $\rho = \dfrac{\rho_0}{1 - \Delta V/V} \approx \rho_0 (1 + \Delta V/V)$
$= 1000 (1 + 0.004) = 1004$ kg/m³.

6. Elastic Potential Energy

Work done in stretching a wire is stored as elastic potential energy.

Derivation: Energy in Stretched Wire 3 Marks

Step 1: Work Done
Average force during stretching from 0 to $\Delta L$ is $\dfrac{0+F}{2} = \dfrac{F}{2}$ .
Work $W = \text{Avg Force} \times \text{Displacement} = \dfrac{1}{2}F \Delta L$ .

Step 2: Express in Stress/Strain
$W = \dfrac{1}{2} (\text{Stress} \times A) \times (\text{Strain} \times L)$
$W = \dfrac{1}{2} \times \text{Stress} \times \text{Strain} \times (AL)$

Step 3: Energy Density ( $u$ )
Energy per unit volume $u = \dfrac{W}{V} = \dfrac{1}{2} \times \text{Stress} \times \text{Strain}$
$u = \dfrac{1}{2} Y (\text{Strain})^2$ .

7. Poisson’s Ratio ( $\sigma$ )

When a wire is stretched, it becomes longer but thinner. The ratio of lateral strain to longitudinal strain is Poisson’s Ratio.

\sigma = \dfrac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = \dfrac{-\Delta d/d}{\Delta L/L}

Theoretical limits: $-1$ to $0.5$ . For most solids: $0.2$ to $0.4$ .

Problem 3: A steel wire is extended by 0.1%. Find the percentage change in its volume if Poisson’s ratio is 0.3.

Show Answer

Longitudinal strain $\dfrac{\Delta L}{L} = 0.001$ .
Lateral strain $\dfrac{\Delta r}{r} = -\sigma \dfrac{\Delta L}{L} = -0.3(0.001) = -0.0003$ .
Volume $V = \pi r^2 L \Rightarrow \dfrac{\Delta V}{V} = 2\dfrac{\Delta r}{r} + \dfrac{\Delta L}{L}$
$= 2(-0.0003) + 0.001 = -0.0006 + 0.001 = 0.0004$ .
% Change = $0.04\%$ .

8. Applications of Elastic Behaviour

The elastic properties of materials are crucial in engineering design:

8.1 Design of Cranes and Ropes

A crane lifting 10 tonnes must use a steel rope thick enough so that stress does not exceed the yield strength ( $\sigma_y \approx 300 \times 10^6$ Pa for mild steel). The required cross-sectional area is:

A \geq \dfrac{Mg}{\sigma_y}

For safety, a factor of 10 is often used, leading to thicker ropes made of braided wires for flexibility.

8.2 I-Shaped Beams

In bridges and buildings, beams are designed with an I-shaped cross-section. This provides large depth $d$ (which reduces bending $\delta \propto 1/d^3$ ) while minimizing material and weight.

Comparison of I-beam and rectangular beam under load. — I-beams resist bending more efficiently than solid rectangular beams.

8.3 Maximum Height of Mountains

The maximum height of a mountain (~10 km) is limited by the elastic strength of rocks. At the base, the shear stress due to the weight of the mountain is $h \rho g$ . When this exceeds the critical shear stress of rock ( $\sim 3 \times 10^7$ Pa), the rock flows. Solving $h \rho g = 3 \times 10^7$ gives $h \approx 10$ km.