This MCQ module is based on: Stress Strain Hookes Law
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Stress Strain Hookes Law
🎓 Class 11
Physics
CBSE
Theory
Ch 8 – Mechanical Properties of Solids
⏱ ~14 min
🌐 Language: [gtranslate]
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Stress Strain Hookes Law
8.1 Introduction: Elastic and Plastic Bodies
In earlier chapters we treated objects as rigid bodies — objects whose shape and size remain unchanged when forces act on them. In practice, every real body deforms when an external force is applied. The study of how materials respond to forces and deformations is the subject of this chapter.
When you stretch a rubber band and release it, it returns to its original shape. When you press a lump of clay, it stays deformed. These two contrasting behaviours define the two fundamental categories of material response:
Elasticity: The property of a body by which it regains its original shape and size after the deforming force is removed. Examples: steel springs, rubber bands, quartz fibres.
Plasticity: The property of a body by which it does not regain its original shape and size after the deforming force is removed. The deformation is permanent. Examples: clay, putty, mud, dough.
No material is perfectly elastic or perfectly plastic. Steel is highly elastic (not perfectly), while putty is nearly plastic. Even a rubber ball, though elastic, shows slight permanent deformation after repeated use.
8.2 Stress and Strain
Stress
When an external force is applied to a body, the body gets deformed. Internal restoring forces develop within the body that oppose the deformation. At equilibrium, the restoring force equals the applied force in magnitude.
Stress is defined as the internal restoring force per unit cross-sectional area of the body:
\[\text{Stress} = \frac{F}{A}\]
SI unit: N/m² = pascal (Pa). Dimensions: \([ML^{-1}T^{-2}]\)
Stress depends on both the magnitude of the force and the area over which it acts. A small force on a thin wire produces greater stress than the same force on a thick rod.
Types of Stress
| Type | Description | Force Direction |
|---|---|---|
| Tensile Stress | Developed when a body is stretched (elongated) by two equal and opposite forces | Perpendicular to cross-section, pulling outward |
| Compressive Stress | Developed when a body is compressed by two equal and opposite forces pushing inward | Perpendicular to cross-section, pushing inward |
| Shearing (Tangential) Stress | Developed when a tangential force acts parallel to the surface of the body | Parallel to the cross-section |
Strain
Strain is defined as the fractional change in dimension (length, volume, or shape) of a body. It is the ratio of change in dimension to the original dimension. Strain has no unit — it is a dimensionless quantity.
| Type of Strain | Formula | Associated Stress |
|---|---|---|
| Longitudinal (Linear) Strain | \(\displaystyle\frac{\Delta L}{L}\) | Tensile or Compressive |
| Volumetric Strain | \(\displaystyle\frac{\Delta V}{V}\) | Hydraulic (all-round pressure) |
| Shearing Strain | \(\displaystyle\frac{\Delta x}{L} = \tan\theta \approx \theta\) (for small θ) | Shearing Stress |
Key Point: For shearing strain, \(\Delta x\) is the lateral displacement of one face relative to the opposite fixed face, and \(L\) is the distance between the two faces. When the angle \(\theta\) is small, \(\tan\theta \approx \theta\) (in radians).
8.3 Hooke's Law
In 1676, Robert Hooke discovered that for most solid materials, stress and strain are directly proportional to each other within a certain limit called the elastic limit.
Hooke's Law: Within the elastic limit, the stress developed in a body is directly proportional to the strain produced.
\[\text{Stress} \propto \text{Strain}\]
\[\text{Stress} = E \times \text{Strain}\]
where \(E\) is called the modulus of elasticity (or elastic modulus). Its unit is Pa (same as stress) since strain is dimensionless.
Stress-Strain Curve
A stress-strain curve is obtained by gradually increasing the load on a test specimen (typically a metallic wire) and plotting the resulting stress against strain. The curve reveals several important features of material behaviour:
Interactive: Stress-Strain Curve Explorer L2 Understand
Hover over each region of the curve below to learn about the material behaviour in that zone.
Click a button above to explore each region of the stress-strain curve.
Ductile vs Brittle Materials
Ductile materials (like copper, aluminium, mild steel) exhibit a large plastic region before fracture. They can be drawn into wires. The stress-strain curve shows a significant gap between the yield point and the fracture point.
Brittle materials (like cast iron, glass, ceramics) fracture soon after the elastic limit with little or no plastic deformation. Their stress-strain curves show almost no plastic region.
Elastomers: Materials like rubber have no well-defined plastic region but can sustain very large strains (up to 800%). They do not obey Hooke's law. The stress-strain curve is non-linear and shows a large elastic region, but on unloading the curve does not retrace the loading curve — this is called elastic hysteresis.
Worked Examples
Example 1: Calculating Tensile Stress
A steel wire of cross-sectional area \(3.0 \times 10^{-5}\) m² is subjected to a tensile force of 600 N. Calculate the tensile stress developed in the wire.
Given: \(F = 600\) N, \(A = 3.0 \times 10^{-5}\) m²
Formula: \(\text{Stress} = \frac{F}{A}\)
Calculation: \[\text{Stress} = \frac{600}{3.0 \times 10^{-5}} = 2.0 \times 10^{7} \text{ Pa} = 20 \text{ MPa}\] Dimensional check: \(\frac{[\text{N}]}{[\text{m}^2]} = \frac{[\text{kg}\cdot\text{m/s}^2]}{[\text{m}^2]} = [\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}] = [\text{Pa}]\) ✓
Answer: The tensile stress is \(\boxed{2.0 \times 10^{7} \text{ Pa}}\) or 20 MPa.
Formula: \(\text{Stress} = \frac{F}{A}\)
Calculation: \[\text{Stress} = \frac{600}{3.0 \times 10^{-5}} = 2.0 \times 10^{7} \text{ Pa} = 20 \text{ MPa}\] Dimensional check: \(\frac{[\text{N}]}{[\text{m}^2]} = \frac{[\text{kg}\cdot\text{m/s}^2]}{[\text{m}^2]} = [\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}] = [\text{Pa}]\) ✓
Answer: The tensile stress is \(\boxed{2.0 \times 10^{7} \text{ Pa}}\) or 20 MPa.
Example 2: Longitudinal Strain in a Rod
A copper rod of length 2.5 m is stretched by 0.5 mm under a tensile load. Calculate the longitudinal strain.
Given: \(L = 2.5\) m, \(\Delta L = 0.5\) mm \(= 0.5 \times 10^{-3}\) m \(= 5 \times 10^{-4}\) m
Formula: \(\text{Longitudinal Strain} = \frac{\Delta L}{L}\)
Calculation: \[\text{Strain} = \frac{5 \times 10^{-4}}{2.5} = 2.0 \times 10^{-4}\] Note: Strain is dimensionless — no unit.
Answer: The longitudinal strain is \(\boxed{2.0 \times 10^{-4}}\).
Formula: \(\text{Longitudinal Strain} = \frac{\Delta L}{L}\)
Calculation: \[\text{Strain} = \frac{5 \times 10^{-4}}{2.5} = 2.0 \times 10^{-4}\] Note: Strain is dimensionless — no unit.
Answer: The longitudinal strain is \(\boxed{2.0 \times 10^{-4}}\).
Example 3: Stress and Modulus of Elasticity
A wire of length 1.5 m and diameter 0.60 mm is stretched by 0.30 mm when a load of 8.0 kg is suspended from it. Calculate: (a) the stress, (b) the strain, and (c) the modulus of elasticity of the wire material. Take \(g = 9.8\) m/s².
Given:
Length: \(L = 1.5\) m
Diameter: \(d = 0.60\) mm \(= 6.0 \times 10^{-4}\) m, so radius \(r = 3.0 \times 10^{-4}\) m
Extension: \(\Delta L = 0.30\) mm \(= 3.0 \times 10^{-4}\) m
Mass: \(m = 8.0\) kg
(a) Stress:
Force: \(F = mg = 8.0 \times 9.8 = 78.4\) N
Cross-sectional area: \(A = \pi r^2 = \pi \times (3.0 \times 10^{-4})^2 = \pi \times 9.0 \times 10^{-8} = 2.827 \times 10^{-7}\) m²
\[\text{Stress} = \frac{F}{A} = \frac{78.4}{2.827 \times 10^{-7}} = 2.77 \times 10^{8} \text{ Pa} \approx 277 \text{ MPa}\] (b) Strain: \[\text{Strain} = \frac{\Delta L}{L} = \frac{3.0 \times 10^{-4}}{1.5} = 2.0 \times 10^{-4}\] (c) Modulus of Elasticity: \[E = \frac{\text{Stress}}{\text{Strain}} = \frac{2.77 \times 10^{8}}{2.0 \times 10^{-4}} = 1.39 \times 10^{12} \text{ Pa} \approx 1.39 \text{ TPa}\] Dimensional check: \(E\) has the same dimensions as stress: \([ML^{-1}T^{-2}]\) ✓
Answers: (a) Stress = \(\boxed{2.77 \times 10^8 \text{ Pa}}\), (b) Strain = \(\boxed{2.0 \times 10^{-4}}\), (c) \(E = \boxed{1.39 \times 10^{12} \text{ Pa}}\)
Length: \(L = 1.5\) m
Diameter: \(d = 0.60\) mm \(= 6.0 \times 10^{-4}\) m, so radius \(r = 3.0 \times 10^{-4}\) m
Extension: \(\Delta L = 0.30\) mm \(= 3.0 \times 10^{-4}\) m
Mass: \(m = 8.0\) kg
(a) Stress:
Force: \(F = mg = 8.0 \times 9.8 = 78.4\) N
Cross-sectional area: \(A = \pi r^2 = \pi \times (3.0 \times 10^{-4})^2 = \pi \times 9.0 \times 10^{-8} = 2.827 \times 10^{-7}\) m²
\[\text{Stress} = \frac{F}{A} = \frac{78.4}{2.827 \times 10^{-7}} = 2.77 \times 10^{8} \text{ Pa} \approx 277 \text{ MPa}\] (b) Strain: \[\text{Strain} = \frac{\Delta L}{L} = \frac{3.0 \times 10^{-4}}{1.5} = 2.0 \times 10^{-4}\] (c) Modulus of Elasticity: \[E = \frac{\text{Stress}}{\text{Strain}} = \frac{2.77 \times 10^{8}}{2.0 \times 10^{-4}} = 1.39 \times 10^{12} \text{ Pa} \approx 1.39 \text{ TPa}\] Dimensional check: \(E\) has the same dimensions as stress: \([ML^{-1}T^{-2}]\) ✓
Answers: (a) Stress = \(\boxed{2.77 \times 10^8 \text{ Pa}}\), (b) Strain = \(\boxed{2.0 \times 10^{-4}}\), (c) \(E = \boxed{1.39 \times 10^{12} \text{ Pa}}\)
Example 4: Shearing Strain
A metallic cube of side 5.0 cm has its upper face displaced by 0.020 cm relative to the lower fixed face when a tangential force is applied. Calculate the shearing strain and the angle of shear.
Given: Side \(L = 5.0\) cm \(= 0.050\) m, Displacement \(\Delta x = 0.020\) cm \(= 2.0 \times 10^{-4}\) m
Shearing strain: \[\text{Shearing strain} = \frac{\Delta x}{L} = \frac{2.0 \times 10^{-4}}{0.050} = 4.0 \times 10^{-3}\] Angle of shear: \[\theta = \tan^{-1}\left(\frac{\Delta x}{L}\right) \approx \frac{\Delta x}{L} = 4.0 \times 10^{-3} \text{ rad}\] (Since the strain is very small, \(\tan\theta \approx \theta\))
Converting to degrees: \(\theta = 4.0 \times 10^{-3} \times \frac{180}{\pi} \approx 0.23°\)
Answer: Shearing strain = \(\boxed{4.0 \times 10^{-3}}\), Angle of shear \(\approx 0.23°\)
Shearing strain: \[\text{Shearing strain} = \frac{\Delta x}{L} = \frac{2.0 \times 10^{-4}}{0.050} = 4.0 \times 10^{-3}\] Angle of shear: \[\theta = \tan^{-1}\left(\frac{\Delta x}{L}\right) \approx \frac{\Delta x}{L} = 4.0 \times 10^{-3} \text{ rad}\] (Since the strain is very small, \(\tan\theta \approx \theta\))
Converting to degrees: \(\theta = 4.0 \times 10^{-3} \times \frac{180}{\pi} \approx 0.23°\)
Answer: Shearing strain = \(\boxed{4.0 \times 10^{-3}}\), Angle of shear \(\approx 0.23°\)
Activity — Exploring Elasticity of Different Materials
L3 Apply
Predict first: If you hang a weight from a rubber band and a steel wire of similar dimensions, which one will stretch more? Which one will return to its original length more precisely?
- Take a thin rubber band, a steel spring (from a pen), and a piece of modelling clay.
- Hang a small weight (say 100 g) from each and measure the extension.
- Remove the weight and observe whether each returns to its original length.
- Record which material showed the most extension and which showed permanent deformation.
Observation:
The rubber band stretches the most but returns to its original length (elastic). The steel spring stretches very little but also returns perfectly (elastic with high modulus). The clay stretches and stays deformed (plastic).
Conclusion: Steel has the highest elastic modulus (stiffest), rubber has the lowest (most compliant), and clay is plastic. Elasticity is not about how much something stretches — it is about whether it returns to its original shape. Steel is more elastic than rubber in the physics sense because it recovers its shape more perfectly.
The rubber band stretches the most but returns to its original length (elastic). The steel spring stretches very little but also returns perfectly (elastic with high modulus). The clay stretches and stays deformed (plastic).
Conclusion: Steel has the highest elastic modulus (stiffest), rubber has the lowest (most compliant), and clay is plastic. Elasticity is not about how much something stretches — it is about whether it returns to its original shape. Steel is more elastic than rubber in the physics sense because it recovers its shape more perfectly.
Competency-Based Questions
An engineer is testing a new alloy wire for use in suspension bridges. A wire sample of length 2.0 m and diameter 1.0 mm is subjected to increasing tensile loads. The wire stretches by 0.40 mm under a load of 50 N. At a load of 200 N, the wire shows permanent deformation. At 350 N, the wire snaps.
Q1. L1 Remember What is the SI unit of stress?
Answer: B. Stress = Force/Area. Its SI unit is N/m², also called the pascal (Pa).
Q2. L3 Apply Calculate the stress in the wire under the 50 N load. (2 marks)
Answer:
Diameter \(d = 1.0\) mm \(= 1.0 \times 10^{-3}\) m, so \(r = 5.0 \times 10^{-4}\) m
\(A = \pi r^2 = \pi \times (5.0 \times 10^{-4})^2 = 7.854 \times 10^{-7}\) m²
\(\text{Stress} = \frac{50}{7.854 \times 10^{-7}} = 6.37 \times 10^{7}\) Pa \(\approx 63.7\) MPa
Diameter \(d = 1.0\) mm \(= 1.0 \times 10^{-3}\) m, so \(r = 5.0 \times 10^{-4}\) m
\(A = \pi r^2 = \pi \times (5.0 \times 10^{-4})^2 = 7.854 \times 10^{-7}\) m²
\(\text{Stress} = \frac{50}{7.854 \times 10^{-7}} = 6.37 \times 10^{7}\) Pa \(\approx 63.7\) MPa
Q3. L3 Apply Calculate the longitudinal strain and the modulus of elasticity of the alloy. (3 marks)
Answer:
Strain \(= \frac{\Delta L}{L} = \frac{0.40 \times 10^{-3}}{2.0} = 2.0 \times 10^{-4}\)
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{6.37 \times 10^{7}}{2.0 \times 10^{-4}} = 3.18 \times 10^{11}\) Pa \(\approx 318\) GPa
Strain \(= \frac{\Delta L}{L} = \frac{0.40 \times 10^{-3}}{2.0} = 2.0 \times 10^{-4}\)
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{6.37 \times 10^{7}}{2.0 \times 10^{-4}} = 3.18 \times 10^{11}\) Pa \(\approx 318\) GPa
Q4. L4 Analyse Identify the elastic limit and the breaking stress of the wire. Is this material ductile or brittle? Justify your answer. (3 marks)
Answer:
The elastic limit corresponds to the load at which permanent deformation begins = 200 N.
Elastic limit stress \(= \frac{200}{7.854 \times 10^{-7}} = 2.55 \times 10^{8}\) Pa \(\approx 255\) MPa.
Breaking stress \(= \frac{350}{7.854 \times 10^{-7}} = 4.46 \times 10^{8}\) Pa \(\approx 446\) MPa.
The wire is ductile because there is a significant difference between the elastic limit (200 N) and breaking point (350 N), meaning the material undergoes substantial plastic deformation before fracture.
The elastic limit corresponds to the load at which permanent deformation begins = 200 N.
Elastic limit stress \(= \frac{200}{7.854 \times 10^{-7}} = 2.55 \times 10^{8}\) Pa \(\approx 255\) MPa.
Breaking stress \(= \frac{350}{7.854 \times 10^{-7}} = 4.46 \times 10^{8}\) Pa \(\approx 446\) MPa.
The wire is ductile because there is a significant difference between the elastic limit (200 N) and breaking point (350 N), meaning the material undergoes substantial plastic deformation before fracture.
Q5. L5 Evaluate The engineer claims this alloy is suitable for suspension bridge cables that must support a maximum load of 100 N per wire without permanent deformation. Evaluate this claim. (2 marks)
Answer:
The elastic limit of the wire is at 200 N. Since the maximum working load is 100 N (which is half the elastic limit), the wire operates well within its elastic region with a safety factor of 2. The claim is valid — the wire will not undergo permanent deformation under 100 N. However, for critical applications like bridges, a safety factor of at least 3-5 is typically recommended, so a more thorough evaluation would be needed.
The elastic limit of the wire is at 200 N. Since the maximum working load is 100 N (which is half the elastic limit), the wire operates well within its elastic region with a safety factor of 2. The claim is valid — the wire will not undergo permanent deformation under 100 N. However, for critical applications like bridges, a safety factor of at least 3-5 is typically recommended, so a more thorough evaluation would be needed.
Assertion-Reason Questions
Assertion (A): Steel is more elastic than rubber.
Reason (R): Steel has a much larger Young's modulus than rubber, meaning it deforms far less for the same applied stress.
Answer: A. Steel has a Young's modulus of about 200 GPa compared to rubber's ~0.01 GPa. A more elastic material is one that resists deformation better and returns to its original shape more completely — steel does this far better than rubber. The larger modulus of elasticity makes steel more elastic, and R correctly explains this.
Assertion (A): Strain is a dimensionless quantity.
Reason (R): Strain is the ratio of two quantities having the same dimensions (change in dimension / original dimension).
Answer: A. Strain = \(\Delta L / L\) = [L]/[L] = dimensionless. Since it is a ratio of two quantities with the same dimension, the dimensions cancel out. R correctly explains A.
Assertion (A): The stress-strain curve for rubber does not have a well-defined straight-line (linear) region.
Reason (R): Rubber does not obey Hooke's law.
Answer: A. Rubber's stress-strain curve is non-linear from the start because the polymer chains uncoil progressively. Since stress is not proportional to strain, Hooke's law does not apply. The non-linear curve is a direct consequence of not following Hooke's law, so R correctly explains A.
Frequently Asked Questions - Stress Strain Hookes Law
What is the main concept covered in Stress Strain Hookes Law?
In NCERT Class 11 Physics Chapter 8 (Mechanical Properties of Solids), "Stress Strain Hookes Law" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Stress Strain Hookes Law useful in real-life applications?
Real-life applications of Stress Strain Hookes Law from NCERT Class 11 Physics Chapter 8 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Stress Strain Hookes Law?
Key formulas in Stress Strain Hookes Law (NCERT Class 11 Physics Chapter 8 Mechanical Properties of Solids) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 8?
NCERT Class 11 Physics Chapter 8 (Mechanical Properties of Solids) is structured so each part builds on the previous one. Stress Strain Hookes Law connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Stress Strain Hookes Law?
CBSE board questions from Stress Strain Hookes Law typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Stress Strain Hookes Law lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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