This MCQ module is based on: Elastic Moduli
TOPIC 2 OF 33
Elastic Moduli
🎓 Class 11
Physics
CBSE
Theory
Ch 8 – Mechanical Properties of Solids
⏱ ~14 min
🌐 Language: [gtranslate]
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Elastic Moduli
8.4 Young's Modulus (Y)
The ratio of longitudinal (tensile or compressive) stress to longitudinal strain is called Young's modulus, denoted by \(Y\):
Young's Modulus:
\[Y = \frac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}} = \frac{F/A}{\Delta L/L} = \frac{F \cdot L}{A \cdot \Delta L}\]
SI unit: Pa (or N/m²). Common practical units: GPa (\(10^9\) Pa).
Young's modulus applies only to solids — fluids do not have a definite shape to resist longitudinal deformation.
Young's modulus applies only to solids — fluids do not have a definite shape to resist longitudinal deformation.
Table 8.1 — Young's Modulus of Common Materials
| Material | Young's Modulus Y (GPa) |
|---|---|
| Steel | 200 |
| Copper | 120 |
| Aluminium | 70 |
| Brass | 91 |
| Iron (wrought) | 190 |
| Glass | 65 |
| Bone | 9 – 16 |
| Rubber | 0.01 – 0.10 |
| Tungsten | 410 |
Experimental Determination: Searle's apparatus uses two identical long wires — one reference and one test wire — suspended from a rigid support. The test wire is loaded with slotted weights and the extension is measured using a spirit level and micrometer screw. Using the formula \(Y = \frac{FL}{A \cdot \Delta L}\), the Young's modulus is calculated.
Worked Examples — Young's Modulus
Example 1 (NCERT 8.1): Steel Beam Elongation
A structural steel rod has a radius of 10 mm and a length of 1.0 m. A force of \(100 \times 10^3\) N (100 kN) stretches it along its length. Calculate the elongation. Given: \(Y_{\text{steel}} = 2.0 \times 10^{11}\) Pa.
Given:
\(r = 10\) mm \(= 0.01\) m, \(L = 1.0\) m, \(F = 100 \times 10^3 = 10^5\) N
\(Y = 2.0 \times 10^{11}\) Pa
Step 1: Cross-sectional area \[A = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} = 3.14 \times 10^{-4} \text{ m}^2\] Step 2: Using \(Y = \frac{FL}{A \cdot \Delta L}\), solve for \(\Delta L\): \[\Delta L = \frac{FL}{AY} = \frac{10^5 \times 1.0}{3.14 \times 10^{-4} \times 2.0 \times 10^{11}}\] \[\Delta L = \frac{10^5}{6.28 \times 10^{7}} = 1.59 \times 10^{-3} \text{ m} \approx 1.59 \text{ mm}\] Dimensional check: \(\frac{[\text{N}][\text{m}]}{[\text{m}^2][\text{Pa}]} = \frac{[\text{N}\cdot\text{m}]}{[\text{m}^2]\cdot[\text{N/m}^2]} = [\text{m}]\) ✓
Answer: The elongation is \(\boxed{1.59 \text{ mm}}\).
\(r = 10\) mm \(= 0.01\) m, \(L = 1.0\) m, \(F = 100 \times 10^3 = 10^5\) N
\(Y = 2.0 \times 10^{11}\) Pa
Step 1: Cross-sectional area \[A = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} = 3.14 \times 10^{-4} \text{ m}^2\] Step 2: Using \(Y = \frac{FL}{A \cdot \Delta L}\), solve for \(\Delta L\): \[\Delta L = \frac{FL}{AY} = \frac{10^5 \times 1.0}{3.14 \times 10^{-4} \times 2.0 \times 10^{11}}\] \[\Delta L = \frac{10^5}{6.28 \times 10^{7}} = 1.59 \times 10^{-3} \text{ m} \approx 1.59 \text{ mm}\] Dimensional check: \(\frac{[\text{N}][\text{m}]}{[\text{m}^2][\text{Pa}]} = \frac{[\text{N}\cdot\text{m}]}{[\text{m}^2]\cdot[\text{N/m}^2]} = [\text{m}]\) ✓
Answer: The elongation is \(\boxed{1.59 \text{ mm}}\).
Example 2: Copper Wire Under Load
A copper wire of length 3.0 m and diameter 0.80 mm is stretched by a force of 50 N. Find the extension produced. Given: \(Y_{\text{Cu}} = 1.2 \times 10^{11}\) Pa.
Given:
\(L = 3.0\) m, \(d = 0.80\) mm \(\Rightarrow r = 0.40\) mm \(= 4.0 \times 10^{-4}\) m
\(F = 50\) N, \(Y = 1.2 \times 10^{11}\) Pa
Step 1: \(A = \pi r^2 = \pi \times (4.0 \times 10^{-4})^2 = \pi \times 1.6 \times 10^{-7} = 5.027 \times 10^{-7}\) m²
Step 2: \[\Delta L = \frac{FL}{AY} = \frac{50 \times 3.0}{5.027 \times 10^{-7} \times 1.2 \times 10^{11}} = \frac{150}{6.032 \times 10^{4}} = 2.49 \times 10^{-3} \text{ m}\] Answer: \(\Delta L = \boxed{2.49 \text{ mm}}\)
\(L = 3.0\) m, \(d = 0.80\) mm \(\Rightarrow r = 0.40\) mm \(= 4.0 \times 10^{-4}\) m
\(F = 50\) N, \(Y = 1.2 \times 10^{11}\) Pa
Step 1: \(A = \pi r^2 = \pi \times (4.0 \times 10^{-4})^2 = \pi \times 1.6 \times 10^{-7} = 5.027 \times 10^{-7}\) m²
Step 2: \[\Delta L = \frac{FL}{AY} = \frac{50 \times 3.0}{5.027 \times 10^{-7} \times 1.2 \times 10^{11}} = \frac{150}{6.032 \times 10^{4}} = 2.49 \times 10^{-3} \text{ m}\] Answer: \(\Delta L = \boxed{2.49 \text{ mm}}\)
Example 3: Comparing Two Wires
Two wires of the same material have lengths in the ratio 1:2 and diameters in the ratio 2:1. If the same force is applied to both, find the ratio of their extensions.
Let: Wire 1: length \(L\), diameter \(2d\); Wire 2: length \(2L\), diameter \(d\)
Since the material is the same, \(Y\) is the same for both.
\[\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y} = \frac{FL}{\pi (d/2)^2 Y} = \frac{4FL}{\pi d^2 Y}\] For Wire 1: \(\Delta L_1 = \frac{4F \cdot L}{\pi (2d)^2 Y} = \frac{4FL}{4\pi d^2 Y} = \frac{FL}{\pi d^2 Y}\)
For Wire 2: \(\Delta L_2 = \frac{4F \cdot 2L}{\pi d^2 Y} = \frac{8FL}{\pi d^2 Y}\)
\[\frac{\Delta L_1}{\Delta L_2} = \frac{FL/(\pi d^2 Y)}{8FL/(\pi d^2 Y)} = \frac{1}{8}\] Answer: The ratio of extensions is \(\boxed{1:8}\). The thinner, longer wire stretches 8 times more.
Since the material is the same, \(Y\) is the same for both.
\[\Delta L = \frac{FL}{AY} = \frac{FL}{\pi r^2 Y} = \frac{FL}{\pi (d/2)^2 Y} = \frac{4FL}{\pi d^2 Y}\] For Wire 1: \(\Delta L_1 = \frac{4F \cdot L}{\pi (2d)^2 Y} = \frac{4FL}{4\pi d^2 Y} = \frac{FL}{\pi d^2 Y}\)
For Wire 2: \(\Delta L_2 = \frac{4F \cdot 2L}{\pi d^2 Y} = \frac{8FL}{\pi d^2 Y}\)
\[\frac{\Delta L_1}{\Delta L_2} = \frac{FL/(\pi d^2 Y)}{8FL/(\pi d^2 Y)} = \frac{1}{8}\] Answer: The ratio of extensions is \(\boxed{1:8}\). The thinner, longer wire stretches 8 times more.
8.5 Bulk Modulus (B)
When a body is subjected to uniform pressure from all sides (hydraulic stress), its volume changes. The bulk modulus relates the applied pressure to the fractional volume change.
Bulk Modulus:
\[B = \frac{-p}{\Delta V / V} = \frac{-pV}{\Delta V}\]
The negative sign ensures \(B\) is positive (since \(\Delta V\) is negative when pressure increases — volume decreases).
SI unit: Pa. Applies to solids, liquids, and gases.
SI unit: Pa. Applies to solids, liquids, and gases.
Compressibility: The reciprocal of bulk modulus is called compressibility (\(k\)):
\[k = \frac{1}{B}\]
Unit: Pa¹ (inverse pascal). Materials with low compressibility (high B) resist volume change — solids have the highest B, gases the lowest.
| Material | Bulk Modulus B (GPa) |
|---|---|
| Steel | 160 |
| Copper | 140 |
| Aluminium | 72 |
| Glass | 37 |
| Water | 2.2 |
| Air (at STP) | 1.0 × 10¹&sup4; Pa |
Example 4: Volume Change Under Pressure
A solid copper sphere of volume \(0.50\) m³ is placed at the bottom of a lake where the pressure is \(2.0 \times 10^{6}\) Pa above atmospheric pressure. Calculate the change in volume. Given: \(B_{\text{Cu}} = 140\) GPa.
Given: \(V = 0.50\) m³, \(p = 2.0 \times 10^{6}\) Pa, \(B = 140 \times 10^{9}\) Pa
Using: \(B = \frac{-pV}{\Delta V}\) so \(\Delta V = \frac{-pV}{B}\) \[\Delta V = \frac{-2.0 \times 10^{6} \times 0.50}{140 \times 10^{9}} = \frac{-1.0 \times 10^{6}}{1.4 \times 10^{11}} = -7.14 \times 10^{-6} \text{ m}^3\] The negative sign indicates a decrease in volume.
Answer: The volume decreases by \(\boxed{7.14 \times 10^{-6} \text{ m}^3}\) or about \(7.14\) cm³.
Using: \(B = \frac{-pV}{\Delta V}\) so \(\Delta V = \frac{-pV}{B}\) \[\Delta V = \frac{-2.0 \times 10^{6} \times 0.50}{140 \times 10^{9}} = \frac{-1.0 \times 10^{6}}{1.4 \times 10^{11}} = -7.14 \times 10^{-6} \text{ m}^3\] The negative sign indicates a decrease in volume.
Answer: The volume decreases by \(\boxed{7.14 \times 10^{-6} \text{ m}^3}\) or about \(7.14\) cm³.
8.5.3 Shear Modulus / Modulus of Rigidity (G)
The shear modulus (also called modulus of rigidity) relates the tangential stress to the angular deformation (shearing strain).
Shear Modulus:
\[G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} = \frac{F/A}{\Delta x / L} = \frac{F/A}{\theta}\]
SI unit: Pa. Applies only to solids.
| Material | Shear Modulus G (GPa) |
|---|---|
| Steel | 84 |
| Copper | 42 |
| Aluminium | 25 |
| Iron | 70 |
| Glass | 23 |
| Lead | 5.6 |
Example 5: Metal Cube Under Shear
A metal cube of side 10 cm is subjected to a tangential force of \(5.0 \times 10^4\) N on its top face while the bottom face is fixed. If the shear modulus of the metal is 80 GPa, find the lateral displacement of the top face.
Given:
Side \(L = 10\) cm \(= 0.10\) m, \(F = 5.0 \times 10^4\) N
Area of top face: \(A = (0.10)^2 = 0.01\) m²
\(G = 80 \times 10^9\) Pa
Using: \(G = \frac{F/A}{\Delta x / L}\), so \(\Delta x = \frac{FL}{AG}\) \[\Delta x = \frac{5.0 \times 10^4 \times 0.10}{0.01 \times 80 \times 10^9} = \frac{5.0 \times 10^3}{8.0 \times 10^8} = 6.25 \times 10^{-6} \text{ m}\] Answer: The top face is displaced by \(\boxed{6.25 \,\mu\text{m}}\) (about 6 micrometres).
Side \(L = 10\) cm \(= 0.10\) m, \(F = 5.0 \times 10^4\) N
Area of top face: \(A = (0.10)^2 = 0.01\) m²
\(G = 80 \times 10^9\) Pa
Using: \(G = \frac{F/A}{\Delta x / L}\), so \(\Delta x = \frac{FL}{AG}\) \[\Delta x = \frac{5.0 \times 10^4 \times 0.10}{0.01 \times 80 \times 10^9} = \frac{5.0 \times 10^3}{8.0 \times 10^8} = 6.25 \times 10^{-6} \text{ m}\] Answer: The top face is displaced by \(\boxed{6.25 \,\mu\text{m}}\) (about 6 micrometres).
Example 6: Comparing Shear and Tensile Stress
A steel rod of diameter 2.0 cm and length 50 cm is subjected to: (a) a tensile force of 10 kN, (b) a shear force of 10 kN. Calculate the deformation in each case. Given: \(Y = 200\) GPa, \(G = 84\) GPa.
Given: \(d = 2.0\) cm, \(r = 0.01\) m, \(L = 0.50\) m, \(F = 10 \times 10^3 = 10^4\) N
\(A = \pi r^2 = \pi \times (0.01)^2 = 3.14 \times 10^{-4}\) m²
(a) Tensile elongation: \[\Delta L = \frac{FL}{AY} = \frac{10^4 \times 0.50}{3.14 \times 10^{-4} \times 200 \times 10^9} = \frac{5 \times 10^3}{6.28 \times 10^7} = 7.96 \times 10^{-5} \text{ m}\] \(\Delta L \approx 0.080\) mm
(b) Shear displacement: \[\Delta x = \frac{FL}{AG} = \frac{10^4 \times 0.50}{3.14 \times 10^{-4} \times 84 \times 10^9} = \frac{5 \times 10^3}{2.64 \times 10^7} = 1.89 \times 10^{-4} \text{ m}\] \(\Delta x \approx 0.189\) mm
Answer: (a) Tensile: \(\boxed{0.080 \text{ mm}}\), (b) Shear: \(\boxed{0.189 \text{ mm}}\). Shear deformation is larger because \(G < Y\).
\(A = \pi r^2 = \pi \times (0.01)^2 = 3.14 \times 10^{-4}\) m²
(a) Tensile elongation: \[\Delta L = \frac{FL}{AY} = \frac{10^4 \times 0.50}{3.14 \times 10^{-4} \times 200 \times 10^9} = \frac{5 \times 10^3}{6.28 \times 10^7} = 7.96 \times 10^{-5} \text{ m}\] \(\Delta L \approx 0.080\) mm
(b) Shear displacement: \[\Delta x = \frac{FL}{AG} = \frac{10^4 \times 0.50}{3.14 \times 10^{-4} \times 84 \times 10^9} = \frac{5 \times 10^3}{2.64 \times 10^7} = 1.89 \times 10^{-4} \text{ m}\] \(\Delta x \approx 0.189\) mm
Answer: (a) Tensile: \(\boxed{0.080 \text{ mm}}\), (b) Shear: \(\boxed{0.189 \text{ mm}}\). Shear deformation is larger because \(G < Y\).
8.5.4 Poisson's Ratio (σ)
When a wire is stretched along its length, it becomes thinner — its diameter decreases. This lateral contraction accompanies every longitudinal extension. Poisson's ratio quantifies this effect.
Poisson's Ratio:
\[\sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} = \frac{-\Delta d / d}{\Delta L / L}\]
It is dimensionless. The negative sign makes \(\sigma\) positive since lateral and longitudinal strains have opposite signs.
Range: Theoretically, \(-1 \leq \sigma \leq 0.5\). For most materials, \(0 < \sigma < 0.5\).
Range: Theoretically, \(-1 \leq \sigma \leq 0.5\). For most materials, \(0 < \sigma < 0.5\).
| Material | Poisson's Ratio (σ) |
|---|---|
| Steel | 0.28 |
| Copper | 0.33 |
| Aluminium | 0.33 |
| Glass | 0.24 |
| Rubber | 0.49 |
| Cork | ≈ 0 |
Activity — Observing Poisson's Effect
L3 Apply
Predict first: When you stretch a rubber band, what happens to its width? When you try to push a cork into a bottle, does it expand sideways? How about a rubber stopper?
- Take a wide rubber band and mark two dots across its width.
- Stretch the band along its length and observe the distance between the dots — it decreases.
- Now try pushing a rubber stopper into a bottle — it resists because it expands laterally.
- Try the same with a cork stopper — it slides in easily because it barely expands sideways.
Observation: Rubber has a high Poisson's ratio (~0.49), so it contracts significantly in width when stretched and expands when compressed. Cork has Poisson's ratio near zero, so it shows almost no lateral deformation. This is precisely why cork is ideal for bottle stoppers — it can be pushed in without expanding to jam the neck.
Key Takeaway: Poisson's ratio determines whether a material is suitable for applications involving compression into tight spaces.
Key Takeaway: Poisson's ratio determines whether a material is suitable for applications involving compression into tight spaces.
Competency-Based Questions
A civil engineer needs to select materials for a building column that will bear a compressive load of \(5.0 \times 10^5\) N. The column has a cross-section of 0.02 m² and a height of 3.0 m. The maximum allowable compression is 0.50 mm.
Q1. L1 Remember Which elastic modulus determines a material's resistance to longitudinal compression?
Answer: C. Young's modulus governs both tensile and compressive longitudinal deformations.
Q2. L3 Apply What is the minimum Young's modulus required for the column material? (3 marks)
Answer:
\(Y = \frac{FL}{A \cdot \Delta L} = \frac{5.0 \times 10^5 \times 3.0}{0.02 \times 0.50 \times 10^{-3}} = \frac{1.5 \times 10^6}{1.0 \times 10^{-5}} = 1.5 \times 10^{11}\) Pa = 150 GPa.
Steel (Y = 200 GPa) or iron (Y = 190 GPa) would work. Aluminium (70 GPa) would not.
\(Y = \frac{FL}{A \cdot \Delta L} = \frac{5.0 \times 10^5 \times 3.0}{0.02 \times 0.50 \times 10^{-3}} = \frac{1.5 \times 10^6}{1.0 \times 10^{-5}} = 1.5 \times 10^{11}\) Pa = 150 GPa.
Steel (Y = 200 GPa) or iron (Y = 190 GPa) would work. Aluminium (70 GPa) would not.
Q3. L2 Understand Explain why bulk modulus applies to all states of matter but shear modulus applies only to solids. (2 marks)
Answer: Bulk modulus measures resistance to volume change under uniform pressure. All states of matter (solids, liquids, gases) have a definite volume and resist compression, so B applies to all. Shear modulus measures resistance to shape change (angular deformation). Liquids and gases do not have a definite shape — they flow freely when tangential forces are applied — so shear modulus is zero for fluids and meaningful only for solids.
Q4. L4 Analyse A steel wire and a copper wire have the same length and diameter. The same tensile force is applied to both. Which wire elongates more and by what factor? Use: \(Y_{\text{steel}} = 200\) GPa, \(Y_{\text{Cu}} = 120\) GPa. (3 marks)
Answer:
\(\Delta L = \frac{FL}{AY}\). Since \(F\), \(L\), and \(A\) are the same:
\(\frac{\Delta L_{\text{Cu}}}{\Delta L_{\text{steel}}} = \frac{Y_{\text{steel}}}{Y_{\text{Cu}}} = \frac{200}{120} = \frac{5}{3} \approx 1.67\)
The copper wire elongates \(\frac{5}{3}\) times more than the steel wire.
\(\Delta L = \frac{FL}{AY}\). Since \(F\), \(L\), and \(A\) are the same:
\(\frac{\Delta L_{\text{Cu}}}{\Delta L_{\text{steel}}} = \frac{Y_{\text{steel}}}{Y_{\text{Cu}}} = \frac{200}{120} = \frac{5}{3} \approx 1.67\)
The copper wire elongates \(\frac{5}{3}\) times more than the steel wire.
Q5. L5 Evaluate Cork has Poisson's ratio close to zero. Explain why this makes it the ideal material for sealing wine bottles. What problems would arise if rubber (σ ≈ 0.49) were used instead? (3 marks)
Answer: When a stopper is pushed into a bottle neck, it undergoes longitudinal compression. With σ ≈ 0, cork does not expand laterally when compressed, so it slides smoothly into the bottle neck. If rubber were used, its high Poisson's ratio means it would expand significantly in diameter when compressed — making it extremely difficult to push into the bottle and potentially breaking the glass neck. Additionally, when removing the stopper, rubber would grip the neck far too tightly, while cork can be extracted with moderate force.
Assertion-Reason Questions
Assertion (A): Young's modulus is defined only for solids, not for liquids or gases.
Reason (R): Liquids and gases do not have a fixed shape, so the concept of longitudinal strain is not applicable to them.
Answer: A. Young's modulus = longitudinal stress / longitudinal strain. Since fluids do not have fixed length or shape, applying a force does not produce a well-defined longitudinal strain — the fluid simply flows. R correctly explains why Y is defined only for solids.
Assertion (A): The bulk modulus of a gas is much smaller than that of a solid.
Reason (R): Gas molecules are far apart and easily compressed, whereas solid molecules are closely packed and strongly resist compression.
Answer: A. B(steel) ~ 160 GPa while B(air) ~ 10&sup5; Pa — a factor of about 10&sup6;. The large intermolecular spacing in gases means they can be compressed easily (high compressibility, low B). Solids with tightly packed atoms resist compression strongly (low compressibility, high B). R correctly explains A.
Assertion (A): For most materials, Young's modulus is greater than the shear modulus.
Reason (R): It is easier to change the shape of a body than to change its length.
Answer: A. For steel: Y = 200 GPa, G = 84 GPa. Since Y > G, a material resists longitudinal deformation more strongly than angular (shape) deformation. R correctly explains this — changing shape (shearing) requires less stress per unit strain than changing length (stretching).
Frequently Asked Questions - Elastic Moduli
What is the main concept covered in Elastic Moduli?
In NCERT Class 11 Physics Chapter 8 (Mechanical Properties of Solids), "Elastic Moduli" 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 Elastic Moduli useful in real-life applications?
Real-life applications of Elastic Moduli 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 Elastic Moduli?
Key formulas in Elastic Moduli (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. Elastic Moduli 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 Elastic Moduli?
CBSE board questions from Elastic Moduli 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 Elastic Moduli 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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