Class 11 : Physics (In English) – Lesson Chapter 8: Mechanical Properties of Solids
EXPLANATION & SUMMARY
⚙️ EXPLANATION SECTION
🔵 1. Introduction
When external forces act on a solid, it may deform — changing shape or size. The study of how solids respond to such forces is called mechanical properties of solids. It explains how materials stretch, compress, bend, and resist deformation.
✔️ Elastic behavior helps engineers and scientists design safe structures, bridges, and machines.
🟢 2. Elasticity and Plasticity
💡 Elasticity: The ability of a body to regain its original shape and size after the deforming force is removed.
💡 Plasticity: The inability of a body to regain its original shape once the deforming force is removed.
✔️ Rubber is highly elastic.
✔️ Clay and wax are plastic.
🔴 3. Stress and Strain
When a force is applied on a body, internal restoring forces are developed to resist deformation. These are described through stress and strain.
✏️ Stress (σ):
The internal restoring force per unit area.
➡️ Formula: σ = F / A
Unit: N/m² (Pascal)
Dimension: [ML⁻¹T⁻²]
✏️ Types of Stress:
🔹 Tensile stress → due to stretching
🔹 Compressive stress → due to squeezing
🔹 Shearing stress → due to tangential forces
✏️ Strain (ε):
It is the fractional change in dimension produced by stress.
➡️ Formula: ε = ΔL / L
It is dimensionless (no unit).
✏️ Types of Strain:
🔹 Longitudinal strain → change in length / original length
🔹 Volumetric strain → change in volume / original volume
🔹 Shear strain → angular deformation
🟡 4. Hooke’s Law
💡 Within elastic limit, stress is directly proportional to strain.
➡️ σ ∝ ε
or, σ = E × ε
Here E is the modulus of elasticity (elastic constant).
✔️ Hooke’s law is valid only within the elastic limit.
Beyond that, deformation becomes permanent.
🔵 5. Elastic Moduli (Elastic Constants)
They measure how strongly a material resists deformation.
🟢 (a) Young’s Modulus (Y)
Ratio of tensile stress to longitudinal strain.
➡️ Y = (F × L) / (A × ΔL)
Unit: N/m² (Pa)
Dimension: [ML⁻¹T⁻²]
✔️ Higher Y means greater rigidity (steel > copper > rubber).
🔴 (b) Bulk Modulus (K)
Resistance to uniform compression.
➡️ K = −(Pressure change)/(Fractional change in volume)
or, K = −ΔP / (ΔV/V)
✔️ Negative sign shows that pressure increase reduces volume.
✔️ For solids, K is large; for liquids, smaller.
🟠 (c) Shear Modulus or Rigidity Modulus (η)
Resistance to change in shape (without volume change).
➡️ η = (Tangential stress)/(Shear strain)
✔️ Fluids have η = 0 since they cannot resist shape change.
🟣 6. Relation Between Elastic Moduli
For isotropic solids (same properties in all directions):
➡️ Y = 3K(1 − 2σ)
➡️ Y = 2η(1 + σ)
where σ = Poisson’s ratio.
💡 7. Poisson’s Ratio (σ)
When a wire is stretched, its length increases and diameter decreases.
➡️ σ = (Lateral strain) / (Longitudinal strain)
✔️ It has no unit.
✔️ For most solids: 0.25 ≤ σ ≤ 0.5
✔️ For incompressible material, σ = 0.5.
🧠 8. Stress–Strain Curve
As stress increases:
🔹 Proportional limit: Stress ∝ Strain (Hooke’s law valid).
🔹 Elastic limit or Yield point: Slight permanent deformation begins.
🔹 Plastic region: Large strain for small stress.
🔹 Breaking point: Fracture occurs.
✔️ Elastic limit marks the end of reversible deformation.
🟢 9. Elastic Fatigue and Elastic Limit
✔️ Elastic limit: Maximum stress without permanent deformation.
✔️ Elastic fatigue: Loss of elasticity after repeated loading and unloading (e.g., weakened springs).
🔴 10. Behavior of Common Materials
✔️ Steel: Very rigid, high Y, ideal for bridges.
✔️ Copper: Ductile, moderate Y, used in wires.
✔️ Rubber: Flexible, low Y, large strain range.
✔️ Glass: Brittle, high Y, breaks easily.
🧩 Steel is the most elastic material because it returns almost completely to its original shape.
💡 11. Applications of Elasticity
🔹 Designing bridges, buildings, and cranes (to avoid structural failure).
🔹 Manufacturing springs and shock absorbers.
🔹 Wire experiments for measuring Y.
🔹 Understanding pressure in hydraulic systems (bulk modulus).
🔹 Choosing materials based on required elasticity.
🔵 12. Determination of Young’s Modulus (Searle’s Method)
💡 Principle:
When a known weight is attached to a wire, elongation is measured.
➡️ Y = (mgL) / (πr²ΔL)
where
m = load mass,
L = wire length,
r = wire radius,
ΔL = elongation.
✔️ Micrometer screw gauge and spirit level are used for accurate measurement.
🟢 13. Elastic Potential Energy
When a wire is stretched, work done is stored as potential energy per unit volume.
➡️ U = ½ × stress × strain
or, U = ½ × (FΔL)/Volume
✔️ Maximum energy stored just before elastic limit.
🟠 14. Modulus of Resilience
💡 The energy stored per unit volume within the elastic limit.
➡️ Ur = ½ × (σ² / Y)
✔️ High modulus of resilience = better energy absorption before deformation.
🔴 15. Factors Affecting Elasticity
🔹 Temperature: Elasticity decreases with rise in temperature.
🔹 Impurities: May increase or decrease elasticity depending on impurity type.
🔹 Hammering and annealing: Hammering increases, annealing decreases elasticity.
💡 16. Comparison of Elastic Constants
For most solids:
Y > K > η
✔️ Resistance to length change > resistance to volume change > resistance to shape change.
⚡ 17. Compressibility
It is the reciprocal of bulk modulus.
➡️ β = 1 / K
✔️ Higher β means easier compression.
✔️ Gases have very high compressibility.
🟣 18. Importance in Engineering
Knowledge of elasticity ensures that stress never exceeds elastic limit.
✔️ Used in designing bridges, aircrafts, cranes, and cables.
✔️ Helps predict safe load and failure points.
🧩 19. Relation Between Stress, Strain, and Energy
The area under stress–strain curve = strain energy per unit volume.
➡️ U = ½ × σ × ε
✔️ Represents energy stored during elastic deformation.
🟢 20. Practical Examples
🔹 Steel wire stretches slightly but returns to original → highly elastic.
🔹 Rubber band stretches a lot but imperfectly → not truly elastic.
🔹 Building materials must have large Y and moderate σ to avoid breakage.
🌿 SUMMARY SECTION (~300 words)
✔️ Elasticity: Ability to regain shape after deforming force is removed.
✔️ Stress (σ): Restoring force per unit area.
✔️ Strain (ε): Fractional change in dimension.
✔️ Hooke’s Law: Stress ∝ Strain (within elastic limit).
✔️ Elastic Moduli:
🔸 Y (Young’s modulus): Measures rigidity.
🔸 K (Bulk modulus): Measures compressibility.
🔸 η (Shear modulus): Measures resistance to shape change.
✔️ Poisson’s Ratio (σ): Ratio of lateral to longitudinal strain.
✔️ Elastic Potential Energy: U = ½ σε.
✔️ Relation: Y = 3K(1 − 2σ) = 2η(1 + σ).
✔️ Stress–Strain Curve: Explains elastic, yield, plastic, and breaking regions.
✔️ Elastic Limit: Maximum stress without permanent deformation.
✔️ Elastic Fatigue: Reduction of elasticity after repeated use.
✔️ Steel: Most elastic; rubber: most flexible.
✔️ Applications: Bridge design, spring making, material selection, and stress analysis.
📝 QUICK RECAP
🔹 Stress = Force/Area
🔹 Strain = ΔL/L
🔹 Y = σ/ε, K = −ΔP/(ΔV/V), η = stress/shear strain
🔹 Hooke’s Law valid only within elastic limit
🔹 Poisson’s Ratio has no unit
🔹 Energy stored per unit volume = ½ σε
🔹 Steel → most elastic, Rubber → most stretchable
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QUESTIONS FROM TEXTBOOK
🔷 Question 8.1
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10⁻⁵ m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
✅ Answer:
📘 For same elongation under same load:
Δl = (F × L)/(A × Y)
⇒ (L/A × Y) is constant
📘 So,
L₁ / (A₁Y₁) = L₂ / (A₂Y₂)
⇒ Y₁ / Y₂ = (L₁ / A₁) / (L₂ / A₂)
📌 L₁ = 4.7 m, A₁ = 3.0 × 10⁻⁵ m²
📌 L₂ = 3.5 m, A₂ = 4.0 × 10⁻⁵ m²
Y₁ / Y₂ = (4.7 / 3.0 × 10⁻⁵) / (3.5 / 4.0 × 10⁻⁵)
= (4.7 / 3.0) × (4.0 / 3.5) ≈ 1.5667 × 1.1429 ≈ 1.79
✅ Final Answer: 1.79
🔷 Question 8.2
Figure 8.9 shows the strain-stress curve for a given material. What are
(a) Young’s modulus and
(b) approximate yield strength for this material?
✅ Answer:
📘 From graph (read visually):
At point with stress = 300 × 10⁶ N/m², strain = 0.002
(a) Young’s modulus (Y) = Stress / Strain
= (300 × 10⁶) / 0.002 = 1.5 × 10¹¹ N/m²
(b) Yield strength ≈ maximum stress at elastic limit
= ~300 × 10⁶ N/m²
✅ Final Answers:
✔️ Young’s modulus = 1.5 × 10¹¹ N/m²
✔️ Yield strength ≈ 3.0 × 10⁸ N/m²
🔷 Question 8.3
The stress-strain graphs for materials A and B are shown in Fig. 8.10.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
✅ Answer:
(a) Greater Young’s modulus → steeper slope
⇒ Material A has higher Young’s modulus
(b) Stronger material → can withstand more stress before breaking
⇒ Material B is stronger (higher ultimate stress)
✅ Final Answers:
✔️ Young’s modulus: A > B
✔️ Strength: B > A
🔷 Question 8.4
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young’s modulus of rubber is greater than that of steel.
(b) The stretching of a coil is determined by its shear modulus.
✅ Answer:
(a) ❌ False – Rubber stretches more for same force, hence has lower Young’s modulus.
Steel resists elongation → higher Young’s modulus.
(b) ✅ True – Coils undergo deformation due to twisting (shearing), hence shear modulus is the relevant quantity.
🔷 Question 8.5
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
✅ Answer:
📌 Given:
Diameter = 0.25 cm = 0.0025 m ⇒ A = πr² = π(0.00125)² ≈ 4.91 × 10⁻⁶ m²
g = 9.8 m/s²
Mass on steel = 4.0 kg → Force F = 4 × 9.8 = 39.2 N
Length L = 1.5 m, Y (steel) = 2 × 10¹¹ N/m²
Δl (steel) = (F × L) / (A × Y)
= (39.2 × 1.5) / (4.91 × 10⁻⁶ × 2 × 10¹¹)
≈ 6.0 × 10⁻⁴ m = 0.60 mm
Mass on brass = 6.0 kg → F = 6 × 9.8 = 58.8 N
Length L = 1.0 m, Y (brass) = 0.91 × 10¹¹ N/m²
Δl (brass) = (58.8 × 1.0) / (4.91 × 10⁻⁶ × 0.91 × 10¹¹)
≈ 1.33 × 10⁻³ m = 1.33 mm
✅ Final Answers:
✔️ Elongation of steel wire = 0.60 mm
✔️ Elongation of brass wire = 1.33 mm
🔷 Question 8.6
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
✅ Answer:
📌 Given:
Edge length, L = 10 cm = 0.1 m
Force, F = mg = 100 × 9.8 = 980 N
Shear modulus, G = 25 GPa = 25 × 10⁹ N/m²
Area, A = (0.1 m)² = 0.01 m²
Shear strain = Δx / L = F / (A × G)
Δx = (F × L) / (A × G)
= (980 × 0.1) / (0.01 × 25 × 10⁹)
= 98 / (2.5 × 10⁸)
= 3.92 × 10⁻⁷ m = 0.392 µm
✅ Final Answer: 0.392 micrometres
🔷 Question 8.7
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 mm and 60 mm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
✅ Answer:
📌 Given:
Mass = 50,000 kg → F = mg = 50,000 × 9.8 = 4.9 × 10⁵ N
Each column supports F/4 = 1.225 × 10⁵ N
Inner radius = 30 mm = 0.03 m
Outer radius = 60 mm = 0.06 m
Area, A = π(R² − r²)
= π[(0.06)² − (0.03)²] = π(0.0036 − 0.0009) = π(0.0027) ≈ 8.48 × 10⁻³ m²
Young’s modulus for steel, Y = 2 × 10¹¹ N/m²
Strain = Stress / Y
Stress = F / A = (1.225 × 10⁵) / (8.48 × 10⁻³) ≈ 1.44 × 10⁷ N/m²
Strain = (1.44 × 10⁷) / (2 × 10¹¹) = 7.2 × 10⁻⁵
✅ Final Answer: 7.2 × 10⁻⁵
🔷 Question 8.8
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
✅ Answer:
📌 Area = 15.2 mm × 19.1 mm = (15.2 × 10⁻³) × (19.1 × 10⁻³) = 2.9032 × 10⁻⁴ m²
Force = 44,500 N
Young’s modulus for copper, Y = 1.1 × 10¹¹ N/m²
Stress = F / A = 44500 / 2.9032 × 10⁻⁴ ≈ 1.53 × 10⁸ N/m²
Strain = Stress / Y = (1.53 × 10⁸) / (1.1 × 10¹¹) ≈ 1.39 × 10⁻³
✅ Final Answer: 1.39 × 10⁻³
🔷 Question 8.9
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N/m², what is the maximum load the cable can support?
✅ Answer:
📌 Radius = 1.5 cm = 0.015 m → A = πr² = π(0.015)² ≈ 7.07 × 10⁻⁴ m²
Maximum Stress = 10⁸ N/m²
Stress = Force / Area ⇒ F = Stress × Area
F = 10⁸ × 7.07 × 10⁻⁴ = 7.07 × 10⁴ N
✅ Final Answer: 70,700 N
🔷 Question 8.10
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
✅ Answer:
📌 For same tension, extensions must be equal.
Δl = (F × L) / (A × Y) ⇒ Δl ∝ 1 / (A × Y)
Let diameters be d₁ (Cu), d₂ (Fe)
Since length and force are same,
1 / (πd₁²/4 × Y_Cu) = 1 / (πd₂²/4 × Y_Fe)
⇒ d₁² / Y_Cu = d₂² / Y_Fe
⇒ (d₁ / d₂)² = Y_Cu / Y_Fe
Y_Cu = 1.1 × 10¹¹ N/m²
Y_Fe = 2.0 × 10¹¹ N/m²
⇒ (d₁ / d₂)² = 1.1 / 2.0 = 0.55
⇒ d₁ / d₂ = √0.55 ≈ 0.741
✅ Final Answer: d_Cu / d_Fe ≈ 0.741
🔷 Question 8.11
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². Calculate the elongation of the wire when the mass is at the lowest point of its path.
✅ Answer:
📌 Mass = 14.5 kg, ω = 2 rev/s = 4π rad/s
L = 1.0 m
Area A = 0.065 cm² = 6.5 × 10⁻⁶ m²
Y = 2 × 10¹¹ N/m²
Centripetal Force at bottom:
T = mg + mω²R = 14.5 × 9.8 + 14.5 × (4π)² × 1
= 142.1 + 14.5 × 157.9 = 142.1 + 2299.6 ≈ 2441.7 N
Elongation = (F × L) / (A × Y)
= (2441.7 × 1) / (6.5 × 10⁻⁶ × 2 × 10¹¹)
= 2441.7 / (1.3 × 10⁶) ≈ 1.88 × 10⁻³ m = 1.88 mm
✅ Final Answer: 1.88 mm
🔷 Question 8.12
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10⁶ Pa. Bulk modulus of glass = 5 × 10⁹ Pa.
✅ Answer:
📌 Fractional change in volume:
ΔV / V = – (P / B)
Where,
P = 10⁶ Pa
B = 5 × 10⁹ Pa
⇒ ΔV / V = – (10⁶ / 5 × 10⁹) = – 2 × 10⁻⁴
✅ Final Answer: –2 × 10⁻⁴
🔷 Question 8.13
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7 × 10⁶ Pa. Bulk modulus of copper = 1.4 × 10¹¹ Pa.
✅ Answer:
📌 Volume of cube = (0.10 m)³ = 1.0 × 10⁻³ m³
Pressure, P = 7 × 10⁶ Pa
Bulk modulus, B = 1.4 × 10¹¹ Pa
Volume contraction:
ΔV = V × (P / B)
= 1.0 × 10⁻³ × (7 × 10⁶ / 1.4 × 10¹¹)
= 1.0 × 10⁻³ × 5 × 10⁻⁵ = 5.0 × 10⁻⁸ m³
✅ Final Answer: 5.0 × 10⁻⁸ m³
🔷 Question 8.14
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. A force of 100 N is applied to this composite rod. Estimate the change in length of the composite rod.
(Y_brass = 0.9 × 10¹¹ Pa, Y_steel = 2.0 × 10¹¹ Pa)
✅ Answer:
📌 Length of each = 0.5 m
Diameter = 3.0 mm = 0.003 m
Area A = πr² = π × (0.0015)² ≈ 7.07 × 10⁻⁶ m²
Force F = 100 N
Change in length = Δl = (F × L) / (A × Y)
For brass:
Δl₁ = (100 × 0.5) / (7.07 × 10⁻⁶ × 0.9 × 10¹¹) ≈ 7.86 × 10⁻⁵ m
For steel:
Δl₂ = (100 × 0.5) / (7.07 × 10⁻⁶ × 2.0 × 10¹¹) ≈ 3.53 × 10⁻⁵ m
Total elongation = Δl₁ + Δl₂ = (7.86 + 3.53) × 10⁻⁵ = 1.139 × 10⁻⁴ m = 0.114 mm
✅ Final Answer: 0.114 mm
🔷 Question 8.15
A uniform wire of length 3 m and diameter 0.5 mm is stretched by a force of 100 N. Calculate the elongation produced. Young’s modulus of wire = 2 × 10¹¹ Pa.
✅ Answer:
📌 L = 3 m, d = 0.5 mm = 0.0005 m
r = 0.00025 m ⇒ A = πr² = π × (0.00025)² ≈ 1.96 × 10⁻⁷ m²
Y = 2 × 10¹¹ N/m², F = 100 N
Δl = (F × L) / (A × Y)
= (100 × 3) / (1.96 × 10⁻⁷ × 2 × 10¹¹)
= 300 / (3.92 × 10⁴) = 7.65 × 10⁻³ m = 7.65 mm
✅ Final Answer: 7.65 mm
🔷 Question 8.16
The Young’s modulus of steel is twice that of brass. Two wires of same length and same area of cross-section are suspended from a rigid support. One is of brass and the other is of steel. At the lower ends of the wires, a platform is suspended such that the platform remains horizontal when both wires are elongated. If the weight on the platform is 300 N, find the individual weights supported by the steel and brass wires.
✅ Answer:
📌 Y_steel = 2 × Y_brass
Let elongation = Δl (same for both)
⇒ F ∝ Y × A / L ⇒ Force ∝ Y (since A and L same)
Let force in brass = F_B, in steel = F_S
Then, F_S / F_B = Y_S / Y_B = 2
⇒ F_S = 2 × F_B
Also, total force = 300 N = F_B + F_S = F_B + 2F_B = 3F_B
⇒ F_B = 100 N, F_S = 200 N
✅ Final Answers:
✔️ Brass wire supports = 100 N
✔️ Steel wire supports = 200 N
🔷 Question 8.17
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ N/m², what is the maximum load the cable can support?
✅ Answer:
📌 Radius, r = 1.5 cm = 0.015 m
Area, A = πr² = π × (0.015)² = π × 2.25 × 10⁻⁴ ≈ 7.07 × 10⁻⁴ m²
Maximum stress, σ = F / A
⇒ F = σ × A = 10⁸ × 7.07 × 10⁻⁴ = 7.07 × 10⁴ N
✅ Final Answer: Maximum load = 70,700 N
🔷 Question 8.18
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
✅ Answer:
📌 Given: Length = same (L = 2.0 m), Tension = same
For equal tension, all wires must stretch by same amount
⇒ ΔL_copper = ΔL_iron
Using elongation formula:
ΔL = (F × L) / (A × Y) ⇒ ΔL ∝ 1 / (A × Y) ⇒ For same F and L:
1 / (πr² × Y) must be same ⇒ (1 / d²Y) same
Let diameters be d_C (copper) and d_I (iron):
⇒ (1 / d_C²Y_C) = (1 / d_I²Y_I)
Taking ratio:
d_C² / d_I² = Y_C / Y_I
Given:
Y_C = 1.1 × 10¹¹ Pa
Y_I = 2.0 × 10¹¹ Pa
⇒ d_C² / d_I² = 1.1 / 2.0 = 11 / 20
⇒ d_C / d_I = √(11/20) ≈ 0.7416
✅ Final Answer:
✔️ Ratio of diameters (d_C : d_I) ≈ 0.7416 : 1
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OTHER IMPORTANT QUESTIONS FOR EXAMS
(CBSE MODEL QUESTIONS PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
⚙️ SECTION A: Multiple Choice Questions (Q1–Q18)
Question 1:
Which of the following quantities is dimensionless?
🔵 (A) Stress
🟢 (B) Strain
🟠 (C) Modulus of elasticity
🔴 (D) Force
Answer: (B) Strain
Question 2:
The unit of Young’s modulus is
🔵 (A) N
🟢 (B) N/m
🟠 (C) N/m²
🔴 (D) m/N
Answer: (C) N/m²
Question 3:
If a wire is stretched and its length increases by 1%, the strain produced is
🔵 (A) 1
🟢 (B) 0.1
🟠 (C) 0.01
🔴 (D) 0.001
Answer: (C) 0.01
Question 4:
Hooke’s law is valid
🔵 (A) Beyond elastic limit
🟢 (B) Up to proportional limit
🟠 (C) For all values of stress
🔴 (D) Only for plastic materials
Answer: (B) Up to proportional limit
Question 5:
The ratio of lateral strain to longitudinal strain is
🔵 (A) Elastic limit
🟢 (B) Poisson’s ratio
🟠 (C) Stress
🔴 (D) Yield ratio
Answer: (B) Poisson’s ratio
Question 6:
If stress is doubled, within elastic limit strain will
🔵 (A) Double
🟢 (B) Halve
🟠 (C) Remain same
🔴 (D) Become four times
Answer: (A) Double
Question 7:
Which of the following is most elastic?
🔵 (A) Rubber
🟢 (B) Copper
🟠 (C) Steel
🔴 (D) Glass
Answer: (C) Steel
Question 8:
The bulk modulus of a fluid is
🔵 (A) Infinite
🟢 (B) Zero
🟠 (C) Finite
🔴 (D) Cannot be defined
Answer: (B) Zero
Question 9:
The SI unit of stress is
🔵 (A) J
🟢 (B) N/m²
🟠 (C) N·m
🔴 (D) W
Answer: (B) N/m²
Question 10:
The reciprocal of bulk modulus is called
🔵 (A) Rigidity
🟢 (B) Compressibility
🟠 (C) Flexibility
🔴 (D) Plasticity
Answer: (B) Compressibility
Question 11:
Which of the following has maximum Young’s modulus?
🔵 (A) Rubber
🟢 (B) Glass
🟠 (C) Copper
🔴 (D) Steel
Answer: (D) Steel
Question 12:
The dimensional formula of stress is
🔵 (A) [MLT⁻²]
🟢 (B) [ML⁻¹T⁻²]
🟠 (C) [M⁻¹L³T⁻²]
🔴 (D) [ML²T⁻²]
Answer: (B) [ML⁻¹T⁻²]
Question 13:
Which statement is correct about elastic limit?
🔵 (A) It is the maximum strain possible
🟢 (B) It is the maximum stress that can be applied without permanent deformation
🟠 (C) It depends on length only
🔴 (D) It is same for all materials
Answer: (B) It is the maximum stress that can be applied without permanent deformation
Question 14:
For an ideal solid, Poisson’s ratio can have a maximum value of
🔵 (A) 0.5
🟢 (B) 1
🟠 (C) 0
🔴 (D) ∞
Answer: (A) 0.5
Question 15:
Elastic potential energy per unit volume is
🔵 (A) σε
🟢 (B) ½ σε
🟠 (C) σ²ε
🔴 (D) σε²
Answer: (B) ½ σε
Question 16:
The wire elongates by 1 mm under a load of 100 N. If area = 2×10⁻⁶ m² and length = 2 m, find Young’s modulus.
🔵 (A) 1×10¹⁰ Pa
🟢 (B) 2×10¹⁰ Pa
🟠 (C) 5×10⁹ Pa
🔴 (D) 10¹¹ Pa
Answer: (A) 1×10¹⁰ Pa
Question 17:
The slope of the linear portion of stress–strain curve gives
🔵 (A) Bulk modulus
🟢 (B) Young’s modulus
🟠 (C) Rigidity modulus
🔴 (D) Poisson’s ratio
Answer: (B) Young’s modulus
Question 18:
If a steel wire and copper wire of same length and area are stretched by same load, which elongates more?
🔵 (A) Steel
🟢 (B) Copper
🟠 (C) Both equally
🔴 (D) None
Answer: (B) Copper
🌿 SECTION B: Very Short/Short Answers (Q19–Q23)
Question 19:
Define stress and strain.
Answer:
✔️ Stress: Force per unit area, σ = F/A.
✔️ Strain: Fractional change in length, ε = ΔL/L.
Both measure deformation effects under applied force.
Question 20:
State Hooke’s law.
Answer:
💡 Within elastic limit, stress is directly proportional to strain.
σ ∝ ε → σ = Yε.
Here Y = modulus of elasticity.
Question 21:
What is Poisson’s ratio?
Answer:
✔️ The ratio of lateral strain to longitudinal strain.
σ = lateral strain / longitudinal strain.
It has no unit and is less than 0.5 for solids.
Question 22:
Define elastic limit.
Answer:
💡 Elastic limit is the maximum stress up to which the material returns to its original shape on removal of force.
Beyond this, permanent deformation occurs.
Question 23:
What is meant by elastic fatigue?
Answer:
✔️ Elastic fatigue is the loss of elasticity due to repeated loading and unloading cycles.
Example: weakening of a spring after long use.
⚡ SECTION C: Mid-Length Numericals/Theory (Q24–Q28)
Question 24:
Derive the expression for Young’s modulus.
Answer:
✏️ Let a wire of length L and area A be stretched by force F.
Elongation = ΔL.
Stress = F/A, Strain = ΔL/L.
Therefore,
➡️ Y = Stress / Strain = (F/A)/(ΔL/L) = FL / (AΔL).
✔️ Hence, Y = (F×L)/(A×ΔL).
Question 25:
A wire 2 m long and 1 mm² cross-section is stretched by 1 mm under a load of 2 N. Find Young’s modulus.
Answer:
✏️ Given: L = 2 m, A = 1×10⁻⁶ m², F = 2 N, ΔL = 1×10⁻³ m.
Y = (F×L)/(A×ΔL)
= (2×2)/(1×10⁻⁶ × 1×10⁻³)
= 4 / 1×10⁻⁹ = 4×10⁹ Pa.
✔️ Y = 4 × 10⁹ Pa.
Question 26:
Derive the relation between Y, K and Poisson’s ratio (σ).
Answer:
💡 For isotropic material:
Y = 3K(1 − 2σ).
✔️ This is derived from the relation between longitudinal and volumetric strains under uniform stress.
Question 27:
Explain stress–strain curve for a typical ductile material.
Answer:
✔️ Proportional limit: Stress ∝ strain (Hooke’s law valid).
✔️ Elastic limit / yield point: Permanent deformation starts.
✔️ Plastic region: Large strain occurs with small increase in stress.
✔️ Breaking point: Material fractures.
💡 The area under curve = energy stored per unit volume.
Question 28:
What is elastic potential energy per unit volume? Derive expression.
Answer:
✏️ Work done = average stress × strain = (½σ) × ε.
Thus, energy per unit volume,
➡️ U = ½ σε.
✔️ It represents the energy stored in a body under deformation.
⚙️ SECTION D: Long Answer Questions (Q29–Q31)
Question 29:
Derive the relation between stress and strain for an elastic wire and hence explain Hooke’s Law.
Answer:
💡 Concept:
When a force acts on a solid, it produces deformation (strain). The ratio of stress to strain determines how elastic a material is.
✏️ Step 1: Definition of Stress
If a wire of cross-sectional area A is stretched by a force F,
then,
➡️ Stress (σ) = F / A
It represents internal restoring force per unit area.
✏️ Step 2: Definition of Strain
If the original length of the wire is L and its increase in length is ΔL,
➡️ Strain (ε) = ΔL / L
It is a dimensionless quantity (no unit).
✏️ Step 3: Hooke’s Law
Within elastic limit, the stress is directly proportional to strain.
➡️ σ ∝ ε
or, σ = Yε
where Y = Young’s Modulus (a constant for the material).
✏️ Step 4: Derivation of Y
From the above relation,
➡️ Y = σ / ε = (F/A) / (ΔL/L)
Therefore,
➡️ Y = (F × L) / (A × ΔL)
✔️ Interpretation:
Y measures the stiffness or rigidity of the material.
Higher Y → more elastic (e.g., steel).
Hooke’s law holds true only up to proportional limit.
💡 Result:
Y = (F × L) / (A × ΔL) and Stress ∝ Strain (Hooke’s Law).
Question 30:
Derive an expression for the energy stored per unit volume in a stretched wire.
Answer:
💡 Concept:
When a wire is stretched under a load, work is done on it, which is stored as elastic potential energy within the elastic limit.
✏️ Step 1:
Let the wire be stretched by a small length ΔL under force F.
Work done (W) = average force × extension
= ½ F × ΔL
✏️ Step 2:
Divide by the original volume of wire (A × L):
➡️ Energy per unit volume (U) = (½ F × ΔL) / (A × L)
✏️ Step 3:
From definitions,
F/A = σ (stress), ΔL/L = ε (strain)
Substitute:
➡️ U = ½ × σ × ε
✔️ Final Expression:
Elastic potential energy per unit volume:
➡️ U = ½ σε = ½ (Yε²) = ½ (σ²/Y)
💡 Interpretation:
The energy is proportional to both stress and strain.
The area under the linear portion of the stress–strain graph represents this energy.
Question 31:
Explain the Stress–Strain Curve for a ductile material (like mild steel).
Answer:
💡 Concept:
The stress–strain curve shows the behavior of a material under gradually increasing stress.
✏️ Step 1: Proportional Limit (O → A)
Stress ∝ strain (Hooke’s law valid).
The slope gives Young’s modulus (Y).
✏️ Step 2: Elastic Limit / Yield Point (A → B)
Beyond A, proportionality fails but material still returns to original shape.
Point B is elastic limit — end of perfect elasticity.
✏️ Step 3: Plastic Region (B → D)
Material deforms permanently.
Between B and C: plastic flow begins (yield point).
Between C and D: necking occurs (cross-section reduces).
✏️ Step 4: Breaking Point (D)
Material fractures; this is ultimate strength.
After D, wire breaks and stress falls to zero.
💡 Key Points from Curve:
✔️ Up to A → Elastic behavior (reversible).
✔️ Beyond A → Plastic behavior (irreversible).
✔️ Area under curve → Strain energy per unit volume.
🧠 Result:
Stress–strain curve clearly distinguishes elastic and plastic deformation, helping determine safe working stress of materials.
🌿 SECTION E: Case/Application Based Questions (Q32–Q33)
Question 32:
A brass wire of length 1.5 m and diameter 0.8 mm is subjected to a load of 2 kg.
Find the extension produced. (Y = 1×10¹¹ N/m², g = 9.8 m/s²)
Answer:
✏️ Step 1:
Given:
L = 1.5 m, r = 0.4×10⁻³ m, F = 2×9.8 = 19.6 N, Y = 1×10¹¹ N/m²
✏️ Step 2:
Formula: ΔL = (F × L) / (A × Y)
A = πr² = 3.14 × (0.4×10⁻³)² = 5.02×10⁻⁷ m²
✏️ Step 3:
ΔL = (19.6 × 1.5) / (5.02×10⁻⁷ × 1×10¹¹)
= 29.4 / 5.02×10⁴
= 5.86×10⁻⁴ m
✔️ Result:
Extension = 0.586 mm
Question 33:
A steel wire and a copper wire of same length and area are subjected to the same load.
If Yₛ = 2×10¹¹ N/m² and Y꜀ = 1.1×10¹¹ N/m², compare their extensions.
Answer:
✏️ Step 1:
For same L, A, and F:
ΔL ∝ 1/Y
✏️ Step 2:
ΔL꜀ / ΔLₛ = Yₛ / Y꜀ = (2×10¹¹)/(1.1×10¹¹) = 1.82
✔️ Result:
Extension in copper wire = 1.82 times that in steel wire.
💡 Hence, copper stretches more than steel (less rigid).
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