StudyOSStudyOS CBSE Study Material, Class 5–12

Mechanical Properties of Solids

NCERT Class 11 · Physics Based on NCERT Class 11 Physics textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter introduces the concepts of elasticity and plasticity in solids, explaining how materials deform under various types of forces and return to their original shape when the force is removed. It is essential in the syllabus because it lays the foundation for understanding material behaviour in engineering and structural design, from buildings to bridges to artificial limbs. The one big idea a student must walk away with is that for small deformations, stress is proportional to strain (Hooke's law), and the proportionality constant (modulus of elasticity) is a characteristic property of the material that determines its stiffness and strength.
Key Concepts (26)
Elasticity
The property of a body by virtue of which it tends to regain its original size and shape when the applied force is removed.
e.g. A helical spring returns to its original length after being gently pulled and released.
Plasticity
The property of a substance by which it gets permanently deformed and has no gross tendency to regain its previous shape when the applied force is removed.
e.g. A lump of putty or mud, when pressed, retains its new shape permanently.
Stress
The restoring force per unit area developed in a body when it is subjected to a deforming force.
e.g. When a steel wire is stretched, the internal restoring force per unit cross-sectional area is the stress.
Tensile stress
The restoring force per unit area when a body is stretched by two equal forces applied normal to its cross-sectional area.
e.g. A cylinder being pulled from both ends experiences tensile stress.
Compressive stress
The restoring force per unit area when a body is compressed under the action of applied forces.
e.g. A pillar supporting a heavy load experiences compressive stress.
Longitudinal stress
Another term for tensile or compressive stress, as both involve a change in length.
e.g. Both stretching and compressing a rod produce longitudinal stress.
Shearing stress
The restoring force per unit area developed due to an applied tangential force, parallel to the cross-sectional area.
e.g. When a book is pressed with the hand and pushed horizontally, the force per unit area parallel to the cover is shearing stress.
Hydraulic stress
The internal restoring force per unit area when a body is compressed uniformly on all sides by a fluid, equal in magnitude to the hydraulic pressure.
e.g. A solid sphere submerged in a high-pressure fluid experiences hydraulic stress.
Strain
The fractional change in dimension (length, shape, or volume) of a body due to stress.
e.g. The ratio of change in length to original length is longitudinal strain.
Longitudinal strain
The ratio of change in length (ΔL) to the original length (L) of the body.
e.g. If a wire of 1 m length stretches by 1 mm, longitudinal strain is 0.001.
Shearing strain
The ratio of relative displacement of the faces (Δx) to the length of the cylinder (L), equal to tan θ, and for small angles, approximately equal to θ.
e.g. When a book is pushed sideways, the angle through which it tilts is the shearing strain.
Volume strain
The ratio of change in volume (ΔV) to the original volume (V) of the body.
e.g. When a solid sphere is compressed under high pressure, the fractional decrease in its volume is volume strain.
Hooke's law
For small deformations, stress is directly proportional to strain.
e.g. In the linear region of a stress-strain curve, doubling the stress doubles the strain.
Modulus of elasticity
The proportionality constant in Hooke's law, equal to stress divided by strain.
e.g. Young's modulus, shear modulus, and bulk modulus are all types of modulus of elasticity.
Yield point (elastic limit)
The point on a stress-strain curve beyond which the material does not return to its original dimensions when the load is removed.
e.g. Point B in a typical stress-strain curve for a metal is the yield point.
Yield strength (σy)
The stress corresponding to the yield point.
e.g. For mild steel, yield strength is about 300 × 10^6 N m^-2.
Ultimate tensile strength (σu)
The maximum stress a material can withstand before fracture.
e.g. Point D on a stress-strain curve is the ultimate tensile strength.
Brittle material
A material in which the ultimate strength and fracture points are close together.
e.g. Glass is brittle because it breaks soon after reaching its ultimate strength.
Ductile material
A material in which the ultimate strength and fracture points are far apart.
e.g. Copper is ductile as it can be drawn into wires before breaking.
Elastomers
Substances that can be stretched to cause large strains and do not have a well-defined plastic region.
e.g. Rubber and the elastic tissue of aorta are elastomers.
Young's modulus (Y)
The ratio of tensile (or compressive) stress to longitudinal strain.
e.g. For steel, Y = 2.0 × 10^11 N m^-2.
Shear modulus (G) or Modulus of rigidity
The ratio of shearing stress to shearing strain.
e.g. For steel, G = 84 × 10^9 N m^-2.
Bulk modulus (B)
The ratio of hydraulic stress (pressure) to the corresponding volume strain.
e.g. For water, B = 2.2 × 10^9 N m^-2.
Compressibility (k)
The reciprocal of bulk modulus, defined as the fractional change in volume per unit increase in pressure.
e.g. Gases have high compressibility, solids have very low compressibility.
Poisson's ratio
The ratio of lateral strain to longitudinal strain within the elastic limit.
e.g. For steel, Poisson's ratio is between 0.28 and 0.30.
Elastic potential energy
The work done against inter-atomic forces stored in a stretched wire.
e.g. When a wire is stretched, the energy stored per unit volume is (1/2) × stress × strain.
Formulas & Equations (15)
Stress = F/A
Reads as: Stress equals force per unit area
Variables: F = applied force normal to cross-section (N), A = area of cross-section (m^2)
Use when: Calculating stress in any type of deformation (tensile, compressive, shearing, hydraulic)
⚠ Using the wrong area — for tensile stress, use the cross-sectional area perpendicular to the force, not the surface area.
Longitudinal strain = ΔL / L
Reads as: Longitudinal strain equals change in length divided by original length
Variables: ΔL = change in length (m), L = original length (m)
Use when: Finding strain when a rod or wire is stretched or compressed
⚠ Forgetting that strain is dimensionless and has no units.
Shearing strain = Δx / L = tan θ ≈ θ
Reads as: Shearing strain equals relative displacement divided by length, approximately equal to the angle in radians
Variables: Δx = relative displacement between opposite faces (m), L = length of cylinder (m), θ = angular displacement (rad)
Use when: Calculating strain in shear deformation
⚠ Using degrees instead of radians for θ when θ is small.
Volume strain = ΔV / V
Reads as: Volume strain equals change in volume divided by original volume
Variables: ΔV = change in volume (m^3), V = original volume (m^3)
Use when: Finding strain under hydraulic compression
⚠ Forgetting the negative sign in bulk modulus formula — volume decreases when pressure increases.
Hooke's law: stress ∝ strain, or stress = k × strain
Reads as: Stress is proportional to strain, or stress equals modulus of elasticity times strain
Variables: k = modulus of elasticity (N m^-2 or Pa)
Use when: Relating stress and strain within the elastic limit
⚠ Applying Hooke's law beyond the proportional limit (beyond point A on the stress-strain curve).
Young's modulus: Y = (F/A) / (ΔL/L) = (F × L) / (A × ΔL)
Reads as: Young's modulus equals tensile stress divided by longitudinal strain
Variables: F = force (N), A = cross-sectional area (m^2), L = original length (m), ΔL = change in length (m)
Use when: Calculating elongation or force in wires/rods under tension or compression
⚠ Using diameter instead of radius in area calculation, or mixing units (e.g., cm instead of m).
Shear modulus: G = (F/A) / (Δx/L) = (F × L) / (A × Δx) = F / (A × θ)
Reads as: Shear modulus equals shearing stress divided by shearing strain
Variables: F = tangential force (N), A = area parallel to force (m^2), L = height (m), Δx = displacement (m), θ = shear angle (rad)
Use when: Calculating displacement or force in shear deformation
⚠ Using the cross-sectional area perpendicular to the force instead of the area parallel to the force.
Bulk modulus: B = -p / (ΔV/V)
Reads as: Bulk modulus equals negative pressure divided by volume strain
Variables: p = hydraulic pressure (Pa), ΔV = change in volume (m^3), V = original volume (m^3)
Use when: Calculating volume change under uniform pressure
⚠ Omitting the negative sign — B is always positive, so the negative sign accounts for volume decrease when pressure increases.
Compressibility: k = 1/B = -(1/Δp) × (ΔV/V)
Reads as: Compressibility equals reciprocal of bulk modulus
Variables: B = bulk modulus (Pa), Δp = change in pressure (Pa), ΔV/V = volume strain
Use when: Comparing how easily materials are compressed
⚠ Confusing compressibility with bulk modulus — they are reciprocals.
Poisson's ratio = (Δd/d) / (ΔL/L) = (Δd/ΔL) × (L/d)
Reads as: Poisson's ratio equals lateral strain divided by longitudinal strain
Variables: Δd = change in diameter (m), d = original diameter (m), ΔL = change in length (m), L = original length (m)
Use when: Finding lateral contraction when a wire is stretched
⚠ Forgetting that Poisson's ratio is dimensionless and its value is typically between 0 and 0.5.
Elastic potential energy per unit volume: u = (1/2) × stress × strain = (1/2) × σ × ε
Reads as: Elastic potential energy per unit volume equals half times stress times strain
Variables: σ = stress (N m^-2), ε = strain (dimensionless)
Use when: Calculating energy stored in a deformed body
⚠ Forgetting the factor of 1/2 — students often write stress × strain directly.
Work done in stretching a wire: W = (1/2) × Y × (strain)^2 × volume = (1/2) × stress × strain × volume
Reads as: Work done equals half times Young's modulus times strain squared times volume
Variables: Y = Young's modulus (N m^-2), strain = ΔL/L, volume = A × L (m^3)
Use when: Calculating total elastic potential energy stored in a wire
⚠ Using the wrong volume — it is the original volume of the wire, not the change in volume.
Bending of a beam: δ = W l^3 / (4 b d^3 Y)
Reads as: Sag (deflection) equals W times l cubed divided by 4 times b times d cubed times Y
Variables: δ = sag (m), W = load at centre (N), l = length of beam (m), b = breadth (m), d = depth (m), Y = Young's modulus (N m^-2)
Use when: Calculating deflection of a beam supported at ends and loaded at centre
⚠ Confusing breadth (b) and depth (d) — depth has a much larger effect (d^-3) than breadth (b^-1).
Minimum cross-sectional area of rope: A ≥ W/σy = Mg/σy
Reads as: Area must be greater than or equal to weight divided by yield strength
Variables: W = load (N), M = mass (kg), g = acceleration due to gravity (m s^-2), σy = yield strength (N m^-2)
Use when: Designing a crane rope or any load-bearing cable
⚠ Forgetting to include a factor of safety — the calculated area is the minimum, but a larger area is used in practice.
Maximum height of a mountain: h = σy / (ρ g)
Reads as: Height equals yield strength divided by density times g
Variables: σy = critical shearing stress (N m^-2), ρ = density (kg m^-3), g = acceleration due to gravity (m s^-2)
Use when: Estimating the maximum possible height of a mountain on Earth
⚠ Using bulk modulus instead of shear stress — the limiting factor is shearing stress at the base.
How It Works
Stress and strain generation under different forces
  1. Step 1: Apply an external force to a solid body while keeping it in static equilibrium.
  2. Step 2: The body deforms — its length, shape, or volume changes depending on the direction and nature of the force.
  3. Step 3: A restoring force develops inside the body, equal in magnitude but opposite in direction to the applied force.
  4. Step 4: The restoring force per unit area is defined as stress.
  5. Step 5: The fractional change in dimension (length, angle, or volume) is defined as strain.
💡 Stress is the cause (applied force per area), strain is the effect (fractional change). They are proportional only for small deformations (Hooke's law).
Obtaining a stress-strain curve experimentally
  1. Step 1: Take a test cylinder or wire and apply a gradually increasing tensile force.
  2. Step 2: Record the applied force and the corresponding fractional change in length (strain) at each step.
  3. Step 3: Plot stress (force per unit area) on the y-axis and strain on the x-axis.
  4. Step 4: Identify the linear region (O to A) where Hooke's law is obeyed.
  5. Step 5: Identify the yield point (B) beyond which permanent deformation occurs.
  6. Step 6: Note the ultimate tensile strength (D) and the fracture point (E).
💡 The curve varies from material to material. For elastomers like rubber, there is no well-defined plastic region and Hooke's law is not obeyed over most of the region.
Calculating elongation of a wire under tension
  1. Step 1: Identify the force (F) applied along the length of the wire.
  2. Step 2: Measure the original length (L) and cross-sectional area (A) of the wire.
  3. Step 3: Use Young's modulus (Y) of the material (from tables).
  4. Step 4: Apply the formula: ΔL = (F × L) / (A × Y).
  5. Step 5: Ensure all units are in SI (N, m, m^2, Pa).
💡 For wires connected end to end, the same force acts on both, but elongations are different if lengths or Young's moduli differ.
Calculating shear displacement
  1. Step 1: Identify the tangential force (F) applied parallel to the face.
  2. Step 2: Determine the area (A) of the face parallel to the force.
  3. Step 3: Measure the height (L) perpendicular to the force.
  4. Step 4: Use shear modulus (G) of the material.
  5. Step 5: Apply the formula: Δx = (F × L) / (A × G).
💡 The area used is the area of the face parallel to the applied force, not perpendicular.
Calculating volume change under hydraulic pressure
  1. Step 1: Determine the hydraulic pressure (p) applied uniformly on all sides.
  2. Step 2: Measure the original volume (V) of the body.
  3. Step 3: Use bulk modulus (B) of the material.
  4. Step 4: Apply the formula: ΔV = - (p × V) / B.
  5. Step 5: The negative sign indicates a decrease in volume when pressure increases.
💡 Bulk modulus is always positive. The negative sign in the formula ensures that ΔV is negative when p is positive.
Storing elastic potential energy in a stretched wire
  1. Step 1: Apply a force F to stretch a wire of original length L and area A.
  2. Step 2: The wire elongates by l. The force at any elongation is F = YA × (l/L).
  3. Step 3: The work done for an infinitesimal elongation dl is dW = F dl = (YAl/L) dl.
  4. Step 4: Integrate from l = 0 to l = l to get total work: W = (1/2) × (YA/L) × l^2.
  5. Step 5: Express W as (1/2) × stress × strain × volume of the wire.
  6. Step 6: This work is stored as elastic potential energy (U).
💡 The energy per unit volume is (1/2) × stress × strain. Do not forget the factor of 1/2.
Designing a crane rope
  1. Step 1: Determine the maximum load (W = Mg) the rope must lift.
  2. Step 2: Find the yield strength (σy) of the material from tables.
  3. Step 3: Calculate the minimum cross-sectional area: A ≥ W/σy.
  4. Step 4: Apply a factor of safety (typically about 10) to get a practical area.
  5. Step 5: Use multiple thin wires braided together for flexibility and strength.
💡 The calculated area is the minimum; a factor of safety is always applied in real engineering designs.
Estimating maximum height of a mountain
  1. Step 1: Consider the force per unit area at the base due to the weight of the mountain: hρg.
  2. Step 2: This force has a shear component approximately equal to hρg.
  3. Step 3: Equate this to the critical shearing stress (elastic limit) of rock: hρg = σy.
  4. Step 4: Solve for h: h = σy / (ρg).
  5. Step 5: Using typical values (σy ≈ 30 × 10^7 N m^-2, ρ ≈ 3 × 10^3 kg m^-3, g = 10 m s^-2), h ≈ 10 km.
💡 The limiting factor is shearing stress, not compressive stress. The sides of the mountain are free, so the stress is not purely compressive.
Exam Traps — Don't Make These Mistakes
⚠ Confusing stress and pressure
✗ Stress and pressure are the same thing because both are force per unit area.
→ Pressure is always normal to the surface and acts equally in all directions in a fluid. Stress can be normal (tensile/compressive) or tangential (shear), and it is a restoring force developed inside a solid, not an externally applied force.
✓ Stress is the internal restoring force per unit area developed within a solid when deformed. Pressure is an external force per unit area applied by a fluid. Stress can be tensile, compressive, or shear; pressure is always compressive and normal.
⚠ Using the wrong area for shear modulus
✗ For shear stress, area = cross-sectional area perpendicular to the force.
→ In shear, the force is applied parallel to the face. The area used is the area of the face parallel to the applied force, not the cross-sectional area perpendicular to it.
✓ For shear stress, area = area of the face parallel to the applied tangential force. For example, in a square lead slab, if force is applied on the narrow face, area = side × thickness.
⚠ Omitting the negative sign in bulk modulus
✗ B = p / (ΔV/V)
→ When pressure increases (p positive), volume decreases (ΔV negative). Without the negative sign, B would be negative, which is incorrect. Bulk modulus is always positive.
✓ B = -p / (ΔV/V). The negative sign ensures B is positive. For example, if p = 3 × 10^7 Pa and ΔV/V = -0.0136, then B = -3×10^7 / (-0.0136) = 2.2 × 10^9 Pa.
⚠ Confusing 'more elastic' with 'stretches more'
✗ Rubber is more elastic than steel because it stretches more.
→ Elasticity is defined by the ability to regain original shape, not by how much it stretches. A material that stretches less for a given load is actually more elastic (has a higher modulus). Steel has a much higher Young's modulus than rubber.
✓ Steel is more elastic than rubber because for the same stress, steel undergoes less strain. A material with a larger Young's modulus requires a larger force to produce the same strain, meaning it is stiffer and more elastic.
⚠ Forgetting the factor of 1/2 in elastic potential energy
✗ Elastic potential energy per unit volume = stress × strain.
→ The work done in stretching a wire is the area under the stress-strain graph, which for a linear elastic material is a triangle, not a rectangle. The area of a triangle is (1/2) × base × height.
✓ Elastic potential energy per unit volume = (1/2) × stress × strain. This is derived from integrating F dl, where F varies linearly with l.
⚠ Applying Hooke's law beyond the elastic limit
✗ Hooke's law is valid for all deformations until fracture.
→ Hooke's law is only valid in the linear region (O to A) of the stress-strain curve. Beyond point A, stress and strain are not proportional, and beyond the yield point (B), the deformation becomes plastic and permanent.
✓ Hooke's law is valid only for small deformations within the elastic limit, specifically in the linear region of the stress-strain curve (region O to A). Beyond this, the material may yield and undergo plastic deformation.
Compare & Contrast
Direction of applied force relative to the surface
Tensile stress
Shearing stress
Tensile stress involves forces applied normal (perpendicular) to the cross-sectional area, causing a change in length. Shearing stress involves forces applied parallel to the surface, causing a change in shape (angular distortion) without a change in volume.
Type of deformation and state of matter applicable
Young's modulus
Bulk modulus
Young's modulus applies to solids under tensile or compressive stress, involving change in length. Bulk modulus applies to solids, liquids, and gases under hydraulic pressure, involving change in volume. Young's modulus is relevant only for solids, while bulk modulus is relevant for all three states of matter.
Behaviour beyond ultimate tensile strength
Brittle material
Ductile material
In a brittle material, the ultimate tensile strength and fracture points are close together, so the material breaks soon after reaching its maximum strength. In a ductile material, these points are far apart, allowing significant plastic deformation before fracture.
Ability to regain original shape after deformation
Elasticity
Plasticity
Elasticity is the property of a body to regain its original size and shape when the applied force is removed. Plasticity is the property of a body to retain the deformed shape permanently after the force is removed.
Exam Focus
Marks pattern: 1 MCQ (1 mark) on definitions or units, 1 SA-I (2 marks) on stress-strain curve or Hooke's law, 1 SA-II (3 marks) on numerical problem (e.g., elongation of wire), 1 LA (5 marks) on combined numerical or derivation of elastic potential energy
Most asked: Numerical problems on Young's modulus, shear modulus, and bulk modulus · Stress-strain curve interpretation (yield point, ultimate strength, fracture point) · Hooke's law and its application · Elastic potential energy stored in a stretched wire · Applications of elastic behaviour (crane rope design, beam bending, mountain height)
Don't skip: Numerical problems combining Young's modulus, force, elongation, and cross-sectional area; Derivation of elastic potential energy per unit volume; Stress-strain curve analysis with labels and explanations
One-Line Revision (read aloud before the exam)
  1. Stress is the restoring force per unit area developed inside a solid when deformed.
  2. Strain is the fractional change in dimension (length, shape, or volume).
  3. Hooke's law states that for small deformations, stress is directly proportional to strain.
  4. Young's modulus Y = (F × L) / (A × ΔL) applies to tensile or compressive stress in solids.
  5. Shear modulus G = (F × L) / (A × Δx) applies to tangential forces causing shape change.
  6. Bulk modulus B = -p / (ΔV/V) applies to uniform pressure causing volume change.
  7. Compressibility is the reciprocal of bulk modulus: k = 1/B.
  8. Poisson's ratio is the ratio of lateral strain to longitudinal strain.
  9. Elastic potential energy per unit volume = (1/2) × stress × strain.
  10. The yield point is where permanent (plastic) deformation begins.
  11. Ultimate tensile strength is the maximum stress a material can withstand.
  12. Steel is more elastic than rubber because it has a higher Young's modulus.
  13. For a beam, increasing depth (d) reduces bending more effectively than increasing breadth (b).
  14. The maximum height of a mountain on Earth is about 10 km, limited by the shear strength of rock.
  15. Stress and strain are proportional only in the linear region of the stress-strain curve.
  16. Shearing strain for small angles is approximately equal to the angle in radians.
  17. Bulk modulus of solids is much larger than that of liquids, which is much larger than that of gases.
  18. A factor of safety (about 10) is used in designing crane ropes to avoid permanent deformation.
  19. Elastomers like rubber do not obey Hooke's law over most of their stress-strain curve.
  20. Brittle materials break soon after reaching ultimate strength; ductile materials undergo significant plastic deformation first.

MCQs — 10 Questions with Answers

Q1. A rod of original length 1 m is stretched to 1.005 m. What is the longitudinal strain?

  • A. 0.005 ✓
  • B. 0.05
  • C. 5
  • D. 0.5

Answer: A — Longitudinal strain = ΔL/L = (1.005 - 1)/1 = 0.005/1 = 0.005 (dimensionless).

Q2. Which statement about elasticity and plasticity is correct?

  • A. Elasticity and plasticity are the same property with different names
  • B. An elastic material returns to original shape after force removal; a plastic material does not ✓
  • C. Plastic materials are always stronger than elastic materials
  • D. Steel is purely plastic and mud is purely elastic

Answer: B — Elasticity is recovery of shape after force removal; plasticity is permanent deformation—they are opposite properties.

Q3. A copper wire of length L and cross-sectional area A is stretched by a force F. The stress on the wire is:

  • A. F × A
  • B. F/A ✓
  • C. L/A
  • D. F × L

Answer: B — Stress is defined as force per unit area: σ = F/A, measured in Pa or N m⁻².

Q4. A cylinder is compressed uniformly from all sides by hydraulic pressure. Which quantity remains unchanged?

  • A. Volume
  • B. Geometric shape (spherical remains spherical) ✓
  • C. Density
  • D. Length

Answer: B — Under hydraulic compression, the body is squeezed uniformly; volume decreases but shape is preserved (sphere stays sphere, cube stays cube).

Q5. In a stress-strain graph for a metal, Hooke's law is valid in the region marked as:

  • A. O to A (linear region only) ✓
  • B. A to B (curved region)
  • C. Beyond B (plastic region)
  • D. Entire curve from O onwards

Answer: A — Hooke's law (stress ∝ strain) holds only in the linear elastic region O to A; beyond A, the relationship becomes nonlinear.

Q6. A wire experiences shearing strain of 0.02 rad when subjected to a shear force. If the original length is 2 m, the relative displacement between opposite faces is:

  • A. 0.01 m
  • B. 0.02 m
  • C. 0.04 m ✓
  • D. 0.1 m

Answer: C — Shearing strain = Δx/L; therefore Δx = strain × L = 0.02 × 2 = 0.04 m.

Q7. A material obeys Hooke's law. If stress increases from 2 × 10⁸ Pa to 4 × 10⁸ Pa, the strain will:

  • A. Decrease by half
  • B. Double (increase by factor of 2) ✓
  • C. Remain constant
  • D. Become zero

Answer: B — Under Hooke's law, stress ∝ strain; doubling stress directly doubles strain within the elastic limit.

Q8. Which of the following is NOT a correct statement? Assertion (A): Strain is a dimensionless quantity. Reason (R): Strain is the ratio of two quantities with the same dimension.

  • A. Both A and R are true, and R is the correct reason for A ✓
  • B. Both A and R are true, but R is not the correct reason for A
  • C. A is true, R is false
  • D. A is false, R is true

Answer: A — Strain (ΔL/L or ΔV/V) is indeed dimensionless because it's a ratio of identical quantities, making both assertion and reason correct.

Q9. A steel rod of diameter 1 cm and length 1 m experiences a tensile stress of 10⁸ Pa. Calculate the tensile force applied (use π = 3.14). Which force is closest?

  • A. 785 N
  • B. 7850 N ✓
  • C. 78500 N
  • D. 785000 N

Answer: B — Stress = F/A; F = stress × A = 10⁸ × π(0.005)² = 10⁸ × 7.85 × 10⁻⁵ ≈ 7850 N.

Q10. A solid sphere of bulk modulus K is subjected to an external pressure P. Derive an expression for the fractional change in volume (ΔV/V). Which relationship is correct?

  • A. ΔV/V = K/P
  • B. ΔV/V = P/K ✓
  • C. ΔV/V = K × P
  • D. ΔV/V = K + P

Answer: B — Bulk modulus K = P/(ΔV/V); rearranging gives ΔV/V = P/K, showing volume strain inversely proportional to bulk modulus.

Flashcards

What is elasticity?

The property of a body to regain its original size and shape when the applied deforming force is removed.

Define stress and state its SI unit.

Stress is the restoring force per unit area (F/A); SI unit is pascal (Pa) or N m⁻².

What is longitudinal strain?

The ratio of change in length (ΔL) to the original length (L) of the body: strain = ΔL/L.

Distinguish tensile stress from shearing stress.

Tensile stress acts perpendicular to the cross-section and changes length; shearing stress acts parallel and causes relative displacement between opposite faces.

State Hooke's law.

For small deformations, stress is directly proportional to strain: stress = k × strain, where k is the modulus of elasticity.

What is volume strain?

The ratio of change in volume (ΔV) to the original volume (V) produced by hydraulic pressure: volume strain = ΔV/V.

What does the linear region of a stress-strain curve represent?

The elastic region (O to A) where Hooke's law is obeyed and the body regains its original dimensions when the force is removed.

Why does strain have no units?

Strain is a dimensionless ratio of two lengths (or volumes), so units cancel out.

What is plasticity?

The property of a material to undergo permanent deformation without regaining its original shape when the applied force is removed.

Why is the I-shaped cross-section used in railway tracks?

The I-shape maximizes the moment of inertia and bending resistance while minimizing material use, providing strength with low weight.

Important Board Questions

Define stress and strain. Give one example each for tensile stress and shearing stress in everyday life. [2 marks]

State stress = F/A (with units Pa) and strain = ΔL/L (dimensionless). Example: tension in a rope pulling a load (tensile); pushing a book horizontally on a table (shearing).

A brass wire of length 2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by a force of 50 N. If the extension produced is 0.01 m, calculate the (a) tensile stress, (b) longitudinal strain, and (c) Young's modulus of brass. [5 marks]

Use stress = F/A = 50/(2×10⁻⁶); strain = ΔL/L = 0.01/2; Young's modulus E = stress/strain. Show working for all three parts separately.

Explain with a stress-strain diagram why the I-shaped cross-section is preferred in railway tracks over a square or circular cross-section. Derive the relationship between Young's modulus, stress, and strain, and explain why engineers must ensure materials operate within the elastic region. [6 marks]

Describe the linear elastic region (O to A) where Hooke's law applies and material recovers. Derive E = (F/A)/(ΔL/L). Explain that beyond elastic limit, permanent deformation occurs; I-shape maximizes moment of inertia (resistance to bending) while minimizing material—connect this to elastic properties and safety limits.

Assertion-Reason Questions for Mechanical Properties of Solids

Q1. Assertion (A):

Strain is a dimensionless quantity.

Reason (R):

Strain is defined as the ratio of change in dimension to the original dimension, and the ratio of two quantities with the same dimension is dimensionless.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why strain is dimensionless. Since strain (ΔL/L, ΔV/V, or Δx/L) is a ratio of quantities with identical dimensions, the dimensional units cancel out, making strain dimensionless with formula [M⁰L⁰T⁰].

Q2. Assertion (A):

A body under hydraulic pressure experiences volume strain but no change in its shape.

Reason (R):

Hydraulic pressure is applied uniformly on all surfaces of the body in perpendicular directions.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. Because hydraulic pressure acts equally and perpendicularly at every point on the surface, it compresses the volume uniformly without distorting the shape. This is fundamentally different from unidirectional tensile stress which changes shape through Poisson effect.

Q3. Assertion (A):

Shearing strain is defined as the ratio of relative displacement between opposite faces to the length of the object.

Reason (R):

When tangential forces are applied parallel to the cross-section, they cause relative displacement between opposite faces of the object.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true, but the reason does not explain the assertion. While both statements are factually correct, the reason merely describes what causes shearing strain rather than explaining why shearing strain is defined specifically as Δx/L. The assertion is a definition, and the reason describes the mechanism, but they are not in a causative relationship.

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for Mechanical Properties of Solids

Case Study:

A construction company is designing a steel suspension bridge. The main support cables are made of steel with Young's modulus of 2 × 10¹¹ Pa. Each cable has a cross-sectional area of 0.5 m² and a total length of 1000 m. During the design phase, engineers need to calculate the extension of these cables when they support a total tension force of 5 × 10⁸ N due to the weight of the bridge and expected traffic load. The engineers must ensure the extension is within safe limits (typically not more than 0.5% of the original length for structural integrity).

Q1. Calculate the tensile stress developed in the cable.

  • A. 1 × 10⁹ Pa ✓
  • B. 1 × 10⁸ Pa
  • C. 2.5 × 10⁸ Pa
  • D. 5 × 10⁸ Pa

Answer: A

Q2. Calculate the tensile strain produced in the cable.

  • A. 0.005 ✓
  • B. 5 × 10⁻³
  • C. 5 × 10⁻⁴
  • D. 0.5 × 10⁻³

Answer: A

Q3. What is the actual extension of the cable in meters?

  • A. 50 m
  • B. 5 m ✓
  • C. 500 m
  • D. 5000 m

Answer: B

Q4. Is the extension within the safe limit of 0.5% of the original length?

  • A. Cannot be determined from the given information
  • B. No, because the extension exceeds the safe limit
  • C. Yes, but the extension is much less than 0.5%, providing extra safety margin
  • D. Yes, because the extension (5 m) is equal to 0.5% of 1000 m, meeting the safety requirement exactly ✓

Answer: D

Unlock more case studies in Premium — unlock →

Class 11 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

View the formula sheet →

Next chapterMechanical Properties of Fluids →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.