Mechanical Properties of Solids | ISC Class 11 Physics Notes

This note covers elastic and plastic behaviour, stress and strain, Hooke’s law, stress-strain curves, Young’s modulus and its determination, shear modulus, bulk modulus, Poisson’s ratio, and the work stored as elastic energy in a stretched wire.

What makes a solid elastic or plastic?

A solid has a definite shape and size, but this does not make it perfectly rigid. Deformation means a change in shape or size under an applied force. A sufficiently large external force can deform even an appreciably rigid steel bar.

Elasticity is the property by which a body tends to regain its original size and shape when the applied force is removed. The associated recoverable change is called elastic deformation. Elastic behaviour concerns recovery, rather than simply how far an object stretches.

Definition: Plasticity is the property associated with permanent deformation: the body has no gross tendency to regain its previous shape after the deforming force is removed.

A helical spring pulled gently at its ends increases slightly in length and regains its original size and shape when released. Putty and mud, by contrast, retain a changed shape. They are close to ideal plastics, rather than examples of perfect elastic recovery.

Why does the distinction matter?

Elastic properties help determine whether a material can support a load, meaning an applied force, without acquiring a permanent change in dimensions. Building columns, bridge supports and machine parts require knowledge of both the material’s resistance to deformation and the range over which it recovers.

A restoring force is the internal force developed in response to deformation. In static equilibrium, meaning equilibrium with the body at rest, this force balances the applied deforming force in magnitude and opposes it in direction.

The same material need not behave elastically under every load. A small deformation can be recoverable, while a sufficiently increased load can leave a permanent change. Describing a material therefore requires both the applied loading conditions and the response after unloading.

How are stress and strain defined and classified?

Stress is restoring force per unit area. Write its magnitude as σ, the Greek letter sigma. A cross-section is a slice perpendicular to the body’s length. For a uniform cross-section, let F be the magnitude of the force normal to the section and A its area. Then:

σ = F/A

The SI unit of stress is the pascal, symbol Pa: 1 Pa = 1 N m⁻². The SI unit of force is the newton, symbol N, with 1 N = 1 kg m s⁻². Here kg, m and s denote kilogram, metre and second.

Dimensions express a physical quantity in terms of base quantities. Stress has dimensions [ML⁻¹T⁻²], where M, L and T inside square brackets denote the dimensions of mass, length and time. Force has dimensions [MLT⁻²], while area has dimensions [L²]. Dimensional L is distinct from a particular wire length.

Strain is the fractional change in a dimension. Let L be original length and ΔL its change; Δ means “change in”. The longitudinal strain, denoted ε, the Greek letter epsilon, is:

ε = ΔL/L

Both lengths must use the same unit. Their units cancel, so strain has no unit and is dimensionless. The denominator is the original length, rather than the extended length. A strain expressed as a percentage is the fractional strain multiplied by 100.

Tensile and compressive stresses are both longitudinal stresses. “Normal” means perpendicular to the surface; “tangential” means along it. Hydraulic pressure is force per unit area exerted by the surrounding fluid and has the same unit and dimensions as stress.

The drawing shows a cylinder pulled lengthwise, a cylinder displaced sideways, a book sheared by a hand, and a round body with inward arrows normal to its surface. The stretched cylinder is labelled with its original length and increase in length.

See Fig. 8.1 in your NCERT textbook

Note: A suspended wire pulled by a load F has tension, or internal pulling force, equal to F, not 2F. The equal and opposite force at its support does not double the tensile stress, which remains F/A.

When does Hooke’s law apply?

Hooke’s law states that stress is proportional to strain for small deformations. It is an empirical law, meaning a relationship established experimentally. It is found to be valid for most materials, but some materials do not exhibit this linear relationship.

The constant ratio of stress to corresponding strain is an elastic modulus. The kind of modulus depends on whether the deformation changes length, produces shear, or changes volume under uniform pressure. Each modulus has the dimensions and unit of stress because strain is dimensionless.

How does proportionality appear on a graph?

Plot stress vertically and strain horizontally. Provided the observations remain in the proportional region, the graph is a straight line through the origin. The slope, meaning change in vertical coordinate divided by change in horizontal coordinate, gives the relevant elastic modulus.

A steeper initial stress-strain line means more stress is required for the same strain. This interpretation compares material properties. Comparing extensions alone is insufficient unless the applied force, original length and cross-sectional area are also considered.

Elastic behaviour and linear behaviour are distinct. A material can return to its original dimensions even in a region where stress is no longer proportional to strain. Hooke’s law applies only to the linear part of the stress-strain curve.

Elastomers are substances, such as rubber and elastic tissue of the aorta, that can be stretched to produce large strains. The aorta is the large vessel carrying blood from the heart. Its tissue has a large elastic region but does not obey Hooke’s law over most of that region.

Stress is plotted vertically and strain horizontally. The curve begins near the origin, rises gently at first, then becomes increasingly steep. It is curved rather than a single straight proportional line.

See Fig. 8.3 in your NCERT textbook

Rubber can be pulled to several times its original length and still returns to its original shape. Such a large recoverable deformation does not establish a constant stress-to-strain ratio. For these elastomers there is also no well-defined plastic region.

What does a metal’s stress-strain curve reveal?

A stress-strain curve records the response to increasing load. A wire or test cylinder is stretched, the force is increased in steps, and the change in length is recorded. Dividing force by cross-sectional area gives stress; dividing extension by original length gives strain.

Stress is on the vertical axis and strain on the horizontal axis. The curve passes from the origin O through A and B, rises towards a maximum at D, then falls towards fracture point E. A dashed unloading line from C meets the strain axis at a non-zero permanent set.

See Fig. 8.2 in your NCERT textbook

How should the labelled regions be read?

1. O to A: Stress and strain are proportional, so Hooke’s law applies. On removal of the load, the body regains its original dimensions. A marks the proportional limit, the end of this linear region.
2. A to B: Stress and strain are not proportional, but the body still returns to its original dimensions when unloaded. B is the yield point, also called the elastic limit in this curve.
3. Beyond B: Strain increases rapidly even for a small change in stress. Unloading from C between B and D leaves a permanent set, meaning residual strain when the applied stress becomes zero.
4. D to E: D corresponds to the ultimate tensile strength, the maximum tensile stress on the curve. Beyond D, additional strain occurs even with reduced applied force, and E is the fracture point where the specimen breaks.

The stress at B is the yield strength; it has the same unit and dimensions as other stresses. Deformation that leaves a permanent set is plastic. This differs from the recoverable, though non-proportional, behaviour between A and B.

If the ultimate-strength point D and fracture point E are close, the material is described as brittle. If they are far apart, it is described as ductile. Stress-strain curves vary from material to material, so these labels must be interpreted for the curve being considered.

How does Young’s modulus describe a change in length?

Young’s modulus, denoted Y, is the ratio of longitudinal stress to longitudinal strain in the proportional region. It applies to tensile or compressive loading. For a given material, the strain magnitude is the same for equal magnitudes of tensile and compressive stress in this treatment.

Y = σ/ε

The SI unit of Young’s modulus is Pa. Its dimensions are [ML⁻¹T⁻²]. A large Y means a large longitudinal stress is required to produce a small strain. For the same stress, a material with greater Y has smaller longitudinal strain.

Derivation: Extension of a uniform wire

1. For axial force F, meaning force along the wire’s length, acting over cross-sectional area A, the longitudinal stress is F/A.
2. For original length L and extension ΔL, the longitudinal strain is ΔL/L.
3. Divide stress by strain to obtain Y = FL/(AΔL), then rearrange for the extension while remaining in the proportional region.

ΔL = FL/(AY)

This expression separates the effects of material and geometry. For a fixed material and cross-section, a longer wire extends more under the same force. For a fixed material and length, a greater cross-sectional area gives a smaller extension under that force.

For a circular wire, let r be its radius, d its diameter and π the ratio of a circle’s circumference to its diameter. Then A = πr² and r = d/2. Radius and diameter must not be interchanged in the area calculation.

In this comparison, steel is more elastic than copper and aluminium: it strains less under the same stress. The everyday description of something easily stretched as “more elastic” can therefore be misleading when comparing Young’s moduli.

How can Young’s modulus be determined experimentally?

The determination requires the original wire length, its cross-sectional area, the added stretching force and the corresponding extension. These measurements give stress and strain separately. Their ratio gives Young’s modulus while the measurements remain within the proportional region.

What is the principle of Searle’s method?

Searle’s apparatus compares an experimental wire with a reference wire suspended alongside it from a common rigid support. The reference wire provides a comparison for the change in length of the experimental wire. The two wires have the same material, length and diameter.

A spirit level, a bubble device that indicates horizontal alignment, links the lower frames. A micrometer screw, which measures small displacements through a calibrated screw movement, is adjusted to restore the bubble to its reference position after the experimental wire is loaded.

1. Apply initial loads to straighten the wires. Keep the reference-wire load unchanged during the measurements.
2. Measure the experimental wire’s length and diameter. A screw gauge measures small diameters through calibrated screw movement. Measure diameter at several positions in perpendicular directions, calculate area from the mean diameter, then establish the initial level and micrometer reading.
3. Add known loads gradually to the experimental wire. After each addition, restore the spirit-level bubble to its initial position and record the micrometer reading.
4. Remove the added loads in steps and record unloading readings. Average loading and unloading readings for each load. Use the measured extension corresponding to each added load to calculate Young’s modulus.

Let m be the added mass in kilograms and g the acceleration due to gravity, in m s⁻² with dimensions [LT⁻²]. The added stretching force is F = mg. Use this force with the extension measured relative to the initial loaded position.

Y = mgL/(AΔL)

Keep the load within the elastic limit. Allow the apparatus to settle before reading. Accurate diameter measurement matters because cross-sectional area depends on the square of diameter. The reference wire remains taut under its constant load while the experimental-wire load changes.

How are longitudinal-extension calculations worked out?

Begin by identifying the force acting in each wire and converting all lengths to metres. Calculate area before using Young’s modulus. Then distinguish extension, an actual length change, from strain, the dimensionless ratio of that change to the original length.

Here mm means millimetre, kN means kilonewton, and their conversions are 1 mm = 10⁻³ m and 1 kN = 10³ N. In each calculation the same unit system must be used throughout, including for the area.

Worked example 1. A structural steel rod has radius 10 mm and length 1.0 m. An axial force of 100 kN stretches it. Find stress, extension and strain, given Y = 2.0 × 10¹¹ Pa. Use π = 3.14 and assume proportional elastic behaviour.

Formula: A = πr²; σ = F/A; ΔL = σL/Y; ε = ΔL/L.

Substitute: r = 10⁻² m and F = 100 × 10³ N. Thus A = 3.14 × 10⁻⁴ m² and σ = (100 × 10³)/(3.14 × 10⁻⁴) = 3.18 × 10⁸ Pa.

Answer: ΔL = (3.18 × 10⁸ × 1.0)/(2.0 × 10¹¹) = 1.59 × 10⁻³ m = 1.59 mm. The strain is 1.59 × 10⁻³, approximately 0.16%.

What changes when wires are joined end to end?

Wires connected in series, meaning end to end, transmit the same tensile load. Their extensions add to give the total extension. Equal tension does not imply equal extension because each wire’s original length, area and Young’s modulus enter its extension formula.

Worked example 2. Copper and steel wires of lengths 2.2 m and 1.6 m, each 3.0 mm in diameter, are joined end to end. Their combined extension is 0.70 mm. Find the tensile load using Y for copper = 1.1 × 10¹¹ Pa and Y for steel = 2.0 × 10¹¹ Pa; take π = 3.14.

Formula: A = πr²; ΔL = FL/(AY). Subscripts c and s below identify copper and steel respectively. Equal force and area give ΔL꜀/ΔLₛ = (Yₛ/Y꜀)(L꜀/Lₛ).

Substitute: The extension ratio is (2.0/1.1)(2.2/1.6) = 2.5. With ΔL꜀ + ΔLₛ = 7.0 × 10⁻⁴ m, the individual extensions are 5.0 × 10⁻⁴ m and 2.0 × 10⁻⁴ m.

Answer: F = AY꜀ΔL꜀/L꜀ = 3.14 × (1.5 × 10⁻³)² × (1.1 × 10¹¹) × (5.0 × 10⁻⁴)/2.2 = approximately 1.8 × 10² N, or 180 N.

How does shear modulus describe a change in shape?

Shear is the relative sideways displacement of opposite faces. Let Δx be that displacement and L the original perpendicular separation of the faces. The shearing strain, denoted γ, the Greek letter gamma, is their ratio.

γ = Δx/L

If θ, the Greek letter theta, is the angular displacement from the original direction, then γ = tan θ. Usually θ is very small, so tan θ is nearly equal to θ when the angle is expressed in radians. A radian measures angle as arc length divided by radius.

Shear modulus, G, is the ratio of shearing stress to shearing strain. It is also called the modulus of rigidity. The SI unit of shear modulus is Pa, and its dimensions are [ML⁻¹T⁻²]. It measures resistance to shear deformation.

Derivation: Sideways displacement under shear

1. For a tangential force F over face area A, the shearing stress magnitude is F/A.
2. The corresponding shearing strain is Δx/L, where L is the distance between the displaced faces.
3. Divide shearing stress by shearing strain to obtain G = FL/(AΔx), and rearrange to find the displacement.

Δx = FL/(AG)

Shear modulus is generally less than Young’s modulus. For most materials, G is approximately Y/3. This is an approximate comparison, not a universal exact relation to substitute for a given material’s measured shear modulus.

Worked example 3. A square lead slab has side 50 cm and thickness 10 cm, where cm means centimetre (1 cm = 10⁻² m). Its lower edge is fixed. A tangential force of 9.0 × 10⁴ N acts over its narrow upper face. Find the upper-edge displacement using G = 5.6 × 10⁹ Pa.

Formula: A = side × thickness; Δx = FL/(AG).

Substitute: A = 0.50 × 0.10 = 0.05 m²; L = 0.50 m. Hence Δx = (9.0 × 10⁴ × 0.50)/(0.05 × 5.6 × 10⁹).

Answer: Δx = 1.6 × 10⁻⁴ m = 0.16 mm. Use the narrow face area, not the square front face area.

Worked example 4. An aluminium cube has edge 10 cm. One face is fixed to a vertical wall and a 100 kg mass is attached to the opposite face. Find that face’s vertical deflection using G = 25 × 10⁹ Pa and g = 9.8 m s⁻².

Formula: F = mg; A = edge²; Δx = FL/(AG).

Substitute: F = 980 N, A = 0.01 m² and L = 0.10 m. Thus Δx = (980 × 0.10)/(0.01 × 25 × 10⁹).

Answer: The downward deflection is 3.92 × 10⁻⁷ m, or 0.000000392 m. The face shifts tangentially, so this calculation uses shear modulus rather than Young’s modulus.

How are bulk modulus and compressibility related?

Bulk modulus, B, describes resistance to volume change under uniform hydraulic pressure. Let V be original volume and ΔV the signed change, final volume minus original volume. Both volumes have SI unit m³ and dimensions [L³]. The volume strain, denoted εᵥ, is:

εᵥ = ΔV/V

Let p be the increase in pressure causing compression. The bulk modulus is the negative ratio of this pressure increase to the corresponding volume strain:

B = −p/(ΔV/V)

For compression, p is positive while ΔV is negative. The minus sign therefore gives a positive bulk modulus for a system in equilibrium. The SI unit of bulk modulus is Pa, and its dimensions are [ML⁻¹T⁻²].

Compressibility, k, is the reciprocal of bulk modulus. It measures the fractional volume decrease per unit pressure increase. Its SI unit is Pa⁻¹ and its dimensions are [M⁻¹LT²]. Thus:

k = 1/B

How does the response differ between states of matter?

Bulk modulus applies to solids, liquids and gases. Solids are the least compressible and gases the most compressible in the material comparison. Gas compressibilities vary with pressure and temperature. Tight coupling between neighbouring atoms primarily accounts for the small compressibility of solids.

Use volume contraction for the positive magnitude of the volume decrease. If C denotes this magnitude, then C = −ΔV and C = pV/B. Distinguishing C from signed ΔV avoids reporting a positive signed change for a compression.

Worked example 5. A solid copper cube has edge 10 cm and is subjected to a hydraulic pressure increase of 7.0 × 10⁶ Pa. Calculate its volume contraction, given B = 140 × 10⁹ Pa.

Formula: V = edge³; C = pV/B.

Substitute: V = (0.10)³ = 1.0 × 10⁻³ m³. Thus C = (7.0 × 10⁶ × 1.0 × 10⁻³)/(140 × 10⁹).

Answer: The volume contraction is 5.0 × 10⁻⁸ m³, or 0.000000050 m³; the signed change ΔV is −5.0 × 10⁻⁸ m³.

Worked example 6. How much should the pressure on one litre of water be increased to compress it by 0.10%? Use B = 2.2 × 10⁹ Pa. A litre is a volume unit equal to 10⁻³ m³.

Formula: p = B(C/V).

Substitute: The fractional compression C/V = 0.10/100 = 0.0010, so p = (2.2 × 10⁹) × 0.0010.

Answer: Increase the pressure by 2.2 × 10⁶ Pa, or 2,200,000 Pa. Because the fractional compression is already given, the initial volume cancels from the calculation.

What does Poisson’s ratio measure?

A wire stretched along its length also undergoes a change in its transverse dimensions. Lateral strain is the strain perpendicular to the applied force. For a stretched wire, it describes the fractional decrease in diameter accompanying longitudinal extension.

Let d be the original diameter and Δd the positive magnitude of its contraction. Then the magnitude of lateral strain is Δd/d. With original length L and extension ΔL, the longitudinal strain is ΔL/L.

Poisson’s ratio, denoted ν, the Greek letter nu, is the ratio of lateral contraction strain to longitudinal extension strain, using these positive magnitudes:

ν = (Δd/d)/(ΔL/L)

Why must the convention be stated?

The length increases while the diameter decreases. Here Δd means the amount of contraction, not final diameter minus original diameter. Using that definition keeps the displayed ratio positive for the stretched wires discussed. It should not be mixed with the signed-volume convention used for bulk compression.

Within the elastic limit, lateral strain is directly proportional to longitudinal strain. Poisson’s ratio depends on the nature of the material. Because it divides one strain by another, it is a pure number with no dimensions or units.

For steels its value is between 0.28 and 0.30, while for aluminium alloys it is about 0.33. The aluminium-alloy value is approximate and need not be identical for every alloy.

Young’s modulus relates longitudinal stress to longitudinal strain. Poisson’s ratio instead connects two strains in perpendicular directions. They therefore describe different aspects of the same stretching experiment, and their numerical values cannot be directly compared as if they measured the same quantity.

How is work stored as elastic energy in a wire?

Work is energy transferred by a force acting through a displacement. Stretching a wire requires work against its interatomic forces, meaning forces between its atoms. In elastic deformation, this work is stored in the wire as elastic potential energy, also called strain energy.

The SI unit of energy is the joule, symbol J: 1 J = 1 N m. Energy and work have dimensions [ML²T⁻²]. They are different from stress or elastic modulus, whose unit is force per unit area.

Why does the changing force matter?

As the extension grows in the proportional region, the stretching force also grows. The force during the whole stretching process is therefore not simply its final value. A description of the work must account for the progressive increase in force as the wire elongates.

The stored energy belongs to the deformed wire. Removing the load allows elastic recovery towards its original length. This is the connection between doing work on the wire and its tendency to restore its original dimensions after the force is removed.

For the same wire within the proportional region, a greater extension involves more work and a greater store of elastic energy. This conclusion concerns recoverable deformation. Once a permanent set is produced, the assumption of complete elastic recovery no longer describes the deformation.

Keep three questions separate: stress measures force per area, strain measures fractional deformation, and elastic energy describes stored work. Young’s modulus connects stress and strain; it is not itself the energy stored in the wire.

Glossary

• Elasticity — Property by which a body tends to regain its original size and shape after the deforming force is removed.
• Plasticity — Property associated with permanent deformation and no gross tendency to recover the previous shape after unloading.
• Stress — Internal restoring force per unit area developed when a body is subjected to a deforming force.
• Longitudinal strain — Change in a body’s length divided by its original length, expressed as a dimensionless ratio.
• Shearing strain — Relative sideways displacement of opposite faces divided by their original perpendicular separation.
• Volume strain — Signed change in volume divided by original volume, negative for compression under the stated convention.
• Hooke’s law — Empirical proportionality between stress and strain for small deformations in the linear response region.
• Young’s modulus — Ratio of longitudinal stress to longitudinal strain within the proportional region of elastic response.
• Shear modulus — Ratio of shearing stress to corresponding shearing strain, also known as the modulus of rigidity.
• Bulk modulus — Positive measure of resistance to uniform compression, given by pressure increase divided by fractional volume contraction.
• Compressibility — Reciprocal of bulk modulus, measuring fractional volume decrease per unit increase in pressure.
• Poisson’s ratio — Ratio of lateral contraction strain to longitudinal extension strain, with both strains taken as positive magnitudes here.
• Permanent set — Residual strain that remains after the applied stress has been removed from a plastically deformed body.
• Elastic potential energy — Work stored in a deformed body as a result of its elastic deformation.

Common errors and misconceptions

• Misconception: All recoverable deformation obeys Hooke’s law. Correct: Hooke’s law describes proportional stress and strain. The typical metal curve includes a recoverable region where the relation is no longer proportional.
• Misconception: Strain is measured in metres. Correct: Extension is measured in metres, but strain divides extension by original length. It is dimensionless and has no unit.
• Misconception: Equal forces at both ends give stress 2F/A. Correct: The tension across the wire is F. Equal and opposite external forces maintain equilibrium without doubling that internal tension.
• Misconception: A larger Young’s modulus means greater extension under the same conditions. Correct: With force, area and original length fixed, extension is inversely proportional to Young’s modulus.
• Misconception: Diameter can replace radius in πr². Correct: Radius is half the diameter. Convert the given diameter to radius before calculating the circular cross-sectional area.
• Misconception: Compression makes bulk modulus negative. Correct: The signed volume change is negative. The minus sign in the defining formula makes the equilibrium bulk modulus positive.
• Misconception: Poisson’s ratio has the unit Pa. Correct: It is the ratio of two strains and has no unit. Young’s, shear and bulk moduli have the unit Pa.

Exam-style questions with model answers

Q1. Define elasticity and plasticity, giving one example of each. [2 marks]

1. Elasticity is the tendency to regain original size and shape after removal of the deforming force, as a gently stretched helical spring does.
2. Plasticity is associated with permanent deformation and no gross tendency to recover the previous shape. Putty is close to an ideal plastic.

Q2. State Hooke’s law, specify its range on a stress-strain curve, and explain why elastic recovery alone does not prove that the law is obeyed. [3 marks]

1. Hooke’s law states that stress is proportional to strain for small deformations. It is an empirical relationship found to hold for most materials.
2. It applies to the linear part of the stress-strain curve, where the stress-to-strain ratio remains constant.
3. A body can recover its dimensions in a non-proportional region, as between A and B on the typical metal curve. Recovery alone therefore does not establish Hooke’s law.

Q3. A structural steel rod of radius 10 mm and length 1.0 m is stretched axially by 100 kN. Given Young’s modulus 2.0 × 10¹¹ Pa and π = 3.14, calculate its area, stress, extension and strain, assuming proportional elastic behaviour. [4 marks]

1. The radius is 10⁻² m, so the cross-sectional area is πr² = 3.14 × 10⁻⁴ m². The diameter is not used in place of the radius.
2. The force is 100 × 10³ N. Stress equals force divided by area, giving 3.18 × 10⁸ Pa.
3. Extension equals stress multiplied by original length and divided by Young’s modulus: ΔL = 1.59 × 10⁻³ m.
4. Longitudinal strain is extension divided by original length, giving 1.59 × 10⁻³. This ratio has no unit.

Q4. Describe the arrangement and procedure for determining Young’s modulus of a wire by Searle’s method. Explain how the observations give the modulus and state a loading precaution. [5 marks]

1. Suspend an experimental wire and a reference wire of the same material, length and diameter from a common rigid support. Apply initial loads to straighten the wires, keeping the reference-wire load unchanged during measurements.
2. Measure the experimental wire’s original length and diameter, then calculate its cross-sectional area. Establish the initial level and micrometer reading.
3. Add known loads gradually to the experimental wire. Restore the spirit-level bubble to its initial position with the micrometer screw and record the reading after each addition.
4. Record readings during unloading too. The difference from the initial reading gives extension; added mass multiplied by gravitational acceleration gives the additional stretching force.
5. Calculate stress divided by strain, or Y = FL/(AΔL). Keep loading within the elastic limit and allow sufficient time after changing the load before adjusting the spirit level.

Q5. A copper cube of edge 10 cm experiences a uniform pressure increase of 7.0 × 10⁶ Pa. Its bulk modulus is 140 × 10⁹ Pa. Find its initial volume, volume contraction and signed volume change. [3 marks]

1. Convert the edge to 0.10 m. The initial volume is the cube of the edge, giving V = 1.0 × 10⁻³ m³.
2. The positive volume contraction is pressure increase multiplied by original volume and divided by bulk modulus: (7.0 × 10⁶ × 1.0 × 10⁻³)/(140 × 10⁹) = 5.0 × 10⁻⁸ m³.
3. The signed volume change is final volume minus original volume. Since the cube contracts, this change is negative: ΔV = −5.0 × 10⁻⁸ m³.

Q6. Define Poisson’s ratio using a stretched wire, explain the contraction convention, and state its units and dimensions. [3 marks]

1. Let the original wire length and diameter be L and d. If extension is ΔL and diameter contraction is Δd, longitudinal strain is ΔL/L and lateral contraction strain is Δd/d.
2. Poisson’s ratio is (Δd/d)/(ΔL/L). Here Δd is the positive amount of contraction, rather than the signed final-minus-original diameter change.
3. Both strains are dimensionless ratios. Consequently Poisson’s ratio is a pure number with no units or dimensions; within the elastic limit it relates lateral and longitudinal strain.

Q7. Explain qualitatively how work done in stretching a wire becomes strain energy and why the increasing force matters. [2 marks]

1. Stretching does work against interatomic forces. During elastic deformation, this work is stored in the wire as elastic potential energy, also called strain energy.
2. The force grows as extension increases in the proportional region. Therefore the work must account for the changing force, rather than treating the final force as constant throughout.

Key takeaways

• Elasticity describes a tendency to recover original dimensions after unloading; plastic deformation leaves a permanent change.
• Stress is restoring force per unit area, while strain is a dimensionless fractional change relative to the original dimension.
• Hooke’s law applies to the proportional region; some recoverable deformation occurs without a linear stress-strain relationship.
• Young’s modulus describes longitudinal response, shear modulus describes shearing response, and bulk modulus describes response to uniform pressure.
• For a wire, extension depends on force, original length, cross-sectional area and Young’s modulus, so geometry matters alongside material.
• Uniform compression gives a negative signed volume change; the defining minus sign makes the equilibrium bulk modulus positive.
• Poisson’s ratio connects lateral contraction strain with longitudinal extension strain and has neither units nor dimensions.
• Work done against interatomic forces during elastic stretching is stored in the wire as elastic potential energy.

Test yourself

Why does a suspended wire under load F have stress F/A rather than 2F/A?

The equal and opposite end forces balance the wire. The internal tension across its cross-section is F, so tensile stress is F/A.

Can a body behave elastically without obeying Hooke’s law?

Yes. It may regain its original dimensions even where stress and strain are not proportional, as in the A-to-B region of the typical metal curve.

What does a steeper initial stress-strain line indicate?

With stress vertical and longitudinal strain horizontal, a steeper proportional line indicates greater Young’s modulus and more stress needed for the same strain.

Why must the original length be used when calculating longitudinal strain?

Longitudinal strain is defined as change in length divided by original length. Using the final stretched length would calculate a different ratio.

Why is there a minus sign in the signed-volume formula for bulk modulus?

Increasing pressure compresses the body, giving a negative volume change. The minus sign makes the equilibrium bulk modulus positive.

How does compressibility change when bulk modulus increases?

Compressibility decreases because it is the reciprocal of bulk modulus. A larger bulk modulus means a smaller fractional compression for the same pressure increase.

What distinguishes Poisson’s ratio from Young’s modulus?

Poisson’s ratio divides lateral contraction strain by longitudinal strain. Young’s modulus divides longitudinal stress by longitudinal strain, so it has pressure units while Poisson’s ratio has none.

What is the physical meaning of strain energy in a stretched wire?

It is elastic potential energy stored through the work done against interatomic forces during elastic deformation. The stored energy accompanies the wire’s recoverable change in length.