PhysicsISC Class 11 · Mechanical Properties of Solids

Bulk Modulus, Rigidity, Poisson's Ratio and Elastic Energy

Updated 17 October 2026 · Fact-checked

Bulk modulus B = −ΔP/(ΔV/V) measures resistance to volume change; modulus of rigidity η = shearing stress/shear strain; Poisson's ratio σ = lateral strain/longitudinal strain. The work stored in a stretched wire is ½FΔL, which is ½ × stress × strain per unit volume.

Understand Elastic Moduli, Poisson's Ratio and Elastic Energy

Bulk modulus. When a uniform pressure change ΔP acts on all sides, the volumetric strain is ΔV/V and B = −ΔP/(ΔV/V), the minus sign showing that a pressure increase reduces the volume. The reciprocal 1/B is the compressibility. All substances (solids, liquids, gases) have a bulk modulus.

Modulus of rigidity. A tangential force F on the face of area A shears a solid by an angle θ. Shearing stress = F/A; shear strain = θ (in radian), so η = F/(Aθ). Only solids have a rigidity modulus, because liquids and gases flow.

Poisson's ratio. When a wire is stretched, its diameter shrinks. σ = lateral strain/longitudinal strain, with lateral strain = Δd/d. It is a pure number; for most solids it lies between 0 and 0.5.

Elastic energy. The work done in stretching a wire is stored as elastic potential energy: U = ½FΔL = ½ × stress × strain × volume = (YAΔL²)/(2L). The energy per unit volume is u = ½ × stress × strain = ½Y(strain)² = stress²/2Y. This is the area under the stress-strain graph.

Dependence. Doubling the load doubles the extension and quadruples the stored energy (as U ∝ ΔL² within the elastic limit). The energy is recovered when the load is removed slowly, but the energy stored per unit volume is not the same for every wire: it depends on the material.

Elastic Moduli, Poisson's Ratio and Elastic Energy: formulas and key facts

Bulk modulus: $B = - Δ P / (Δ V / V) ∣ C o m p r ess ibi l i t y = 1/ B$; ΔV is negative when ΔP is positive.
Modulus of rigidity: $η = (F / A) / θ$; θ = angle of shear in radian.
Poisson's ratio: $σ = (Δ d / d) / (Δ L / L)$; A pure number.
Elastic energy: $U = ½ F Δ L ∣ u = ½ \times s t r ess \times s t r ain = s t r es s^{2} /2 Y$; u is the energy per unit volume.

Elastic Moduli, Poisson's Ratio and Elastic Energy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Energy stored in a loaded wire

Suppose a steel wire (Y = 2 × 10¹¹ Pa) 2 m long and of area 1 mm² supports a 100 N load.

Find the elongation, the stored energy and the energy per unit volume.

Try it first, then show the solutionSolution

ΔL = FL/(AY) = 100 × 2/(10⁻⁶ × 2 × 10¹¹) = 1.0 × 10⁻³ m.
U = ½FΔL = ½ × 100 × 10⁻³ = 0.05 J.
u = U/V = 0.05/(2 × 10⁻⁶) = 2.5 × 10⁴ J m⁻³; check ½ × 10⁸ × 5 × 10⁻⁴ = 2.5 × 10⁴.

Answer: ΔL = 1 mm; U = 0.05 J; u = 2.5 × 10⁴ J m⁻³.

Memory hook: The ½ comes from the force growing steadily from 0 to F.

Tricks to remember Elastic Moduli, Poisson's Ratio and Elastic Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Double load, quadruple energy
Within the elastic limit U ∝ ΔL² ∝ F².
Use it when: Energy ratio questions.
Only solids have rigidity
Bulk modulus exists for all states; rigidity modulus only for solids.
Use it when: MCQs on moduli.

Elastic Moduli, Poisson's Ratio and Elastic Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Reciprocal of bulk modulus · 1 mark · MCQBoard level

The reciprocal of the bulk modulus of a substance is called its:
(a) compressibility
(b) rigidity
(c) Poisson's ratio
(d) elasticity

Try it first, then show the solutionSolution

Compressibility = 1/B = −(ΔV/V)/ΔP.

Answer: (a) compressibility

Example 2: Compressing a liquid · 3 marks · Short answerBoard level

A pressure of 2 × 10⁷ Pa is applied to 1000 cm³ of water (B = 2.2 × 10⁹ Pa). Find the decrease in volume.

Try it first, then show the solutionSolution

ΔV/V = ΔP/B = 2 × 10⁷/2.2 × 10⁹ = 9.09 × 10⁻³. (1 mark)
V = 1000 cm³ = 10⁻³ m³. (1 mark)
ΔV = 9.09 × 10⁻³ × 1000 = 9.09 cm³. (1 mark)

Answer: The volume decreases by about 9.1 cm³.

Example 3: Shear of a block · 4 marks · Short answerHarder board level

A tangential force of 4 × 10⁵ N acts on the top face (area 0.05 m²) of a steel block of height 0.2 m, whose modulus of rigidity is 8 × 10¹⁰ Pa. Find the shear stress, the shear strain and the lateral displacement of the top face.

Try it first, then show the solutionSolution

Shear stress = F/A = 4 × 10⁵/0.05 = 8 × 10⁶ Pa. (1 mark)
Shear strain θ = stress/η = 8 × 10⁶/8 × 10¹⁰ = 1 × 10⁻⁴ rad. (1 mark)
Displacement x = θ h = 10⁻⁴ × 0.2. (1 mark)
x = 2 × 10⁻⁵ m. (1 mark)

Answer: Stress 8 × 10⁶ Pa; strain 10⁻⁴ rad; displacement 2 × 10⁻⁵ m.

Example 4: Poisson's ratio · 4 marks · Short answerHarder board level

A wire 2 m long and of radius 1 mm is stretched by 1 mm. If Poisson's ratio is 0.3, find the longitudinal strain, the lateral strain and the decrease in radius.

Try it first, then show the solutionSolution

Longitudinal strain = 1 mm/2000 mm = 5 × 10⁻⁴. (1 mark)
Lateral strain = σ × longitudinal = 0.3 × 5 × 10⁻⁴ = 1.5 × 10⁻⁴. (1 mark)
Δr = 1.5 × 10⁻⁴ × 1 mm = 1.5 × 10⁻⁴ mm. (1 mark)
= 1.5 × 10⁻⁷ m. (1 mark)

Answer: Longitudinal 5 × 10⁻⁴; lateral 1.5 × 10⁻⁴; the radius decreases by 1.5 × 10⁻⁷ m.

Example 5: Elastic energy · 5 marks · Long answerHOTS / competency

A steel wire (Y = 2 × 10¹¹ Pa) of length 2 m and area 1 mm² carries a 100 N load. Find (i) the extension, (ii) the elastic energy stored, (iii) the energy per unit volume, and (iv) the energy stored if the load were doubled.

Try it first, then show the solutionSolution

ΔL = FL/(AY) = 100 × 2/(10⁻⁶ × 2 × 10¹¹) = 1 × 10⁻³ m. (1 mark)
U = ½FΔL = ½ × 100 × 10⁻³ = 0.05 J. (1 mark)
V = 10⁻⁶ × 2 = 2 × 10⁻⁶ m³; u = 0.05/2 × 10⁻⁶ = 2.5 × 10⁴ J m⁻³. (1 mark)
Check: stress 10⁸ Pa, strain 5 × 10⁻⁴: ½ × 10⁸ × 5 × 10⁻⁴ = 2.5 × 10⁴ J m⁻³. (1 mark)
Doubling the load doubles ΔL, so U becomes 4 × 0.05 = 0.2 J. (1 mark)

Answer: ΔL = 1 mm; U = 0.05 J; u = 2.5 × 10⁴ J m⁻³; 0.2 J with the doubled load.

Where students lose marks in Elastic Moduli, Poisson's Ratio and Elastic Energy

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the ½ in elastic energy
Writing U = FΔL.
Do this: The force grows from 0 to F, so U = ½FΔL.
Usually costs: 1 to 2 marks.
Mistake: Applying rigidity modulus to liquids
Saying liquids have a modulus of rigidity.
Do this: Liquids cannot sustain a shear; only a bulk modulus exists.
Usually costs: 1 mark.
Mistake: Ignoring the minus sign in B
Reporting a negative bulk modulus.
Do this: ΔV is negative for an increase in pressure, so B = −ΔP/(ΔV/V) is positive.
Usually costs: 1 mark.

Exam tips

Name the type of strain (volume, shear, lateral) before choosing the formula.
Express the energy per unit volume as ½ × stress × strain and check it against U/V.
Use radians for shear angles; they are very small, so θ ≈ tanθ.
Say "within the elastic limit" in every energy derivation.

Practice questions on Elastic Moduli, Poisson's Ratio and Elastic Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which modulus of elasticity exists for liquids as well as solids?
(a) Young's modulus
(b) modulus of rigidity
(c) bulk modulus
(d) none of these
Hint
Liquids resist volume change but not shear.
Answer
(c) bulk modulus
Q2 · 2 marks · Short answerBoard level
The bulk modulus of a material is 1.6 × 10¹¹ Pa. Find the fractional volume change under an extra pressure of 8 × 10⁷ Pa.
Hint
ΔV/V = ΔP/B in magnitude.
Answer
8 × 10⁷/1.6 × 10¹¹ = 5 × 10⁻⁴ (a decrease).
Q3 · 3 marks · Short answerHarder board level
A steel wire (Y = 2 × 10¹¹ Pa), 1.5 m long and of area 2 × 10⁻⁶ m², is stretched by 1 mm. Find the force applied and the work done.
Hint
F = YAΔL/L; W = ½FΔL.
Answer
F = 2 × 10¹¹ × 2 × 10⁻⁶ × 10⁻³/1.5 = 267 N. W = ½ × 267 × 10⁻³ = 0.133 J.
Q4 · 3 marks · Short answerHarder board level
A rod of diameter 10 mm is stretched by 0.2% of its length and Poisson's ratio is 0.25. By how much does its diameter change?
Hint
Lateral strain = σ × longitudinal strain.
Answer
Lateral strain = 0.25 × 0.002 = 5 × 10⁻⁴. Δd = 5 × 10⁻⁴ × 10 mm = 5 × 10⁻³ mm (a decrease).
Q5 · 3 marks · Short answerHarder board level
A brass cube of edge 10 cm is lowered in the sea to where the extra pressure is 4 × 10⁷ Pa (B = 1.4 × 10¹¹ Pa for brass). Find the change in volume and the change in the edge.
Hint
ΔV/V = 3 Δa/a for a cube.
Answer
ΔV/V = 4 × 10⁷/1.4 × 10¹¹ = 2.86 × 10⁻⁴. V = 10⁻³ m³ so ΔV = 2.86 × 10⁻⁷ m³ = 0.286 cm³. Δa/a = 9.5 × 10⁻⁵, so Δa = 9.5 × 10⁻⁶ m.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a lift of 800 kg hangs from a steel cable of cross-section 4 × 10⁻⁴ m² and length 40 m (Y = 2 × 10¹¹ Pa, g = 9.8 m s⁻²).
(i) Find the stress when the lift is at rest. (1 mark)
(ii) Find the strain and the extension. (1 mark)
(iii) Find the tension when the lift accelerates upward at 2 m s⁻². (1 mark)
(iv) Find the new extension. (1 mark)
Hint
During upward acceleration T = m(g + a).
Answer
(i) F = 7840 N; stress = 7840/4 × 10⁻⁴ = 1.96 × 10⁷ Pa. (ii) Strain = 9.8 × 10⁻⁵; extension = 3.92 mm. (iii) T = 800 × 11.8 = 9440 N. (iv) Extension = 3.92 × 9440/7840 = 4.72 mm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define Poisson's ratio.Show answer
Lateral strain divided by longitudinal strain.
2.Elastic energy per unit volume?Show answer
½ × stress × strain.
3.Which materials have a modulus of rigidity?Show answer
Solids only.
4.What is compressibility?Show answer
The reciprocal of the bulk modulus.

Elastic Moduli, Poisson's Ratio and Elastic Energy: frequently asked questions

Why is the stored energy ½FΔL and not FΔL?

Because the stretching force rises from zero to its final value F as the wire extends, so the average force is F/2. The work is the area of the triangle under the F-ΔL graph.

What happens to the stored energy when a wire breaks?

The stored energy is released as heat and sound, and as kinetic energy of the broken ends.

Why can't Poisson's ratio exceed 0.5?

A value above 0.5 would mean the volume decreases on stretching, which is not found in ordinary solids; 0.5 corresponds to a constant volume.