PhysicsCBSE Class 11 · Mechanical Properties of Solids

Bulk Modulus and Elastic Potential Energy Class 11

Updated 17 October 2026 · Fact-checked

The bulk modulus B = −ΔP/(ΔV/V) measures resistance to uniform compression, and its inverse is the compressibility. The work done in stretching a wire within the elastic limit is stored as elastic potential energy U = ½FΔL = ½kΔL², so the energy per unit volume is ½ × stress × strain = σ²/2Y.

The force grows from zero, so the stored energy is the triangle's area, ½FΔL, not FΔL; divide by volume to get σ²/2Y.

Illustration drawn for QuestionBuddy.

Understand Bulk Modulus and Elastic Potential Energy

Volume strain and bulk modulus. When a uniform pressure increase $Δ P$ acts on a body, its volume changes by $Δ V$ . The volume strain is $Δ V / V$ and the bulk modulus is
$ $B = - \frac{Δ P}{Δ V / V} = - \frac{V Δ P}{Δ V} .$ $
The negative sign shows the volume decreases when the pressure increases; use magnitudes in numericals. The reciprocal $1/ B$ is the compressibility. Solids and liquids have large $B$ (hard to compress); gases have a small $B$ . Because a compressed body gets denser, $ρ^{'} \approx ρ (1 + Δ P / B)$ for small compressions.

Hooke's law as a spring. Within the elastic limit a stretched wire behaves like a spring of constant $k = A Y / L$ . The force rises from zero to $F$ while the extension rises from zero to $Δ L$ , so the work done is the area of the F- $Δ L$ triangle:
$ $U = \frac{1}{2} F Δ L = \frac{1}{2} k (Δ L)^{2} = \frac{F ^{2} L}{2 A Y} .$ $
The work is not $F Δ L$ because the force is not constant. Energy per unit volume (energy density):
$ $u = \frac{U}{A L} = \frac{1}{2} stress \times strain = \frac{σ ^{2}}{2 Y} .$ $

Work between two states. Going from extension $x_{1}$ to $x_{2}$ , $W = \frac{1}{2} k (x_{2}^{2} - x_{1}^{2})$ ; equivalently, the average force times the extra extension.

Sudden loading. If a mass $m$ is released suddenly from the unstretched position of a wire (or spring), the maximum extension is *twice* the static extension: $m g x = \frac{1}{2} k x^{2}$ gives $x = 2 m g / k$ , and the maximum force is $2 m g$ (the wire momentarily overshoots the equilibrium).

Series and parallel. Wires joined end to end carry the same force; their extensions add, so $1/ k = 1/ k_{1} + 1/ k_{2}$ . Wires side by side stretch equally; $k = k_{1} + k_{2}$ .

Bulk Modulus and Elastic Potential Energy: formulas and key facts

Volume strain: $\frac{Δ V}{V}$; Fractional change in volume.
Bulk modulus: $B = - \frac{V Δ P}{Δ V}$; Pa.
Compressibility: $κ = \frac{1}{B}$; Pa⁻¹.
Elastic energy: $U = \frac{1}{2} F Δ L = \frac{1}{2} k (Δ L)^{2}$; Within the elastic limit.
Energy density: $u = \frac{1}{2} σ ε = \frac{σ ^{2}}{2 Y}$; Per unit volume.
Sudden loading: $x_{ma x} = \frac{2 m g}{k} = 2 x_{s t a t i c}$; Released from the natural length.

How to solve Bulk Modulus and Elastic Potential Energy questions

For volume changes use B with the fractional volume change; for energy use the area under the force-extension graph, never F times extension.

1Identify whether the question is about a change of volume (use B) or of length (use Y).
2For B problems, compute ΔV/V = ΔP/B and then ΔV from V.
3For stored energy, use ½FΔL, ½kx² or u × volume.
4For energy between two loads, use the average force times the extra extension.
5For series wires, add the extensions; for parallel wires, add the spring constants.
6For sudden loading, equate the loss of potential energy to the elastic energy.

Bulk Modulus and Elastic Potential 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.

A diver's equipment

Suppose a sealed liquid-filled gauge of volume 20 cm³ has a bulk modulus of 2.0 × 10⁹ Pa and is carried down by a diver where the extra pressure is 4.0 × 10⁵ Pa.

Find the change in volume.

Try it first, then show the solutionSolution

ΔV/V = 4.0 × 10⁵/2.0 × 10⁹ = 2.0 × 10⁻⁴.
ΔV = 2.0 × 10⁻⁴ × 20 = 4.0 × 10⁻³ cm³.

Answer: 4.0 × 10⁻³ cm³ decrease.

Memory hook: Liquids hardly compress.

Energy in a stretched cable

Suppose a cable of spring constant 5.0 × 10⁴ N/m is stretched by 2.0 mm.

Find the elastic energy.

Try it first, then show the solutionSolution

U = ½kx² = ½ × 5.0 × 10⁴ × (2.0 × 10⁻³)² = 0.10 J.

Answer: 0.10 J.

Memory hook: Energy grows with the square of the stretch.

Tricks to remember Bulk Modulus and Elastic Potential Energy

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

Half F x
Energy = ½ × final force × extension, not full F x.
Use it when: Work in stretching.
Energy density = ½ σ ε
Per unit volume, stress × strain over two.
Use it when: Energy density questions.
Twice for sudden
Released suddenly, the extension (and force) double.
Use it when: Dynamic loading.
Series adds extensions
Same force; sum the stretches.
Use it when: Composite wires.

Bulk Modulus and Elastic Potential Energy: solved examples

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

Example 1: Compressing a liquid · 2 marks · Very short answerBoard level

A liquid of bulk modulus 1.6 × 10⁹ Pa is subjected to an additional pressure of 4.0 × 10⁶ Pa. Find the fractional change in volume, and the change in volume of 2.0 litres of it.

Try it first, then show the solutionSolution

ΔV/V = ΔP/B = 4.0 × 10⁶/1.6 × 10⁹ = 2.5 × 10⁻³. (1)
ΔV = 2.5 × 10⁻³ × 2000 mL = 5 mL decrease. (1)

Answer: 2.5 × 10⁻³; 5 mL.

Example 2: Energy density in a wire · 3 marks · Short answerHarder board level

A steel wire (Y = 2 × 10¹¹ Pa) of length 2 m and cross-section 1.5 mm² is under a tensile stress of 2 × 10⁸ Pa. Find the strain, the energy per unit volume and the total elastic energy stored.

Try it first, then show the solutionSolution

Strain = σ/Y = 2 × 10⁸/2 × 10¹¹ = 1 × 10⁻³. (1)
u = ½ σ ε = ½ × 2 × 10⁸ × 10⁻³ = 1 × 10⁵ J/m³. (1)
Volume = 2 × 1.5 × 10⁻⁶ = 3 × 10⁻⁶ m³, so U = 1 × 10⁵ × 3 × 10⁻⁶ = 0.3 J. (1)

Answer: 10⁻³; 10⁵ J/m³; 0.3 J.

Example 3: Work between two loads · 3 marks · Short answerHarder board level

A wire stretches by 0.4 mm when a force of 80 N is applied and by 0.6 mm when the force is 120 N. Find the energy stored at 80 N and the extra work done in raising the load from 80 N to 120 N.

Try it first, then show the solutionSolution

U₁ = ½ × 80 × 0.4 × 10⁻³ = 0.016 J. (1)
Extra work = average force × extra extension = ½(80 + 120) × 0.2 × 10⁻³ = 0.02 J. (1)
Check: U₂ = ½ × 120 × 0.6 × 10⁻³ = 0.036 J = 0.016 + 0.02. (1)

Answer: 0.016 J; 0.02 J.

Example 4: Density of a compressed solid · 4 marks · Long answerHarder board level

A metallic block of volume 500 cm³, bulk modulus 1.5 × 10¹¹ Pa and density 9000 kg/m³ is lowered to a depth where the extra pressure is 3.0 × 10⁷ Pa. Find the decrease in volume, the new density and the percentage change in density.

Try it first, then show the solutionSolution

ΔV/V = ΔP/B = 3.0 × 10⁷/1.5 × 10¹¹ = 2.0 × 10⁻⁴. (1)
ΔV = 2.0 × 10⁻⁴ × 500 = 0.10 cm³. (1)
New density ≈ ρ(1 + 2.0 × 10⁻⁴) = 9001.8 kg/m³. (1)
Increase = 0.02%. (1)

Answer: 0.10 cm³; 9001.8 kg/m³; 0.02% increase.

Example 5: Wires in series and a suddenly applied load · 4 marks · Long answerHOTS / competency

Two wires, each of length 1 m and area 1.0 mm², are joined end to end; one has Y = 2.0 × 10¹¹ Pa and the other Y = 1.0 × 10¹¹ Pa. A 100 N load is hung from the lower end. (a) Find the total extension. (b) Find the elastic energy stored and its division between the wires. (c) If the same 100 N weight were instead released suddenly from the unstretched position, what would the maximum extension be?

Try it first, then show the solutionSolution

(a) Same force: ΔL₁ = 100/(10⁻⁶ × 2 × 10¹¹) = 0.5 mm; ΔL₂ = 100/(10⁻⁶ × 10¹¹) = 1.0 mm; total 1.5 mm. (1)
(b) U = ½ × 100 × 1.5 × 10⁻³ = 0.075 J. (1)
Energy ∝ extension at the same force: 0.025 J and 0.050 J. (1)
(c) Sudden release gives twice the static value: 3.0 mm. (1)

Answer: (a) 1.5 mm; (b) 0.075 J (0.025 J + 0.050 J); (c) 3.0 mm.

Where students lose marks in Bulk Modulus and Elastic Potential Energy

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

Mistake: Using work = F × extension for a stretched wire.
Treating force as constant.
Do this: Work = ½Fx (area of the triangle).
Usually costs: The answer mark.
Mistake: Forgetting the volume in energy density problems.
Stopping at u.
Do this: Total U = u × A × L.
Usually costs: 1 mark.
Mistake: Mixing Y and B.
Using length strain for a volume change.
Do this: Y for stretching, B for uniform compression.
Usually costs: Concept mark.
Mistake: Ignoring the negative sign meaning in B.
Quoting a negative modulus.
Do this: B is positive; the sign only shows volume decreases.
Usually costs: Reasoning mark.

Exam tips

Write the energy as ½ × F × ΔL before substituting.
For energy density, use σ²/2Y when stress and Y are given.
Mention "within the elastic limit" in energy answers.
State the compressibility as 1/B when it is asked.
For series wires, show the equal force first.

Practice questions on Bulk Modulus and Elastic Potential 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
A wire stretched by x stores elastic energy U. Within the elastic limit, stretching it by 3x stores
(a) 3U (b) 6U (c) 9U (d) 27U
Hint
U ∝ x².
Answer
(c) 9U.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The work done in stretching a wire by force F through extension x is ½Fx and not Fx.
Reason (R): The stretching force increases from zero to F as the extension grows.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Average force is F/2.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A wire with Y = 2 × 10¹¹ Pa carries a stress of 10⁸ Pa. Find the energy stored per unit volume.
Hint
u = σ²/2Y.
Answer
u = (10⁸)²/(2 × 2 × 10¹¹) = 2.5 × 10⁴ J/m³.
Q4 · 3 marks · Short answerHarder board level
A wire of length 1.5 m, area 3 mm² and Y = 2 × 10¹¹ Pa is treated as a spring. Find its spring constant, the force for an extension of 1 mm and the energy stored then.
Hint
k = AY/L.
Answer
k = 3 × 10⁻⁶ × 2 × 10¹¹/1.5 = 4 × 10⁵ N/m; F = 400 N; U = ½ × 400 × 10⁻³ = 0.2 J.
Q5 · 4 marks · Long answerHOTS / competency
A 5 kg mass is attached to the lower end of a vertical wire of length 2 m, area 1 mm² and Y = 2 × 10¹¹ Pa and is lowered slowly until it is in equilibrium. (a) Find the static extension. (b) If the mass is instead released suddenly from the unstretched position, find the maximum extension and the maximum stress (g = 10 m/s²). (c) Why is it dangerous to apply loads suddenly?
Hint
k = AY/L; energy balance for sudden release.
Answer
(a) k = 10⁻⁶ × 2 × 10¹¹/2 = 10⁵ N/m; x = 50/10⁵ = 0.5 mm. (b) x_max = 2 × 0.5 = 1 mm; force 100 N, stress 100/10⁻⁶ = 10⁸ Pa. (c) The peak force is twice the weight, so the stress can exceed the safe value.

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.Bulk modulus formula?Show answer
B = −VΔP/ΔV.
2.Compressibility?Show answer
1/B.
3.Elastic energy of a wire?Show answer
½FΔL = ½kx².
4.Energy per unit volume?Show answer
½ × stress × strain.
5.Sudden loading extension?Show answer
Twice the static value.
6.Spring constant of a wire?Show answer
AY/L.

Bulk Modulus and Elastic Potential Energy: frequently asked questions

Why are gases easily compressed?

Their bulk modulus is small because the molecules are far apart.

Where does the stored elastic energy go when the wire is released?

It is returned as kinetic energy and then dissipated as heat after the oscillations die out.

What is the difference between Y and B?

Young's modulus relates stress and strain for a change in length; the bulk modulus relates pressure and volume strain for uniform compression.

Does the energy stored depend on the shape of the load-extension graph?

Yes, it is the area under the graph; for a straight line it is a triangle.

What is a perfectly rigid body?

A body whose Young's modulus is infinite; it does not deform.