NEB Class 11 • Physics • Properties of Matter

Elasticity: NEB Class 11 Physics Guide

Explain deformation with stress and strain, distinguish material stiffness from strength, and use Young, bulk and shear moduli only inside stated model limits.

Stress, strain and Hooke’s-law region
Young, bulk and shear modulus models
Material curves, energy, examples and exam checks

Modulus links stress to strain inside the linear elastic region.

From rigid model to real material

Where Elasticity fits in Class 11 Physics

Earlier Mechanics often treats bodies as rigid. Elasticity asks how a body’s size or shape changes under force and whether it returns after unloading. The CDC Physics Grade 11 page is the Nepal scope checkpoint. The OpenStax stress–strain reference defines tensile, bulk and shear models and their linear moduli.

Review Dynamics if force balance is weak and Work and Energy before elastic-energy questions.

Separate four ideas

Elasticity, plasticity, stiffness and strength

Elastic deformation disappears after the load is removed; plastic deformation remains. Stiffness means a large stress is required for a given strain, so a stiffer material has a larger relevant modulus. Strength concerns stress before failure. A material can be stiff but brittle, or flexible and able to sustain large elastic strain.

Term	Operational meaning	Do not confuse with
Elasticity	Recovery after unloading	Large extension
Stiffness	High stress/strain ratio	Strength
Strength	Ability to withstand stress before failure	Modulus
Plasticity	Permanent deformation	Elastic limit

The OpenStax elasticity and plasticity section uses characteristic stress–strain curves to show why proportional, elastic and fracture limits matter.

Load intensity

Stress and strain

Normal stress is σ = F/A, measured in pascals. Tensile strain is ε = ΔL/L₀ and is dimensionless. Stress depends on how force is distributed over cross-sectional area; strain compares deformation with original size. Use metres consistently when area is in m².

Worked example: tensile stress and strain

A 2.0 m wire of area 1.0×10⁻⁶ m² extends 1.0 mm under 100 N. Stress is 100/(1.0×10⁻⁶) = 1.0×10⁸ Pa. Strain is 0.001/2.0 = 5.0×10⁻⁴.

Strain has no unit because it is a length ratio. Writing metres beside strain is a quick signal that the definitions have been mixed.

Length deformation

Young’s modulus and wire extension

Within the linear elastic range, Y = σ/ε = (F/A)/(ΔL/L₀). Rearranging gives ΔL = FL₀/(AY). Extension increases with force and original length, and decreases with cross-sectional area and modulus.

Double force

Extension doubles while linear behaviour remains valid.

Double length

Extension doubles for the same force, area and material.

Double radius

Area becomes four times larger, so extension becomes one quarter.

Worked example: find extension

A 1.5 m steel wire of area 2.0×10⁻⁶ m² carries 200 N. With Y = 2.0×10¹¹ Pa, ΔL = 200×1.5/(2.0×10⁻⁶×2.0×10¹¹) = 7.5×10⁻⁴ m = 0.75 mm.

Volume deformation

Bulk modulus and compressibility

Bulk stress is pressure change and volume strain is ΔV/V₀. With magnitude convention, B = −Δp/(ΔV/V₀); the minus sign records that increasing external pressure normally reduces volume. Large B means the material is difficult to compress. Compressibility is the reciprocal 1/B.

Worked example: volume decrease

A liquid of volume 0.020 m³ has B = 2.0×10⁹ Pa and pressure rises 5.0×10⁶ Pa. ΔV/V = −Δp/B = −0.0025, so ΔV = −5.0×10⁻⁵ m³. The negative sign indicates a decrease.

Shape deformation

Shear modulus

Shear stress is tangential force per area, while small shear strain can be written Δx/L or approximately the shear angle in radians. The shear modulus S = (F/A)/(Δx/L). Fluids at rest cannot sustain static shear stress, whereas solids can.

Worked example: shear displacement

A block of height 0.10 m and top area 0.020 m² experiences tangential force 400 N. If S = 5.0×10⁸ Pa, shear stress is 2.0×10⁴ Pa, strain is 4.0×10⁻⁵, and Δx = 4.0×10⁻⁶ m.

Respect material limits

Reading a stress–strain curve

The initial straight region obeys proportional stress and strain.
Its slope in a tensile plot is Young’s modulus.
Up to the elastic limit, unloading can return the sample close to its original form.
Beyond yield, substantial plastic deformation occurs.
The maximum stress and fracture point describe other material behaviours, not the modulus alone.

A straight-line calculation is invalid after the material leaves its relevant linear region. Do not extrapolate Hooke’s law through yielding. Modulus values also depend on material condition and temperature; use the data supplied in the problem.

Curve feature	Meaning	Useful evidence
Initial slope	Stiffness/modulus	Stress divided by strain
Elastic limit	Recovery boundary	Unloading behaviour
Yield/plastic region	Permanent deformation	Residual strain
Fracture	Material separation	End of curve

Work stored during loading

Elastic energy density

In a linear loading process, force rises from zero to its final value, so work is ½FΔL. Dividing by volume AL gives energy per volume = ½σε = σ²/(2Y) = ½Yε². The triangular area under a linear stress–strain graph therefore represents elastic energy density.

Worked example: energy density

If stress is 8.0×10⁷ Pa and strain is 4.0×10⁻⁴, energy density is ½σε = 1.6×10⁴ J m⁻³.

This expression assumes reversible linear elasticity. Once plastic deformation occurs, loading and unloading can follow different paths and energy is dissipated.

Use concepts responsibly

Nepal-relevant examples and model limits

Cables, building components, suspension systems, climbing ropes and reinforced concrete all involve deformation, but real design requires safety factors, material testing, temperature effects, fatigue and engineering codes. Classroom equations explain trends; they are not structural design instructions.

A long thin cable extends more than a short thick cable of the same material under equal tension. A bridge expansion joint reduces thermally generated stress. Rubber may stretch visibly with low modulus yet remain elastic over a wide strain range, while glass is stiff but can fracture with little plastic deformation.

Exam method

A reliable Elasticity solution

Name tensile, compressive, bulk or shear deformation.
Convert area and length to SI units.
Calculate stress and strain separately.
Select the corresponding modulus.
State the linear elastic assumption.
Solve symbolically before numbers.
Check units, sign and scaling with area or length.

Use the Elasticity study guide for a spaced learning sequence. For online or physical NEB tuition, call 9846662070 with one attempted graph question.

Closed-book checkpoint

Practice tasks

Distinguish stiffness, strength and elasticity.
Find stress for 500 N on 2 mm².
Find strain for 0.4 mm extension of a 2 m wire.
Predict extension if wire radius doubles.
Explain the minus sign in bulk modulus.
Label a generic stress–strain curve.
Derive elastic energy density in the linear region.

Checks: (2) 2.5×10⁸ Pa; (3) 2×10⁻⁴; (4) one quarter.

Frequently asked questions

Questions students ask about Elasticity

What is the difference between stress and strain?

Stress is deforming force per area and has pascal units; strain is fractional deformation and is dimensionless.

Does a larger Young’s modulus mean more elastic?

It means stiffer in the linear tensile model: more stress is required for the same strain. Elastic recovery range is a separate property.

When can I use Hooke’s law?

Use proportional stress–strain relations only inside the relevant linear elastic range.

Why does wire radius strongly affect extension?

Area is proportional to radius squared, and extension is inversely proportional to area.

What does area under a stress–strain graph represent?

It represents work per unit volume; in the linear elastic region it is ½σε.

Where can I get help with NEB Elasticity?

For current online or physical tuition options, call 9846662070 and confirm timetable, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Continue with the Elasticity Study Guide for a diagnostic and revision plan. Scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

Measurement and uncertainty

Why diameter measurement controls a wire result

For a circular wire A = πd²/4, so a small fractional diameter error approximately doubles in the area. Since Y = FL/(AΔL), an overestimated diameter produces an overestimated area and therefore an underestimated modulus. Measure diameter with a micrometer at several positions and two perpendicular orientations, check zero error and average the readings.

Extension can be very small, so use a long wire and a sensitive pointer or optical scale where available. Read at eye level, add load in controlled steps and let oscillations settle. Plot stress against strain and use a best-fit gradient over the linear region; two adjacent points amplify reading noise.

A complete conclusion reports the modulus with appropriate significant figures, compares it with a suitable reference only after considering material identity and temperature, and states whether unloading returned close to the original length. Safety and apparatus limits take priority over obtaining a larger extension.

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