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NEET UG · Physics

Mechanical properties of solids

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NEET UG · Physics

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Convert area, calculate stress and extension, and distinguish Young modulus from sample stiffness before NEET chapter MCQs. Stress, strain and stiffness: keep material and geometry separate

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Stress, strain and stiffness: keep material and geometry separate

Convert area, calculate stress and extension, and distinguish Young modulus from sample stiffness before NEET chapter MCQs.

What you will understand

For NEET UG students: distinguish stress, strain and extension, convert a wire’s area correctly, and decide whether a change concerns the material modulus or the sample’s geometry. This focused lesson explains the reasoning before chapter practice.

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Quick answer: keep the material and the sample separate

For a uniform wire under a small axial load in its linear elastic range, engineering stress is σ=F/A0\sigma=F/A_0, longitudinal strain is ε=ΔL/L0\varepsilon=\Delta L/L_0 and Young modulus is Y=σ/εY=\sigma/\varepsilon. Here A0A_0 and L0L_0 are the original area and length. Thus ΔL=FL0/(A0Y)\Delta L=FL_0/(A_0Y).

Stress and YY have unit Pa; strain is dimensionless. The sample’s effective axial stiffness is k=F/ΔL=YA0/L0k=F/\Delta L=YA_0/L_0, with unit N/m. Changing length or radius can change kk without changing the material’s YY.

Do not add the two end loads when calculating stress

A uniform wire of original length 2.0 metres and original cross-sectional area 1.0 square millimetre is pulled axially by equal and opposite 100 N forces. Its tension is 100 N, not 200 N. With Young modulus 2.0 times ten to the eleventh pascals in the small linear elastic range, stress is 100 MPa, strain is 0.00050, and extension is 1.0 millimetre. Strain is dimensionless. Wire and force arrows are schematic.
A uniform wire of original length 2.0 metres and original cross-sectional area 1.0 square millimetre is pulled axially by equal and opposite 100 N forces. Its tension is 100 N, not 200 N. With Young modulus 2.0 times ten to the eleventh pascals in the small linear elastic range, stress is 100 MPa, strain is 0.00050, and extension is 1.0 millimetre. Strain is dimensionless. Wire and force arrows are schematic.

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Worked example: convert area before dividing

Take L0=2.0 mL_0=2.0\,\mathrm m, A0=1.0 mm2A_0=1.0\,\mathrm{mm^2}, F=100 NF=100\,\mathrm N and Y=2.0×1011 PaY=2.0\times10^{11}\,\mathrm{Pa}. Treat the wire as uniform and neglect its weight. Two opposite 100 N end loads produce tension 100 N across a section, not 200 N.

Because 1 mm2=(10−3 m)2=10−6 m21\,\mathrm{mm^2}=(10^{-3}\,\mathrm m)^2=10^{-6}\,\mathrm{m^2}, stress is 108 Pa=100 MPa10^8\,\mathrm{Pa}=100\,\mathrm{MPa}. Strain is 108/(2.0×1011)=5.0×10−410^8/(2.0\times10^{11})=5.0\times10^{-4}, or 0.050%. Extension is 2.0(5.0×10−4)=0.0010 m=1.0 mm2.0(5.0\times10^{-4})=0.0010\,\mathrm m=1.0\,\mathrm{mm}. A strain value is not itself a length.

Choose the deformation and respect the limits

Young modulus describes axial length change. Shear modulus relates tangential stress to shear strain. Bulk modulus is B=−Δp/(ΔV/V0)B=-\Delta p/(\Delta V/V_0) for small volume changes; the minus sign makes BB positive when a pressure increase compresses the volume.

The proportionality limit marks the end of a linear stress–strain relation. An elastic response can still be nonlinear beyond that limit; an elastic limit concerns recovery on unloading. Neither the modulus nor the linear slope alone specifies breaking strength. Do not extend the small-strain formula to arbitrary loads or materials.

Compare geometry at the same load and material

Three uniform wires have the same Young modulus and carry the same 100 N axial load in the linear range. A has length 2 metres, area 1 square millimetre, extension 1.00 millimetre and stiffness 100000 N per metre. B has the same length but twice the radius, hence four times the area: extension is 0.25 millimetre and stiffness 400000 N per metre. C has twice A’s length and the same area: extension is 2.00 millimetres and stiffness 50000 N per metre. Drawn wire lengths and thicknesses compare geometry schematically; the numerical values define the examples.
Three uniform wires have the same Young modulus and carry the same 100 N axial load in the linear range. A has length 2 metres, area 1 square millimetre, extension 1.00 millimetre and stiffness 100000 N per metre. B has the same length but twice the radius, hence four times the area: extension is 0.25 millimetre and stiffness 400000 N per metre. C has twice A’s length and the same area: extension is 2.00 millimetres and stiffness 50000 N per metre. Drawn wire lengths and thicknesses compare geometry schematically; the numerical values define the examples.

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Worked example: doubling radius does not double extension

Keep the original length, material and 100 N load, but double the circular wire radius. Its area becomes four times as large. Since ΔL∝1/A0\Delta L\propto1/A_0 under these conditions, extension becomes 1.0/4=0.25 mm1.0/4=0.25\,\mathrm{mm}. Young modulus stays the same; the sample stiffness becomes four times larger.

For the original wire, k=1.0×105 N m−1k=1.0\times10^5\,\mathrm{N\,m^{-1}}. Slow linear loading from zero stores U=12FΔL=0.050 JU=\tfrac12F\Delta L=0.050\,\mathrm J. Using FΔLF\Delta L would incorrectly assume the load stayed at its final value throughout the extension.

Common mistakes to catch

Square the millimetre-to-metre conversion when converting an area. Use the force carried across one section, not the sum of opposite end-force magnitudes. Keep fractional strain distinct from percentage strain and extension. A doubled radius means four times the area. The material modulus, a sample’s stiffness, its elastic limit and its breaking strength describe different properties.

Try it yourself 1: choose the section force

The original 1.0 mm² wire is pulled by two opposite 100 N end forces. What is the axial stress? A. 200 MPa; B. 100 MPa; C. zero.

Reveal the answer

B. The force transmitted across a cross-section is 100 N. Dividing by 10−6 m210^{-6}\,\mathrm{m^2} gives 108 Pa10^8\,\mathrm{Pa}. Opposite end loads balance the wire as a whole; that does not make the internal stress zero or double the section force.

Try it yourself 2: shorten the sample

Cut an otherwise identical sample to half the original length. Keep area, material and the same 100 N load. What is its extension in the linear model? A. 2.0 mm; B. 1.0 mm; C. 0.50 mm.

Reveal the answer

C. Extension is proportional to original length at fixed force, area and modulus. The material’s Young modulus does not halve; the shorter sample is twice as stiff.

A useful check on your next visit

Explain stress, strain and extension in one sentence each, then recover the effect of changing length or radius without substituting numbers. On your next practice question, state which quantities stay fixed before choosing a proportionality.

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Explain the distinction you missed, then apply it to a fresh chapter question.

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Sources and review

Technical review: Codex (automated). Primary textbook sources checked on 23 October 2026. Diagrams, numerical examples and self-checks are original teaching material. Human subject review is not claimed.

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