Mechanical Properties of Solids

Mechanical Properties of SolidsNEET Physics · Class 11 · NCERT Chapter 8

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Mechanical Properties of Solids. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Stress-strain curve regions

Young's modulus solver

Elastic potential energy

Bulk modulus and compressibility

Two wires in series and parallel

Poisson's ratio visualiser

Elastic moduli reference table

Stress, strain and Young's modulus

The proportional region, the elastic limit, the yield point — and how Young's modulus quantifies stiffness.

Stress-strain

Stress-strain curve regions

Slide the marker along the strain axis. The shaded region tells you which part of the curve you are in.

Strain marker: 15% of fracture strain

Current region

Proportional (Hooke's law)

Try this

Stay below A → Hooke's law holds, the body returns fully on unloading.
Between A and B → still elastic (returns), but stress-strain is no longer linear.
Beyond B → permanent deformation. Unloading leaves the wire longer than its original length.
Brittle materials (glass) have a tiny region after A — they crack instead of yielding.

Young's modulus

Young's modulus solver

Leave one field blank — the widget fills it in via Y = FL/(A·ΔL). Use scientific notation (1e-6) for small values.

F (N)

L (m)

A (m²)

ΔL (m)

Y (Pa)

Answer

Y = 2.0000e+11 Pa

Y = \frac{F L}{A Δ L} = \frac{100 \times 2}{0.000001 \times 0.001} = 2.000 e + 11

Try this

Defaults give Y = 2 × 10¹¹ Pa — the canonical value for steel.
Try F = 100 N, L = 1 m, A = 1e-6 m², Y = 2e11 → ΔL = 5e-4 m = 0.5 mm.
Common units: 1 mm² = 10⁻⁶ m². 1 GPa = 10⁹ Pa.

Elastic energy and bulk modulus

Energy stored in a stretched material, and how bulk modulus measures resistance to compression.

Elastic PE

Elastic potential energy in a stretched material

Energy stored per unit volume = ½·Y·ε². Multiply by volume for the total energy.

Young's modulus Y: 200 GPa

Rubber ≈ 0.01 GPa · Bone ≈ 14 GPa · Aluminium ≈ 70 GPa · Steel ≈ 200 GPa

Strain ε: 0.100%

Volume: 10 cm³

Energy density u

1.000e+5 J/m³

Total energy U

1.0000 J

Stress σ

2.00e+8 Pa

u = \frac{1}{2} Y ε^{2} = \frac{1}{2} (200 \times 1 0^{9}) (1.00 e - 3)^{2}

U = u V = 1.00 e + 5 \times 10 \times 1 0^{- 6} = 1.0000 J

Try this

Doubling the strain quadruples the energy density (u ∝ ε²).
Same strain on a stiffer material (larger Y) → more energy stored. That is why a steel spring stores more energy than a rubber band of the same volume at the same elongation.
For ε = 0.1% in steel: u ≈ 10⁵ J/m³ = 0.1 J/cm³. A small but real amount.

Bulk modulus

Bulk modulus and compressibility

Apply hydrostatic pressure to a material. Higher B (stiffer) means smaller fractional volume change.

Bulk modulus B: 2.20e+9 Pa

Pressure increase ΔP: 10 MPa

Quick presets

Fractional volume change

ΔV/V = 4.545e-1%

\frac{Δ V}{V} = \frac{Δ P}{B} = \frac{1.00 e + 7}{2.20 e + 9}

Compressibility (1/B)

4.545e-10 Pa⁻¹

Try this

Click "Air" then ΔP = 1 atm (~0.1 MPa) → ΔV/V = 100%. A gas can be compressed to almost any extent.
Click "Water" then ΔP = 100 MPa → ΔV/V ≈ 4.5%. Water is nearly incompressible.
Click "Steel" → even hundreds of MPa cause less than 0.1% volume change. Solids are very stiff under hydrostatic pressure.

Wire combinations and Poisson's ratio

Series and parallel wires, plus the lateral contraction that comes with longitudinal stretch.

Wires

Two wires in series and parallel

Series: same force, elongations add. Parallel: same elongation, forces add. Stiffer wire (Y·A/L) carries more in parallel.

Total force F: 100 N

Wire 1 (steel-ish)

L₁: 1.0 m

r₁: 1.0 mm

Y₁: 200 GPa

Wire 2 (copper-ish)

L₂: 1.0 m

r₂: 1.0 mm

Y₂: 110 GPa

Total elongation

ΔL = 0.449 mm

ΔL₁

0.159 mm

ΔL₂

0.289 mm

Δ L = \frac{F L _{1}}{A _{1} Y _{1}} + \frac{F L _{2}}{A _{2} Y _{2}}

Try this

Series mode: each wire has the same force, but elongations add. The thinner / less-stiff wire stretches more.
Parallel mode: both wires stretch by the same amount; the stiffer wire (larger Y A / L) carries more force.
In parallel, set Y₁ = Y₂, r₁ = r₂ → equal sharing F₁ = F₂ = F/2.

Poisson's ratio

Poisson's ratio visualiser

Stretch a block longitudinally — it gets thinner sideways. Adjust Poisson's ratio to see different materials.

Longitudinal strain (stretch): 2.0%

Poisson's ratio σ: 0.30

Steel ≈ 0.3 · Concrete ≈ 0.2 · Aluminium ≈ 0.33 · Rubber ≈ 0.5

Lateral strain

-0.600%

Volume change ΔV/V

0.800%

σ = - \frac{ε _{lat}}{ε _{long}} \Rightarrow ε_{lat} = - σ ε_{long}

Try this

σ = 0 → no lateral contraction. Cork is close to this.
σ = 0.5 → lateral strain exactly cancels the volume change. Rubber and many soft tissues behave this way (incompressible).
σ = 0.3 (typical metal) → ΔV/V ≈ 0.4 × ε_long. The wire gets thinner and slightly larger in volume.

Reference

Elastic moduli reference table

Approximate Young's modulus, shear modulus, bulk modulus and Poisson's ratio for common NEET materials.

Approximate values at room temperature. All in GPa (10⁹ Pa). Use these to sanity-check NEET problems.

Material

Y (GPa)

G (GPa)

B (GPa)

σ (Poisson)

Steel

200

160

0.3

Copper

110

140

0.34

Brass

100

110

0.34

Aluminium

0.33

Bone

—

0.3

Granite

—

0.25

Concrete

—

0.2

Wood (oak)

—

Rubber

0.05

—

0.5

Try this

Steel's Y is about 4000× larger than rubber's. That's why rubber stretches so easily.
For metals, G ≈ Y/3 holds approximately — check it against this table.
Rubber has σ ≈ 0.5 → nearly incompressible. That's why it bulges sideways visibly when squeezed.

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