Mechanical Properties of Solids — NEET / JEE
Class 11 Physics · Chapter 9

Mechanical
Properties
of Solids

Unlock your full potential — today we turn one of the toughest chapters into your scoring weapon. Concepts, graphs, numericals and exam traps all in one place.

NEET Ready JEE Focus CBSE Class 11 Interactive
Physics
01

Core Concepts

Tap any card to expand the full explanation, formula and exam trap for that concept.

02

Elastic Moduli — YBS

A solid cube can deform in three fundamentally different ways. Each has its own modulus. Mnemonic: Y B S → Young, Bulk, Shear.

Y
Young's Modulus
Y
Stretching or compressing a wire/rod along one axis. Measures how stiff the material is under tension or compression.
Y = (F · L₀) / (A · ΔL)
B
Bulk Modulus
B
Uniform pressure on all sides → volume decreases. Applies to solids, liquids and gases alike.
B = −P · V / ΔV
G
Shear Modulus
G
Layers sliding over each other — like pushing the top of a book stack sideways. Only solids have this.
G = (F / A) / θ
02b

Which Modulus for Which State?

🔩
Solids (steel, copper, glass, rubber) — possess all three moduli: Y, B and G. They resist stretching, volume compression and shearing.
💧
Liquids — Bulk modulus only. Liquids cannot support tensile or shear stress — they flow. Only volume compression (B) is meaningful.
💨
Gases — Bulk modulus only (very small B). Gases are highly compressible. They have absolutely no Y or G.
Common misconception: "Steel is less elastic than rubber." Wrong. Steel's Y ≈ 200 GPa vs rubber's ≈ 0.01 GPa. Higher Y = harder to deform = MORE elastic. "Elastic" means "returns to shape," not "easy to stretch."
Thick vs thin wire: ΔL = FL/(AY). Double the diameter → 4× the area → elongation becomes ¼. Thick wires always stretch less under the same force.
Compressibility: K = 1/B. Small B (gases) → highly compressible. Large B (solids, liquids) → nearly incompressible.
03

Stress–Strain Curve

This single graph tells the complete life story of a material — from perfect elasticity to catastrophic fracture. Mastering it is worth 4–5 marks in every exam.

Strain → Stress ↑ O A B C C' D E slope = Y Area = energy / volume
Hooke's law / elastic (O→B)
Yield / plastic flow (C→C')
Strain hardening (C'→D)
Necking & fracture (D→E)
A
Proportional Limit
Hooke's law is perfectly obeyed — stress is exactly proportional to strain. Slope of this straight line = Young's Modulus Y. The most useful region for all calculations.
B
Elastic Limit
Last safe point. Remove the force here and the material fully recovers. Beyond B, permanent (plastic) deformation begins — the material has been "damaged."
C — C'
Upper & Lower Yield Points
The material suddenly "gives" — large strain with no increase in stress. Permanent deformation sets in. In mild steel this is a dramatic drop; in other materials it may be gradual.
D
Ultimate Tensile Strength (UTS)
Maximum stress the material can withstand. After D, the specimen thins locally (necking). Even decreasing the load will not prevent fracture now.
E
Fracture / Breaking Point
The wire snaps. The fracture stress appears lower than UTS on the graph because the neck's cross-section has reduced drastically — actual local stress is enormous.
04

Solved Numericals

Universal 5-Step Method — Apply to every problem

① IdentifyTensile/compressive → Y  |  Pressure on all sides → B  |  Layer sliding → G
② List DataWrite F, A, L₀, ΔL — or P, V, ΔV — or F, A, θ with SI units
③ Write FormulaWrite the modulus equation before substituting anything
④ SubstituteUse SI units throughout. Simplify step-by-step; don't skip.
⑤ VerifyStress → Pa · Strain → dimensionless · Modulus → Pa or GPa
Level 1 — BasicWire Elongation using Young's Modulus
A steel wire of length 2 m and cross-sectional area 4 × 10⁻⁶ m² is stretched by a force of 200 N. Young's modulus of steel = 2 × 10¹¹ Pa. Find the elongation.
1
Type: Tensile stress along one axis → use Young's Modulus Y.
2
Given: F = 200 N  |  L₀ = 2 m  |  A = 4 × 10⁻⁶ m²  |  Y = 2 × 10¹¹ Pa
3
Formula: Y = FL₀ / (A · ΔL)  →  rearrange: ΔL = FL₀ / (A · Y)
4
Substitute: ΔL = (200 × 2) / (4 × 10⁻⁶ × 2 × 10¹¹) = 400 / (8 × 10⁵)
ΔL = 5 × 10⁻⁴ m = 0.5 mm
Level 2 — ModerateElastic Potential Energy Stored in a Wire
Using the same wire (F = 200 N, ΔL = 5 × 10⁻⁴ m, L₀ = 2 m, A = 4 × 10⁻⁶ m²), calculate the elastic energy stored and the energy density.
1
Formula for wire under tension: U = ½ × F × ΔL
2
Substitute: U = ½ × 200 × 5 × 10⁻⁴ = ½ × 0.1
U = 0.05 J = 50 mJ
3
Volume = A × L₀ = 4 × 10⁻⁶ × 2 = 8 × 10⁻⁶ m³
4
Energy density = U / V = 0.05 / (8 × 10⁻⁶)
= 6250 J/m³
Level 3 — NEET/JEEVolume Change using Bulk Modulus
The bulk modulus of water is 2 × 10⁹ Pa. Find (a) the change in volume of 1 m³ of water when pressure is increased by 10⁵ Pa, and (b) the compressibility of water.
1
Type: Uniform pressure on all sides → Bulk modulus. ΔV is negative (compression).
2
Given: B = 2 × 10⁹ Pa  |  P = 10⁵ Pa  |  V = 1 m³
3
Formula: B = −PV / ΔV  →  ΔV = −PV / B
4
Substitute: ΔV = −(10⁵ × 1) / (2 × 10⁹) = −5 × 10⁻⁵ m³
|ΔV| = 5 × 10⁻⁵ m³ (decrease)
5
Compressibility K = 1/B = 1/(2 × 10⁹)
K = 5 × 10⁻¹⁰ Pa⁻¹
Level 3 — JEE AdvancedComposite Wire — Total Elongation
A steel wire (Y = 2 × 10¹¹ Pa, L = 1 m, A = 1 × 10⁻⁶ m²) and a copper wire (Y = 1.2 × 10¹¹ Pa, L = 2 m, A = 2 × 10⁻⁶ m²) are connected end-to-end and a 4 kg mass is hung from the bottom. Find total elongation. (g = 10 m/s²)
1
Key insight — series: Both wires carry the same tension F = mg = 4 × 10 = 40 N.
2
Steel: ΔL₁ = FL / (AY) = (40 × 1) / (1 × 10⁻⁶ × 2 × 10¹¹) = 40 / (2 × 10⁵) = 2 × 10⁻⁴ m
3
Copper: ΔL₂ = (40 × 2) / (2 × 10⁻⁶ × 1.2 × 10¹¹) = 80 / (2.4 × 10⁵) = 3.33 × 10⁻⁴ m
4
Total: ΔL = ΔL₁ + ΔL₂ = (2.00 + 3.33) × 10⁻⁴
ΔL_total ≈ 5.33 × 10⁻⁴ m ≈ 0.53 mm
05

Quick Quiz — NEET / JEE Style

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06

Mnemonics

YBS
Young → Bulk → Shear. The three elastic moduli in order of exam frequency.
SSS
Stress → Strain → Slope = Young's Modulus. The core logic chain behind Hooke's Law.
PEYUD
Proportional → Elastic → Yield → Ultimate → (Break) Down. The five key curve points in order.
06b

Common Exam Traps

06c

Final Mastery Checklist

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