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State Board · Class 11 · Physics

Lesson 6: Mechanical Properties of Solids

7 concepts · 56 practice questions with answers · Free to read

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Elasticity and plasticity

From daily life

A rubber band returns when released; pressed clay stays pressed. Link to earlier classes : pressure and force of classes 9-10 now extend to stress and strain.

Simple definition

Elasticity is the property of regaining shape when the deforming force is removed.

Elastic and plasticLoriginalstretchedreturns after removal = elasticstays deformed = plastic

In the figure : The green part is the stretch; it returns when the force is removed.

Rule 1: Elasticity — A body regains its shape after the force is removed : elasticity. Steel is more elastic than rubber (larger Y).

  • Easy : A stretched spring returns when released.
  • Medium : Steel Y = 2 × 10¹¹, rubber ≈ 10⁶-10⁸ Pa.
  • Exam (CET) : Most elastic? (1) steel (2) rubber (3) clay (4) lead → answer 1)

Rule 2: Elastic limit — The maximum stress up to which a body returns is the elastic limit; beyond it, permanent deformation.

  • Easy : A wire pulled slightly returns; pulled more, it stays bent.
  • Medium : Plastic state = permanent deformation.
  • Exam (CET) : A plastic body? (1) clay (2) steel (3) glass (4) spring → answer 1)

Rule 3: More terms — Restoring force = the internal force pulling back to the original shape. Fatigue = weakening from repeated loading.

  • Easy : Bridges tire from repeated vibrations.
  • Medium : Restoring force arises from interatomic forces.
  • Exam (CET) : Fatigue means? (1) repeated loading (2) one-time break (3) heating (4) shrinking → answer 1)

Derivation : Origin of the restoring force

  • Step 1 : At equilibrium atoms are at r₀; net force 0.
  • Step 2 : Stretching gives r > r₀ ⇒ net attraction.
  • Step 3 : Compression gives r < r₀ ⇒ net repulsion.
  • Step 4 : For small displacements F ∝ −x (spring-like).
Remember

Within the elastic limit it returns; beyond it, plastic.

Common mistake / CET trap

Elasticity ≠ ease of stretching; steel is more elastic than rubber.

Shortcut

Memory clue : "Back within the limit, permanent beyond".

Spot the difference

PropertyElasticPlastic
After removalreturnspermanent change
Examplespringclay
Limitwithinbeyond

Practice questions

Easy
  1. 1What is elasticity?
  2. 2What is the elastic limit?
  3. 3Name a plastic body.
Medium
  1. 4Rubber or steel more elastic? Why?
  2. 5What is fatigue?
  3. 6What causes the restoring force?
Challenging
  1. 7Force when r < r₀?
  2. 8Why is Y small for a rubber band?
Show answers (try them yourself first)
  1. 1Regains shape on removal
  2. 2Max stress up to which it returns
  3. 3Clay
  4. 4Steel; larger Y
  5. 5Weakening from repeated load
  6. 6Interatomic force
  7. 7Repulsive
  8. 8Large strain for small stress

Stress, strain and Hooke's law

From daily life

A thicker wire bears more load — the same force gives less stress.

Simple definition

Stress is restoring force per unit area; strain is the relative change in length.

Stress and strainFLΔLstress = F ÷ Astrain = ΔL ÷ Lstress = E × strainunit : N/m² (Pa)

In the figure : F stretches; ΔL is the extension.

Rule 1: Stress — Stress = F ÷ A (N/m² = Pa); dimension [M L⁻¹ T⁻²]. Types : tensile/compressive, volume (pressure), shear.

  • Easy : F = 500 N, A = 2 × 10⁻⁴ m² ⇒ 2.5 × 10⁶ Pa.
  • Medium : A halved ⇒ stress doubles.
  • Exam (CET) : Unit of stress? (1) Pa (2) N (3) J (4) kg → answer 1)

Rule 2: Strain — Strain = ΔL ÷ L (longitudinal); no unit or dimension. Types : longitudinal, volume (ΔV/V), shear (θ).

  • Easy : L = 2 m, ΔL = 1 mm ⇒ 5 × 10⁻⁴.
  • Medium : Strain has no unit.
  • Exam (CET) : Dimension of strain? (1) none (2) M (3) L (4) T → answer 1)

Rule 3: Hooke's law — Within the elastic limit stress ∝ strain; the constant is the modulus of elasticity E (Pa).

  • Easy : E = stress ÷ strain = (2.5 × 10⁶) ÷ (5 × 10⁻⁴) = 5 × 10⁹ Pa.
  • Medium : Strain doubled ⇒ stress doubled.
  • Exam (CET) : Hooke's law holds? (1) within the elastic limit (2) everywhere (3) at fracture (4) never → answer 1) Trick : 'straight inside the limit'.

Derivation : Unit and dimension of E

  • Step 1 : E = stress ÷ strain.
  • Step 2 : Strain is dimensionless ⇒ E has the dimension of stress.
  • Step 3 : [F/A] = [M L T⁻²]/[L²] = M L⁻¹ T⁻².
  • Step 4 : E : N/m² (Pa), [M L⁻¹ T⁻²].
Remember

Stress = F/A; strain = ΔL/L; stress = E × strain.

Common mistake / CET trap

Strain is unitless; Hooke's law holds only within the elastic limit.

Shortcut

Memory clue : "Stress = force over area".

Spot the difference

QuantityFormulaUnitDimension
StressF/APaM L⁻¹ T⁻²
StrainΔL/L——
Estress/strainPaM L⁻¹ T⁻²

Practice questions

Easy
  1. 1Formula for stress?
  2. 2What is strain?
  3. 3State Hooke's law.
Medium
  1. 4F = 1000 N, A = 5 × 10⁻⁴ : stress?
  2. 5L = 4 m, ΔL = 2 mm : strain?
  3. 6Stress 4 × 10⁶, strain 2 × 10⁻⁴ : E?
Challenging
  1. 7Prove the dimension of E.
  2. 8Difference between pressure and stress?
Show answers (try them yourself first)
  1. 1F/A
  2. 2ΔL/L
  3. 3stress ∝ strain
  4. 42 × 10⁶ Pa
  5. 55 × 10⁻⁴
  6. 62 × 10¹⁰ Pa
  7. 7M L⁻¹ T⁻²
  8. 8Both F/A; pressure is external normal, stress is internal

Young's modulus

From daily life

A crane's cable is thick steel — it stretches little.

Simple definition

Young's modulus Y = longitudinal stress ÷ longitudinal strain.

Young's modulus : wireML, AF = MgΔLY = FL ÷ AΔLΔL ∝ 1/Ysteel 2 × 10¹¹ Pa

In the figure : Load M stretches the wire by ΔL.

Rule 1: Formula — Y = (F ÷ A) ÷ (ΔL ÷ L) = F L ÷ (A ΔL); unit Pa; wire : A = π d² ÷ 4. ΔL = F L ÷ (A Y).

  • Easy : F = 200 N, L = 2 m, A = 10⁻⁶ m², Y = 2 × 10¹¹ ⇒ ΔL = 2 mm.
  • Medium : M = 10 kg (g = 10), A = 10⁻⁶, L = 1 m, Y = 2 × 10¹¹ ⇒ ΔL = 0.5 mm.
  • Exam (CET) : L doubled ⇒ ΔL? (1) double (2) half (3) four times (4) same → answer 1) Trick : ΔL ∝ L.

Rule 2: Comparison : ΔL — ΔL ∝ F L ÷ (A Y) ∝ F L ÷ (r² Y). Radius doubled ⇒ ΔL becomes ¼. Larger Y ⇒ smaller ΔL.

  • Easy : r doubled ⇒ ΔL ¼.
  • Medium : Same F, L, A : steel and copper (Y = 1.2 × 10¹¹) : ΔL_copper/ΔL_steel = 5/3.
  • Exam (CET) : r doubled ⇒ ΔL? (1) ¼ (2) ½ (3) 2 (4) 4 → answer 1)

Rule 3: Safe load — Breaking stress σ_b; safe load : F = σ_b A ÷ factor of safety.

  • Easy : σ_b = 5 × 10⁸, A = 2 × 10⁻⁶ ⇒ F_max = 1000 N.
  • Medium : Safety factor 5 ⇒ 200 N.
  • Exam (CET) : Safe load? (1) σ_b A / factor (2) σ_b (3) A (4) E → answer 1)

Derivation : ΔL = F L ÷ (A Y)

  • Step 1 : Stress = F/A.
  • Step 2 : Strain = ΔL/L.
  • Step 3 : Y = (F/A) ÷ (ΔL/L).
  • Step 4 : Rearranging : ΔL = F L ÷ (A Y).
Remember

Y = FL/(AΔL); ΔL = FL/(AY).

Common mistake / CET trap

ΔL ∝ 1/r²; Y depends on the material, not the shape; use F = Mg.

Shortcut

Memory clue : "F L over A delta L".

Spot the difference

MaterialY (Pa)
Steel2 × 10¹¹
Copper1.2 × 10¹¹
Aluminium7 × 10¹⁰
Rubber≈ 10⁷

Practice questions

Easy
  1. 1Formula for Y?
  2. 2Unit of Y?
  3. 3Highest Y?
Medium
  1. 4F = 400, L = 1.5 m, A = 2 × 10⁻⁶, Y = 2 × 10¹¹ : ΔL?
  2. 5ΔL = 1 mm; L tripled : ΔL?
  3. 6d doubled : ΔL?
Challenging
  1. 7σ_b = 6 × 10⁸, d = 2 mm : F_max (π = 3.14)?
  2. 8Two wires, same F, L, A : Y₁ = 2Y₂ ⇒ ΔL₁/ΔL₂?
Show answers (try them yourself first)
  1. 1FL/AΔL
  2. 2Pa
  3. 3Steel (of those given)
  4. 41.5 mm
  5. 53 mm
  6. 6ΔL becomes ¼ of the original
  7. 71884 N
  8. 8½

Bulk modulus, shear modulus and Poisson's ratio

From daily life

Deep water is squeezed a little; push a book sideways and it shears — three kinds of strain.

Simple definition

K measures volume change, η shape change, σ lateral contraction.

Three moduliKΔV/Vηθσlat/longK = −ΔP V/ΔV ; η = (F/A)/θ ; σ = −(Δd/d)/(ΔL/L)Y = 3K (1 − 2σ) = 2η (1 + σ)

In the figure : K, η, σ and the relations between them.

Rule 1: Bulk modulus — K = −ΔP ÷ (ΔV/V) = −ΔP V ÷ ΔV; compressibility = 1/K. Small for gases, large for solids.

  • Easy : Water K = 2.2 × 10⁹, ΔP = 10⁷ ⇒ ΔV/V = 4.5 × 10⁻³.
  • Medium : ΔP doubled ⇒ ΔV/V doubled.
  • Exam (CET) : Compressibility? (1) 1/K (2) K (3) K² (4) 0 → answer 1)

Rule 2: Shear (rigidity) modulus — η = (F/A) ÷ θ (θ in radians); η = 0 for liquids and gases.

  • Easy : F = 10⁴ N, A = 0.01 m², η = 8 × 10¹⁰ ⇒ θ = 1.25 × 10⁻⁵ rad.
  • Medium : Larger η ⇒ smaller θ.
  • Exam (CET) : η in a liquid? (1) 0 (2) K (3) Y (4) infinite → answer 1)

Rule 3: Poisson's ratio — σ = lateral strain ÷ longitudinal strain (ignoring the sign); dimensionless; between −1 and 0.5; typically 0.2–0.4.

  • Easy : σ = 0.25, longitudinal 2 × 10⁻³ ⇒ lateral 5 × 10⁻⁴.
  • Medium : K = 10¹¹, σ = 0.25 ⇒ Y = 3K(1 − 2σ) = 1.5 × 10¹¹.
  • Exam (CET) : Unit of σ? (1) none (2) Pa (3) N (4) m → answer 1)

Derivation : Bulk modulus : K

  • Step 1 : Volume stress = ΔP (increase in pressure).
  • Step 2 : Volume strain = −ΔV/V.
  • Step 3 : K = stress ÷ strain.
  • Step 4 : K = −ΔP V ÷ ΔV; the minus sign means V decreases with pressure.
Remember

K = −ΔP V/ΔV; η = (F/A)/θ; Y = 3K(1 − 2σ) = 2η(1 + σ).

Common mistake / CET trap

θ in radians; the sign in K; σ has no unit.

Shortcut

Memory clue : "K = volume, eta = shape".

Spot the difference

ModulusStressStrain
YF/A (normal)ΔL/L
KΔP−ΔV/V
ηF/A (tangential)θ

Practice questions

Easy
  1. 1Formula for K?
  2. 2Formula for η?
  3. 3What is σ?
Medium
  1. 4ΔP = 2 × 10⁷, K = 2 × 10⁹ : ΔV/V?
  2. 5F = 2 × 10³, A = 0.02, η = 10¹⁰ : θ?
  3. 6σ = 0.3, longitudinal 10⁻³ : lateral?
Challenging
  1. 7η = 4 × 10¹⁰, σ = 0.25 : Y?
  2. 8Why η = 0 for a liquid?
Show answers (try them yourself first)
  1. 1−ΔP V/ΔV
  2. 2(F/A)/θ
  3. 3lateral ÷ longitudinal strain
  4. 410⁻²
  5. 510⁻⁵ rad
  6. 63 × 10⁻⁴
  7. 7Y = 2η(1 + σ) = 10¹¹
  8. 8A liquid does not resist tangential stress

Stress–strain graph and applications

From daily life

A wire pulled harder first returns, then stays bent, then breaks — the graph shows exactly this.

Simple definition

The stress–strain curve shows how a material behaves.

Stress–strain graphPEYUBstressstrainHooke

In the figure : P, E, Y, U, B = important points.

Live figure

Stress–strain graph and applications

In the app you can move this figure with your finger.

Open in the app

Rule 1: Parts of the graph — OP straight : Hooke (P = proportional limit); E elastic limit; Y yield point; U ultimate stress; B fracture.

  • Easy : Slope of OP = Y.
  • Medium : After U the stress appears to fall (necking).
  • Exam (CET) : The straight part? (1) Hooke (2) yield (3) fracture (4) fatigue → answer 1)

Rule 2: Ductile and brittle — Ductile : long plastic region (copper, gold). Brittle : short plastic region (glass, cast iron) — breaks suddenly.

  • Easy : Wires are drawn ⇒ ductile.
  • Medium : Glass breaks instead of bending.
  • Exam (CET) : Brittle? (1) glass (2) copper (3) gold (4) lead → answer 1)

Rule 3: Applications — Factor of safety = breaking stress ÷ working stress for bridges and cranes. Rubber for shock absorption, steel for stiffness.

  • Easy : σ_b = 4 × 10⁸, working 10⁸ ⇒ factor 4.
  • Medium : Rubber loop area = energy dissipated.
  • Exam (CET) : Factor of safety? (1) σ_b ÷ σ_working (2) σ_working ÷ σ_b (3) σ_b σ (4) 0 → answer 1)

Derivation : Y from the graph

  • Step 1 : In OP, stress = Y × strain.
  • Step 2 : Y = stress ÷ strain = slope.
  • Step 3 : One point : stress 4 × 10⁸, strain 2 × 10⁻³.
  • Step 4 : Y = 2 × 10¹¹ Pa.
Remember

Slope = Y; ductile = long plastic part; factor = σ_b ÷ σ_working.

Common mistake / CET trap

P and E are not the same; brittle ≠ weak.

Shortcut

Memory clue : "P-E-Y-U-B".

Spot the difference

PointMeaning
Pproportional limit
Eelastic limit
Yyield point
Uultimate stress
Bfracture

Practice questions

Easy
  1. 1What is P?
  2. 2Name a ductile material.
  3. 3Name a brittle material.
Medium
  1. 4What does the slope of OP give?
  2. 5σ_b = 6 × 10⁸, working 2 × 10⁸ : factor?
  3. 6A = 4 × 10⁻⁶, σ_b = 5 × 10⁸ : F_max?
Challenging
  1. 7What is necking?
  2. 8Why does the graph fall after U?
Show answers (try them yourself first)
  1. 1Proportional limit
  2. 2Copper
  3. 3Glass
  4. 4Y (Young's modulus)
  5. 53
  6. 62000 N
  7. 7Part of the wire thins out after U
  8. 8The area drops, while stress uses the original area

Strain energy

From daily life

A drawn bow stores energy and releases it when the arrow is let go.

Simple definition

Strain energy is the potential energy stored in a deformed body.

Strain energyΔLF(ΔL, F)U = ½ F ΔLu = ½ × stress × strain= ½ Y × (strain)²

In the figure : The blue area = stored energy.

Rule 1: Stored energy — U = ½ F ΔL = ½ Y A ΔL² ÷ L (area under the F–ΔL graph).

  • Easy : F = 200 N, ΔL = 2 mm ⇒ U = 0.2 J.
  • Medium : ΔL doubled ⇒ U four times.
  • Exam (CET) : ΔL doubled ⇒ U? (1) four times (2) double (3) half (4) same → answer 1) Trick : U ∝ ΔL².

Rule 2: Energy density — u = U ÷ (A L) = ½ × stress × strain = ½ Y × strain² (J/m³).

  • Easy : Stress 2 × 10⁸, strain 10⁻³ ⇒ u = 10⁵ J/m³.
  • Medium : Check : 0.2 J ÷ 2 × 10⁻⁶ m³ = 10⁵ J/m³.
  • Exam (CET) : Unit of u? (1) J/m³ (2) J (3) N (4) Pa s → answer 1)

Rule 3: Stretched wire/spring — Wire/spring : F = k ΔL, k = YA/L; U = ½ k ΔL². Work done = U (when stretched slowly).

  • Easy : k = 10⁵ N/m, ΔL = 2 cm ⇒ U = 20 J.
  • Medium : k = YA/L = 2 × 10¹¹ × 10⁻⁶ ÷ 2 = 10⁵ N/m.
  • Exam (CET) : U = ? (1) ½ k x² (2) k x² (3) k x (4) 2k x → answer 1)

Derivation : U = ½ F ΔL

  • Step 1 : While stretching, F grows from 0 to F (F = kx).
  • Step 2 : dW = F dx = k x dx.
  • Step 3 : W = ∫₀^ΔL k x dx = ½ k ΔL².
  • Step 4 : k ΔL = F ⇒ U = ½ F ΔL.
Remember

U = ½ F ΔL; u = ½ Y × strain².

Common mistake / CET trap

U ≠ F ΔL (do not forget ½); U ∝ ΔL².

Shortcut

Memory clue : "Half force times extension".

Spot the difference

QuantityFormulaUnit
U½ F ΔLJ
u½ × stress × strainJ/m³
kYA/LN/m

Practice questions

Easy
  1. 1Formula for U?
  2. 2Unit of u?
  3. 3U is the area under which graph?
Medium
  1. 4F = 100 N, ΔL = 4 mm : U?
  2. 5Stress 10⁸, strain 10⁻³ : u?
  3. 6k = 2 × 10⁴, ΔL = 1 cm : U?
Challenging
  1. 7Prove U = ½ F ΔL.
  2. 8Steel or rubber stores more energy at equal stress? Why?
Show answers (try them yourself first)
  1. 1½ F ΔL
  2. 2J/m³
  3. 3F–ΔL
  4. 40.2 J
  5. 55 × 10⁴ J/m³
  6. 61 J
  7. 7W = ∫k x dx = ½ k ΔL²
  8. 8Rubber; u = stress²/2Y, small Y ⇒ larger u

Friction and hardness

From daily life

Walking on ice is hard — low friction; on sand it is easy — high friction.

Simple definition

Friction is the force between surfaces that opposes motion.

Friction : f vs Ff_maxf = Ff_k = μ_k Nf_s ≤ μ_s NfF

In the figure : f = F up to f_max; then f_k.

Rule 1: Friction force — f_s ≤ μ_s N (static), f_k = μ_k N (kinetic); μ_k < μ_s; independent of contact area.

  • Easy : m = 10 kg, μ_s = 0.5 ⇒ f_max = 50 N (g = 10).
  • Medium : Pushing with 40 N ⇒ f = 40 N (no motion).
  • Exam (CET) : f depends on? (1) N (2) area (3) speed (4) shape → answer 1)

Rule 2: Incline and angle of repose — tan θ = μ_s (angle of repose). Sliding down an incline : a = g (sin θ − μ_k cos θ).

  • Easy : μ_s = 0.5 ⇒ θ = tan⁻¹ 0.5 ≈ 26.6°.
  • Medium : θ = 37° (sin 0.6, cos 0.8), μ_k = 0.5 ⇒ a = 10(0.6 − 0.4) = 2 m/s².
  • Exam (CET) : Angle of repose? (1) tan⁻¹ μ_s (2) sin⁻¹ μ_s (3) μ_s (4) 45° → answer 1)

Rule 3: Rolling friction and hardness — Rolling friction is far less than sliding (wheels, bearings). Hardness = resistance to scratching; Mohs scale : talc 1 … diamond 10.

  • Easy : Wheels make pulling easier.
  • Medium : Diamond scratches all the others.
  • Exam (CET) : Hardest? (1) diamond (2) talc (3) steel (4) glass → answer 1)

Derivation : Angle of repose : tan θ = μ_s

  • Step 1 : On the incline : N = m g cos θ.
  • Step 2 : Component along the incline : m g sin θ.
  • Step 3 : At the point of slipping m g sin θ = μ_s N = μ_s m g cos θ.
  • Step 4 : tan θ = μ_s.
Remember

f_s ≤ μ_s N; f_k = μ_k N; tan θ = μ_s.

Common mistake / CET trap

f is independent of area; N is not always mg (it is mg cos θ on an incline).

Shortcut

Memory clue : "Static > kinetic > rolling".

Spot the difference

TypeFormulaComparison
Staticf_s ≤ μ_s Nlarger
Kineticf_k = μ_k Nsmaller
Rollingf_r = μ_r Nsmallest

Practice questions

Easy
  1. 1Formula for f_s?
  2. 2What is the angle of repose?
  3. 3How is hardness measured?
Medium
  1. 4m = 20, μ_s = 0.4 : f_max (g = 10)?
  2. 5μ_s = 1 : angle of repose?
  3. 6θ = 30° (sin 0.5, cos 0.87), μ_k = 0.2 : a?
Challenging
  1. 7Prove tan θ = μ_s.
  2. 8Why is μ_k < μ_s?
Show answers (try them yourself first)
  1. 1f_s ≤ μ_s N
  2. 2Maximum slope angle before slipping
  3. 3With the Mohs scale
  4. 480 N
  5. 545°
  6. 63.26 m/s²
  7. 7m g sin θ = μ_s m g cos θ
  8. 8Surface bonds have less time to form once moving

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
StressF/APa; M L⁻¹ T⁻²
StrainΔL/L—
YFL/(AΔL)Pa
K−ΔP V/ΔVPa
η(F/A)/θPa
RelationY = 3K(1 − 2σ) = 2η(1 + σ)—
Strain energyu = ½ × stress × strainJ/m³
Frictionf_k = μ_k N; tan θ = μ_sN

🎯 Most important points for the exam

  • Hooke within the elastic limit : stress ∝ strain.
  • Strain is unitless; E has the dimension of stress.
  • ΔL = FL/(AY); ΔL ∝ 1/r².
  • K, η, Y, σ : Y = 3K(1 − 2σ) = 2η(1 + σ).
  • Graph : P → E → Y → U → B; slope = Y.
  • U = ½ F ΔL; u = ½ Y × strain².
  • f_s > f_k > f_rolling; tan θ = μ_s.
Study mantra

Stress ÷ strain = modulus; ½ F ΔL = energy.