NCERT Plus One (Class 11) Physics • Chapter 8

Mechanical Properties of Solids

Q: Steel is more elastic than rubber. Justify this statement mathematically.

Elasticity is measured by Young's modulus Y = Stress / Strain = (F/A)/(ΔL/L) . For identical dimensions and load F , strain produced in steel ( ΔL_s/L ) is far smaller than strain in rubber ( ΔL_r/L ). Since Y ∝ 1/Strain , Y_steel > Y_rubber — hence steel is more elastic.

Q: Draw a typical stress–strain curve for a ductile metal and mark (a) proportional limit, (b) yield point, (c) ultimate tensile strength, and (d) fracture point. What are elastomers?

Plot Stress (y-axis) vs Strain (x-axis): • O–A: Linear region obeying Hooke's law ( A = Proportional limit). • B: Yield point / elastic limit (yield strength S_y ). • D: Ultimate tensile strength ( S_u ). • E: Fracture point (large D–E plastic zone = ductile). Elastomers: Materials (e.g., rubber, aorta tissue) that can stretch to large strains elastically without obeying Hooke's law (non-linear curve).

Q: State Hooke's law. Name the three elastic moduli and mention which one applies to gases.

Hooke's Law: Within the elastic limit (proportional limit), stress is directly proportional to strain ( Stress ∝ Strain ). Three moduli: Young's modulus ( Y ), Shear modulus ( G ), and Bulk modulus ( B ). Only Bulk modulus (B) applies to gases (and liquids).

Q: A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a given load. Find the ratio Y_steel / Y_copper.

Since F and ΔL are equal for both wires: Y_s = (F L_s)/(A_s ΔL) and Y_c = (F L_c)/(A_c ΔL) . Y_s / Y_c = (L_s / L_c) × (A_c / A_s) = (4.7 / 3.5) × (4.0 / 3.0) = 1.79 : 1 (approx 1.8 ).

Q: Derive an expression for the elastic potential energy stored per unit volume (energy density) in a stretched wire.

For extension dl under force F = (YA/L) l : U = ∫₀^ΔL (YA/L) l dl = ½ (YA/L)(ΔL)² = ½ F ΔL . Dividing by wire volume V = A L : u = U / (AL) = ½ (F/A)(ΔL/L) = ½ × Stress × Strain = ½ Y ε² .

Q: Define Poisson's ratio. What are its SI unit and dimensions?

Poisson's ratio (σ): Ratio of lateral strain ( Δd/d ) to longitudinal strain ( ΔL/L ) within the elastic limit. It is a pure number with NO SI unit and NO dimensions ( [M⁰L⁰T⁰] ).

Q: Why are girders used for bridges and buildings given an I-shaped cross-section rather than a solid square shape?

1. Sag of a loaded beam is δ = Wl³/(4bd³Y) , so maximizing depth d sharply reduces bending. 2. During bending, maximum compressive/tensile stresses concentrate near the top and bottom surfaces while the middle is nearly neutral. 3. An I-section keeps wide top/bottom flanges for strength and anti-buckling while removing unused middle steel, reducing weight and cost.

Q: Explain why the maximum height of a mountain on Earth is limited to roughly 10 km.

At the base of a mountain of height h , the weight of rock produces pressure h ρ g and uneven lateral shearing stress. For rocks, elastic shear limit is ~3 × 10⁸ N/m² and density ρ ≈ 3 × 10³ kg/m³ . Thus h_max ≈ (3 × 10⁸) / (3 × 10³ × 10) = 10 km — taller peaks cause base rock to flow plastically.

Stress, strain, Hooke's law, stress–strain curves, Young/Shear/Bulk moduli, Poisson's ratio, elastic energy & engineering design.

📄 Official NCERT Textbook PDF

Topic 8.2: Stress & Strain (Tensile, Compressive & Shearing)

Pull a rubber catapult from a Thrissur village fair, squeeze a kitchen sponge, and then push only the top cover of a thick Physics textbook sideways on your desk! In all three cases your hands apply a deforming force, and the solid's tiny interatomic springs push right back with an internal restoring force.

STRESS (σ)Stress = F_restoringA | SI unit: N/m² or Pa | Dim: [ML⁻¹T⁻²]

STRAIN (ε)Strain = Δdimensionoriginal dimension | Unitless & Dimensionless!

3 STRESS–STRAIN PAIRSLongitudinal (ΔL/L), Shearing (Δx/L = tan θ), Hydraulic/Volume (ΔV/V)

🧠 Memory Trick: Stress = internal push-back per unit area (like pressure); Strain = fractional change (percentage stretch, NO units!).

💡 Simple Kid-Check: Imagine atoms inside a steel rod as tiny balls connected by invisible coil springs. Pull the rod, and every spring stretches slightly and pulls back! Stress measures how hard each square metre pulls back, while Strain measures what fraction the rod actually changed!

1
Longitudinal (Tensile or Compressive): Force acts perpendicular (normal) to cross-section A. Stretching gives tensile stress; squashing a pillar gives compressive stress. Strain = ΔL / L.
2
Shearing (Tangential) Stress: Equal and opposite forces act parallel to opposite faces, tilting the block sideways by angle θ. Shearing strain = Δx / L = tan θ ≈ θ (in radians).
3
Hydraulic (Volume) Stress: Equal fluid pressure p squeezes from all sides, changing volume without changing shape! Volume strain = ΔV / V.

Tensile/Compressive Stress = FA | Shear Strain = ΔxL = tan θ | Volume Strain = ΔVV

🧒 Super-Easy Example: A steel wire of cross-section 2 × 10⁻⁶ m² and length 2 m holds a 40 N weight and stretches by 0.2 mm = 2 × 10⁻⁴ m.
• Tensile Stress = 40 / (2 × 10⁻⁶) = 2 × 10⁷ N/m².
• Longitudinal Strain = (2 × 10⁻⁴) / 2 = 10⁻⁴ (no unit!).

Two classic NCERT traps! First, Stress is NOT a vector (it cannot be given a single fixed direction — force direction depends on which side of the cut surface you look at). Second, when a wire hangs from the ceiling carrying weight F at its lower end, the tension and stress at every cross-section is F/A, NOT 2F/A!

Topic 8.3 – 8.4: Hooke's Law & Stress–Strain Curve (Yield & Fracture)

Bend a steel paperclip slightly and it springs right back; twist it too far and it stays permanently crooked; wiggle it harder and *snap*! Every metal wire follows a dramatic life-story graph from elastic springiness to permanent bend, necking, and final fracture.

HOOKE'S LAWFor small deformations: Stress ∝ Strain ⇒ Stress = k × Strain (k = modulus of elasticity)

CURVE LANDMARKSO–A: Linear Hooke • B: Yield Point (S_y) • D: Ultimate Strength (S_u) • E: Fracture

DUCTILE vs BRITTLE vs ELASTOMERDuctile: D–E far (Copper) • Brittle: D–E close (Glass) • Elastomer: Non-linear (Rubber)

🧠 Memory Trick: Remember curve points A–B–D–E as "Always Bend, Don't Explode!" → Proportional (A), Yield/Bend (B), Peak Strength (D), Fracture (E)!

💡 Simple Kid-Check: Up to point B (Elastic Limit), a metal behaves like an obedient spring — let go and it snaps back to zero stretch. Cross point B, and the metal enters the Plastic Zone — even after removing the weight, it keeps a permanent stretch!

1
O to A (Proportional Limit): Straight line! Hooke's law holds strictly: doubling stress doubles strain. Slope = Young's modulus.
2
A to B (Yield Point): Still elastic up to Yield Point B (corresponding stress = Yield Strength S_y), though no longer perfectly straight.
3
B to D (Plastic Flow & Ultimate Strength): Permanent deformation occurs! Strain grows rapidly even for small extra load until stress peaks at D (Ultimate Tensile Strength S_u).
4
D to E (Necking & Fracture): The wire thins locally ("necking") and snaps at E. Large D–E gap = Ductile; tiny D–E gap = Brittle.

Hooke's Law: Stress = Modulus × Strain | Elastomers (rubber, aorta tissue) stretch huge amounts non-linearly and do NOT obey Hooke's law!

Watch out: Hooke's law is an empirical law valid ONLY inside the small linear region (O to A) — NOT up to fracture! Also, elastomers like vulcanised rubber can return to original length after 500% strain, yet they do NOT obey Hooke's law because their stress–strain curve is curved, not straight!

Topic 8.5.1: Young's Modulus & Wire Extension

Imagine two swings hanging from a mango tree in Palakkad — one hung with a rubber band, the other with a thin steel cable. Ask a non-science friend which is "more elastic" and they point at the rubber! Ask a Physicist and the answer flips upside down: Steel is thousands of times more elastic than rubber!

YOUNG'S MODULUS (Y)Y = Tensile StressLongitudinal Strain = F LA ΔL

WIRE EXTENSION FORMULAΔL = M g Lπ r² Y | Unit of Y: N/m² or Pa

STEEL vs RUBBERY_steel ≈ 200 × 10⁹ Pa | Y_copper ≈ 110 × 10⁹ Pa | Y_rubber ≈ 10⁶ Pa

🧠 Memory Trick: Elasticity in Physics = Stubbornness to Stretch! Smaller stretch ΔL for the same load = BIGGER Young's Modulus Y = MORE Elastic!

💡 Simple Kid-Check: In daily Malayalam or English, "elastic" means "stretchy." In Physics, elasticity means how strongly a material resists deformation and snaps back! Since steel barely budges (ΔL is tiny) under a heavy truck, Y = Stress / Strain becomes gigantic for steel!

1
Definition: Within the elastic limit, the ratio of tensile (or compressive) stress to longitudinal strain is Young's modulus Y.
2
Valid ONLY for Solids: Liquids and gases cannot support a tensile pull, so Young's modulus Y has meaning only for solids.
3
Material Property: Changing wire length L or radius r changes extension ΔL, yet Y stays constant for a given material at a given temperature!

Y = F / AΔL / L = F Lπ r² ΔL ⇒ For same load & same wire size: Y_steelY_rubber = ΔL_rubberΔL_steel > 1

🧒 NCERT-Style Example: A steel wire of length 4 m and cross-section 2 × 10⁻⁵ m² (Y = 2 × 10¹¹ N/m²) supports a 200 kg load (g = 10 m/s²). Find elongation ΔL:
F = 200 × 10 = 2000 N.
ΔL = F LA Y = 2000 × 4(2 × 10⁻⁵)(2 × 10¹¹) = 80004 × 10⁶ = 2 × 10⁻³ m = 2 mm!

Two giant 1-mark & 2-mark exam traps: (1) If you double a wire's radius (r → 2r), area πr² grows 4×, so elongation ΔL shrinks to ¼ — while Young's modulus Y does NOT change at all! (2) Always state why steel is more elastic than rubber: for identical stress, steel suffers far smaller strain!

Topic 8.5.2 – 8.5.3: Shear Modulus & Bulk Modulus (Compressibility)

Try twisting a heavy steel axle with a wrench (**shear**), or diving 3000 m deep into the Indian Ocean where water squeezes a submarine from every direction (**bulk compression**)! Solids fight both twisting and squeezing, while fluids give up completely on twisting and only resist being squashed.

SHEAR MODULUS (G)G = F / AΔx / L = FA θ | Only for Solids! (G ≈ Y/3 for most metals)

BULK MODULUS (B)B = −pΔV / V | Applies to Solids, Liquids & Gases!

COMPRESSIBILITY (k)k = 1B (unit: Pa⁻¹) | Order: B_solid > B_liquid ≫ B_gas

🧠 Memory Trick: Y & G = Solids only (they have length & shape!). Bulk = Both solids & fluids (every matter has volume!).

💡 Simple Kid-Check: Try pressing an air-filled syringe vs a water-filled syringe with your thumb sealing the tip. Air squishes easily (tiny B, huge compressibility k!), water barely budges (huge B ≈ 2.2 GPa!), and steel refuses to budge (B ≈ 160 GPa)!

Modulus	Formula	Deformation	States of Matter
Young's (`Y`)	`(F/A) / (ΔL/L)`	Length change	Solids only
Shear (`G`)	`(F/A) / θ`	Shape tilt (constant V)	Solids only (ideal fluids `G = 0`)
Bulk (`B`)	`−p / (ΔV/V)`	Volume change (constant shape)	Solids, Liquids & Gases

🧒 NCERT Indian Ocean Example (8.5): Average ocean depth is 3000 m (water ρ = 1000 kg/m³, g = 10 m/s², B_water = 2.2 × 10⁹ Pa).
• Pressure at bottom: p = h ρ g = 3000 × 1000 × 10 = 3 × 10⁷ Pa.
• Fractional compression: |ΔV/V| = p / B = (3 × 10⁷) / (2.2 × 10⁹) = 1.36 × 10⁻² = 1.36%!

Why is there a negative sign in B = −p / (ΔV/V)? Because increasing pressure (+p) shrinks volume (ΔV < 0), so the minus sign ensures Bulk Modulus B and Compressibility k = 1/B are always positive numbers! Remember gases are about a million times more compressible than solids!

Topic 8.5.4 – 8.5.5: Poisson's Ratio & Elastic Potential Energy

Pull a thick rubber band between your fingers and look sideways — as it gets longer, it visibly gets thinner! And as you pull back a slingshot before releasing a pebble, your muscles don't lose that energy: it hides inside the stretched rubber as elastic potential energy.

POISSON'S RATIO (σ)σ = Lateral strainLongitudinal strain = Δd / dΔL / L (No unit, no dimension!)

ELASTIC ENERGY (U)U = 12 F ΔL = 12 Y AL (ΔL)²

ENERGY DENSITY (u)u = UVolume = 12 × Stress × Strain = 12 Y ε² = σ_s²2Y

🧠 Memory Trick: ½ appears in U = ½ F ΔL because force ramps from 0 to F (triangle area under the F–ΔL graph, NOT a rectangle!).

💡 Simple Kid-Check: Stretch a noodle and its waist gets thinner — that ratio of waist-thinning strain to length-stretching strain is Poisson's ratio (σ)! For steels σ ≈ 0.28–0.30; for aluminium ~0.33.

1
Poisson's Ratio σ: Ratio of lateral contraction strain (Δd/d) to longitudinal extension strain (ΔL/L). It is a pure dimensionless number (for real solids 0 < σ < 0.5; theoretical isotropic range −1 to 0.5).
2
Work Done Stretching a Wire: Force starts at 0 and rises linearly to F, so average force is ½ F! Thus stored potential energy U = ½ F ΔL.
3
Energy Per Unit Volume (Energy Density u): Dividing U = ½ F ΔL by wire volume A L gives u = ½ (F/A)(ΔL/L) = ½ × Stress × Strain!

u = 12 × Stress × Strain = 12 Y (Strain)² = (Stress)²2Y

🧒 Super-Easy Example: A wire stretches by 2 mm = 2 × 10⁻³ m when pulled up to final force 100 N.
Elastic potential energy stored = ½ × F × ΔL = ½ × 100 × (2 × 10⁻³) = 0.1 J! (Notice: NOT 100 × 0.002 = 0.2 J!)

Classic 2-mark exam trap: What happens to the other half of the work F ΔL when you gently hang a weight F = Mg on a wire? Weight drops by ΔL, losing gravitational PE Mg ΔL = F ΔL. Only HALF (½ F ΔL) stores as elastic PE in the wire — the other half dissipates as heat/oscillation damping!

Topic 8.6: Engineering Applications (Cranes, I-Beams & Bridges)

Walk beneath a Kochi Metro flyover girder or watch a massive crane at Vallarpadam container terminal! Why are bridge girders shaped like the capital letter I instead of solid square blocks, and why can no mountain on Earth tower higher than roughly 10 km?

CRANE ROPE DESIGNRequire F/A ≤ S_y (with safety factor ~10) + braided thin wires for flexibility!

BEAM SAG (δ)δ = W l³4 b d³ Y ⇒ Increase depth d (cubed!), use I-shape!

MOUNTAIN LIMIT (~10 km)Rock shear stress at base h ρ g ≤ 3 × 10⁸ Pa ⇒ h_max ≈ 10 km (Everest 8.85 km!)

🧠 Memory Trick: Remember δ ∝ 1/(b d³)! Making a girder deeper (d) crushes sag by d³, while scooping out the lazy middle makes a light, stiff I-beam!

💡 Simple Kid-Check: Lay a plastic scale flat on two erasers and press the middle — it bends easily! Now stand the exact same scale edge-up (tall depth d) and press — it barely bends! That is why railway rails and metro bridges use tall I-beams!

1
Crane Rope Radius: For lifting 10 tonnes (10⁵ N) with yield stress σ_y = 300 × 10⁶ N/m² (and 10× safety margin), area A ≥ 10⁵ / (3 × 10⁷) gives radius r ≈ 1–3 cm. Braided thin strands keep the same area while bending cleanly over pulleys!
2
Bending of Beams & I-Shape: A loaded beam of length l, breadth b, depth d sags by δ = W l³ / (4 b d³ Y). Since top layers compress and bottom layers stretch (middle is nearly unstressed!), an I-section concentrates steel at top & bottom flanges while keeping a thin tall web — maximum stiffness, minimal steel & zero buckling!
3
Columns & 10-km Mountains: Pillars use flared/distributed ends to carry heavy loads without buckling. At a mountain's base, uneven rock load produces shear stress ~ h ρ g; rocks start flowing past ~3 × 10⁸ Pa, capping mountain height around 10 km!

Beam Sag: δ = W l³4 b d³ Y | Max Mountain Height: h ≈ σ_shearρ g ≈ 3 × 10⁸3 × 10³ × 10 = 10 km

Watch the d³ vs b exam question! To reduce bridge sag without adding heavy steel, engineers increase depth d (not breadth b), because δ ∝ 1/d³! And why don't bridges use tall thin flat rectangles (huge d, tiny b)? Because tall thin blades twist sideways ("buckling") — the I-shape prevents buckling!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Steel is more elastic than rubber. Justify this statement mathematically.

Elasticity is measured by Young's modulus Y = Stress / Strain = (F/A)/(ΔL/L).
For identical dimensions and load F, strain produced in steel (ΔL_s/L) is far smaller than strain in rubber (ΔL_r/L).
Since Y ∝ 1/Strain, Y_steel > Y_rubber — hence steel is more elastic.

Q2 (Kerala HSE March • 3 Marks): Draw a typical stress–strain curve for a ductile metal and mark (a) proportional limit, (b) yield point, (c) ultimate tensile strength, and (d) fracture point. What are elastomers?

Plot Stress (y-axis) vs Strain (x-axis):
• O–A: Linear region obeying Hooke's law (A = Proportional limit).
• B: Yield point / elastic limit (yield strength S_y).
• D: Ultimate tensile strength (S_u).
• E: Fracture point (large D–E plastic zone = ductile).
Elastomers: Materials (e.g., rubber, aorta tissue) that can stretch to large strains elastically without obeying Hooke's law (non-linear curve).

Q3 (Kerala HSE Say • 2 Marks): State Hooke's law. Name the three elastic moduli and mention which one applies to gases.

Hooke's Law: Within the elastic limit (proportional limit), stress is directly proportional to strain (Stress ∝ Strain).
Three moduli: Young's modulus (Y), Shear modulus (G), and Bulk modulus (B).
Only Bulk modulus (B) applies to gases (and liquids).

Q4 (NCERT / DHSE Model • 3 Marks): A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10⁻⁵ m² under a given load. Find the ratio Y_steel / Y_copper.

Since F and ΔL are equal for both wires:
Y_s = (F L_s)/(A_s ΔL) and Y_c = (F L_c)/(A_c ΔL).
Y_s / Y_c = (L_s / L_c) × (A_c / A_s) = (4.7 / 3.5) × (4.0 / 3.0) = 1.79 : 1 (approx 1.8).

Q5 (Kerala HSE March • 2 Marks): Derive an expression for the elastic potential energy stored per unit volume (energy density) in a stretched wire.

For extension dl under force F = (YA/L) l:
U = ∫₀^ΔL (YA/L) l dl = ½ (YA/L)(ΔL)² = ½ F ΔL.
Dividing by wire volume V = A L:
u = U / (AL) = ½ (F/A)(ΔL/L) = ½ × Stress × Strain = ½ Y ε².

Q6 (Kerala HSE Model • 1 Marks): Define Poisson's ratio. What are its SI unit and dimensions?

Poisson's ratio (σ): Ratio of lateral strain (Δd/d) to longitudinal strain (ΔL/L) within the elastic limit.
It is a pure number with NO SI unit and NO dimensions ([M⁰L⁰T⁰]).

Q7 (Kerala HSE March • 2 Marks): Why are girders used for bridges and buildings given an I-shaped cross-section rather than a solid square shape?

1. Sag of a loaded beam is δ = Wl³/(4bd³Y), so maximizing depth d sharply reduces bending.
2. During bending, maximum compressive/tensile stresses concentrate near the top and bottom surfaces while the middle is nearly neutral.
3. An I-section keeps wide top/bottom flanges for strength and anti-buckling while removing unused middle steel, reducing weight and cost.

Q8 (Kerala HSE Say • 2 Marks): Explain why the maximum height of a mountain on Earth is limited to roughly 10 km.

At the base of a mountain of height h, the weight of rock produces pressure h ρ g and uneven lateral shearing stress.
For rocks, elastic shear limit is ~3 × 10⁸ N/m² and density ρ ≈ 3 × 10³ kg/m³.
Thus h_max ≈ (3 × 10⁸) / (3 × 10³ × 10) = 10 km — taller peaks cause base rock to flow plastically.