Class 11 Physics · Chapter 8
Short answer: Class 11 Physics Chapter 8, "Mechanical Properties of Solids," padhaata hai ki solid materials force lagne par kaise deform hote hain — aur wapas apni shape kaise pakadte hain. Stress, strain, Hooke's Law, aur teen elastic moduli (Young's, shear, bulk) is chapter ki backbone hain. Rationalised (2026-27 session) NCERT syllabus me yeh Class 11 Physics ka Chapter 8 hai — 14-chapter structure me, "Units and Measurements" se shuru hoke.
Jaise class 11 physics chapter 2 motion in a straight line ncert solutions me hum motion ko numbers me dhalna seekhte hain, waise hi Chapter 8 "Mechanical Properties of Solids" me hum deformation ko numbers me dhalna seekhte hain — kitna force, kitna badlaav, aur material kab tak elastic rehta hai, kab permanently deform ho jata hai.
Ye chapter conceptually chhota hai par numerically dense — Young's modulus, shear modulus, bulk modulus teeno alag-alag deformation types (stretching, twisting, uniform pressure) ko handle karte hain, aur exam me inko confuse karna sabse common mistake hai. Agar aap class 11 physics ncert solutions poori tarah revise kar rahe ho, to yaad rakho — is chapter ke concepts (stress-strain, elastic limit) system of particles and rotational motion important formulas aur gravitation class 11 numericals with solutions dono me indirectly background assumption ki tarah kaam aate hain.
Rationalised 2026-27 syllabus me Class 11 Physics ab 14 chapters ka hai (purana "Physical World" standalone chapter poori tarah hata diya gaya hai). Mechanical Properties of Solids is naye structure me Chapter 8 hai — is spec me hum poore NCERT text ke worked examples aur end-of-chapter exercises (8.1 se 8.16 tak) cover karenge, step-by-step marking-scheme style working ke saath.
Chapter 8 Summary — 5 Minute Revision
Chapter 8 me humne dekha ki koi bhi solid deforming force ke against apna shape/size wapas laane ki koshish karta hai — yehi elasticity hai. Elastic limit ke andar, stress strain ke directly proportional hota hai (Hooke's Law), aur ye proportionality constant hi material ka elastic modulus hai.
Teen tarah ke deformation, teen alag moduli: Young's modulus (Y) length ke stretch/compress ke liye, shear modulus (η) shape change (twisting) ke liye, aur bulk modulus (B) volume change (uniform pressure) ke liye. Stress-strain curve pe proportional limit, yield point, ultimate tensile strength aur fracture point — ye chaaron landmark har numerical aur conceptual question ka aadhaar hain.
Practical takeaway: is chapter ke numericals me sabse zyada marks unit-conversion (cm² → m², atm → Pa) aur sahi modulus choose karne se katte hain — is spec ke "mistakes" section me exactly yehi cover kiya gaya hai.
In-Text Questions — Solutions
Ek structural steel rod ki radius 10 mm aur length 1.0 m hai. Isko lambaai ki disha me 100 kN force se khincha jaata hai. Calculate (a) stress, (b) strain, aur (c) elongation on the rod. (Young's modulus of structural steel, Y = 2.0 × 10¹¹ N m⁻²)
Given: r = 10 mm = 1.0 × 10⁻² m, L = 1.0 m, F = 100 kN = 1.0 × 10⁵ N, Y = 2.0 × 10¹¹ Pa
A = πr² = π × (1.0×10⁻²)² = 3.14 × 10⁻⁴ m²
(a) Stress:
Stress = F/A = (1.0×10⁵)/(3.14×10⁻⁴) = 3.18 × 10⁸ N m⁻²
(b) Strain:
Strain = Stress/Y = (3.18×10⁸)/(2.0×10¹¹) = 1.59 × 10⁻³
(c) Elongation:
ΔL = Strain × L = 1.59×10⁻³ × 1.0 = 1.59 × 10⁻³ m ≈ 1.59 mm
Stress-strain curves ke do materials A aur B diye gaye hain, jahan A ka slope B se zyada steep hai lekin B zyada strain ke baad fracture hota hai. In dono ke Young's modulus aur ductility ke baare me kya kaha ja sakta hai?
Young's modulus curve ke slope (stress/strain) ke barabar hota hai (elastic region me). Steeper slope = zyada Young's modulus → Material A ka Y, B se zyada hoga — yaani A stiffer (kam strain me zyada stress) hai.
Jo material zyada strain tak fracture nahi hota (yahaan B), wo zyada ductile kehlata hai — wo fracture se pehle zyada plastic deformation sahta hai. A relatively kam ductile (brittle-leaning) hoga agar wo kam strain pe hi toot jaata hai.
Ek metal cube ke ek face ko fix karke uske opposite (upar wale) face par tangential force lagayi jaati hai. Cube apni shape (rectangular parallelepiped) me badal jaata hai lekin volume same rehta hai. Kaunsa elastic modulus is deformation ko govern karta hai, aur kyun?
Shear modulus (η), kyunki yahan deformation shape ka hai, size/volume ka nahi. Tangential (shearing) force lagne se ek layer doosri layer ke relative slip karti hai, jisse shape ka angle badalta hai (shearing strain, θ) lekin volume constant rehta hai. Iska formula: η = Shearing stress / Shearing strain = (F/A) / θ.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q19)
8.1 — Ek steel wire ki length 4.7 m aur cross-sectional area 3.0 × 10⁻⁵ m² hai. Ek copper wire ki length 3.5 m aur area 4.0 × 10⁻⁵ m² hai. Same load ke under dono wires equal elongation dikhate hain. Steel aur copper ke Young's moduli ka ratio nikalein.
Y = FL/(AΔL) ⇒ Y_steel/Y_copper = (L_steel × A_copper) / (A_steel × L_copper)
= (4.7 × 4.0×10⁻⁵) / (3.0×10⁻⁵ × 3.5) = (18.8×10⁻⁵)/(10.5×10⁻⁵)
Y_steel : Y_copper ≈ 1.79 : 1
8.2(a) — Figure me diye gaye stress-strain graph ke ek point par stress = 150 × 10⁶ Pa aur strain = 0.002 hai. Us point tak material ka Young's modulus nikalein.
Y = Stress/Strain = (150 × 10⁶)/0.002 = 7.5 × 10¹⁰ N m⁻²
8.2(b) — Usi graph se material ka approximate yield strength (yield point pe stress) bataiye.
Graph par jahan linear (elastic) region khatam hokar curve moadna shuru hoti hai — wahi yield point hai. Iss chapter ke typical stress-strain curve me yield strength approx 3 × 10⁸ N m⁻² ke aas-paas padhi jaati hai (exact value diye gaye graph ke exact scale par depend karta hai — answer likhte waqt apni copy me diye graph ki reading use karein).
8.3(a) — Materials A aur B ke stress-strain graphs diye hain. In dono me se kis material ka Young's modulus zyada hai?
Jis material ka stress-strain graph zyada steep (slope zyada) hai, uska Young's modulus zyada hai — kyunki Y = stress/strain = slope. Diye gaye typical graph me Material A ka slope zyada hota hai, isliye Y_A > Y_B.
8.3(b) — In dono materials (A aur B) me se kaunsa zyada "strong" hai?
"Strength" ka matlab hai fracture hone se pehle material kitna maximum stress (ultimate tensile strength) sahn kar sakta hai — sirf steep slope (stiffness) se nahi. Graph par jis material ka fracture point zyada stress value par hai, wo zyada strong hai, chahe uska slope kam ho.
8.4(a) — "Rubber ka Young's modulus steel se zyada hota hai." — Ye statement sahi hai ya galat? Reason dijiye.
Galat. Same stress lagane par rubber steel se kaafi zyada strain (elongation) dikhata hai. Chunki Y = Stress/Strain, aur rubber ka strain bahut bada hota hai same stress ke liye, isliye Y_rubber << Y_steel. Steel bahut stiffer material hai.
8.4(b) — "Coiled spring ka stretching uske shear modulus se determine hota hai, Young's modulus se nahi." — Sahi hai ya galat?
Sahi. Jab spring stretch hoti hai, spring ka material (wire) khud lambaai me nahi khinchta — balki wire apni length ke along twist (shear deformation) hota hai. Isliye spring constant essentially uske material ke shear modulus (η) par depend karta hai, Young's modulus par nahi.
8.5 — 0.25 cm diameter ke steel (length 1.5 m) aur brass (length 1.0 m) wires ek rigid support se is tarah lagaye gaye hain ki steel wire poora load (4.0 kg + 6.0 kg = 10.0 kg) uthaata hai aur brass wire sirf 6.0 kg. Dono wires ki elongation nikalein. (Y_steel = 2.0×10¹¹ Pa, Y_brass = 0.91×10¹¹ Pa)
r = 0.125 cm = 1.25×10⁻³ m ⇒ A = πr² = 4.91 × 10⁻⁶ m²
Steel wire (load = 10.0 kg, L = 1.5 m):
ΔL_steel = FL/(AY) = (10×9.8 × 1.5)/(4.91×10⁻⁶ × 2.0×10¹¹) = 147/(9.82×10⁵) ≈ 1.5 × 10⁻⁴ m
Brass wire (load = 6.0 kg, L = 1.0 m):
ΔL_brass = (6×9.8 × 1.0)/(4.91×10⁻⁶ × 0.91×10¹¹) = 58.8/(4.47×10⁵) ≈ 1.3 × 10⁻⁴ m
8.6 — 10 cm edge ke aluminium cube ka ek face wall se firmly fixed hai. Iske opposite face par 100 kg mass ka tangential force lagaya jaata hai. Vertical deflection of this face nikalein. (Shear modulus of aluminium, η = 25 GPa)
F = mg = 100 × 9.8 = 980 N; A = (0.1)² = 0.01 m²
Shearing stress = F/A = 980/0.01 = 9.8 × 10⁴ Pa
Shearing strain = stress/η = (9.8×10⁴)/(25×10⁹) = 3.92 × 10⁻⁶
Δx = strain × L = 3.92×10⁻⁶ × 0.1 = 3.92 × 10⁻⁷ m
8.7 — 4 identical hollow cylindrical columns (inner radius 30 cm, outer radius 60 cm) 50,000 kg ka structure symmetrically support karte hain. Har column ki compressional strain nikalein. (Y = 2 × 10¹¹ Pa)
Har column par load = 50000/4 = 12500 kg ⇒ F = 12500 × 9.8 = 122500 N
A = π(r₂² − r₁²) = π[(0.6)² − (0.3)²] = π(0.27) = 0.848 m²
Stress = F/A = 122500/0.848 = 1.444 × 10⁵ Pa
Strain = Stress/Y = (1.444×10⁵)/(2×10¹¹) ≈ 7.22 × 10⁻⁷
8.8 — Copper ke ek piece ka rectangular cross-section 15.2 mm × 19.1 mm hai. Ise 44,500 N ki tension force se khincha jaata hai (sirf elastic deformation hoti hai). Resulting strain nikalein. (Y_copper = 1.1 × 10¹¹ Pa)
A = 15.2×10⁻³ × 19.1×10⁻³ = 2.903 × 10⁻⁴ m²
Stress = F/A = 44500/(2.903×10⁻⁴) = 1.533 × 10⁸ Pa
Strain = Stress/Y = (1.533×10⁸)/(1.1×10¹¹) ≈ 1.39 × 10⁻³
8.9 — Ek steel cable ki radius 1.5 cm hai aur is par maximum allowable stress 10⁸ N m⁻² hai. Cable jo maximum load support kar sakta hai wo nikalein.
A = πr² = π × (0.015)² = 7.07 × 10⁻⁴ m²
F_max = Stress × A = 10⁸ × 7.07×10⁻⁴ ≈ 7.07 × 10⁴ N
8.10 — 15 kg ka rigid bar teen wires (har ek 2.0 m length) se symmetrically supported hai — dono ends par copper wire, beech me iron wire. Agar teeno wires me equal tension chahiye, to unke diameters ka ratio nikalein. (Y_copper = 1.1×10¹¹ Pa, Y_iron = 1.9×10¹¹ Pa)
Rigid bar horizontal rehta hai, isliye teeno wires (same length L) ka elongation ΔL same hota hai ⇒ strain same. Diya gaya condition: tension T bhi same hai teeno me.
Strain = T/(AY) ⇒ A_copper × Y_copper = A_iron × Y_iron (kyunki T aur strain dono same)
A_copper/A_iron = Y_iron/Y_copper = 1.9/1.1 = 1.727
(d_copper/d_iron)² = 1.727 ⇒ d_copper/d_iron = √1.727 ≈ 1.31 : 1
8.11 — 14.5 kg ka mass, steel wire (unstretched length 1.0 m, cross-section area 0.065 cm²) ke end me bandh kar vertical circle me 2 rev/s se whirl kiya jaata hai. Circle ke lowest point par wire ki elongation nikalein. (Y_steel = 2×10¹¹ Pa)
ω = 2πn = 2π(2) = 4π rad/s; v = ωr = 4π(1.0) = 12.57 m/s
Lowest point par tension (centripetal + weight dono support karta hai):
T = m(g + v²/r) = 14.5 × (9.8 + 157.9/1.0) = 14.5 × 167.7 ≈ 2431.7 N
A = 0.065 × 10⁻⁴ m² = 6.5 × 10⁻⁶ m²
ΔL = TL/(AY) = (2431.7 × 1.0)/(6.5×10⁻⁶ × 2×10¹¹) ≈ 1.87 × 10⁻³ m
8.12 — 100.0 litre paani ko 100.0 atm pressure badhaane se uska volume 99.5 litre ho jaata hai. Bulk modulus of water nikalein aur ise air ke bulk modulus se compare karein.
ΔV/V = 0.5/100.0 = 5 × 10⁻³; ΔP = 100.0 × 1.013×10⁵ = 1.013×10⁷ Pa
B_water = ΔP/(ΔV/V) = (1.013×10⁷)/(5×10⁻³) ≈ 2.03 × 10⁹ N m⁻²
Air ka isothermal bulk modulus uske pressure ke barabar hota hai, yaani B_air ≈ 1.0 × 10⁵ Pa.
B_water/B_air ≈ 2 × 10⁴ — yaani paani air se karib 20,000 guna kam compressible hai, kyunki liquid ke molecules already bahut close-packed hote hain.
8.13 — Paani ki surface density 1.03 × 10³ kg m⁻³ hai. Us depth par jahan pressure 80.0 atm hai, paani ki density kya hogi? (Compressibility of water = 45.8 × 10⁻¹¹ Pa⁻¹)
ΔP = 80.0 × 1.013×10⁵ = 8.104 × 10⁶ Pa
ΔV/V = compressibility × ΔP = 45.8×10⁻¹¹ × 8.104×10⁶ ≈ 3.71 × 10⁻³
Mass constant rehta hai, isliye density me fractional increase = volume me fractional decrease:
Δρ = ρ₀ × (ΔV/V) = 1.03×10³ × 3.71×10⁻³ ≈ 3.82 kg m⁻³
ρ_new = 1.03×10³ + 3.82 ≈ 1.034 × 10³ kg m⁻³
8.14 — Ek glass slab ko 10 atm hydraulic pressure diya jaata hai. Uska fractional volume change (ΔV/V) nikalein. (Bulk modulus of glass, B ≈ 37 × 10⁹ Pa)
ΔP = 10 × 1.013×10⁵ = 1.013 × 10⁶ Pa
ΔV/V = ΔP/B = (1.013×10⁶)/(37×10⁹) ≈ 2.74 × 10⁻⁵
8.15 — 10 cm edge ke copper cube ko 7.0 × 10⁶ Pa pressure diya jaata hai. Volume contraction (ΔV) nikalein. (Bulk modulus of copper, B ≈ 140 × 10⁹ Pa)
V = (0.1)³ = 1 × 10⁻³ m³
ΔV/V = ΔP/B = (7.0×10⁶)/(140×10⁹) = 5 × 10⁻⁵
ΔV = 5×10⁻⁵ × 1×10⁻³ = 5 × 10⁻⁸ m³
8.16 — 1 litre paani ko 0.10% se compress karne ke liye kitna pressure change chahiye? (Bulk modulus of water, B = 2.2 × 10⁹ Pa)
ΔV/V = 0.10% = 1.0 × 10⁻³
ΔP = B × (ΔV/V) = 2.2×10⁹ × 1.0×10⁻³ = 2.2 × 10⁶ Pa
≈ 2.2 × 10⁶ Pa ≈ 21.7 atm
Important Equations — Ek Nazar Me
| Quantity | Formula | SI Unit |
|---|---|---|
| Stress | Stress = F/A | N m-2 (Pa) |
| Longitudinal (linear) strain | Strain = ΔL/L | No unit |
| Volume strain | Strain = ΔV/V | No unit |
| Shearing strain | Strain = Δx/L ≈ θ | No unit (radian) |
| Hooke's Law | Stress ∝ Strain (elastic limit ke andar) | — |
| Young's modulus | Y = (F/A)/(ΔL/L) = FL/(AΔL) | N m-2 (Pa) |
| Shear modulus (rigidity) | η = Shearing stress/Shearing strain = (F/A)/θ | N m-2 (Pa) |
| Bulk modulus | B = −ΔP/(ΔV/V) | N m-2 (Pa) |
| Compressibility | k = 1/B | Pa-1 |
| Poisson's ratio | σ = Lateral strain / Longitudinal strain | No unit |
| Elastic P.E. per unit volume | u = ½ × Stress × Strain = ½ Y (Strain)² | J m-3 |
| Elastic P.E. in a stretched wire | U = ½ F ΔL | J |
| 1 atm (unit conversion) | 1 atm = 1.013 × 105 Pa | Pa |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Area unit convert karna bhool jaana — cm² ya mm² me diya gaya cross-section seedha formula me daal dena bina m² me convert kiye; is ek galti se final answer ka order of magnitude hi galat ho jaata hai.
- atm ko Pa me convert na karna (1 atm = 1.013 × 10⁵ Pa) — bulk modulus ke questions me pressure directly "atm" me use kar lena bahut common numeric slip hai.
- Young's modulus aur shear modulus ko confuse karna — jaise spring ki stretching ko Y se solve karne ki koshish karna, jabki wahan shear/rigidity modulus (η) lagta hai (deformation ka TYPE dekhna zaroori hai: length change vs shape change vs volume change).
- Bulk modulus ke formula me negative sign ka matlab galat samajhna — B hamesha positive quantity hai (pressure badhne se volume ghatta hai, isliye formula me minus sign hai, lekin B ki value khud negative nahi likhni).
- Vertical circular motion wale tension problems (jaise wire par whirl hota mass) me sirf centripetal force (mv²/r) le lena aur weight (mg) add karna bhool jaana — lowest point par T = m(g + v²/r) hota hai, sirf mv²/r nahi.
- Significant figures aur rounding ka dhyaan na rakhna — intermediate steps me bahut jaldi round-off kar dena, jisse final answer expected range se bahar chala jaata hai; hamesha final step tak poori precision carry karo.
Board-Style Important Questions
- Stress ki definition dijiye aur uska SI unit likhiye.
- Hooke's Law ko statement ke roop me likhiye aur bataiye ye kis condition ke andar hi valid hota hai.
- Ek stretched wire me store hone wali elastic potential energy ke liye expression derive karein.
- Young's modulus, shear modulus, aur bulk modulus — in teeno me difference spasht karein, har ek ka ek real-life example ke saath.
- Ek ductile metal ke typical stress-strain curve ko draw karke us par proportional limit, elastic limit, yield point, ultimate tensile strength, aur fracture point label karein aur har region ka physical significance samjhaein.
- Agar ek wire ko same material ke doosre wire se replace kiya jaaye jiski length double aur diameter bhi double ho, to same load ke liye elongation kaise change hogi — derivation ke saath dikhayein.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 8 Mechanical Properties of Solids start karne se pehle kya revise kar lena chahiye?
Force aur units ka basic sense clear hona chahiye — agar Chapter 1 (Units and Measurements) aur class 11 physics chapter 2 motion in a straight line ncert solutions wale basics (displacement, unit conversion) solid hain, to stress-strain ke numericals me unit-conversion ki galtiyan kaafi kam ho jaati hain. Laws of Motion (force, Newton's laws) ka concept bhi background me kaam aata hai kyunki stress khud F/A hai.
class 11 physics deleted syllabus 2026-27 me kya-kya trim hua hai, aur kya Mechanical Properties of Solids bhi affected hai?
Rationalised 2026-27 session me sabse bada change ye hai ki purana "Physical World" chapter poori tarah standalone chapter ke roop me hata diya gaya — ab Class 11 Physics 14 chapters ka hai, seedha "Units and Measurements" se shuru hokar. Mechanical Properties of Solids (Chapter 8) apne aap me poora retained hai — koi major topic trim nahi hua yahan (trimming zyada tar Gravitation, Thermodynamics, System of Particles jaise chapters me hui hai). Phir bhi, kisi bhi numerically-precise claim se pehle apni copy ke official ncert.nic.in wale latest PDF se ek baar table of contents cross-check kar lena best practice hai.
class 11 physics chapter wise weightage cbse me Mechanical Properties of Solids ka kitna weightage hota hai?
Exact marks-weightage har saal ke sample paper/marking scheme ke saath thoda change ho sakta hai, isliye koi specific number yahan claim nahi kiya ja raha. Sabse reliable tareeka hai apne current academic year ka official sample question paper aur marking scheme dekhna — wahi authoritative source hai, kisi bhi third-party estimate se zyada bharosemand.
Is chapter ke baad revision karte waqt kaunse related chapters saath me cover karne chahiye?
Sabse natural sequel Chapter 9 (Mechanical Properties of Fluids) hai — wahi solid-vs-fluid deformation ka comparison deta hai (mechanical properties of fluids class 11 numericals bhi similar unit-conversion discipline maangte hain). Agar aap class 11 physics all chapters pdf 2026-27 ek saath revise kar rahe ho, to mechanics ka poora set — motion in a plane projectile motion class 11 solutions, gravitation class 11 numericals with solutions, aur system of particles and rotational motion important formulas — saath me karna helpful rehta hai kyunki force/energy concepts overlap karte hain.
Numericals me galti kam karne ke liye kaunsi practice habit sabse zyada help karti hai?
Har numerical me pehle "kaunsa modulus lagega" decide karo (deformation ka type dekhkar), phir units convert karo, tab formula lagao — is order ko follow karna. Ye discipline laws of motion class 11 ncert exemplar level ke tricky numericals me bhi kaam aati hai, jahan galat formula choose karna sabse common mistake hota hai, calculation error nahi.
Mechanical Properties of Solids ke baad Part II (thermal + oscillations + waves) ki taiyaari kaise plan karein?
Book do physical parts me print hoti hai (Part I roughly Ch 1-8 mechanics, Part II Ch 9-14), lekin ye sirf printing split hai, syllabus-unit distinction nahi. Part II me thermodynamics class 11 physics notes, kinetic theory of gases class 11 derivation, oscillations class 11 physics important questions, aur waves class 11 physics doppler effect formula — in sabka foundation bhi wahi stress-strain jaisi step-by-step, unit-disciplined problem-solving habit hai jo Chapter 8 me build hoti hai.
Class 11 Physics — Saare Chapters
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