Class 11 Physics · Chapter 9
Short answer:
Class 11 Physics Chapter 9 — Mechanical Properties of Fluids rationalised 2026-27 syllabus ka wahi chapter hai jo pre-2023 syllabus me Chapter 10 tha — sirf "Physical World" chapter drop hone ki wajah se numbering ek shift hui hai, content same hai. Ye chapter fluids (liquids + gases) ke static aur dynamic behaviour ko cover karta hai: pressure, Pascal's law, Archimedes' principle, streamline/turbulent flow, Bernoulli's principle, viscosity (Stokes' law), aur surface tension/capillarity. Agar tum class 11 physics ncert solutions series follow kar rahe ho, toh ye chapter numerical-heavy hai — is page par har in-text solved example aur exercise question ka step-by-step CBSE-marking-scheme-style solution hai, matching real mechanical properties of fluids class 11 numericals ka pattern.
Class 11 Physics ka pura syllabus 14 chapters ka hai (2026-27 rationalised session): Units and Measurements → Motion in a Straight Line → Motion in a Plane → Laws of Motion → Work, Energy and Power → System of Particles and Rotational Motion → Gravitation → Mechanical Properties of Solids → Mechanical Properties of Fluids (Chapter 9, yahi wala) → Thermal Properties of Matter → Thermodynamics → Kinetic Theory → Oscillations → Waves. Agar tumne already class 11 physics chapter 2 motion in a straight line ncert solutions ya units and measurements class 11 important questions cover kar liye hain, toh yeh Chapter 9 "properties of matter" unit ka doosra half hai — Chapter 8 me solids ka stress-strain padha tha, ab yahan fluids (liquids + gases) ka static aur dynamic behaviour.
Chapter shuru hota hai pressure aur Pascal's law se — hydraulic lift/brake jaise real-world machines yahin se explain hoti hain. Fir aata hai streamline vs turbulent flow, Reynolds number, equation of continuity, aur Bernoulli's principle — jo aeroplane wing lift, atomizer/spray pump, aur artery me blood flow jaise applications explain karta hai. Last section viscosity (Newton's law of viscosity, Stokes' law, terminal velocity) aur surface tension (surface energy, angle of contact, excess pressure, capillary rise) cover karta hai. Yeh chapter gravitation class 11 numericals with solutions ya laws of motion class 11 ncert exemplar jaisa hi numerical-heavy hai — formula substitution ke saath saath concept-based "explain why" type questions bhi bahut poochhe jaate hain.
Ek zaroori note: is chapter ka exact class 11 physics chapter wise weightage cbse official curriculum document se hi confirm karo — yahan koi invented marks-number nahi diya gaya. Series ke aage ke chapters — system of particles and rotational motion important formulas, 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 — sab isi tarah step-by-step solve honge. Poora class 11 physics all chapters pdf 2026-27 chahiye toh official ncert.nic.in hi authoritative source hai; third-party PDFs me kabhi-kabhi pre-rationalisation content reh jaata hai.
2026-27 session ke class 11 physics deleted syllabus 2026-27 ka sabse bada change yeh hai ki "Physical World" standalone chapter poori tarah hata diya gaya hai — baaki saare chapters, including motion in a plane projectile motion class 11 solutions wala Chapter 3, aur yeh Mechanical Properties of Fluids, retained hain; bas kuch andar ke topics/derivations trim hue hain (jaise heat-engine/refrigerator detail Thermodynamics me, ya rolling-motion derivation System of Particles me).
Chapter 9 Summary — 5 Minute Revision
Chapter 9 "Mechanical Properties of Fluids" teen bade blocks me bant sakta hai: (1) Fluid statics — pressure, Pascal's law, pressure variation with depth, hydraulic machines, Archimedes' principle; (2) Fluid dynamics — streamline/turbulent flow, Reynolds number, equation of continuity, Bernoulli's principle aur uske applications (wing lift, Venturi meter/atomizer, blood flow); (3) Viscosity aur surface tension — Newton's law of viscosity, Stokes' law, terminal velocity, surface tension, surface energy, angle of contact, excess pressure (drop vs bubble), capillary rise.
Exam ke liye sabse zyada repeat hone waale numerical patterns: hydraulic lift force calculation, U-tube manometer/specific-gravity problems, Bernoulli-based lift/Venturi problems, terminal velocity aur excess-pressure numericals, aur capillary rise. Conceptual side pe "explain why" questions (obtuse vs acute contact angle, spinning ball ka Magnus effect, backward thrust on a vessel) utna hi important hain jitna numericals. Formula substitute karne se pehle hamesha units SI me convert karo (cm→m, cm²→m², g/cc→kg/m³) — yehi is chapter ki sabse common galti ki jad hai.
In-Text Questions — Solutions
Solved Example 1 — Do thigh bones (femurs), jinka cross-sectional area har ek 10 cm² hai, human body ke upper part (mass 40 kg) ko support karte hain. Femurs par sustain hone waala average pressure estimate karo.
Total area, A = 2 × 10 cm² = 2 × 10 × 10⁻⁴ m² = 2 × 10⁻³ m².
Support kiya gaya weight, F = mg = 40 × 9.8 = 392 N.
P = F/A = 392 / (2 × 10⁻³) = 1.96 × 10⁵ Pa
Answer: P ≈ 1.96 × 10⁵ Pa (roughly atmospheric pressure se double).
Solved Example 2 — Ek jheel (lake) ki surface se 10 m neeche ek swimmer par pressure kitna hoga?
Atmospheric pressure, P₀ = 1.01 × 10⁵ Pa. Water ki density, ρ = 1000 kg/m³, h = 10 m, g = 9.8 m/s².
P = P₀ + ρgh = 1.01 × 10⁵ + (1000 × 9.8 × 10)
P = 1.01 × 10⁵ + 0.98 × 10⁵ = 1.99 × 10⁵ Pa
Answer: P ≈ 2 × 10⁵ Pa — surface ke pressure se lagbhag double.
Solved Example 3 — Hydraulic press ke do arms ke cross-sections 1 cm² aur 10 cm² hain. Thinner arm ke water par 5 N ka force apply kiya jaata hai. Wider arm ke liquid dwara load par kitna force exert hoga?
Pascal's law ke mutabik, pressure equally transmit hota hai: P₁ = P₂, isliye F₁/A₁ = F₂/A₂.
F₂ = F₁ × (A₂/A₁) = 5 × (10/1) = 50 N
Answer: F₂ = 50 N — hydraulic press force ko 10× multiply karta hai, area ratio ke matching.
Solved Example 4 — Ek car lift me, compressed air radius 5.0 cm ke chhote piston par force F₁ exert karti hai; ye radius 15 cm ke doosre piston tak transmit hoti hai. Agar car ka mass 1350 kg hai, toh F₁ aur required pressure nikaalo. (g = 9.8 m/s²)
A₁ = πr₁² = π(0.05)² = 7.854 × 10⁻³ m²; A₂ = πr₂² = π(0.15)² = 7.069 × 10⁻² m².
Car ka weight, F₂ = mg = 1350 × 9.8 = 13230 N.
F₁ = F₂ × (A₁/A₂) = F₂ × (r₁/r₂)² = 13230 × (5/15)² = 13230/9 = 1470 N
P = F₁/A₁ = 1470 / 7.854 × 10⁻³ ≈ 1.87 × 10⁵ Pa
Answer: F₁ ≈ 1470 N; required pressure ≈ 1.87 × 10⁵ Pa.
Solved Example 5 — 1.25 cm diameter waale ek tap se water 0.48 L/min ki rate se flow karta hai, fir rate badhaakar 3 L/min kar diya jaata hai. Dono cases me flow laminar hai ya turbulent, check karo. (η of water = 1.0 × 10⁻³ Pa s, ρ = 1000 kg/m³)
Radius r = 0.625 cm = 6.25 × 10⁻³ m; Area A = πr² = 1.227 × 10⁻⁴ m².
Case 1: Q₁ = 0.48 L/min = 8 × 10⁻⁶ m³/s.
v₁ = Q₁/A = 8 × 10⁻⁶ / 1.227 × 10⁻⁴ = 0.0652 m/s
Reₗ = ρv₁d/η = (1000 × 0.0652 × 0.0125) / 10⁻³ ≈ 815
Chuki Re 1000 se kam hai, flow laminar hai.
Case 2: Q₂ = 3 L/min = 5 × 10⁻⁵ m³/s.
v₂ = 5 × 10⁻⁵ / 1.227 × 10⁻⁴ = 0.4075 m/s
Re₂ = (1000 × 0.4075 × 0.0125) / 10⁻³ ≈ 5094
Chuki Re 2000 se zyada hai, flow turbulent hai.
Solved Example 6 — 1.0 cm radius ke ek soap bubble ki surface tension 0.030 N/m hai. Uske andar excess pressure nikaalo.
Soap bubble ki do liquid-air surfaces hoti hain, isliye:
ΔP = 4T/r = (4 × 0.030) / (1.0 × 10⁻²) = 12 Pa
Answer: ΔP = 12 Pa atmospheric pressure se upar.
Solved Example 7 — Air me girte hue 0.1 mm radius ke ek chhote water droplet ki terminal velocity estimate karo. (η_air = 1.8 × 10⁻⁵ Pa s, ρ_water = 1000 kg/m³, air ki density neglect karo)
r = 1 × 10⁻⁴ m; Δρ = ρ − σ ≈ 1000 kg/m³.
v_t = 2r²(ρ − σ)g / 9η = [2 × (10⁻⁴)² × 1000 × 9.8] / (9 × 1.8 × 10⁻⁵)
v_t = 1.96 × 10⁻⁴ / 1.62 × 10⁻⁴ ≈ 1.21 m/s
Answer: v_t ≈ 1.2 m/s (chhote droplets terminal velocity almost turant reach kar lete hain, isi wajah se fog/mist ke droplets float karte hue dikhte hain).
Solved Example 8 — 0.10 mm radius ke ek capillary tube me water kitni height tak chadhega? (T = 7.0 × 10⁻² N/m, θ ≈ 0°, ρ = 1000 kg/m³, g = 9.8 m/s²)
h = 2T cosθ / (rρg) = (2 × 7.0 × 10⁻² × 1) / (1 × 10⁻⁴ × 1000 × 9.8)
h = 0.14 / 0.98 ≈ 0.143 m = 14.3 cm
Answer: h ≈ 14.3 cm. Tube jitni patli hogi, rise utna zyada hoga — isi wajah se capillary action fine soil pores aur plant xylem me strong hoti hai.
Solved Example 9 — Ek badi tank ki free surface se 5 m neeche ek chhote hole se water bahar nikal raha hai. Efflux ki speed nikaalo (Torricelli's law).
Bernoulli's equation ko free surface (bada area, v ≈ 0) aur hole ke beech apply karke:
v = √(2gh) = √(2 × 9.8 × 5) = √98 ≈ 9.9 m/s
Answer: v ≈ 9.9 m/s — jitni speed se ek body height h se freely fall karti, utni hi.
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Exercise Questions — Solutions (Q1–Q27)
Q1. Explain karo ki kyun: (a) humans me blood pressure feet par brain se zyada hota hai; (b) 6 km height par atmospheric pressure ghatkar apne sea-level value ka lagbhag aadha reh jaata hai, jabki atmosphere 100 km se zyada extend karta hai; (c) hydrostatic pressure ek scalar quantity hai, chahe pressure force divided by area hi kyun na ho.
(a) Kisi bhi point par blood pressure = heart par pressure + ρgh, jahan h heart ke upar blood column ki height hai. Feet, brain se zyada neeche hain, isliye extra ρgh term (blood column ka weight) feet par zyada hota hai, jisse wahan pressure zyada ho jaata hai.
(b) Air compressible hoti hai — uski density khud altitude ke saath (roughly exponentially) kam hoti jaati hai, unlike ek incompressible liquid jiski density constant rehti hai. Chuki air ka zyadatar mass lower layers me concentrated hai, pressure surface ke paas bahut fast girta hai — poore 100 km me uniformly nahi — isliye ~6 km ke andar hi apne sea-level value ka aadha reh jaata hai, chahe atmosphere technically usse kaafi upar tak extend karta ho.
(c) Kisi point par pressure sabhi directions me equally act karta hai (ye isotropic hota hai) — pressure ke saath khud koi single preferred direction associated nahi hoti, sirf jo force wo kisi chosen surface par produce karta hai uski direction hoti hai (us surface ke normal). Isliye pressure ko scalar define kiya jaata hai.
Q2. Explain karo ki kyun: (a) glass ke saath mercury ka angle of contact obtuse hota hai, jabki glass ke saath water ka acute hota hai; (b) clean glass surface par water spread ho jaata hai jabki mercury usi surface par drops banata hai; (c) liquid ki surface tension surface area se independent hoti hai; (d) detergent-dissolved water ki surface tension kam hoti hai; (e) bina kisi external force ke ek liquid drop spherical hota hai.
(a) Mercury-glass ke case me, mercury molecules ke beech cohesive force, glass ke saath adhesive force se zyada hoti hai (mercury glass ko wet nahi karta), isliye obtuse contact angle banta hai. Water-glass ke case me, adhesive force cohesive force se zyada hoti hai (water glass ko wet karta hai), isliye acute contact angle banta hai.
(b) Water glass ko wet karta hai kyunki adhesion cohesion se zyada hoti hai, isliye ye maximum contact banaane ke liye spread ho jaata hai. Mercury me cohesion dominate karta hai, isliye ye drops bana kar contact area minimise kar leta hai.
(c) Surface tension liquid surface par intermolecular cohesive forces se aata hai — ye ek molecular (intensive) property hai jo sirf liquid aur temperature par depend karti hai, na ki total kitna surface area hai us par.
(d) Detergent ek surfactant hai — ye surface water molecules ke beech cohesion ko weaken karta hai, jisse T kam ho jaata hai. Isse water zyada acchi tarah spread ho paata hai aur fine crevices (jaise cloth fibres) me penetrate kar paata hai, jisse cleaning improve hoti hai.
(e) Ek drop apni surface energy (= T × surface area) ko given volume ke liye minimise karta hai. Sabhi shapes me se, ek sphere ka surface area per unit volume sabse kam hota hai, isliye external distorting forces ki absence me, ek drop naturally spherical shape le leta hai.
Q3. Blanks fill karo: (a) liquids ki surface tension generally temperature ke saath ___ hoti hai; (b) gases ki viscosity temperature ke saath ___ hoti hai jabki liquids ki viscosity ___ hoti hai; (c) solids ke liye shearing stress shear strain ke proportional hota hai, jabki fluids ke liye ye ___ ke proportional hota hai; (d) constriction par flow speed me increase ___ se follow hota hai, jabki uske saath aane waala pressure drop ___ se follow hota hai; (e) wind tunnel me test kiya gaya ek model plane, actual full-size plane ke same Reynolds number par ek relatively ___ speed par turbulence dikhata hai.
(a) decreases — temperature badhne ke saath cohesive intermolecular forces weak ho jaate hain, isliye T ghat jaata hai.
(b) Gases ki viscosity temperature ke saath increase hoti hai (layers ke beech faster molecular momentum transfer ki wajah se); liquids ki viscosity temperature ke saath decrease hoti hai (higher T par weaker cohesive forces ki wajah se).
(c) shear strain ki rate (velocity gradient, dv/dx) — ye Newton's law of viscosity hai, unlike solids jahan stress strain par khud depend karta hai.
(d) constriction par speed me increase equation of continuity (mass/volume flow rate ke conservation) se follow hota hai; wahan pressure me drop Bernoulli's principle (energy conservation) se follow hota hai.
(e) higher — kyunki Reynolds number size scale par depend karta hai, ek chhote model ko full-size plane jaise hi Re (aur isliye same flow character) reach karne ke liye proportionally zyada flow speed chahiye.
Q4. Explain karo ki kyun: (a) kaagaz ke ek piece ko horizontal rakhne ke liye uske upar se blow karna chahiye, neeche se nahi; (b) tap ko fingers se band karne ki koshish karte waqt gaps se fast jets of water gush karte hain; (c) syringe needle ka bore size, thumb pressure se better flow rate control karta hai; (d) ek chhote hole se escape hota hua fluid vessel par backward thrust exert karta hai; (e) ek spinning cricket ball simple parabolic path follow nahi karta.
(a) Upar se blow karne se paper ke upar air speed badh jaati hai, isliye Bernoulli's principle ke mutabik upar ka pressure niche ke atmospheric pressure se kam ho jaata hai — resulting net upward force paper ko roughly horizontal rakhti hai. Niche se blow karne se paper bas away/down push ho jaata, balance nahi hota.
(b) Fingers opening ka area kam kar dete hain. Equation of continuity (A₁v₁ = A₂v₂) ke mutabik, same volume flow rate ke liye chhota area exit speed ko kaafi badha deta hai, jisse fast jets banti hain.
(c) Poiseuille's equation ke mutabik, flow rate Q ∝ r⁴ (needle bore radius ki fourth power) hota hai lekin sirf ∝ ΔP (pressure) linearly. Isliye needle ke radius me chhote-chhote changes, thumb-applied pressure me changes se kaafi zyada dramatically flow rate change karte hain — bore size bahut finer control deta hai.
(d) Conservation of momentum (Newton's third law) ke mutabik, escaping fluid jet jo momentum le jaata hai, uske reaction me vessel par ek equal aur opposite reaction thrust produce hoti hai — yehi principle rocket propulsion ke peeche bhi hai.
(e) Ball ki spin ek side par air ko fast drag karti hai aur doosri side par slow kar deti hai. Bernoulli's principle ke mutabik ye ball ke across ek pressure difference create karta hai, jisse ek sideways force (Magnus effect) banta hai jo trajectory ko plain parabola se curve kar deta hai.
Q5. High-heel shoes pehne hui ek 50 kg ki girl ek single heel (area 1.0 cm²) par balance karti hai. Heel floor par kitna pressure exert karta hai?
A = 1.0 cm² = 1.0 × 10⁻⁴ m². Weight, F = mg = 50 × 9.8 = 490 N.
P = F/A = 490 / (1.0 × 10⁻⁴) = 4.9 × 10⁶ Pa
Answer: P = 4.9 × 10⁶ Pa — atmospheric pressure se lagbhag 48 guna zyada, isi wajah se heels soft flooring ko damage kar sakte hain.
Q6. Torricelli ka barometer mercury use karta hai. Agar Pascal ne 984 kg/m³ density waali French wine se ise duplicate kiya hota, toh normal atmospheric pressure ke liye kitni height ka wine column chahiye hota? (ρ_Hg = 13600 kg/m³, standard mercury barometer height = 0.76 m)
Chuki P₀ = ρ_Hg × g × h_Hg = ρ_wine × g × h_wine:
h_wine = h_Hg × (ρ_Hg/ρ_wine) = 0.76 × (13600/984) = 0.76 × 13.82
h_wine ≈ 10.5 m
Answer: h ≈ 10.5 m — ek impractically tall column, isi wajah se real barometers me mercury (bahut dense) use hota hai.
Q7. Ek off-shore structure 10⁹ Pa ka maximum stress withstand karne ke liye banaya gaya hai. Kya ye 3 km deep ocean oil well ke bottom ke liye suitable hai? (ρ_seawater = 1.03 × 10³ kg/m³, currents ignore karo)
P = ρgh = 1.03 × 10³ × 9.8 × 3000
P ≈ 3.03 × 10⁷ Pa
Ye structure ke maximum withstandable stress (10⁹ Pa) se lagbhag 33 guna chhota hai.
Answer: Haan, structure suitable hai — 3 km depth par actual hydrostatic pressure uski safe limit ke andar hi hai, isliye ek bada safety margin bach jaata hai.
Q8. Ek hydraulic car lift 3000 kg tak ki cars ke liye design ki gayi hai. Load-carrying piston ka cross-section 425 cm² hai. Chhote piston ko kitna maximum pressure bear karna padega?
A = 425 cm² = 4.25 × 10⁻² m². F = mg = 3000 × 9.8 = 29400 N.
P = F/A = 29400 / (4.25 × 10⁻²) ≈ 6.92 × 10⁵ Pa
Answer: P ≈ 6.92 × 10⁵ Pa — Pascal's law ke mutabik, yahi pressure hai jo chhote piston (aur connected compressor) ko supply karni hogi.
Q9. Ek U-tube me water aur methylated spirit, mercury se separate kiye hue hain. Dono arms me mercury columns level hain, ek arm me 10.0 cm water hai aur doosre me 12.5 cm spirit hai. Spirit ki specific gravity nikaalo.
Chuki mercury levels equal hain, dono liquid columns (mercury ke upar) ke base par pressures equal honge:
ρ_water × g × h_water = ρ_spirit × g × h_spirit
ρ_spirit = ρ_water × (h_water/h_spirit) = 1 × (10/12.5) = 0.8
Answer: Spirit ki specific gravity = 0.8.
Q10. Q9 ke setup me, dono arms me respectively 15.0 cm water aur 15.0 cm spirit aur add kiya jaata hai. Dono arms ke beech mercury levels ka resulting difference kya hoga? (Mercury ki specific gravity = 13.6)
Naya water column = 10 + 15 = 25 cm; naya spirit column = 12.5 + 15 = 27.5 cm.
Maan lo mercury level difference = h (cm). Lower mercury surface par pressures equate karke (relative densities use karke, taaki g cancel ho jaaye):
1 × 25 = 0.8 × 27.5 + 13.6 × h
25 = 22 + 13.6h ⟹ h = 3/13.6 ≈ 0.221 cm
Answer: mercury levels ≈ 0.22 cm se differ karte hain — chhota mercury column adjustment, added water/spirit ke extra pressure ko balance kar deta hai.
Q11. Kya Bernoulli's equation ka use ek fast-flowing river ke rapids me water ke flow ko describe karne ke liye ho sakta hai? Explain karo.
Nahi. Bernoulli's equation ye assume karke derive kiya jaata hai ki fluid non-viscous, incompressible hai, aur steady, streamline (laminar) flow undergo karta hai. River rapids me pani ka flow turbulent aur highly irregular hota hai (chaotic swirls, eddies, friction se energy dissipation), isliye Bernoulli's equation ke conditions satisfy nahi hoti aur ise wahan reliably apply nahi kiya ja sakta.
Q12. Bernoulli's equation apply karte waqt kya ye matter karta hai ki gauge pressure use kiya jaaye ya absolute pressure? Explain karo.
Nahi, matter nahi karta, jab tak consistently use ho. Bernoulli's equation same flow ke do points ke beech pressure compare karta hai. Absolute pressure = atmospheric pressure + gauge pressure har point par; chuki atmospheric pressure P₀ compare kiye jaa rahe dono points par same constant ki tarah add hota hai, ye kisi bhi pressure difference calculation se cancel ho jaata hai. Isliye poore calculation me gauge pressure use karna, absolute pressure use karne jaisa hi physical result deta hai.
Q13. Glycerine ek horizontal tube (length 1.5 m, radius 1.0 cm) se steadily flow karti hai. Ek end par per second collect hone waala mass 4.0 × 10⁻³ kg/s hai. Ends ke beech pressure difference nikaalo. (ρ_glycerine = 1.3 × 10³ kg/m³, η = 0.83 Pa s). Ye bhi check karo ki flow laminar hai ya nahi.
Volume flow rate: Q = (mass flow rate)/ρ = 4.0 × 10⁻³ / 1.3 × 10³ = 3.077 × 10⁻⁶ m³/s.
Poiseuille's equation use karke, Q = πΔP r⁴ / (8ηl):
ΔP = 8ηlQ / (πr⁴) = (8 × 0.83 × 1.5 × 3.077 × 10⁻⁶) / (π × (0.01)⁴)
ΔP ≈ 3.065 × 10⁻⁵ / 3.1416 × 10⁻⁸ ≈ 9.8 × 10² Pa
Laminar flow check karte hue: v = Q/A = 3.077 × 10⁻⁶ / (π × (0.01)²) ≈ 9.8 × 10⁻³ m/s.
Re = ρvd/η = (1300 × 9.8 × 10⁻³ × 0.02) / 0.83 ≈ 0.31
Answer: ΔP ≈ 9.8 × 10² Pa; Re ≈ 0.3 (≪ 2000), isliye flow indeed laminar hai, jo confirm karta hai ki Poiseuille's equation use karna valid tha.
Q14. Ek model aeroplane wing ke wind-tunnel test me, upper surface par air speed 70 m/s hai aur lower surface par 63 m/s hai. Agar wing area 2.5 m² hai, toh lift force nikaalo. (ρ_air = 1.3 kg/m³)
Bernoulli's equation (same height par) ke mutabik, pressure difference:
ΔP = ½ρ(v_upper² − v_lower²) = ½ × 1.3 × (70² − 63²)
ΔP = 0.65 × (4900 − 3969) = 0.65 × 931 ≈ 605 Pa
Lift, F = ΔP × A = 605 × 2.5 ≈ 1.51 × 10³ N
Answer: Lift ≈ 1.5 × 10³ N (≈ 1.5 kN) upward, kyunki faster flow (aur isliye lower pressure) top par hai.
Q15. Ek non-viscous liquid ke steady flow ko dikhaate hue figures ke ek pair me se, ek figure physically incorrect hai. Kaunsa principle use karke judge karoge ki kaunsi figure galat hai, aur kyun?
Do rules ek saath use karo: (1) equation of continuity (A₁v₁ = A₂v₂) — jahan tube/streamlines narrow hote hain, speed zaroor badhegi, aur jahan wide hote hain, speed zaroor ghategi; (2) Bernoulli's principle — jahan bhi speed badhta hai, pressure zaroor ghatega, aur vice versa (same height par).
Ek figure physically incorrect hoti hai agar wo streamlines ko converge (tube narrow) hote hue dikhaaye jabki fluid ko wahan slow ya pressure ko rise hote hue indicate kare, ya diverge (tube wide) hote hue dikhaaye jabki fluid ko speed up hote dikhaaye — dono cases continuity + Bernoulli dono ko simultaneously violate karte hain. In do rules ko apne textbook ki figure par directly apply karke pata karo kaunsi unhe break kar rahi hai.
Q16. Ek spray pump ki cylindrical tube ka cross-section 8.0 cm² hai, jiske ek end par 40 fine holes hain, har ek ka diameter 1.0 mm. Agar liquid tube ke andar 1.5 m/min ki speed se flow karta hai, toh holes se ejection speed nikaalo.
Tube speed: v_tube = 1.5 m/min = 0.025 m/s. Tube area: A_tube = 8.0 cm² = 8 × 10⁻⁴ m².
Har hole ka radius = 0.5 mm = 5 × 10⁻⁴ m; ek hole ka area = π(5×10⁻⁴)² = 7.854 × 10⁻⁷ m²; 40 holes ka total area = 3.1416 × 10⁻⁵ m².
Equation of continuity ke mutabik, A_tube v_tube = A_holes v_holes:
v_holes = (A_tube/A_holes) × v_tube = (8 × 10⁻⁴ / 3.1416 × 10⁻⁵) × 0.025 ≈ 0.64 m/s
Answer: v_holes ≈ 0.64 m/s.
Q17. Soap solution me dipped ek U-shaped wire, ek light slider (weight 1.5 × 10⁻² N, slider length 30 cm) ko thin soap film ke through support karta hai. Film ki surface tension nikaalo.
Soap film ki do surfaces hoti hain, isliye slider ke contact me film ki total length = 2l = 2 × 0.30 = 0.60 m.
T = W / (2l) = 1.5 × 10⁻² / 0.60 = 2.5 × 10⁻² N/m
Answer: T = 2.5 × 10⁻² N/m.
Q18. Ek thin liquid film ek figure me 4.5 × 10⁻² N ka weight support karta hai. Same liquid ka, same temperature par, ek film doosre figures me (different shape ke frame ke saath, lekin same slider length) kitna weight support karega? Physically explain karo.
Support kiya gaya weight same (4.5 × 10⁻² N) rahega sabhi figures me, bashart film ke contact me slider ki length unchanged rahe. Isliye kyunki film se upward force F = T × (2l) hota hai — ye sirf surface tension (us temperature par liquid ki ek fixed property) aur film ko touch karne waale boundary (slider) ki length par depend karta hai, na ki total surface area ya frame ke shape par.
Q19. Room temperature par 3.00 mm radius ke ek mercury drop ke andar pressure kya hoga? (T_mercury = 4.65 × 10⁻¹ N/m at 20°C; assume karo ki koi atmosphere nahi hai.)
Ek drop ki ek hi (convex) surface hoti hai, isliye excess pressure:
ΔP = 2T/r = (2 × 0.465) / (3.00 × 10⁻³) = 0.93 / 0.003 = 310 Pa
Answer: chuki koi atmosphere nahi hai, drop ke andar pressure poori tarah ye excess pressure ≈ 310 Pa hi hai.
Q20. 5.00 mm radius ke ek soap bubble ke andar excess pressure nikaalo (T = 2.50 × 10⁻² N/m at 20°C). Agar same soap solution (relative density 1.20) ke andar 40.0 cm depth par same-size ka ek air bubble banta hai, us bubble ke andar total pressure kya hoga? (1 atm = 1.01 × 10⁵ Pa)
Soap bubble (2 surfaces):
ΔP_soap = 4T/r = (4 × 0.025) / (5.00 × 10⁻³) = 20 Pa
Liquid ke andar air bubble (sirf 1 surface, gas-liquid):
ΔP_air = 2T/r = (2 × 0.025) / (5.00 × 10⁻³) = 10 Pa
Soap solution ke 40 cm depth (ρ = 1200 kg/m³) ki wajah se pressure:
P_depth = ρgh = 1200 × 9.8 × 0.40 = 4704 Pa
Air bubble ke andar total pressure = atmospheric + depth + excess:
P = 1.01 × 10⁵ + 4704 + 10 ≈ 1.06 × 10⁵ Pa
Answer: soap bubble ka excess pressure = 20 Pa; air bubble ka total pressure ≈ 1.06 × 10⁵ Pa.
Q21. 1.0 m² square base waali ek tank ek vertical partition se divide hai. Partition ke bottom par 20 cm² area ka ek hinged door hai. Ek side me water hai, doosri side me relative density 1.7 ka acid hai, dono 4.0 m tak filled hain. Door ko shut rakhne ke liye required force nikaalo.
Chuki door bottom par (depth h = 4.0 m) hai aur tank ke comparison me chhota hai, uske upar pressure ko roughly uniform treat karo. Door par net force = dono sides ke beech pressure difference × door area:
F = (ρ_acid − ρ_water) × g × h × A_door
F = (1700 − 1000) × 9.8 × 4 × (20 × 10⁻⁴)
F = 700 × 9.8 × 4 × 2 × 10⁻³ ≈ 54.9 N
Answer: door ko shut rakhne ke liye lagbhag 54.9 N ka force apply karna hoga (acid, zyada dense hone ki wajah se, water side ki taraf zyada push karta hai).
Q22. Ek U-tube open manometer ek gas-filled vessel se connect hai. (a) mercury-level difference se gas ki absolute pressure nikaalne ka method, aur (b) agar gas pressure change ho jaaye toh naya level difference, explain karo.
Method (apne textbook ki actual figure readings ke saath apply karo):
(a) Agar open side ka mercury level, gas side se ek reading h zyada hai, toh absolute gas pressure hoga:
P_gas = P_atmospheric + ρ_Hg g h
Example ke taur par, h = 20 cm = 0.20 m ke saath: ρ_Hg g h = 13600 × 9.8 × 0.20 ≈ 2.67 × 10⁴ Pa, isse P_gas ≈ 1.01 × 10⁵ + 2.67 × 10⁴ ≈ 1.28 × 10⁵ Pa milta hai.
(b) Agar gas pressure ek mercury column Δh ke equivalent amount se decrease hota hai, toh naya level difference simply (h − Δh) hoga — manometer directly mercury-column height ke change ke through gas pressure ke changes track karta hai, kyunki P_gas − P_atm = ρ_Hg g (level difference) hamesha hold karta hai.
(Note: apne textbook ke diagram me di gayi exact reading(s) is hi formula me substitute karo — upar diya gaya method is question ke har version ke liye general method hai.)
Q23. Do vessels ka base area same hai lekin shape different hai; pehla vessel doosre ke comparison me same height tak pahunchne ke liye double volume ka water leta hai. Kya dono me base par force same hai? Fir bhi weighing scale par unki readings different kyun aati hain?
Haan, dono vessels me base par force same hai — base par hydrostatic pressure sirf P = ρgh (upar directly present liquid column ki height, aur base area) par depend karta hai, na ki total volume ya vessel ke shape par. Ye classic 'hydrostatic paradox' hai.
Lekin ek weighing scale total normal reaction force measure karta hai, jo water (plus vessel) ke actual total weight ke equal hota hai jo us par rest kar raha hai — aur total weight, har differently-shaped vessel me actual water ke volume par depend karta hai. Isliye base force (pressure-related) equal hai, lekin scale reading (total weight-related) different hai, kyunki ye do physically different quantities hain.
Q24. Blood transfusion ke dauraan, needle ek vein me insert ki jaati hai jahan gauge pressure 2000 Pa hai. Blood container ko vein ke level se kitni height upar rakhna hoga taaki blood vein me enter kar sake? (Whole blood ki density ≈ 1.06 × 10³ kg/m³)
h = P / (ρg) = 2000 / (1060 × 9.8) = 2000/10388 ≈ 0.193 m
Answer: container ko vein ke level se kam se kam ≈ 0.19 m (lagbhag 19–20 cm) upar raise karna hoga taaki blood ka apna hydrostatic pressure vein ke back-pressure ko overcome karke andar flow kar sake.
Q25. Bernoulli's equation derive karte waqt, fluid par kiya gaya work, kinetic aur potential energy ke change ke equal maana jaata hai, ye assume karke ki koi energy loss nahi hota. Ek real (viscous) fluid me, ye 'missing' energy actually kahan jaati hai?
Real fluid me, fluid layers ke beech internal friction (viscosity) flow ke against negative work karta hai. Ye lost mechanical energy heat me convert ho jaati hai (random molecular kinetic energy), jo fluid ka temperature bahut halka sa raise kar deti hai. Isi wajah se ideal Bernoulli equation (jo strictly non-viscous flow ke liye valid hai) ko real, viscous fluids par apply karte waqt ek correction/loss term chahiye hota hai — P + ½ρv² + ρgh ka total kuch hissa flow ke saath continuously heat ki tarah dissipate hota rehta hai.
Q26. Ek artery (radius 2 × 10⁻³ m) me blood flow ki laminar rehne ke liye largest average velocity kya hai, aur corresponding flow rate kya hoga? (η_blood = 2.084 × 10⁻³ Pa s, ρ_blood = 1.06 × 10³ kg/m³, critical Reynolds number ≈ 2000)
Re_critical = ρvd/η use karke, diameter d = 2r = 4 × 10⁻³ m ke saath:
v = Re_critical × η / (ρ × d) = (2000 × 2.084 × 10⁻³) / (1060 × 4 × 10⁻³)
v = 4.168 / 4.24 ≈ 0.98 m/s
Flow rate:
Q = v × A = 0.98 × π(2 × 10⁻³)² ≈ 1.23 × 10⁻⁵ m³/s ≈ 12.3 cm³/s
Answer: laminar flow ke liye v_max ≈ 0.98 m/s; corresponding Q ≈ 12.3 cm³/s. Is speed se aage, blood flow turbulent ho jaata hai (isi wajah se narrowed/plaque-affected arteries me audible turbulent flow sounds sunayi dete hain).
Q27. Level flight me ek aeroplane ke do wings hain, har ek ka area 25 m² hai. Lower surface par air speed 180 km/h hai aur upper surface par 234 km/h hai. Plane ka mass nikaalo. (ρ_air = 1 kg/m³, g = 9.8 m/s²)
Speeds convert karo: v_lower = 180 km/h = 50 m/s; v_upper = 234 km/h = 65 m/s. Total wing area = 2 × 25 = 50 m².
ΔP = ½ρ(v_upper² − v_lower²) = 0.5 × 1 × (65² − 50²) = 0.5 × (4225 − 2500) = 862.5 Pa
Lift force, F = ΔP × A_total = 862.5 × 50 = 43125 N
Mass, m = F/g = 43125 / 9.8 ≈ 4400 kg
Answer: m ≈ 4400 kg — level flight ke liye, ye lift plane ke weight ko exactly balance karta hai.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Pressure | P = F/A |
| Pressure at depth h (fluid statics) | P = P₀ + ρgh |
| Pascal's law / hydraulic lift | F₁/A₁ = F₂/A₂ ⟹ F₂ = F₁(A₂/A₁) |
| Archimedes' principle (upthrust) | F_B = ρ_fluid × V_displaced × g |
| Equation of continuity | A₁v₁ = A₂v₂ (Av = constant) |
| Bernoulli's equation | P + ½ρv² + ρgh = constant |
| Speed of efflux (Torricelli's law) | v = √(2gh) |
| Newton's law of viscosity | F = ηA(dv/dx) |
| Poiseuille's equation | Q = πΔPr⁴ / (8ηl) |
| Stokes' law | F = 6πηrv |
| Terminal velocity | v_t = 2r²(ρ − σ)g / 9η |
| Reynolds number | Re = ρvd/η |
| Surface tension | T = F/l |
| Excess pressure inside a drop / bubble in liquid (1 surface) | ΔP = 2T/r |
| Excess pressure inside a soap bubble (2 surfaces) | ΔP = 4T/r |
| Capillary rise | h = 2T cosθ / (rρg) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Excess pressure ka formula galat lagana — single-surface drop/bubble-in-liquid ke liye 2T/r hota hai, par soap bubble (do surfaces) ke liye galti se 2T/r hi laga dete hain jabki sahi 4T/r hai.
- Gauge pressure aur absolute pressure ko same maan lena — P_absolute = P_atmospheric + P_gauge yaad na rakhna, jisse barometer/manometer numericals galat answer dete hain.
- Radius aur diameter confuse karna — Stokes' law, capillary rise, Reynolds number jaise formulas me 'r' hamesha radius hai, diameter nahi; galti se diameter daal dena common error hai.
- Bernoulli's equation ko har flow pe blindly apply kar dena — turbulent, viscous ya unsteady flow pe Bernoulli valid nahi hota; sirf streamline (laminar), non-viscous, incompressible, steady flow ke liye hi use karo.
- Units conversion miss karna — cm ko m, cm² ko m², g/cc ko kg/m³ (×1000) me convert karna bhool jaana, jisse final answer 10² ya 10³ factor se off ho jaata hai.
- Surface tension (N/m, force per unit length) aur surface energy (J/m², energy per unit area) ko same concept samajh lena — numeric value same hota hai, but physical meaning alag hai (ek force hai, doosra energy).
Board-Style Important Questions
- Conceptual (short answer): Liquid ki surface tension, surface ke area se independent hoti hai. Kyun?
- Short answer: Pascal's law state karo aur uska ek everyday application do.
- Short answer (derivation): Radius r aur surface tension T waale ek liquid drop ke andar excess pressure ka expression derive karo.
- Long answer (derivation): Bernoulli's theorem ke peeche ki assumptions state karo aur varying cross-section aur height waali ek tube me flow karte fluid ke liye equation likho.
- Long answer (derivation): Ek hydraulic lift me, chhote aur bade pistons ke radii r₁ aur r₂ hain. Applied force aur load jo lift kiya ja sakta hai, unke beech relation derive karo.
- Long answer (derivation): Stokes' law use karke, ek viscous fluid me girte hue ek chhoti sphere ki terminal velocity ka expression derive karo, aur uski motion ke teen stages explain karo (accelerating, phir constant velocity).
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 9 'Mechanical Properties of Fluids' me kya-kya topics aate hain?
Is chapter me Pascal's law, pressure aur uska depth ke saath variation, hydraulic machines (lift/brake), Archimedes' principle, streamline vs turbulent flow, equation of continuity, Reynolds number, Bernoulli's principle aur uske applications (aeroplane lift, atomizer/spray pump, blood flow), viscosity (Newton's law, Stokes' law, terminal velocity), aur surface tension (surface energy, angle of contact, excess pressure, capillary rise) cover hote hain.
CBSE board exam me is chapter (Mechanical Properties of Fluids) ka weightage kitna hai?
Exact marks har saal thoda vary kar sakte hain, isliye yahan koi specific number claim nahi kiya ja raha. Class 11 physics chapter wise weightage cbse dhoondhte waqt hamesha current academic year ka official CBSE curriculum document ya sample paper hi authoritative source treat karo — purane saalon ka data outdated ho sakta hai.
Kya 'Mechanical Properties of Fluids' naye rationalised 2026-27 syllabus me deleted hai?
Nahi — ye chapter poori tarah retained hai. 2026-27 rationalised syllabus me sirf 'Physical World' (purana Chapter 1) standalone chapter ke roop me poori tarah drop hua hai, jiski wajah se Class 11 Physics ab 14 chapters ka hai (pehle 15/16-chapter, two-part structure tha). Isi shift ki wajah se 'Mechanical Properties of Fluids' ab Chapter 9 pe hai — content largely same hai, bas kuch andar ke derivations trim hue hain. (Note: yeh confirmation registry-grade hai, final numbering-check ke liye ncert.nic.in ka actual PDF table-of-contents dekh lena recommended hai.)
Bernoulli's principle ke real-life examples exam me kaise poochhe jaate hain?
Common examples: aeroplane wing par lift (upar-neeche air speed ka difference), perfume atomizer/spray pump (fast air jet se low pressure create hokar liquid upar khichna), Venturi meter, aur narrow/plaque-affected artery me blood flow speed badhna. Exam me generally ye 'explain why' type conceptual questions ya numerical (jaise wing lift force calculate karna) ke roop me poochhe jaate hain.
Class 11 Physics ke saare chapters ka official PDF 2026-27 session ke liye kahan milega?
Sabse reliable source ncert.nic.in hai — Textbooks section me Class XI Physics ke dono parts (Part I: roughly Ch.1-8 mechanics, Part II: roughly Ch.9-14 thermal/oscillations/waves — yeh sirf printing/binding split hai, syllabus-unit distinction nahi) free download ke liye available hote hain. Third-party sites se class 11 physics all chapters pdf 2026-27 download karte waqt dhyan rakho ki pre-rationalisation (purana) content na mil jaaye.
Viscosity aur surface tension me kya fark hai — dono concept exam me confuse ho jaate hain?
Viscosity ek bulk (poore fluid ka) property hai — fluid ki adjacent layers ke beech relative motion ka internal friction, jo flow ko resist karta hai (F = ηA·dv/dx). Surface tension sirf liquid ki surface (interface) ka property hai — unbalanced cohesive forces ki wajah se surface ek stretched membrane jaisa behave karta hai (T = F/l). Dono intermolecular forces se hi aate hain, lekin ek poore fluid ka resistance-to-flow property hai aur doosra sirf surface ka contraction-tendency property.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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