Mechanical Properties of Fluids: Class 11 Physics Chapter 9 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

110 min
1 simulation
29 practice questions
27 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki liquids aur gases fluids kyun hain, aur unki compressibility aur shear par response ko solids se compare karo
Pressure aur density define karo, Pascal's law prove karo, P = Pa + ρgh derive karo aur use barometers, manometers aur hydraulic machines par apply karo
Steady, streamline aur turbulent flow ko describe karo, aur equation of continuity use karo
Work–energy theorem se Bernoulli's equation derive karo aur use efflux (Torricelli's law), Venturi flow aur dynamic lift par apply karo
Coefficient of viscosity define karo, Stokes' law use karo aur girte hue sphere ki terminal velocity derive karo
Surface energy, surface tension aur angle of contact samjhao, aur drops aur bubbles mein excess pressure aur capillary rise ki height derive karo

NCERT Physics Class 11, Chapter 9, pp. 180–201 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

9.1 Introduction

Steel ke tumbler ka paani glass mein ulto, aur paani turant glass ka shape le leta hai. Jab pressure cooker ki seeti bajti hai, to uske vent se steam tezi se bahar nikalti hai. Paani aur steam bilkul alag dikhte hain, phir bhi dono mein ek cheez common hai: dono flow karte hain. Jo bhi substance flow kar sakta hai use fluid (तरल) kehte hain, isliye yeh word liquids aur gases dono ke liye use hota hai.

Hawa, tumhara dil jo blood pump karta hai (aaram ki haalat mein lagbhag paanch litre per minute), neem ke ped ka sap aur Western Ghats ke upar monsoon ke baadal, yeh sab chalte hue fluids hain.

Fluid ko solid se alag kya banata hai? Solid apna shape khud rakhta hai; fluid ka koi shape nahi hota aur woh jis bhi cheez mein ho usi ka shape le leta hai. Phir volume ke maamle mein liquids aur gases alag ho jaate hain. Ek litre doodh jug mein ho ya bottle mein, ek litre hi rehta hai, lekin gas apne container ke har kone mein phail jaati hai. Chapter 8 mein tumne dekha ki pressure kisi bhi material ka volume badalta hai. Solids aur liquids ke liye yeh badlaav bahut chhota hota hai, jabki gas aasani se dab jaati hai: gases kahin zyada compressible hoti hain.

Sabse bada farq yeh hai ki material ek shearing force ka jawab kaise deta hai, yaani aisi force jo surface ke along lagti hai. Ek eraser ko desk par chipkao aur uske top ko side mein dhakelo: woh thoda jhukta hai aur phir ruk jaata hai, kyunki solid mein ek shear stress paida hota hai jo tumhare push ko balance karta hai. Liquid ki ek layer ke top ko dhakelo to woh slide karti hi rehti hai, push chahe kitna bhi halka ho. Rest mein fluid lagbhag koi shear stress support nahi kar sakta, aur isi wajah se woh flow karta hai (Fig. 9.1).

3/3Solid apna jhukaav bana kar rakhta hai, lekin liquid tab tak deform hota rehta hai jab tak push lagta rehta hai.

Fig. 9.1Solid shearing push ka virodh karta hai; fluid uske aage jhuk jaata hai.NCERT p. 180

Left mein, ek fixed base aur top plate ke beech solid block; right mein, ek tairti plate ke neeche liquid layer jismein ek vertical dye streak hai. Step 1: kuch bhi push nahi hota. Step 2: dono plates ko side mein push kiya jaata hai; solid thoda tilt hota hai aur dye streak jhuk jaati hai kyunki liquid slide karna shuru karta hai. Step 3: solid apna chhota tilt bana kar rakhta hai, jabki liquid ki plate kaafi aage slide kar chuki hai aur streak bahut zyada jhuk gayi hai, kyunki liquid flow karta rehta hai.

SimulationEk band box mein solid, liquid aur gas

Paani ke particles ka ek band box, jismein ek thermometer aur −20 °C se 120 °C tak ka temperature slider hai. 0 °C se neeche particles sirf fixed jagahon ke aas-paas hilte hain (ice). 0 °C aur 100 °C ke beech woh ek doosre ke paas se slide karte hain lekin box ke neeche saath rehte hain (liquid paani). 100 °C se upar woh bikhar kar poore box mein phail jaate hain (steam).

Fig. 9.2Temperature slider ko drag karo aur dekho ki particles saath rehte hain ya poore box mein bhar jaate hain.

Try karo

25 °C par particles ek doosre ke paas se slide karte hain. Kya woh box bhar dete hain, ya neeche ek fixed volume mein rehte hain?
Box ko 100 °C se upar garam karo. Ab particles box ka kitna hissa gherte hain?
Ise 0 °C se neeche thanda karo. Kya particles ab bhi ek doosre ke paas se move kar sakte hain, ya sirf apni jagah par hilte hain?

Pehle hum rest mein fluids padhenge: pressure, depth ke saath woh kaise badhta hai aur uska use karne wali machines. Phir hum fluids ko chalayenge aur energy conservation se Bernoulli's principle tak pahunchenge. Real fluids ki do properties se chapter khatam hota hai: viscosity, ek tarah ka internal friction, aur surface tension, jiski wajah se liquid ki surface ek khinchi hui skin ki tarah behave karti hai.

Quick check

2 mein se 1

Kaun si property liquids aur gases ko ek class, fluids, mein saath rakhti hai?

Woh hamesha apna container bhar dete hainWoh hamesha solids se kam dense hote hainUnhe compress karna aasan haiRest mein woh shear stress support nahi kar sakte, isliye flow karte hain

Answer dekho

Answer (D) Rest mein woh shear stress support nahi kar sakte, isliye flow karte hain

Fluid kisi bhi shear stress ke aage jhuk jaata hai, isliye flow karta hai. Sirf gases apna container bharti hain aur sirf gases aasani se compress hoti hain, aur kai liquids (jaise mercury) kai solids se zyada dense hote hain.

9.2 Pressure

Ek drawing pin ko anguthe se soft board mein dabao. Nokeela sira andar chala jaata hai, phir bhi flat head tumhe chubhta nahi, jabki dono siron par same force lagti hai. Ek bhaari truck apna load das ya zyada tyres par baant deta hai, aur camel ke chaude gaddedaar pair use registan ki ret mein dhansne se bachate hain. Har case mein effect is baat se tay hota hai ki force ko us area se divide karne par kya milta hai jo use sambhalta hai.

Ab ek patthar ko shaant paani ki baalti mein dheere se utaaro. Paani patthar ke aur baalti ke har hisse par push karta hai, aur har push us surface ke perpendicular hota hai jis par woh lagta hai. Kyun? Rest mein fluid shear stress nahi sambhal sakta (Section 9.1). Agar paani kisi deewar par tircha dabata, to use ek tircha reaction milta jiska deewar ke along wala hissa shear hai, aur paani slide kar jaata. Shaant paani sirf seedha push kar sakta hai (Fig. 9.3, pehla step).

Kisi point par push ko measure karne ke liye, ek chhote sensor ki kalpana karo: ek seal kiya hua, bina hawa ka dabba jo area $A$ ke ek halke piston se band hai, aur jise ek calibrated spring bahar ki taraf thaame hue hai. Fluid mein rakhne par piston tab tak andar jaata hai jab tak spring ki force us par fluid ki normal force $F$ ko balance na kar de. Yeh ratio piston par per unit area average push deta hai:

Average pressure

P_{av} = \frac{F}{A}

(9.1)

Piston ko chhota karte-karte ek point tak le jaane se wahan ka pressure (दाब) define hota hai:

Ek point par pressure

P = \lim_{Δ A \to 0} \frac{Δ F}{Δ A}

(9.2)

Definition

Pressure

Fluid mein kisi point par pressure wahan rakhi ek chhoti surface par per unit area normal force hai. Yeh ek scalar hai, jise pascal mein measure karte hain ( $1 P a = 1 N m^{- 2}$ ).

Example $50 c m^{2}$ base par khadi $1 k g$ paani ki bottle table ko lagbhag $2.0 \times 10^{3} P a$ se dabati hai.

Pressure ki SI unit, $1 N m^{- 2}$ , ko French scientist Blaise Pascal (1623–1662) ke naam par pascal (Pa) kehte hain, jinke fluids par kaam se hum aage milenge; bade pressures aksar standard atmospheres mein bataye jaate hain, $1 a t m = 1.013 \times 10^{5} P a$ , jo sea level par hawa ke pressure ke kareeb hai. Pressure ki dimensions $[M L^{- 1} T^{- 2}]$ hain. Dhyaan do ki pressure ki koi direction nahi hoti: (9.2) chhoti surface ke normal force ko hi use karta hai, aur woh surface kisi bhi taraf face kar sakti hai. Pressure ek scalar hai.

Kyunki fluid ka koi fixed shape nahi hota, isliye woh kisi jagah mein kitna matter bharta hai, yeh hum uski density se batate hain, yaani per unit volume mass:

Density

ρ = \frac{m}{V}

(9.3)

Density ki dimensions $[M L^{- 3}]$ aur unit $k g m^{- 3}$ hai. Ise 4 °C par paani ki density, $1.000 \times 10^{3} k g m^{- 3}$ , se divide karne par relative density milti hai, jo ek pure number hai. Mercury ki relative density lagbhag 13.6 hai, isliye ek litre mercury ka mass 13.6 kg hota hai (Table 9.1). Dabane se liquid ki density mushkil se badalti hai, jabki gas us par lage pressure ke proportion mein denser ho jaati hai.

Table 9.1Kuch fluids ki approximate densities (liquids lagbhag 20 °C par jab tak alag se na bataya ho; gases 20 °C aur 1 atm par)NCERT Table 9.1 · p. 181

Fluid	Density ( $k g m^{- 3}$ )	Relative density
Paani (4 °C)	1000	1.000
Paani (20 °C)	998	0.998
Samudri paani	≈ 1025	≈ 1.03
Doodh	≈ 1030	≈ 1.03
Kerosene	≈ 800	≈ 0.80
Petrol	≈ 740	≈ 0.74
Glycerine	1260	1.26
Mercury	≈ 13 600	13.6
Hawa	1.20	0.0012
Helium	0.17	0.00017

Example 9.1NCERT Example 9.1 · p. 182

Chaar tyres par khadi car

$1500 k g$ mass ki ek car chaar ek jaise tyres par khadi hai. Har tyre road par lagbhag $12 c m \times 15 c m$ ke patch par chapta ho jaata hai. Tyres road par jo average pressure daalte hain uska estimate lagao. $g = 9.8 m s^{- 2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.2.1 Pascal's Law

Pool ki surface se ek metre neeche tairo aur apna sir dheere se ghumao. Tumhare eardrums ko har direction mein same dabaav mehsoos hota hai. Is tarah ke experiments se Pascal rest mein fluid ke baare mein do nateejon par pahunche, jinhe ab Pascal's law kehte hain: kisi point par pressure har direction mein barabar lagta hai, aur ek horizontal level ke saare points par same hota hai. Dono Newton's laws se nikalte hain.

LawPascal's law (rest mein fluid)

Rest mein fluid mein, kisi point par pressure har direction mein same hota hai, aur fluid ki ek hi judi hui body ke andar same horizontal level wale points par pressure same hota hai.

DerivationPressure ki koi direction kyun nahi hoti

Fluid ka ek chhota wedge alag se socho jo right-angled prism ke shape ka ho (Fig. 9.3). Iska cross-section ek right triangle hai jismein area $A_{x}$ ka ek vertical face, area $A_{y}$ ka ek horizontal face aur area $A_{s}$ ka ek dhalaan wala face hai jo horizontal se angle $θ$ banata hai. Aas-paas ka fluid faces par forces $F_{x}$ , $F_{y}$ aur $F_{s}$ se normally push karta hai. Wedge ka weight uske size ke cube ke saath badhta hai aur face forces square ke saath, isliye kaafi chhote wedge ke liye weight negligible hai.

Wedge equilibrium mein hai. Slope par lage push ka ek horizontal hissa $F_{s} \sin θ$ aur ek vertical hissa $F_{s} \cos θ$ hai, jinhe baaki do pushes ko cancel karna hoga. Triangle ki geometry areas ko unhi factors se jodti hai:

\begin{aligned} F_{x} & = F_{s} \sin θ, & A_{x} & = A_{s} \sin θ \\ F_{y} & = F_{s} \cos θ, & A_{y} & = A_{s} \cos θ \end{aligned}

Har force ko uske area se divide karne par,

\frac{F_{x}}{A_{x}} = \frac{F_{y}}{A_{y}} = \frac{F_{s}}{A_{s}}

(9.4)

yaani $P_{x} = P_{y} = P_{s}$ . Angle $θ$ koi bhi ho sakta tha, isliye dhalaan wala face kisi bhi taraf point kar sakta tha: ek point par pressure har direction mein same hota hai. Rest mein fluid ke andar surface ka orientation kuch bhi ho, fluid us par isi ek pressure se normally push karta hai.

4/4 $F_{s} \sin θ$ $F_{x}$ ko balance karta hai aur $F_{s} \cos θ$ $F_{y}$ ko; areas se divide karne par ek hi pressure milta hai.

Fig. 9.3Shaant paani har surface par right angle par push karta hai, aur wedge dikhata hai ki push har direction mein same hai.NCERT Fig. 9.2 · p. 182

Left mein, shaant paani ka tank. Step 1: paani mein ek tircha block hai, jiske har face aur har deewar se arrows right angle par milte hain. Step 2: paani ka ek chhota right-angled wedge right mein bada karke dikhaya gaya hai, faces par x (vertical), y (base) aur s (slope) label hain, saath mein slope angle θ. Step 3: har face par ek force arrow normally dabata hai. Step 4: slope ki force ko horizontal aur vertical hisson mein toda gaya hai jo baaki do forces ko balance karte hain.

Law ke doosre hisse ke liye, fluid ki ek patli horizontal rod lo jiske flat ends ka area $A$ ho. Gravity rod ke aar-paar lagti hai, isliye uske along sirf end pushes $P_{L} A$ aur $P_{R} A$ lagte hain. Rod rest mein rehti hai, isliye $P_{L} A = P_{R} A$ aur $P_{L} = P_{R}$ . Ek level par koi bhi do points, jo same fluid se jude hon, ek hi pressure share karte hain; agar aisa na hota, to fluid high-pressure wale end se doosre ki taraf flow karta.

9.2.2 Depth ke saath Pressure ka Variation

Ek gehre pool ke bottom ki taraf dive karo aur tumhare kaan dukhne lagte hain. Jitna gehre jaate ho, upar utna zyada paani hota hai, aur us paani ko neeche ka pressure hi thaame rakhta hai. Neeche ka argument isse ek formula banata hai.

DerivationEk depth par pressure

Fluid ka hi ek vertical cylinder socho, jiske flat ends ka area $A$ hai: uska top face level 1 par aur bottom face height $h$ neeche, level 2 par (Fig. 9.4). Uski curved surface par side ke pushes jodiyon mein cancel ho jaate hain. Vertically teen forces lagti hain: top par neeche ki taraf $P_{1} A$ , bottom par upar ki taraf $P_{2} A$ aur andar ke fluid ka weight $m g$ . Cylinder rest mein hai, isliye

(P_{2} - P_{1}) A = m g

(9.5)

Andar ka mass $m = ρ V = ρ A h$ hai. Substitute karke aur $A$ cancel karke,

P_{2} - P_{1} = ρ g h

(9.6)

Ab level 1 ko free surface tak utha do, jahan hawa atmospheric pressure $P_{a}$ se dabati hai, aur depth $h$ par pressure ko simply $P$ kaho:

Depth h par pressure

P = P_{a} + ρ g h

(9.7)

Definition

Gauge pressure

Kisi point par gauge pressure woh amount hai jitna wahan ka pressure atmospheric pressure se zyada hai, $P - P_{a}$ . Hawa ke liye khule liquid mein depth $h$ par yeh $ρ g h$ ke barabar hota hai.

ExamplePaani ke do metre neeche gauge pressure lagbhag $2.0 \times 10^{4} P a$ hota hai.

Dekho ki (9.7) kya chhod deta hai. Area $A$ cancel ho gaya, aur result mein kuch bhi vessel ke baare mein nahi batata. Kisi point par pressure sirf is par depend karta hai ki woh free surface se kitna neeche hai, container ki chaudai ya shape par ya usmein kitna liquid hai is par nahi. Paani ki ek lambi patli pipe aur same height tak bhara ek chauda tank, apne base ke har square centimetre par barabar zor se dabate hain. Is chaunkane wale result ko hydrostatic paradox kehte hain (Fig. 9.4, aakhri step).

Alag-alag shape ke vessels ko bottom se jodo aur paani daalo: sabmein paani ek hi level par aa jaata hai, kyunki sirf ek common level hi jodne wali pipe mein har height ko ek hi pressure deta hai. Mistri yahi use karte hain jab woh paani se bhari transparent plastic pipe se do deewaron par level line ka nishan lagate hain; dono siron par paani ki surface hamesha same height par hoti hai.

4/4Same depth, base par same pressure, vessel ka shape ya volume kuch bhi ho.

Fig. 9.4Pressure depth ke saath badhta hai aur container ki kisi aur cheez par depend nahi karta.NCERT Fig. 9.3 · p. 183

Paani ka ek tank. Step 1: level 1 aur usse height h neeche level 2 ke beech paani ka ek imaginary vertical cylinder outline kiya gaya hai. Step 2: arrows dikhate hain uske top par neeche ki taraf P1A, bottom par upar ki taraf P2A, uska weight mg aur side ke barabar pushes. Step 3: cylinder ka top upar khuli surface tak jaata hai, jahan hawa Pa se dabati hai, jisse P = Pa + ρgh milta hai. Step 4: same level tak bhare teen alag shape ke vessels dikhte hain, har ek ke base par ek jaisa pressure dial hai.

Example 9.2NCERT Example 9.2 · p. 183

Swaraj Dweep ke paas ek dive

Ek scuba diver Andamans mein Swaraj Dweep (Havelock Island) ke paas surface se $15 m$ neeche ek reef explore kar rahi hai. Samudri paani ki density $1025 k g m^{- 3}$ , $g = 9.8 m s^{- 2}$ aur $P_{a} = 1.013 \times 10^{5} P a$ lo. Diver par (a) gauge pressure aur (b) absolute pressure nikalo. (c) Samudri paani ki kitni depth ek atmosphere jodti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Ek bottle se teen dhaarein

Tumhe chahiye

ek khaali 1 L plastic bottle
ek moti sui ya compass ki nok
sticky tape
ek tray ya sink

Pehle guess karo

Paani se bhari bottle mein teen heights par chhed kiye gaye hain. Kis chhed se paani sabse tez nikalta hai?

upar wala chhedbeech wala chhedneeche wala chhedteeno barabar

Kisi bade ki madad se, ek khaali plastic bottle ki side mein ek ke upar ek teen chhote chhed karo, lagbhag 5 cm ki doori par.
Chhedon ko tape ki patti se dhako, bottle ko upar tak bharo aur use tray ya sink ke kinaare khada karo.
Tape ko ek jhatke mein kheencho aur dekho ki har dhaar bottle se nikalte waqt kitni zor se nikalti hai.
Level girte samay dekhte raho, aur note karo ki har dhaar ke saath kya hota hai.

Quick check

3 mein se 1

Fresh water (

ρ = 1000 k g m^{- 3}

) mein kis depth par gauge pressure

1.013 \times 10^{5} P a

ke barabar hota hai?

g = 9.8 m s^{- 2}

lo. Depth metres mein do.

Answer dekho

Answer 10.34 m

h = P / (ρ g) = 1.013 \times 10^{5} / (1000 \times 9.8) = 10.3 m

. Paani se bhare barometer ko 10 m se bhi lambi tube chahiye hogi.

9.2.3 Atmospheric Pressure aur Gauge Pressure

Hum hawa ke ek samudra ke tal par rehte hain. Zameen ka har square metre apne upar jama saari hawa ka weight uthata hai, wahan tak jahan atmosphere khatam hota hai, aur per unit area yahi weight atmospheric pressure hai. Sea level ke paas iski value lagbhag $1.013 \times 10^{5} P a$ hai: har square metre ke upar ki hawa ka weight ek lade hue 10-tonne truck jitna hai. Hum kuchle nahi jaate kyunki hamare body fluids bhi lagbhag utni hi zor se bahar ki taraf push karte hain.

Evangelista Torricelli (1608–1647), jo Galileo ke assistant reh chuke the, ne 1643 mein ise measure karne ka tareeka dhoondha. Unhone lagbhag ek metre lambi, ek sire se band glass tube ko mercury se bhara, khule sire ko dhaka, use mercury ki ek dish mein ulta khada kiya aur phir khol diya. Mercury tab tak neeche gira jab tak uska column dish ke level se lagbhag 76 cm upar na ruk gaya, aur upar lagbhag khaali jagah bach gayi. Yeh instrument mercury barometer hai (Fig. 9.5).

Column wahin kyun rukta hai? Uske upar ki jagah mein sirf thodi si mercury vapour hai, jiska pressure negligible hai. Tube mein dish ki surface ke level wala point aur us khuli surface par ek point, same mercury mein ek hi level par hain, isliye unka pressure same hai, jo khuli surface par $P_{a}$ hai. Pehle point ke upar height $h$ ke column ko yahi pressure dena hoga:

Mercury barometer

P_{a} = ρ g h

(9.8)

Yahan $ρ$ mercury ki density hai. $ρ = 1.36 \times 10^{4} k g m^{- 3}$ aur $h = 0.760 m$ ke saath (9.8) se $1.013 \times 10^{5} P a$ milta hai; paani ke barometer ko 10 m se lambi tube chahiye hogi. Doctors aaj bhi blood pressure millimetres of mercury mein batate hain; mercury ke 1 mm column ka pressure, lagbhag $133 P a$ , Torricelli ke samman mein ek torr kehlata hai. Weather scientists bar ( $10^{5} P a$ ) aur millibar use karte hain; weather charts pressure ko hectopascals mein dikhate hain, aur $1 h P a$ ek millibar hai.

Sea level ka pressure mausam ke saath kuch percent upar-neeche hota hai; barometer ka lagataar girna kisi aate hue low-pressure system ki warning deta hai, jaise Bay of Bengal ka cyclone.

Pressure ka difference measure karne ke liye hum open-tube manometer use karte hain: ek U-tube jo kuch hadd tak liquid se bhari hai, ek arm test ki ja rahi gas se juda hai aur doosra hawa ke liye khula hai (Fig. 9.5, aakhri step). Liquid mein ek level ke points ka pressure same hota hai, isliye gas pressure $P$ , $P_{a}$ se khule arm mein extra liquid column $h$ jitna zyada hai: $P - P_{a} = ρ g h$ . Isliye manometer seedha gauge pressure padhta hai. Chhote differences ke liye paani ya oil jaisa halka liquid padhne layak height deta hai; bade differences ke liye dense mercury tube ko chhota rakhta hai.

2/3Column lagbhag 760 mm tak girta hai; dish par hawa ka push use wahin rokta hai.

Fig. 9.5Atmosphere mercury ka 76 cm column thaame rakhta hai; manometer gauge pressure ko height ke difference ki tarah dikhata hai.NCERT Fig. 9.5 · p. 183

Left mein, upar se band ek glass tube mercury ki dish mein ulti khadi hai. Step 1: tube mercury se bhari hai. Step 2: column dish level se 760 mm upar tak gir gaya hai, uske upar near-vacuum hai aur arrows dikhate hain ki hawa dish par dabati hai. Step 3: barometer fade hota hai aur right mein ek U-tube manometer dikhta hai, jiska ek arm gas bulb se juda hai aur doosra khula hai, aur khule arm mein liquid h jitna upar hai.

Example 9.3NCERT Example 9.3 · p. 184

Constant density wala atmosphere

Ek normal din par sea level ki hawa ki density lagbhag $1.20 k g m^{- 3}$ hoti hai. Maan lo, bhale hi yeh realistic nahi hai, ki density upar tak yahi rehti. $1.013 \times 10^{5} P a$ ke sea-level pressure ke liye atmosphere kitni oonchai tak jaata? Apne answer ko Mount Everest ki height, $8849 m$ , se compare karo. $g = 9.8 m s^{- 2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9.4NCERT Example 9.4 · p. 184

Chhe kilometre neeche

India ka Deep Ocean Mission Matsya 6000 bana raha hai, ek submersible jo teen logon ko $6000 m$ ki depth tak le jaane ke liye design kiya gaya hai. Us depth par, samudri paani ki density constant $1030 k g m^{- 3}$ , $g = 9.8 m s^{- 2}$ aur $P_{a} = 1.013 \times 10^{5} P a$ lekar, (a) gauge pressure, (b) absolute pressure aur (c) $10 c m$ diameter ki ek circular window par net force nikalo, agar cabin 1 atm par rakha gaya hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

120 mm of mercury ki blood-pressure reading kitne kilopascals ka gauge pressure hai? (

ρ_{Hg} = 13 600 k g m^{- 3}

g = 9.8 m s^{- 2}

)

kPa

Answer dekho

Answer 16 kPa

P = ρ g h = (13 600) (9.8) (0.120) = 1.60 \times 10^{4} P a = 16.0 k P a

, atmospheric pressure se zyada.

9.2.4 Hydraulic Machines

Band toothpaste tube ke bottom ko dabao to nozzle par paste aa jaata hai; paani ke balloon ko chhuo to woh har jagah ek saath phool jaata hai. Band fluid ke ek hisse par push poore fluid mein mehsoos hota hai. Ise ek plastic bottle se try karo jo paani se upar tak bhari ho, dhakkan band ho, aur kai jagah pinholes hon: use halke se dabao aur har dhaar ek hi pal mein tez ho jaati hai, chhed kisi bhi taraf ho. Tumhare haath ka extra pressure paani ke har point tak barabar pahunch gaya hai.

LawPascal's law (pressure ka transmission)

Rest mein kisi band fluid par lagaya gaya pressure ka koi bhi change fluid ke har hisse tak, aur use rakhne wali deewaron tak, bina kisi loss ke pahunchta hai.

Hydraulic lift isi ka use karta hai (Fig. 9.6). Do pistons, ek chhota area $A_{1}$ ka aur ek bada area $A_{2}$ ka, ek seal kiye hue, liquid se bhare system ke dono ends ko band karte hain. Chhote piston par force $F_{1}$ ek pressure $F_{1} / A_{1}$ jodti hai, jo bade piston tak bina badle pahunchta hai aur use is force se upar push karta hai

Hydraulic lift

F_{2} = \frac{F_{1}}{A_{1}} A_{2} = \frac{A_{2}}{A_{1}} F_{1}

Force $A_{2} / A_{1}$ guna ho jaati hai, jise mechanical advantage kehte hain. Lekin energy create nahi hoti. Liquid practically incompressible hai, isliye chhote cylinder se bahar dhakela gaya volume, $A_{1} d_{1}$ , bade piston ke neeche jude volume, $A_{2} d_{2}$ , ke barabar hai. Bada piston usi factor se kam distance chalta hai, aur $F_{1} d_{1} = F_{2} d_{2}$ : frictionless pistons ke liye, daala gaya work nikle hue work ke barabar hai.

2/3Extra pressure $F_{1} / A_{1}$ har point tak pahunchta hai aur bade piston ko $F_{2} = (A_{2} / A_{1}) F_{1}$ se uthata hai.

Fig. 9.6Chhote piston par chhoti force bada load uthati hai, jo chhote piston ke girne se kam upar uthta hai.NCERT Fig. 9.6 · p. 185

Ek patla aur ek chauda cylinder bottom par liquid se bhare channel se jude hain; chauda piston car uthaye hue hai. Step 1: kuch push nahi hota. Step 2: patle piston par ek force dabati hai, arrows dikhate hain ki pressure liquid mein phailta hai, aur chaude piston par ek badi upward force lagti hai. Step 3: patla piston kaafi neeche chala gaya hai aur car thodi upar uthi hai; dashed bands dikhate hain ki patle piston ke neeche sweep hua volume chaude piston ke neeche judi patli layer ke barabar hai.

Car ke brakes bhi isi idea par chalte hain. Har wheel par ek wheel cylinder ka piston brake pads ya shoes ko ghoomte disc ya drum par dabata hai. Chaaron wheel cylinders brake fluid se bhari patli pipes se pedal ke neeche ek single master cylinder se jude hote hain, jiska piston kahin chhota hota hai. Isliye pedal par thoda sa push har wheel par badi force ban jaata hai, aur kyunki fluid poore system mein ek hi pressure rakhta hai, chaaron wheels par barabar brake lagta hai, jisse car side mein nahi phisalti. Excavator ki arms, barber ki adjustable chair aur mechanic ka bottle jack doosri hydraulic machines hain.

Example 9.5NCERT Example 9.5 · p. 185

Syringe se chalne wala lifting arm

Science fair ke liye Meera paani se bhari tube se jude do syringes se ek lifting arm banati hai. Chhote syringe ke plunger ka diameter $1.2 c m$ hai aur bade wale ka $3.0 c m$ . (a) Jab woh chhote plunger ko $8.0 N$ se push karti hai, to bada plunger kitni upward force lagata hai? (b) Agar woh chhote plunger ko $5.0 c m$ andar push kare, to bada plunger kitna bahar aata hai? Friction ko ignore karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9.6NCERT Example 9.6 · p. 186

Service station par car uthana

Ek car service station par compressed air radius $4.0 c m$ ke ek chhote piston par push karti hai, aur oil yeh pressure radius $20 c m$ ke ek bade piston tak pahunchata hai jis par $1100 k g$ mass ki car hai. (a) Car ko upar thaame rakhne ke liye hawa ko chhote piston par kitni force lagani hogi? (b) Iske liye hawa ka kitna gauge pressure chahiye? $g = 9.8 m s^{- 2}$ lo aur pistons ke weights ignore karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Ek hydraulic press ke pistons ke diameters

2.0 c m

aur

20 c m

hain. Chhote piston par

50 N

lagane par bada piston kitni force, newtons mein, lagata hai?

Answer dekho

Answer 5000 N

Area ratio

(20 / 2.0)^{2} = 100

hai, isliye

F_{2} = 100 \times 50 N = 5000 N

. Area ratio ki jagah diameter ratio (10) use karna aam galti hai.

9.3 Streamline Flow

Ek shaant kamre mein agarbatti jalao aur uska dhuaan dekho. Pehle kuch centimetre tak woh ek smooth, bina toote dhaage ki tarah upar uthta hai; aur upar jaakar dhaaga lehraata hai aur chhallon aur gubbaron mein toot jaata hai. Wahi hawa do bilkul alag tarah se behave karti hai. Chalte fluids ki padhai, fluid dynamics, smooth wale flow ko theek-theek describe karne se shuru hoti hai.

Flow describe karne ke liye hum alag-alag particles ki jagah fixed jagahon ko dekh sakte hain. Socho ki canal mein ek jagah par camera fixed hai, jo us jagah se guzarne wale paani ki velocity record kar raha hai. Agar reading kabhi na badle, us jagah par aur har doosri jagah par, to flow steady hai. Steady ka matlab uniform nahi hai: paani chaude hisse ke mukaable patle hisse mein tez beh sakta hai, isliye koi particle chalte-chalte tez ho sakta hai. Jo fixed rehta hai woh hai har jagah par velocity.

Definition

Streamline

Streamline woh path hai jis par steady flow mein fluid particles chalte hain; kisi bhi point par iski tangent wahan fluid ki velocity ki direction batati hai.

ExampleAgarbatti ke theek upar dhuen ka smooth dhaaga streamlines dikhata hai.

Steady flow mein streamlines ka pattern motion ka ek fixed map hai. Do streamlines kabhi cross nahi kar sakti: crossing par particle ki do possible velocities hoti, aur flow steady nahi ho sakta tha. Paas-paas ki streamlines ka ek bundle tube of flow banata hai; fluid ek end se usmein aata hai aur doosre se nikalta hai, lekin uski sides kabhi cross nahi karta.

DerivationEquation of continuity

Ek tube of flow, ya pipe, lo jiska cross-section $A_{1}$ se $A_{2}$ ho jaata hai, aur wahan speeds $v_{1}$ aur $v_{2}$ hain (Fig. 9.7). Ek chhote time $Δ t$ mein section 1 paar karne wala fluid lambai $v_{1} Δ t$ aur mass $ρ_{1} A_{1} v_{1} Δ t$ ka ek slab bharta hai. Steady flow mein sections ke beech koi fluid jama nahi hota, isliye utna hi mass utne hi time mein section 2 paar karta hai. $Δ t$ cancel karne par,

ρ_{1} A_{1} v_{1} = ρ_{2} A_{2} v_{2}

(9.9)

Liquid practically incompressible hai, isliye $ρ_{1} = ρ_{2}$ aur

Equation of continuity

A_{1} v_{1} = A_{2} v_{2}

(9.10)

Dono sections tube ke along koi bhi do the, isliye

A v = constant

(9.11)

Product $A v$ volume flow rate hai, yaani har second guzarne wala volume, $m^{3} s^{- 1}$ mein. Equation of continuity kehti hai ki yeh poori pipe mein same rehta hai; yeh incompressible fluid ke liye conservation of mass ka law hai. Jahan pipe patli hoti hai wahan fluid ko tez hona padta hai, aur jahan chaudi hoti hai wahan fluid dheema ho jaata hai. Streamline picture mein ghani lines ka matlab tez flow hai. Garden hose ke muh ko anguthe se thoda band karna yahi use karta hai: agar khula area paanchva hissa reh jaaye aur flow rate same rahe, to paani paanch guna tez nikalta hai aur kaafi door tak jaata hai.

Steady flow sirf kaafi kam speeds par hi tikta hai. Ek critical speed se upar, jo fluid, pipe aur kisi rukawat par depend karti hai, flow turbulent (विक्षुब्ध) ho jaata hai: kisi point par velocity be-tarteeb badalti hai aur fluid kai sizes ke eddies mein mathne lagta hai. Tootta hua agarbatti ka dhuaan, Rishikesh ke paas Ganga ke jhaag bhare rapids aur tez bhaagti bus ke peeche ghoomta wake, yeh sab turbulent hain. Smooth flow jismein paas ki layers bina mile ek doosre ke paas se slide karti hain, laminar flow kehlata hai.

Compare

Streamline (laminar) flow aur turbulent flow

Streamline (laminar)

Har point par velocity time ke saath steady
Streamlines kabhi cross nahi karti
Layers bina mile slide karti hain
Dheema flow, patli pipes, gaadhe liquids

Turbulent

Har point par velocity badalti rehti hai
Eddies; koi fixed streamline pattern nahi
Layers zor se mix hoti hain
Tez flow aur rukawaton ke paas flow

2/3Time $Δ t$ mein dono sections se same volume guzarta hai: $A_{1} v_{1} Δ t = A_{2} v_{2} Δ t$ .

Fig. 9.7Jahan streamlines paas-paas hoti hain wahan fluid tez chalta hai; har second har section se same volume guzarta hai.NCERT Fig. 9.7 · p. 187

Ek horizontal pipe left ke chaude section se right ke patle section tak patli hoti hai. Pehle step mein paanch streamlines pipe ke along chalti hain aur patle hisse mein paas aa jaati hain, chaude hisse mein chhota velocity arrow aur patle hisse mein lamba arrow hai. Doosre step mein do shaded slabs us fluid ko dikhate hain jo same chhote time mein har section paar karta hai: left mein lambai v1Δt ka chhota, chauda slab aur right mein lambai v2Δt ka lamba, patla slab, dono mein same volume. Teesre step mein patle hisse ki streamlines ki jagah ghoomte eddies aa jaate hain, jo turbulent flow dikhate hain.

Quick check

3 mein se 1

Paani

12 c m^{2}

cross-section ki pipe mein

0.60 m s^{- 1}

se behta hai, jo phir patli hokar

3.0 c m^{2}

ki ho jaati hai. Patle hisse mein speed kitni hai, m/s mein?

m/s

Answer dekho

Answer 2.4 m/s

v_{2} = v_{1} A_{1} / A_{2} = 0.60 \times 12 / 3.0 = 2.4 m s^{- 1}

. Ek chauthai area matlab chaar guna speed.

9.4 Bernoulli's Principle

Pipe ke patle hisse mein ghusta paani tez ho jaata hai, isliye koi cheez use aage push kar rahi hogi. Yeh sirf aas-paas ka fluid hi kar sakta hai, ek aise pressure se jo paani ke peeche aage se zyada ho. Isliye tez flow ke saath kam pressure hona chahiye. Apni book Hydrodynamica (1738) mein Swiss physicist Daniel Bernoulli ne chalte fluid par energy conservation apply karke is idea ko ek exact relation bana diya.

DerivationWork–energy theorem se Bernoulli's equation

Ek ideal fluid lo: incompressible, viscosity ke bina aur ek aisi pipe mein steady flow mein jiska cross-section aur height dono badalte hain (Fig. 9.8). Neeche wale section 1 par area $A_{1}$ , speed $v_{1}$ , pressure $P_{1}$ aur height $h_{1}$ hai; section 2 par yeh $A_{2}$ , $v_{2}$ , $P_{2}$ aur $h_{2}$ hain. Cross-sections K aur M ke beech ke fluid ko follow karo. Ek chhote time $Δ t$ mein uska peeche ka face K se L tak, $v_{1} Δ t$ distance, aur aage ka face M se N tak, $v_{2} Δ t$ distance, chalta hai. Continuity se dono patle tukdon ke volumes barabar hain, $Δ V = A_{1} v_{1} Δ t = A_{2} v_{2} Δ t$ .

Peeche ka fluid peeche wale face ko $v_{1} Δ t$ aage push karta hai, aur work $P_{1} A_{1} v_{1} Δ t = P_{1} Δ V$ karta hai. Aage ka fluid aage wale face par peeche ki taraf push karta hai jabki woh $v_{2} Δ t$ aage chalta hai, aur work $- P_{2} Δ V$ karta hai. Hamare fluid par net work hai

W = (P_{1} - P_{2}) Δ V

Steady flow mein L se M tak ka hissa pehle aur baad mein same dikhta hai, isliye net change aisa hai jaise mass $Δ m = ρ Δ V$ ka patla tukda section 1 se section 2 tak le jaaya gaya ho. Uski kinetic aur potential energies itni badalti hain

\begin{aligned} Δ K & = \frac{1}{2} Δ m (v_{2}^{2} - v_{1}^{2}) \\ Δ U & = Δ m g (h_{2} - h_{1}) \end{aligned}

Work–energy theorem (Chapter 5) se, $W = Δ K + Δ U$ . Poore ko $Δ V$ se divide karke, $Δ m / Δ V = ρ$ ke saath,

\begin{aligned} P_{1} - P_{2} \\ = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) \\ + ρ g (h_{2} - h_{1}) \end{aligned}

aur har section ke terms ko uski apni side par ikattha karke,

\begin{aligned} P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} \\ = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2} \end{aligned}

(9.12)

Sections 1 aur 2 flow ke along koi bhi do jagah the, isliye

Bernoulli's equation

P + \frac{1}{2} ρ v^{2} + ρ g h = const.

(9.13)

3/3Pressure forces net work $(P_{1} - P_{2}) Δ V$ karti hain, jo tukde ki kinetic aur potential energy badhata hai.

Fig. 9.8Flow ka net effect hai volume ΔV ke ek patle tukde ko neeche ke chaude section se upar ke patle section tak le jaana.NCERT Fig. 9.9 · p. 188

Ek pipe left mein neeche ke chaude section se right mein upar ke patle section tak uthti hai. Step 1: cross-sections K aur M ke beech ka fluid shaded hai, aur K ke paas ek patla tukda doosre rang mein hai. Step 2: fluid ab L aur N ke beech hai; left wala tukda chala gaya hai aur M ke aage ek barabar volume ka, lamba, patla tukda aa gaya hai. Step 3: arrows dikhate hain ki left end par P1A1 aage push karta hai aur right end par P2A2 peeche push karta hai, saath mein reference line se naapi heights h1 aur h2.

PrincipleBernoulli's principle

Incompressible, non-viscous fluid ke steady flow mein, quantity $P + \frac{1}{2} ρ v^{2} + ρ g h$ , yaani pressure plus per unit volume kinetic energy plus per unit volume potential energy, kisi bhi ek streamline ke along ek hi value rakhti hai.

Bernoulli's equation ka har term $J m^{- 3}$ mein measure hota hai, jo pascal ke barabar hai, isliye pressure ko khud per unit volume energy ka store bhi samjha ja sakta hai. Har jagah $v = 0$ rakho to (9.13) wapas $P_{1} - P_{2} = ρ g (h_{2} - h_{1})$ deta hai, yaani hydrostatic result (9.6): fluid statics ek special case hai.

Equation utni hi achhi hai jitne uske assumptions. Ise steady flow chahiye: turbulent flow mein kisi point par speed aur pressure badalte rehte hain aur yeh fail ho jaati hai. Yeh fluid ko incompressible maanti hai, dabane se store hui energy ko ignore karke, isliye gas tabhi chalti hai jab uski density mushkil se badle. Aur yeh no viscosity maanti hai (Section 9.5); real fluid mein kuch mechanical energy heat ban jaati hai, aur aage ka pressure (9.12) ke prediction se kam aata hai. Phir bhi paani aur kam speed wali hawa ke liye yeh kamaal ki useful hai.

ExampleNCERT p. 188

Ek Venturi meter

$6.0 c m$ diameter ka ek horizontal water main smoothly patla hokar $3.0 c m$ diameter ka throat banta hai aur phir chauda hota hai. Pressure taps dikhate hain ki throat mein pressure main pipe se $4.5 k P a$ kam hai. Main pipe mein speed aur flow rate nikalo. Paani ko density $1000 k g m^{- 3}$ ka ideal fluid maano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Scent atomiser ya paint spray gun liquid mein dubi ek patli tube ke top ke upar se tezi se hawa phenkta hai; wahan pressure gir jaata hai, isliye liquid upar dhakela jaata hai aur boondon mein toot jaata hai. Bunsen burner bhi isi wajah se apne base par hawa kheenchta hai.

InteractiveVenturi tube lab

Ek horizontal pipe jiske beech mein patla throat hai aur do vertical manometer tubes hain, ek chaude hisse ke upar aur ek throat ke upar. Sliders chaude hisse ki speed aur pipe ke fraction ki tarah throat ka area set karte hain. Readouts continuity se throat speed, Bernoulli's equation se pressure drop aur dono column heights dete hain; throat patla hone par uske upar ka column girta hai, aur agar throat pressure atmospheric se neeche jaaye to ek warning dikhti hai. Play pipe mein tracer dots bhejta hai, jo throat mein tez chalte hain.

Fig. 9.9Flow speed aur throat ka size set karo, phir Play dabao. Speeds, pressures aur dono manometer columns ko compare karo.

Try karke dekho

Do balloons ke beech phoonko

Tumhe chahiye

do phule hue balloons
dhaaga aur sticky tape
ek drinking straw
do khaali cans aur kuch aur straws

Pehle guess karo

Do balloons kuch centimetre ki doori par saath-saath latke hain. Tum unke beech ke gap mein hawa ki dhaar phoonkte ho. Woh kya karte hain?

door hat jaate hainek doosre ki taraf aate hainjahan hain wahin rehte hain

Do phule hue balloons ko barabar lambai ke dhaagon se kisi shelf ya ruler se latkao, lagbhag 5 cm ki doori par.
Straw se unke beech ke gap mein hawa ki ek steady dhaar phoonko, straw ko seedha rakhte hue.
Aur zor se phoonko aur phir dekho. Phir do khaali soft-drink cans ko drinking straws ki ek line par 3 cm ki doori par lita do, aur unke beech phoonko.

9.4.1 Efflux ki Speed: Torricelli's Law

Efflux ka matlab hai bahar nikalna. Torricelli ne notice kiya ki tank ke chhed se nikalta liquid utni speed se nikalta hai jitni use liquid ki surface ke level se chhed tak free fall karne par milti. Bernoulli's equation dikhati hai ki aisa kyun hai.

DerivationEfflux ki speed

Density $ρ$ ke liquid ke ek tank ki side mein base se height $y_{1}$ par area $A_{1}$ ka ek chhota chhed hai, aur area $A_{2}$ ki free surface height $y_{2}$ par hai (Fig. 9.10). Surface ke upar gas pressure $P$ par hai, aur dhaar $P_{a}$ par atmosphere mein nikalti hai. Continuity se $v_{2} = (A_{1} / A_{2}) v_{1}$ milta hai. Jab tank chhed se bahut chauda ho ( $A_{2} ≫ A_{1}$ ), to surface itni dheere girti hai ki hum $v_{2} \approx 0$ rakh sakte hain. Chhed (1) aur surface (2) ke beech (9.12) apply karne par,

P_{a} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = P + ρ g y_{2}

$h = y_{2} - y_{1}$ ke saath, jo surface se chhed ki depth hai,

v_{1} = \sqrt{2 g h + \frac{2 (P - P_{a})}{ρ}}

(9.14)

Agar tank hawa ke liye khula hai, to $P = P_{a}$ aur

Torricelli's law

v_{1} = \sqrt{2 g h}

(9.15)

Yeh bilkul utni hi speed hai jitni rest se height $h$ tak giraye gaye patthar ki, aur wajah bhi wahi hai: per unit volume potential energy $ρ g h$ kinetic energy $\frac{1}{2} ρ v^{2}$ ban jaati hai. Paani ka ek drum jiska tap surface se $0.80 m$ neeche hai, paani $\sqrt{2 (9.8) (0.80)} \approx 4.0 m s^{- 1}$ par deta hai. Doosri extreme mein, jab liquid ke upar gas ka pressure atmospheric se bahut zyada ho, to $2 g h$ term ka mushkil se farq padta hai aur sirf container pressure hi dhaar ki speed tay karta hai. Fire extinguishers, spray cans aur, zyada complicated tareeke se, rocket engines isi regime mein kaam karte hain.

2/3Dhaar $v_{1} = \sqrt{2 g h}$ se nikalti hai, yaani $h$ tak giraye patthar ki speed.

Fig. 9.10Depth h par chhed se paani utni speed se nikalta hai jitni h tak free fall kiye body ki hoti hai.NCERT Fig. 9.10 · p. 189

Paani ka ek tank legs par khada hai, jiski side mein paani ki surface se depth h neeche ek chhota chhed hai. Pehle step mein chhed par 1 aur surface par 2 likha hai, aur unke beech h dikhaya gaya hai. Doosre step mein chhed se ek dhaar mudti hui nikalti hai aur, tank ke bagal mein, ek patthar paani ki surface ke level se chhed ke level tak girta hai aur dhaar jitni hi speed tak pahunchta hai. Teesre step mein ek dhakkan paani ke upar pressure P par gas band karta hai, aur ab dhaar tank se kaafi door girti hai.

Quick check

3 mein se 1

Ek bade khule tank mein paani ki surface se

1.25 m

neeche ek chhota chhed hai. Paani chhed se kis speed se, m/s mein, nikalta hai? (

g = 9.8 m s^{- 2}

)

m/s

Answer dekho

Answer 4.95 m/s

v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 1.25} = 4.95 m s^{- 1}

, yaani 1.25 m tak giraye patthar ki speed.

9.4.2 Dynamic Lift

Kisi chalti body ke paas se behta fluid use side mein, uski motion ke right angle par, push kar sakta hai. Yahi dynamic lift udte eagle ko bina zyada pankh maare hawa mein rakhta hai, aircraft ko thaame rakhta hai aur ghoomti ball ka raasta mod deta hai. Bernoulli's principle iska kam se kam ek hissa samjhata hai.

Ek aisi ball se shuru karo jo sirf aage chalti hai. Ball ki nazar se dekho to hawa uske paas se behti hai, aur ball ke upar streamlines ka pattern neeche wale pattern ki mirror image hai (Fig. 9.11). Barabar speeds matlab barabar pressures, isliye hawa na upar push karti hai na neeche; woh sirf ball ko peeche kheenchti hai.

Spin symmetry tod deta hai. Ghoomti ball apne saath hawa ki ek patli skin ghumati hai. Backspin mein yeh skin ball ke upar saamne se aati hawa ke saath chalti hai aur neeche uske khilaaf, isliye upar hawa tez ho jaati hai aur neeche dheemi; streamlines upar paas aa jaati hain aur neeche phail jaati hain. Phir Bernoulli's principle neeche ke mukaable upar kam pressure deta hai, aur ball upar push hoti hai. Spin se hone wale lift ko Magnus effect kehte hain. Spin bowler ka drift aur footballer ki mudti free kick ise dikhate hain. (Fast bowler ki swing ki wajah alag hai: seam ki wajah se hawa dono sides par alag tarah chipakti hai.)

Aircraft wing ka ek special cross-section hota hai, aerofoil: aage se gol, upar se curved aur peeche ek nukile kinaare tak patla hota hua. Saamne se aati hawa mein thoda upar jhuka hone par, yeh apni upar wali surface par hawa ko neeche wali hawa se tez behata hai. Upar kam pressure se upward lift milta hai jo level flight mein aircraft ke weight ko balance karta hai. Doosri tarah dekho to wing hawa ko neeche mod deta hai aur, Newton's third law se, hawa wing ko upar push karti hai; dono pictures ek hi flow ko describe karti hain.

3/3Wing apne upar ki hawa ko neeche ki hawa se tez behata hai, isliye pressure difference use uthata hai.

Fig. 9.11Ek side tez hawa matlab wahan kam pressure aur us side ki taraf lift force.NCERT Fig. 9.11 · p. 190

Hawa right se left behti hai. Step 1: bina spin ki ball streamlines ke beech hai jo uske upar aur neeche symmetric tarah mudti hain; koi lift nahi hai. Step 2: ball mein backspin hai, jo ek curved arrow se dikhaya gaya hai; streamlines uske upar paas aa jaati hain aur neeche phail jaati hain, aur ek arrow upward lift dikhata hai. Step 3: right mein ek aircraft wing dikhta hai, uske curved top par streamlines ghani hain, saath mein lift arrow aur labels: upar tez, low-pressure hawa aur neeche dheemi, higher-pressure hawa.

Example 9.7NCERT Example 9.7 · p. 190

Cargo aircraft ko thaame rakhna

$1.5 \times 10^{5} k g$ mass ke ek cargo aircraft ka total wing area $600 m^{2}$ hai. Yeh $720 k m h^{- 1}$ par level udta hai, us height par jahan hawa ki density $0.90 k g m^{- 3}$ hai. (a) Estimate karo ki wings ke neeche average pressure upar se kitna zyada hai. (b) Estimate karo ki wings ke upar ki hawa ko neeche ki hawa se kitne fraction zyada tez chalna hoga. $g = 9.8 m s^{- 2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Ek table-tennis ball ko backspin ke saath right ki taraf maara jaata hai, taaki uski top surface left ki taraf chale. Hawa use push karti hai

upar ki tarafneeche ki tarafright ki taraf, use tez karte huekahin nahi: spin ka koi effect nahi

Answer dekho

Answer (A) upar ki taraf

Ball ke relative hawa left ki taraf behti hai. Backspin top ke upar hawa ko saath le jaata hai, isliye upar ki hawa tez chalti hai aur uska pressure kam hota hai: Magnus force upar point karti hai, aur ball zyada der hawa mein rehti hai.

9.5 Viscosity

Shahad ke jar aur paani ke glass ko same chammach se same speed par hilao: shahad kahin zyada virodh karta hai. Real fluids ek layer ke doosri par slide hone ka virodh karte hain, jo ek tarah ka internal friction hai aur use viscosity (श्यानता) kehte hain. Yeh sirf tab lagta hai jab layers ek doosre ke relative chalti hain; rest mein fluid par yeh nahi lagta.

Viscosity dekhne ka sabse saaf tareeka hai liquid ki ek patli film ko steadily shear karna. Do flat plates ke beech, jo ek chhoti distance $l$ par hain, oil phailao, ek plate ko fixed rakho aur doosri ko uske parallel constant speed $v$ se slide karo (Fig. 9.12). Liquid dono plates se chipakta hai (har plate ko chhoota oil uske saath chalta hai), isliye gap ke aar-paar velocity zero se $v$ tak ek seedhi line mein badhti hai. Plate ko chalate rehne ke liye ek steady force $F$ chahiye, kyunki har layer apne padosiyon ko kheenchti hai. Oil ki jagah shahad lo to same $v$ ke liye $F$ bahut badh jaata hai: shahad zyada viscous hai. Is tarah ka smooth, layered flow laminar hai.

Yahi chipakna pipe mein flow ko shape deta hai. Wall par liquid ruka rehta hai, isliye velocity profile ek gol curve hai, wall par zero aur axis par sabse zyada: liquid ek ke andar ek cylindrical sleeves mein chalta hai, har ek agle ke andar slide karta hai (Fig. 9.12 ka aakhri step).

Steady shear stress mein solid ek fixed strain le leta hai. Fluid aisa nahi kar sakta: liquid ka ek chhota block jo shuru mein rectangular hai, shear hokar ek parallelogram ban jaata hai jo tab tak aur jhukta jaata hai jab tak plate chalti hai. Time $Δ t$ mein uska top bottom ke relative $Δ x = v Δ t$ khisak jaata hai, isliye shear strain $Δ x / l$ badhta rehta hai. Paani, oils aur kai doosre liquids ke liye experiment dikhata hai ki area $A$ ki plate par stress $F / A$ strain ke nahi balki uske rate, $Δ x / (l Δ t) = v / l$ , ke proportional hai. Proportionality ka constant viscosity define karta hai:

1/3Speed fixed plate par zero se moving plate par $v$ tak barabar badhti hai.

Fig. 9.12Laminar flow mein layers ek doosre par slide karti hain; fixed wall par speed zero hoti hai.NCERT Fig. 9.12 · p. 191

Left mein, neeche ki fixed plate aur right ki taraf chalti upar ki plate ke beech oil. Step 1: oil mein arrows neeche se upar barabar lambe hote jaate hain, jo dikhate hain ki speed zero se v tak barabar badhti hai. Step 2: oil ka ek chhota rectangular block jhuk kar parallelogram ban jaata hai, thickness l ke aar-paar uska top Δx khisak jaata hai. Step 3: plates fade hoti hain aur ek pipe dikhti hai, jismein speed arrows axis par sabse lambe aur walls par zero hain, unki nokein ek curved profile par hain.

Coefficient of viscosity

η = \frac{F / A}{v / l} = \frac{F l}{v A}

(9.16)

Definition

Coefficient of viscosity

Kisi fluid ka coefficient of viscosity $η$ (eta) usmein shear stress ko rate of shear strain se divide karne par milta hai. Iski SI unit pascal second hai, $P a s = N s m^{- 2}$ , jise poiseuille (Pl) bhi kehte hain.

Example20 °C par paani ka $η \approx 1.0 \times 10^{- 3} P a s$ hai; glycerine ek hazaar guna se bhi zyada viscous hai.

Viscosity ki dimensions $[M L^{- 1} T^{- 1}]$ hain. Patle liquids jaise paani, petrol aur alcohol ki viscosity kam hoti hai; glycerine, shahad, engine oil aur tar ki viscosity zyada hoti hai. Blood, kuch millipascal seconds ke saath, paani se gaadha hai. Table 9.2 kuch values deta hai.

Table 9.2Kuch fluids ki viscosities (approximate; 1 mPa s = 0.001 Pa s)NCERT Table 9.2 · p. 192

Fluid	Temperature (°C)	Viscosity (mPa s)
Paani	0	1.79
Paani	20	1.00
Paani	40	0.65
Paani	100	0.28
Castor oil	20	≈ 990
Glycerine	20	≈ 1400
Hawa	20	0.018
Hawa	100	0.022

Temperature ka bahut asar hota hai. Liquid ko garam karo to uske molecules apne padosiyon se aasani se chhoot jaate hain, isliye uski viscosity girti hai: garam ghee aasani se behta hai jabki thanda ghee mushkil se hilta hai, aur 100 °C par paani 20 °C ke mukaable teen se chaar guna kam viscous hota hai. Gas mein trend ulta ho jaata hai. Gas molecules ek doosre ko mushkil se attract karte hain; gas ki viscosity molecules ke layers ke beech bhaagne aur apne saath momentum le jaane se aati hai. Garam karne se woh tez hote hain, isliye garam gas zyada viscous ho jaati hai.

Example 9.8NCERT Example 9.8 · p. 191

Oil film par slide karti plate

Area $0.040 m^{2}$ ki ek flat steel plate ek level table par phaili $0.40 m m$ moti oil film par rakhi hai. Plate se bandhi ek halki string table ke kinaare ek ideal pulley ke upar se $50 g$ ke latakte mass tak jaati hai. Chhodne ke baad plate ek steady $2.0 c m s^{- 1}$ se slide karti hai. Oil ka coefficient of viscosity nikalo. ( $g = 9.8 m s^{- 2}$ )

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.5.1 Stokes' Law

Shampoo ki bottle mein doobti marble, shaant hawa mein baithta dhool ka kan aur aquarium ke paani mein dheere-dheere neeche jaata food pellet, sab apne paas ke fluid ko saath kheenchte hain, isliye fluid unhe rokta hai. Dheere chalte chhote sphere ke liye George Gabriel Stokes (1819–1903) ne 1851 mein dikhaya ki yeh drag sphere ki speed, uske radius aur fluid ki viscosity ke proportional hota hai:

Stokes' law

F = 6 π η a v

(9.17)

jahan

$a$: sphere ka radius
$v$: fluid ke relative uski speed

Hum ise derive nahi karenge, lekin iski dimensions sahi baithti hain: $[η] [a] [v] = [M L^{- 1} T^{- 1}] [L] [L T^{- 1}] = [M L T^{- 2}]$ , yaani ek force. Yeh law tab tak chalta hai jab tak sphere ke aas-paas flow smooth rahe, matlab chhote, dheeme spheres ya bahut viscous fluids. Chapter 5 mein tumne resistive forces dwara kiya gaya work calculate kiya tha; Stokes' law batata hai ki chhote sphere ke liye aisi force kya hoti hai.

DerivationTerminal velocity

Radius $a$ aur density $ρ$ ka ek sphere density $σ$ ke fluid mein girao. Teen forces lagti hain: neeche ki taraf uska weight $\frac{4}{3} π a^{3} ρ g$ , upar ki taraf buoyant force $\frac{4}{3} π a^{3} σ g$ aur upar ki taraf viscous drag $6 π η a v$ . Shuru mein $v = 0$ hai aur sphere accelerate karta hai. Speed badhne ke saath drag badhta hai, jab tak drag plus buoyancy weight ke barabar na ho jaaye. Uske baad net force zero hai aur sphere ek constant terminal velocity $v_{t}$ se girta hai:

6 π η a v_{t} = \frac{4}{3} π a^{3} (ρ - σ) g

Terminal velocity

v_{t} = \frac{2 a^{2} (ρ - σ) g}{9 η}

(9.18)

Terminal velocity radius ke square ke saath badhti hai aur viscosity badhne par ghatti hai. $10 μ m$ ki cloud droplet sirf lagbhag $3 m m s^{- 1}$ se baithti hai, isliye sabse halka updraft bhi baadalon ko tairta rakhta hai. Badi baarish ki boondein kai metre per second se girti hain; yeh Stokes' law ke liye bahut tez hai, lekin woh bhi ek steady terminal speed tak pahunchti hain. Agar $ρ < σ$ ho, to (9.18) negative $v_{t}$ deta hai: sphere steady speed se upar uthta hai, jaise shahad ki bottle mein hawa ka bubble.

Example 9.9NCERT Example 9.9 · p. 192

Glycerine mein ball ka time naapna

Ek falling-ball experiment mein radius $1.5 m m$ ki steel ball bearing ko 20 °C par glycerine ke ek lambe jar mein giraya jaata hai. Steady chalne lagne ke baad, use $20.0 c m$ door do nishaanon ke beech girne mein $8.8 s$ lagte hain. Glycerine ki viscosity nikalo. Densities: steel $7.8 \times 10^{3} k g m^{- 3}$ , glycerine $1.26 \times 10^{3} k g m^{- 3}$ ; $g = 9.8 m s^{- 2}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveFalling-ball lab

Liquid ka ek lamba jar, tumhari chuni hui ball (steel, glass ya nylon) aur ek radius slider; liquid glycerine ya castor oil ho sakta hai. Play ball ko girata hai aur pehle kuch milliseconds ke graph par uski speed ko time ke against plot karta hai, jahan speed badhti hai aur terminal value par tik jaati hai. Readouts Stokes' law se terminal velocity, 20 cm girne ka time, aur terminal speed par weight, buoyancy aur drag dete hain. Glycerine mein nylon ball doobne ki jagah upar uthti hai, aur jab ball Stokes' law ke accurate rehne ke liye bahut tez chale to ek note warning deta hai.

Fig. 9.13Ek ball aur ek liquid chuno, radius set karo aur Play dabao. Dekho ki speed terminal velocity par jaakar kaise tik jaati hai.

Quick check

3 mein se 1

Coefficient of viscosity ki SI unit kaun si hai?

N m^{- 1}

P a s^{- 1}

N m^{- 2}

P a s

Answer dekho

Answer (D) $P a s$

η

stress divided by strain rate hai: pascal divided by

s^{- 1}

, jo

P a s

hai.

N m^{- 1}

surface tension ki unit hai aur

N m^{- 2}

khud pascal hai.

9.6 Surface Tension

Pond skater talaab par bina doobe chalta hai. Makdi ke jaale par os ki boondein chhote spheres hoti hain. Baalti mein ek kona dooba towel dheere-dheere paani ki line se kaafi upar tak geela ho jaata hai. Paani par dheere se rakhi steel ki paper clip tairti hai, jabki steel paani se lagbhag aath guna dense hai. Yeh sab liquid ki free surface par hota hai, jo kuch-kuch ek khinchi hui elastic skin ki tarah behave karti hai jismein apni extra energy hoti hai. Is effect ko surface tension (पृष्ठ तनाव) kehte hain. Gases ki koi free surface nahi hoti, isliye yeh liquids ki property hai.

Try karke dekho

Paper clip tairao, phir use dubao

Tumhe chahiye

saaf paani ka ek bowl
ek steel paper clip
tissue paper ka ek chhota tukda
ek toothpick aur thoda liquid detergent

Pehle guess karo

Saaf paani par ek paper clip tair rahi hai. Paani mein zara sa detergent daalne par kya hota hai?

woh aur upar tairti haiwoh doob jaati haikuch nahi badalta

Tissue paper ka ek chhota square paani par seedha bichhao aur us par ek sookhi paper clip dheere se rakho.
Tissue ke bheeg kar doobne ka intezaar karo, jisse clip surface par tiki reh jaaye. Uske aas-paas paani mein bane dent ko dhyaan se dekho.
Toothpick ko liquid detergent mein dubao aur bowl ke kinaare ke paas paani ko chhuo.

9.6.1 Surface Energy

Surface banane mein energy kyun lagni chahiye? Same liquid ke do molecules compare karo (Fig. 9.14): ek andar gehre mein daba hua, doosra surface par jiske upar hawa hai. Dono apne padosiyon se attract hote hain, aur har attraction uski potential energy kam karta hai, kuch waise hi jaise gaddhe mein padi ball ki energy kinaare par tiki ball se kam hoti hai. Andar wale molecule ke chaaron taraf padosi hain; unke kheenchav cancel ho jaate hain aur uski energy jitni kam ho sakti hai utni hai. Surface wale molecule ke padosi sirf bagal mein aur neeche hain, lagbhag aadhe, isliye woh kam mazbooti se bandha hai aur liquid ke andar ki taraf ek net kheenchav mehsoos karta hai.

Kitni energy shaamil hai? Saare molecules ko alag karne mein, yaani liquid ko poora ubaal dene mein, paani ke boiling point par lagbhag 41 kJ per mole lagte hain. Surface molecule ne apne lagbhag aadhe bonds kho diye hain, isliye uske paas, bahut mote taur par, us per-molecule energy ka aadha hissa extra hota hai. Isliye nayi surface banane mein energy lagti hai, aur akela chhoda gaya liquid sabse chhota area apnata hai jitna uska maahaul allow kare. Zyadatar surface effects isi ek fact se nikalte hain.

Molecular scale par dekho to liquid ka kinaara dhundhla hota hai: sirf kuch molecules ki thickness mein density apni poori value se lagbhag zero tak gir jaati hai.

4/4Wire ko $d$ bahar kheenchne se area $2 l d$ badhta hai, kyunki film ke do faces hain.

Fig. 9.14Surface molecules mein extra energy hoti hai, isliye film ko khinchne ke liye ek steady force chahiye.NCERT Fig. 9.15 · p. 193

Left mein, surface ke neeche liquid molecules ka ek grid. Step 1: andar gehre ka ek molecule highlight hai, jise chaaron taraf ke padosi barabar kheenchte hain. Step 2: surface ka ek molecule highlight hai; uske padosi sirf bagal mein aur neeche hain, aur ek mota arrow uska andar ki taraf net kheenchav dikhata hai. Step 3: right mein ek soap film dikhti hai, jo ek wire frame par khinchi hai aur ek slide hone wale wire se band hai. Step 4: slider ko force F se distance d bahar kheencha gaya hai, jisse film ki ek patti jud gayi hai.

9.6.2 Surface Energy aur Surface Tension

Extra energy measure karne ke liye, ek wire frame par soap film khincho jiski ek side slide kar sakti hai: lambai $l$ ka ek seedha wire (Fig. 9.14, aakhri steps). Film slider ko andar kheenchti hai, isliye tumhe use force $F$ se pakadna padta hai. Use dheere se distance $d$ tak bahar kheencho; work $F d$ nayi film ki surface energy ban jaata hai. Film ki do surfaces hoti hain, aage aur peeche, isliye naya area $2 l d$ hai. Agar $S$ per unit area surface energy hai,

S (2 l d) = F d

(9.19)

Surface tension

S = \frac{F}{2 l}

(9.20)

Ek tarah padho to $S$ per unit area surface energy hai, $J m^{- 2}$ mein; doosri tarah padho to yeh per unit length woh force hai jisse surface apne kinaare ko kheenchti hai, $N m^{- 1}$ mein. Dono units same hain.

Definition

Surface tension

Kisi liquid ka surface tension $S$ uski surface ke plane mein, usmein kheenchi kisi bhi line ke right angle par lagne wali per unit length force hai. Yeh surface ki per unit area extra energy ke barabar hai.

Example20 °C par paani: $S \approx 0.073 N m^{- 1}$ .

Do baatein picture ko aur saaf karti hain. Pehli, mann mein surface ko kisi bhi line ke along kaato, jaise khinchi hui drum skin ko kaatoge. Har side kate kinaare ko per unit length force $S$ se kheenchegi, line ke right angle par aur surface ke plane mein. Surface ke andar yeh kheenchav balance ho jaate hain; sirf ek free kinaare par, jaise slider, andar ki taraf net kheenchav hota hai. Doosri, yeh energy do substances ke beech ki boundary ki property hai aur kisi ek ke badalne par badal jaati hai, isliye hum liquid–air surface ke liye $S_{l a}$ , solid–liquid ke liye $S_{s l}$ aur solid–air ke liye $S_{s a}$ likhte hain.

Table 9.3Kuch liquids ke surface tensions (approximate)NCERT Table 9.3 · p. 194

Liquid	Temperature (°C)	Surface tension ( $N m^{- 1}$ )
Paani	20	0.073
Paani	100	0.059
Ethanol	20	0.022
Glycerine	20	0.063
Soap solution	20	≈ 0.03
Mercury	20	≈ 0.48

Surface tension, liquid ki viscosity ki tarah, aam taur par liquid garam hone par girta hai. Liquid solid se chipkega ya nahi, yeh interface energies par depend karta hai. Drop tab chipakta hai jab solid–air surface ke ek patch aur liquid–air surface ke ek patch ki jagah ek single solid–liquid interface aane se energy kam ho, yaani jab $S_{s l} < S_{s a} + S_{l a}$ .

Liquid ke kheenchav ko tola bhi ja sakta hai. Agar ek patli glass plate ka lambai $l$ ka seedha neeche ka kinaara liquid ko bas chhoo raha ho, to surface plate ke dono faces ko kul force $2 S_{l a} l$ se neeche kheenchti hai. To plate ko balance ke ek arm se latkao, use counterpoise karo, aur liquid ko upar laakar uske kinaare se chhuao; phir doosre pan mein mass $m$ tab tak jodo jab tak plate bas chhoot na jaaye. Tab $m g = 2 S_{l a} l$ , yaani

S_{l a} = \frac{m g}{2 l}

(9.21)

9.6.3 Angle of Contact

Jahan liquid ki surface solid se milti hai wahan woh aam taur par curved hoti hai. Angle of contact (स्पर्श कोण) $θ$ woh angle hai jo solid surface aur line of contact par liquid surface ki tangent ke beech hota hai, liquid ke andar se naapa jaata hai. Taro (arbi) ke patte par baarish ki boondein moti ban kar tikti hain jahan $θ$ $90^{\circ}$ se kaafi zyada hai, jabki saaf glass par paani ki boond chhota $θ$ bana kar ek patle lens mein phail jaati hai (Fig. 9.15).

Jis line par drop ka kinaara solid se milta hai, wahan teen interfaces milte hain: liquid–air, solid–liquid aur solid–air, tensions $S_{l a}$ , $S_{s l}$ aur $S_{s a}$ ke saath. Har ek line ko apni surface ke along kheenchta hai. Solid ke along, $S_{s a}$ bahar ki taraf, drop se door, kheenchta hai, jabki $S_{s l}$ aur $S_{l a}$ ka solid ke parallel hissa andar ki taraf kheenchte hain. (Solid $S_{l a}$ ke apne right angle wale hisse ko balance karta hai.) Line apni jagah tiki rahe, iske liye

S_{l a} \cos θ + S_{s l} = S_{s a}

(9.22)

Ise $\cos θ = (S_{s a} - S_{s l}) / S_{l a}$ ki tarah padho. Jo liquid khud se zyada solid ki taraf attract hota hai, woh $S_{s l}$ ko chhota banata hai, isliye $\cos θ$ bada aur $θ$ acute hota hai: liquid solid ko wet karta hai, jaise paani saaf glass ko wet karta hai aur cooking oil steel ki plate par phail jaata hai. Jab liquid ke apne molecules zyada mazbooti se jude hon, to solid–liquid interface mehenga hota hai ( $S_{s l} > S_{s a}$ ), $\cos θ < 0$ aur $θ$ obtuse hota hai: liquid solid ko wet nahi karta aur moti jaisi boondon mein simat jaata hai, jaise nayi wax lagi car par baarish ya glass par mercury.

Washing powders aur soaps contact angle kam karte hain, taaki paani kapde ke fibres mein ghus jaaye, unke upar boondon ki tarah na baithe, aur dirt ko saath le jaaye. Raincoat ya chhatri par lagi water-repellent coating ulta karti hai: woh contact angle badhati hai taaki boondein seedhe lud jaayein.

3/3Solid ke along, $S_{l a} \cos θ + S_{s l} = S_{s a}$ (arrows scale par nahi).

Fig. 9.15Angle of contact liquid ke andar se naapa jaata hai; jahan liquid solid ko wet nahi karta wahan yeh obtuse hota hai.NCERT Fig. 9.17 · p. 195

Paani ki do boondein. Step 1: left wali boond, waxy patte par, lagbhag ek gend jaisi hai; uske kinaare par ek dashed tangent patte ke saath obtuse angle θ banati hai, jo paani ke andar se naapa gaya hai. Step 2: right wali boond, saaf glass par, ek chapti cap hai jiski tangent acute angle θ banati hai. Step 3: right wali boond ke kinaare par teen arrows lagte hain: tangent ke along liquid–air tension, glass ke along boond ki taraf solid–liquid tension, aur usse door solid–air tension.

Quick check

3 mein se 1

Units

N m^{- 1}

aur

J m^{- 2}

dono describe karti hain

coefficient of viscositypressuresurface tensiondensity

Answer dekho

Answer (C) surface tension

Surface tension per unit length kheenchav ko naapta hai aur, barabar taur par, per unit area energy ko;

1 J m^{- 2} = 1 N m m^{- 2} = 1 N m^{- 1}

9.6.4 Drops aur Bubbles

Chhoti boondein, jaise makdi ke jaale par os ya plant sprayer ki dhundh, lagbhag perfect spheres hoti hain, aur hawa mein tairta soap bubble bhi. Wajah hai surface ki energy cost: diye gaye volume ko rakhne wale saare shapes mein sphere ka area sabse kam hota hai. ( $1 c m^{3}$ volume ke sphere ki surface lagbhag $4.84 c m^{2}$ hoti hai; same volume ke cube ki $6 c m^{2}$ .) Sirf gravity aur air resistance hi badi boondon ko gol hone se rokte hain.

Surface tension drop ke andar ka pressure bahar se zyada bhi kar deta hai. Equilibrium mein radius $r$ ka ek spherical drop lo, jiske andar pressure $P_{i}$ aur bahar $P_{o}$ hai, aur socho ki woh thoda phool kar radius $r + Δ r$ ka ho jaata hai.

DerivationDrop ke andar excess pressure

Surface area itna badhta hai

\begin{aligned} Δ A & = 4 π (r + Δ r)^{2} - 4 π r^{2} \\ \approx 8 π r Δ r \end{aligned}

$(Δ r)^{2}$ ko neglect karke, isliye extra surface energy hai

Δ E = 8 π r Δ r S_{l a}

(9.23)

Equilibrium par yeh energy surface ko bahar dhakelne wala pressure difference deta hai, jab volume $4 π r^{2} Δ r$ badhta hai:

W = (P_{i} - P_{o}) 4 π r^{2} Δ r

(9.24)

$W = Δ E$ rakhne par,

Drop ya cavity

P_{i} - P_{o} = \frac{2 S_{l a}}{r}

(9.25)

General rule yeh hai ki curved liquid surface ke aar-paar concave side par pressure zyada hota hai. Liquid ke andar hawa ke bubble (cavity) ki ek surface hoti hai, isliye uska excess pressure bhi $2 S / r$ hai. Soap bubble alag hai: yeh ek patli film hai jiski do surfaces hain, bahar ki aur andar ki, aur har ek $2 S / r$ jodti hai:

Soap bubble

P_{i} - P_{o} = \frac{4 S_{l a}}{r}

(9.26)

Chhote bubbles ka excess pressure zyada hota hai. Isiliye soap bubble phoonkne mein shuru mein zor ki phoonk lagti hai aur bada hone par aasan ho jaata hai, aur isiliye agar ek chhote aur ek bade soap bubble ko tube se joda jaaye, to chhota bubble bade mein khaali ho jaata hai.

3/3Do surfaces: $P_{i} - P_{o} = 4 S / r$ .

Fig. 9.16Liquid surface ki concave side par pressure zyada hota hai; soap bubble ki do surfaces hoti hain.NCERT Fig. 9.18 · p. 196

Same radius r ke teen spheres saath-saath, har ek ke neeche uska excess pressure likha hai. Pehle step mein hawa mein liquid drop highlight hai: uski ek curved surface hai, aur uska andar ka pressure bahar se 2S/r zyada hai. Doosre step mein liquid ke andar hawa ka bubble highlight hai: phir se ek surface, aur excess pressure 2S/r. Teesre step mein soap bubble highlight hai: bahar aur andar ki surface wali ek patli film, isliye uska excess pressure 4S/r hai.

Example 9.10NCERT Example 9.10 · p. 197

Aquarium ka bubbler

Ek aquarium air pump hawa ko ek patli tube mein neeche dhakelta hai jiska khula end, internal radius $0.50 m m$ ka, paani ki surface se $12.0 c m$ neeche hai. Tube ke muh par theek hemisphere jaisa bubble bahar nikalne ke liye pump ko kitna air pressure dena hoga, aur us bubble ke andar excess pressure kitna hai? $S = 7.28 \times 10^{- 2} N m^{- 1}$ , $ρ = 1000 k g m^{- 3}$ , $g = 9.8 m s^{- 2}$ aur $P_{a} = 1.013 \times 10^{5} P a$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

9.6.5 Capillary Rise

Ek patli glass tube ka ek end paani mein dubao aur paani uske andar bahar ke level se upar chadh jaata hai; tube jitni patli, paani utna upar. Yeh capillary rise apna naam baal ke Latin word se leta hai, kyunki baal jaisi patli tube mein rise bahut zyada hota hai. Iski wajah tube mein curved paani ki surface, yaani meniscus, ke aar-paar pressure difference hai.

DerivationCapillary rise ki height

Maan lo tube ka radius $a$ hai aur paani glass se contact angle $θ$ (acute) par milta hai. Tab meniscus upar ki taraf concave hota hai, radius $R$ ke ek sphere ka hissa, aur Fig. 9.17 ki geometry se $R = a / \cos θ$ milta hai. Concave side, yaani upar ki hawa, zyada pressure par hai, isliye meniscus ke theek neeche paani ka pressure $P_{in}$ atmospheric se itna kam hai

\begin{aligned} P_{a} - P_{in} & = \frac{2 S}{R} \\ = \frac{2 S \cos θ}{a} \end{aligned}

(9.27)

Ab tube mein bahar ke paani ki surface ke level par ek point lo. Yeh same judi hui paani mein us khuli surface ke points ke saath ek hi level par hai, isliye iska pressure $P_{a}$ hai. Yeh meniscus se height $h$ neeche hai, isliye

P_{in} + ρ g h = P_{a}

(9.28)

(9.27) aur (9.28) ko milakar,

Capillary rise

h = \frac{2 S \cos θ}{ρ g a}

(9.29)

Rise tube ke radius ke inversely proportional hai aur poori tarah surface tension se aata hai. Saaf glass mein paani ke liye $θ \approx 0$ . Jab $\cos θ$ negative ho, jaise glass mein mercury ke liye ( $θ \approx 140^{\circ}$ ), to (9.29) negative $h$ deta hai: meniscus convex hota hai aur liquid bahar ke level se neeche rehta hai.

2/3Meniscus radius $R = a / \cos θ$ ke sphere ka hissa hai; uske theek neeche pressure $P_{a}$ se $2 S / R$ kam hai.

Fig. 9.17Paani patli tube mein chadhta hai; mercury, jo glass ko wet nahi karta, neeche dhakela jaata hai.NCERT Fig. 9.19 · p. 197

Left mein, paani ke beaker mein ek patli glass tube khadi hai; andar ka paani height h tak chadh gaya hai, aur ek concave meniscus par khatam hota hai. Step 2: ek bada view meniscus ko radius R ke circle ka hissa dikhata hai, jiska centre uske upar hai, aur jo tube ki wall (radius a) se contact angle θ par milta hai, isliye R = a/cos θ. Step 3: bade view ki jagah mercury mein ek tube aati hai, jahan andar ka level bahar ki surface se neeche hai aur top convex hai.

ExampleNCERT p. 197

Kitna upar, aur kitna neeche?

Internal radius $0.25 m m$ ki ek saaf glass capillary ek dish mein seedhi khadi hai. Uske andar level ka change nikalo (a) paani ke liye, $S = 0.073 N m^{- 1}$ , $θ = 0$ , $ρ = 1000 k g m^{- 3}$ ke saath; (b) mercury ke liye, $S = 0.48 N m^{- 1}$ , $θ = 140^{\circ}$ , $ρ = 1.36 \times 10^{4} k g m^{- 3}$ ke saath. $g = 9.8 m s^{- 2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Capillarity diye ki cotton wick mein oil ko upar kheenchti hai aur blotting paper aur towels ko paani sokhne deti hai. Yeh plants ki patli vessels mein paani ko upar chadhne mein madad karti hai, haalanki lambe ped mein zyadatar lift patton par evaporation se aata hai.

Quick check

3 mein se 1

Ek saaf glass capillary mein paani

3.0 c m

chadhta hai. Usse aadhe radius ki tube mein woh kitna, cm mein, chadhega?

Answer dekho

Answer 6 cm

h \propto 1 / a

, isliye radius aadha karne se rise double hokar

6.0 c m

ho jaata hai.

Recap

Summary aur key points

Liquids aur gases flow karte hain kyunki rest mein woh shear stress support nahi kar sakte; liquids apna volume rakhte hain aur lagbhag incompressible hain, gases apna container bhar deti hain (Section 9.1).
Pressure per unit area normal force hai, $P = \lim Δ F / Δ A$ : ek scalar jo pascal mein measure hota hai, jahan $1 a t m = 1.013 \times 10^{5} P a$ ; density $ρ = m / V$ hai (Section 9.2).
Pascal's law: rest mein fluid mein kisi point par pressure har direction mein barabar lagta hai aur ek level ke saare points par same hota hai (Section 9.2.1).
$P = P_{a} + ρ g h$ : pressure sirf depth par depend karta hai, vessel ke shape par nahi; $P - P_{a}$ gauge pressure hai (Section 9.2.2).
Barometer $P_{a}$ ko lagbhag 76 cm mercury se balance karta hai; manometer gauge pressure padhta hai; $1 t o r r = 1$ mm of mercury $\approx 133 P a$ aur $1 b a r = 10^{5} P a$ (Section 9.2.3).
Band fluid pressure ko bina kam kiye transmit karta hai; hydraulic lift force ko $A_{2} / A_{1}$ guna karta hai jabki daala gaya work nikle hue work ke barabar rehta hai (Section 9.2.4).
Steady flow mein streamlines kabhi cross nahi karti, aur $A v$ constant rehta hai (continuity, mass conservation); critical speed se aage flow turbulent ho jaata hai (Section 9.3).
Bernoulli: steady flow mein ideal fluid ki streamline ke along $P + \frac{1}{2} ρ v^{2} + ρ g h$ constant rehta hai; yeh per unit volume energy conservation hai (Section 9.4).
Torricelli's law $v = \sqrt{2 g h}$ khule tank se efflux speed deta hai (Section 9.4.1); body ki ek side tez flow dynamic lift paida karta hai, jaise Magnus effect aur aerofoils mein (Section 9.4.2).
Viscosity: $η$ shear stress divided by rate of shear strain hai, $P a s$ mein; temperature badhne par liquids ke liye yeh girti hai aur gases ke liye badhti hai (Section 9.5).
Stokes' law $F = 6 π η a v$ se terminal velocity $v_{t} = 2 a^{2} (ρ - σ) g / 9 η$ milti hai (Section 9.5.1).
Surface tension $S$ ko surface ki per unit length kheenchav ya per unit area extra energy ki tarah padh sakte hain; film se $S = F / 2 l$ milta hai, aur contact line par $S_{l a} \cos θ + S_{s l} = S_{s a}$ (Section 9.6.2, Section 9.6.3).
Excess pressure drop ya cavity mein $2 S / r$ aur soap bubble mein $4 S / r$ hai; capillary rise $h = 2 S \cos θ / ρ g a$ hai (Section 9.6.4, Section 9.6.5).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises Q9.1	Practice Q1
NCERT Exercises Q9.2	Practice Q2
NCERT Exercises Q9.3 (a)–(d)	Practice Q3
NCERT Exercises Q9.3 (e)	Practice Q4
NCERT Exercises Q9.4 (a)–(d)	Practice Q5
NCERT Exercises Q9.5	Practice Q6
NCERT Exercises Q9.6	Practice Q7
NCERT Exercises Q9.7	Practice Q8
NCERT Exercises Q9.8	Practice Q9
NCERT Exercises Q9.9	Practice Q10
NCERT Exercises Q9.10	Practice Q11
NCERT Exercises Q9.11	Practice Q12
NCERT Exercises Q9.12	Practice Q13
NCERT Exercises Q9.13	Practice Q14
NCERT Exercises Q9.14	Practice Q15
NCERT Exercises Q9.15	Practice Q16
NCERT Exercises Q9.16	Practice Q17
NCERT Exercises Q9.17	Practice Q18
NCERT Exercises Q9.18	Practice Q19
NCERT Exercises Q9.19	Practice Q20
NCERT Exercises Q9.20	Practice Q21
NCERT Exercises Q9.4 (e)	Practice Q29

Quick check: MCQs

9.1 Introduction2 questions
9.2.2 Depth ke saath Pressure ka Variation3 questions
9.2.3 Atmospheric Pressure aur Gauge Pressure3 questions
9.2.4 Hydraulic Machines2 questions
9.3 Streamline Flow3 questions
9.4.1 Efflux ki Speed: Torricelli's Law3 questions
9.4.2 Dynamic Lift2 questions
9.5.1 Stokes' Law3 questions
9.6.3 Angle of Contact3 questions
9.6.5 Capillary Rise3 questions

Glossary · 14

Key terms (English + Hindi)

angle of contactस्पर्श कोणsparsh kon: Jahan solid surface aur liquid surface milte hain wahan solid surface aur liquid surface ki tangent ke beech ka angle, liquid ke andar se naapa gaya.Chapter mein pehli baar yahan aaya →

Bernoulli's principleबरनौली का सिद्धांतbarnauli ka siddhaant: Steady flow mein ideal fluid ki streamline ke along $P + \frac{1}{2} ρ v^{2} + ρ g h$ constant rehta hai.Chapter mein pehli baar yahan aaya →

capillary riseकेशिकत्वkeshikatva: Surface tension ki wajah se patli tube mein liquid ka chadhna ya girna: $h = 2 S \cos θ / ρ g a$ .Chapter mein pehli baar yahan aaya →

equation of continuityसांतत्य समीकरणsaantatya sameekaran: Incompressible fluid ke steady flow mein $A v$ har cross-section par same hota hai; yeh mass conservation batata hai.Chapter mein pehli baar yahan aaya →

fluidतरलtaral: Aisa substance jo flow kar sake: liquid ya gas. Rest mein yeh shear stress support nahi kar sakta.Chapter mein pehli baar yahan aaya →

gauge pressureप्रमापी दाबpramaapi daab: Atmospheric pressure se upar ka pressure, $P - P_{a}$ ; tyre gauges aur manometers yahi padhte hain.Chapter mein pehli baar yahan aaya →

Pascal's lawपास्कल का नियमpaaskal ka niyam: Rest mein fluid mein pressure har direction mein barabar lagta hai, ek level par same hota hai, aur usmein change bina kam hue aage pahunchta hai.Chapter mein pehli baar yahan aaya →
pressureदाबdaab: Fluid mein kisi point par per unit area normal force; ek scalar, jo pascal (Pa) mein measure hota hai.Chapter mein pehli baar yahan aaya →

relative densityआपेक्षिक घनत्वaapekshik ghanatva: Kisi substance ki density divided by 4 °C par paani ki density; iski koi unit nahi hoti.Chapter mein pehli baar yahan aaya →

streamlineधारारेखाdhaararekha: Steady flow mein fluid particles ka path; iski tangent har point par fluid ki velocity batati hai.
surface tensionपृष्ठ तनावprishth tanaav: Liquid surface mein per unit length force, jo uski per unit area extra energy ke barabar hai; unit $N m^{- 1}$ .Chapter mein pehli baar yahan aaya →

terminal velocityसीमांत वेगseemaant veg: Fluid mein girti body ki woh steady speed jab drag aur buoyancy uske weight ko balance kar dete hain.Chapter mein pehli baar yahan aaya →
turbulent flowविक्षुब्ध प्रवाहvikshubdh pravaah: Critical speed se aage be-tarteeb flow, jismein kisi point par velocity badalti rehti hai aur eddies bante hain.Chapter mein pehli baar yahan aaya →

viscosityश्यानताshyaanata: Relative motion mein fluid layers ke beech internal friction; $η$ shear stress divided by strain rate hai, Pa s mein.Chapter mein pehli baar yahan aaya →

Mechanical Properties of Fluids: Class 11 Physics Chapter 9 Hinglish mein

Is chapter mein aap kya seekhenge

9.1 Introduction

9.2 Pressure

9.2.1 Pascal's Law

9.2.2 Depth ke saath Pressure ka Variation

9.2.3 Atmospheric Pressure aur Gauge Pressure

9.2.4 Hydraulic Machines

9.3 Streamline Flow

9.4 Bernoulli's Principle

9.4.1 Efflux ki Speed: Torricelli's Law

9.4.2 Dynamic Lift

9.5 Viscosity

9.5.1 Stokes' Law

9.6 Surface Tension

9.6.1 Surface Energy

9.6.2 Surface Energy aur Surface Tension

9.6.3 Angle of Contact

9.6.4 Drops aur Bubbles

9.6.5 Capillary Rise

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye