Surface tension — origin, Young-Laplace equation: Worked examples

2.2.4 · D3 · Physics › Fluid Mechanics › Surface tension — origin, Young-Laplace equation

Shuru karne se pehle, un tools ko pin down karte hain jo hum baar baar use karte rahenge, taaki koi bhi symbol use karne se pehle uska matlab earn ho chuka ho.

"Two principal radii" ka matlab kya hai (pehle $R_{1}, R_{2}$ use karte hain)

Kisi bhi curved skin ke ek point pe khade ho jao. Surface ko ek plane se slice karo aur ek curved line milegi; woh line kisi circle ko hug karti hai kisi radius ke saath. Ab apna slicing plane rotate karo — hugging circle ka radius badalta jaata hai jaise tum ghoomate ho. Pata chalta hai ki do special, perpendicular slicing directions hote hain jahan radius sabse chota aur sabse bada hota hai. Un dono radii ko principal radii of curvature kehte hain, likha jaata hai $R_{1}$ aur $R_{2}$ .

General formula kahan se aata hai (ek careful sketch)

Hum $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ pe baar baar lean karte hain, toh chalte hain isse actually build karte hain, un $R_{1}, R_{2}$ ko use karke jo humne abhi define kiye. Figure 1 dekhte raho parhte waqt.

Skin ka ek tiny rectangular patch lo jiske sides $d x$ ( $R_{1}$ direction mein) aur $d y$ ( $R_{2}$ direction mein) hain, jo gently outward curve kar raha hai.

Tension har edge ke saath pull karta hai. Do $d y$ -length edges pe (woh edges jo $d x$ span ke across ek doosre ke samne hain), surface tension $γ$ har ek pe $γ d y$ force se pull karta hai — force per length times edge length. Ye dono pulls surface ke saath outward, tangent direction mein point karte hain.
- $γ d y$ kyun? Kyunki $γ$ force per unit length hai us line ki jiske along woh act karta hai, aur us line ki length $d y$ hai.
Kyunki patch curved hai, woh dono pulls parallel nahi hain. $R_{1}$ direction mein surface ek chote angle se bend karta hai. Agar patch $R_{1}$ radius ke circle pe arc-length $d x$ span karta hai, toh do tangent directions mein angle ka difference $d θ_{1} = \frac{d x}{R _{1}}$ hai (arc = radius × angle).
- $d x / R_{1}$ kyun? Yahi arc se subtended angle ki definition hai: arc $d x$ ko radius $R_{1}$ se divide karo.
Isliye har tilted pull ka ek tiny inward component hota hai. $γ d y$ force jo inward half-angle $d θ_{1} /2$ se tilted hai, contribute karta hai $γ d y \cdot sin (d θ_{1} /2) \approx γ d y \cdot \frac{d θ _{1}}{2}$ inward. Pair (dono edges) deta hai $2 \times γ d y \cdot \frac{d θ _{1}}{2} = γ d y d θ_{1} = γ d y \cdot \frac{d x}{R _{1}} = \frac{γ d x d y}{R _{1}}$ .
- Small-angle step $sin x \approx x$ kyun? Kyunki $d θ_{1}$ infinitesimal hai; yahi reason hai ki hum tiny patch lete hain — geometry limit mein exact ho jaati hai.
Perpendicular direction mein bhi same karo ( $R_{2}$ direction, $d x$ length ki edges): woh identical argument se $\frac{γ d x d y}{R _{2}}$ inward contribute karte hain.
Pressure push ke saath balance karo. Excess pressure patch ko outward push karta hai $Δ P \cdot (d x d y)$ force se. Equilibrium pe outward push total inward pull ke barabar hota hai: $Δ P (d x d y) = \frac{γ d x d y}{R _{1}} + \frac{γ d x d y}{R _{2}} .$
Area $d x d y$ cancel karo: $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) .$

Sphere ke liye dono radii $r$ ke barabar hain, isliye $\frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{2}{r}$ — yahi woh jagah hai jahan $2 γ / r$ mein "2" paida hota hai. Notice karo ki sketch jo key link concrete banata hai: tension ka inward component exactly $γ d x d y / R$ hai kyunki edge arc-angle $d x / R$ se tilt hoti hai.

Definition

$R_{1}$ aur $R_{2}$ ke liye sign convention (yeh skip mat karo) Curvature ka radius positive count hota hai jab surface inside ki taraf bulge karta hai (woh region jise hum "in" kehte hain, jahan $P_{in}$ rehta hai) — yaani hugging circle ka centre inside pe hota hai. Yeh negative hota hai jab surface doosri taraf curve karta hai (saddle ya dimple, centre outside pe).

Dono radii positive (drop, bubble): dono terms add hote hain ⇒ $Δ P > 0$ , inside squeeze hota hai.
Ek radius negative (saddle-shaped soap film, jaise do bubbles ke beech ka neck): terms partly cancel hote hain, aur $Δ P$ zero bhi ho sakta hai — exactly yahi wajah hai ki ek soap-film neck ek minimal surface ho sakta hai jahan dono sides pe pressure equal ho (dekho Minimal surfaces & soap films). Hum yeh case fully Ex 9 mein work karte hain.

The scenario matrix

Is topic ka har problem inhi cells mein se kisi ek mein hoga. Example column batata hai ki kaunsa worked example us cell ko nail karta hai.

Cell	Kya cheez isko distinct banati hai	Example
A. One interface (drop)	Single liquid–air skin ⇒ factor 2	Ex 1
B. Two interfaces (soap bubble)	Film mein inside + outside skin ⇒ factor 4	Ex 2
C. Gas cavity in liquid	Paani ke andar gas ka bubble — phir bhi one skin	Ex 3
D. Cylinder (one radius $\infty$ )	$R_{1} = r, R_{2} = \infty$ ⇒ factor 1	Ex 4
E. Flat / degenerate limit	$r \to \infty$ ⇒ $Δ P \to 0$	Ex 4 (aside)
F. Direction of flow (sign of $1/ r$ )	Do connected bubbles; kaunsa zyada $P$ pe hai?	Ex 5
G. Limiting behaviour $r \to 0$	$Δ P \to \infty$ : kyun tiny drops "resist" karte hain	Ex 6
H. Energy method / word problem	Ek drop ko kai drops mein todna; energy released	Ex 7
I. Exam twist — coalescence	Air (volume) + surfaces conserve karo naya radius nikaalte hue	Ex 8
J. Negative curvature (saddle)	Ek radius negative ⇒ terms cancel, $Δ P$ ho sakta hai 0	Ex 9

Neeche har numeric answer machine-checked hai.

Cell A — one interface

Worked example Ex 1 · Raindrop ke andar excess pressure

Ek spherical water raindrop ka radius $r = 2 mm$ hai. Water ki surface tension $γ = 0.072 N/m$ . Surrounding air ke comparison mein andar ka pressure nikalo.

Forecast: pehle guess karo — $Δ P$ kiske kareeb hai: $1 Pa$ , $70 Pa$ , ya $7000 Pa$ ? (Atmospheric pressure $\sim 1 0^{5} Pa$ hai, toh yeh tiny hona chahiye.)

Interfaces ki sankhya pehchano. Ek liquid drop mein ek liquid–air skin hoti hai.
- Yeh step kyun? Formula mein factor sirf skins count karke decide hota hai. Ek skin ⇒ $2 γ / r$ use karo, $4 γ / r$ nahi.
Radius ko metres mein convert karo: $r = 2 mm = 2 \times 1 0^{- 3} m$ .
- Yeh step kyun? $γ$ N/m mein hai; millimetres mix karne se power of ten galat ho jaayegi.
Substitute karo: $Δ P = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 72 Pa .$
- Yeh step kyun? Sphere ke liye dono principal radii $r$ ke barabar hain, isliye general formula deta hai $γ (1/ r + 1/ r) = 2 γ / r$ — outward pressure push single skin ke inward pull ke saath balance karta hai.

Verify: Units: $\frac{N/m}{m} = \frac{N}{m ^{2}} = Pa$ ✓. Magnitude $72 Pa$ atmospheric $1 0^{5} Pa$ ke aage kuch bhi nahi — soft-skinned drop ke liye sensible hai.

Cell B — two interfaces

Neeche figure dono ka side by side crucial difference dikhata hai: drop (left) mein ek lavender boundary hai, jabki soap bubble (right) mein do coral rings hain — inner aur outer skins jo milkar factor 4 earn karte hain. Dekho kaise bubble liquid ka ek shell hai jiske dono faces pe air hai.

Figure — Surface tension — origin, Young-Laplace equation

Figure 1 — Ek skin (drop) versus do skins (soap bubble): boundaries count karna decide karta hai factor 2 vs 4.

Cell C — gas cavity in liquid

Worked example Ex 3 · Paani ke andar ek air bubble

Ek tiny air bubble radius $r = 0.5 mm$ ka paani ke andar hai, $γ = 0.072 N/m$ . Andar ki air ka surrounding water ke upar excess pressure nikalo.

Forecast: kya yeh "drop" case hai (one skin, factor 2) ya "bubble" case (two skins, factor 4)? Decide karne se pehle picture dekho.

Interfaces gino. Bahar paani hai aur andar air, ek single liquid–air surface se separated. Koi thin film nahi hai — paani ek bulk ocean hai.
- Yeh step kyun? "Soap bubble" mein do skins isliye hain kyunki woh thin film hai. Bulk liquid mein gas cavity mein sirf ek skin hoti hai, bilkul drop ki tarah.
One-skin formula apply karo: $Δ P = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{0.5 \times 1 0 ^{- 3}} = 288 Pa .$
- Yeh step kyun? Physics same hai raindrop jaisi — curved skin enclosed gas ko squeeze karti hai.

Verify: Units Pa dete hain ✓. Ex 1 se chota radius (0.5 mm vs 2 mm), isliye pressure bada hona chahiye: $288 > 72$ ✓. "Bubble" word ka matlab yahan factor 4 nahi tha — geometry vocabulary ko beat karta hai.

Figure mein, air cavity mint-green bulk water ke samundar mein baithe ek single lavender skin ke saath hai. Coral arrows andar ki taraf point karte hain — woh hai tension jo trapped air ko squeeze kar rahi hai. Kahi bhi doosri skin nahi hai, isliye factor 2 hai, exactly raindrop ki tarah.

Figure 2 — Bulk water mein ek air cavity: ek interface (factor 2), two-skinned film nahi.

Cell D — cylinder (aur Cell E — flat limit)

Worked example Ex 4 · Ek liquid jet (cylinder) aur uska flat limit

Paani ka ek cylindrical column (ek thin jet) ka radius $r = 1 mm$ hai, $γ = 0.072 N/m$ . Uski curved side ke across $Δ P$ nikalo. Phir pucho: jab $r \to \infty$ toh kya hota hai?

Forecast: ek cylinder apni width ke across circle ki tarah curve karta hai lekin length ke saath straight hota hai. Uska pressure $2 γ / r$ , $γ / r$ , ya kuch beech mein hona chahiye?

Do principal radii nikalo. Circular cross-section ke across curvature ka radius $R_{1} = r$ hai (positive — yeh inside ki taraf bulge karta hai). Axis ke saath surface ek straight line hai, isliye woh curve nahi karta: $R_{2} = \infty$ .
- Yeh step kyun? General Young–Laplace formula ko dono curvatures chahiye, aur cylinder woh case hai jahan unme se ek straight line tak vanish ho jaata hai ( $1/\infty = 0$ ).
General form mein substitute karo: $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = γ (\frac{1}{r} + \frac{1}{\infty}) = \frac{γ}{r} .$
- Yeh step kyun? $1/\infty = 0$ axial term ko kill kar deta hai, ek single factor 1 chhod ke — drop ke factor 2 ka half.
Numbers: $Δ P = \frac{0.072}{1 \times 1 0 ^{- 3}} = 72 Pa .$

Cell E — flat limit: ab radius blow up karo, $r \to \infty$ . Dono terms $1/\infty = 0$ ban jaate hain, isliye $Δ P \to 0$ .

Yeh step kyun? Perfectly flat skin mein tension ka koi inward-pointing component nahi hota, isliye ise hold karne ke liye koi pressure difference nahi chahiye. Yeh degenerate case hai jo pure formula ko anchor karta hai.

Verify: Units Pa ✓. Cylinder factor 1 exactly drop ke factor 2 ka half hai, aur flat surface $0$ deta hai jaise parent ki table demand karti hai ✓.

Figure dono curvatures visible banata hai: coral arrow finite radius $R_{1} = r$ hai round cross-section ke across, jabki mint double-arrow axis ke saath straight chalti hai — woh direction hai jahan $R_{2} = \infty$ hai. Sirf curved direction contribute karta hai, isliye factor 1 hai.

Figure 3 — Cylinder mein ek finite radius ( $R_{1} = r$ ) aur ek infinite ( $R_{2} = \infty$ ) hota hai, jo deta hai $Δ P = γ / r$ .

Cell F — flow ki direction ( $1/ r$ ka sign)

Worked example Ex 5 · Do connected soap bubbles — air kis taraf jaayegi?

Bubble A ka radius $r_{A} = 3 cm$ hai, bubble B ka radius $r_{B} = 1 cm$ . Unhe ek choti tube se valve ke saath joda gaya hai. $γ = 0.030 N/m$ . Valve kholo — air kis taraf jaayegi, aur dono pressures kya hain?

Forecast: intuition chillata hai "bada bubble = zyada air = zyada pressure, isliye air bade wale se niklegi." Formula pe trust karo, gut pe nahi.

Har ka pressure likho. Dono soap bubbles hain (do skins), isliye $Δ P = \frac{4 γ}{r}$ . $Δ P_{A} = \frac{4 ( 0.030 )}{0.03} = 4 Pa, Δ P_{B} = \frac{4 ( 0.030 )}{0.01} = 12 Pa .$
- Yeh step kyun? Har bubble apna internal excess pressure apne radius se set karta hai; tube unhe compare karne deti hai.
Compare karo. $Δ P \propto 1/ r$ , isliye chota bubble (B) zyada pressure pe hai.
- Yeh step kyun? Tighter curve gas ko zyada squeeze karta hai — "Petite = Pressurised" mnemonic.
Direction conclude karo. Air high pressure se low pressure ki taraf jaati hai ⇒ B (small) se A (big) ki taraf. B shrink hoga aur A grow karega.
- Yeh step kyun? Gas hamesha pressure gradient down drift karta hai; yahan gradient chote bubble se bade ki taraf point karta hai.

Verify: $Δ P_{B} /Δ P_{A} = 12/4 = 3 = r_{A} / r_{B} = 3/1$ ✓ — ratio inverse radius ratio ke barabar hona chahiye. Counter-intuitive small-to-large flow confirm hua.

Figure dono bubbles ko tube se joined dikhata hai, har ek apne radius aur pressure ke saath labelled hai. Lavender flow arrow chote, high-pressure bubble se bade, low-pressure wale ki taraf point karta hai — trace karo aur dekho kaise air tightly curved bubble ko chhod deti hai.

Figure 4 — Chota bubble (12 Pa) bade mein drain hota hai (4 Pa): flow pressure follow karta hai, jo $1/ r$ follow karta hai.

Cell G — limiting behaviour $r \to 0$

Worked example Ex 6 · Droplet ka radius half karna

$r$ radius ke drop ka excess pressure $Δ P$ hai. Ise squeeze karke half radius $r /2$ kar diya. $Δ P$ kitne factor se badlega? $r = 1 μ m = 1 0^{- 6} m$ aur $γ = 0.072 N/m$ se illustrate karo.

Forecast: kya pressure double, half, ya quadruple hoga?

$Δ P = 2 γ / r$ use karo. $r$ ki jagah $r /2$ dalo: $Δ P^{'} = \frac{2 γ}{r /2} = \frac{4 γ}{r} = 2 Δ P .$
- Yeh step kyun? Pressure $1/ r$ se scale karta hai, isliye radius half karne se pressure double ho jaata hai.
Tiny limit tak le jao. $r = 1 0^{- 6} m$ ke liye: $Δ P = \frac{2 ( 0.072 )}{1 0 ^{- 6}} = 1.44 \times 1 0^{5} Pa .$
- Yeh step kyun? Micrometre scale pe excess pressure already ek full atmosphere compete karta hai — yahi wajah hai ki mist droplets aur nucleating bubbles form hone mein resist karte hain.
Formal limit $r \to 0$ lo: $Δ P = 2 γ / r \to \infty$ .
- Yeh step kyun? Matrix ka degenerate small-radius end: infinitely tight curvature ke liye infinite pressure chahiye. Bahut chota drop ya bubble banana mushkil hai — "nucleation barriers" ka janam.

Verify: $Δ P^{'} /Δ P = 2$ exactly ✓, aur $1.44 \times 1 0^{5} Pa \approx 1.4$ atm — promised "rivals atmosphere" ✓.

Cell H — energy method / word problem

Worked example Ex 7 · Ek drop ko hazar mein todna

Ek single water drop ka radius $R = 1 mm$ hai, use spray karke $n = 1000$ identical tiny droplets banaye jaate hain. $γ = 0.072 N/m$ . Kitni surface energy supply karni padegi?

Forecast: kya spraying energy release karti hai ya energy cost karti hai? (Kaun si shape mein zyada total surface hoti hai — ek bada drop ya hazar chote wale?)

Chota radius $r$ nikaalane ke liye volume conserve karo. Volume fixed hai: $\frac{4}{3} π R^{3} = n \cdot \frac{4}{3} π r^{3} \Rightarrow r = \frac{R}{n ^{1/3}} = \frac{1 0 ^{- 3}}{100 0 ^{1/3}} = \frac{1 0 ^{- 3}}{10} = 1 0^{- 4} m .$
- Yeh step kyun? Koi paani create ya destroy nahi hota; sirf surface change hoti hai. $r$ nikaalne ke baad hi naya total area nikaal sakte hain.
Pehle aur baad ki total surface area calculate karo. $A_{before} = 4 π R^{2}, A_{after} = n \cdot 4 π r^{2} = 1000 \cdot 4 π (1 0^{- 4})^{2} .$ Numerically $A_{before} = 4 π (1 0^{- 3})^{2} = 4 π \times 1 0^{- 6} m^{2}$ aur $A_{after} = 4 π \times 1 0^{- 5} m^{2}$ .
- Yeh step kyun? Surface energy = $γ$ times surface area ( $γ$ ki energy definition se). Area woh cheez hai jo badli.
Energy required = $γ Δ A$ : $Δ E = γ (A_{after} - A_{before}) = 0.072 (4 π \times 1 0^{- 5} - 4 π \times 1 0^{- 6}) .$ $= 0.072 \cdot 4 π \cdot 1 0^{- 6} (10 - 1) = 0.072 \cdot 4 π \cdot 9 \times 1 0^{- 6} \approx 8.14 \times 1 0^{- 6} J .$
- Yeh step kyun? Positive matlab energy daalni padegi — todne se area badhta hai, isliye free mein spray nahi kar sakte.

Verify: $A_{after} = 10 A_{before}$ (area $n^{1/3} = 10$ se badhti hai) ✓. Sign positive hai ⇒ energy supplied, "bahut chote drops mein zyada surface hoti hai" se match karta hai ✓. Numeric $Δ E \approx 8.14 μ J$ . (General "energy = $γ \times$ area change" bookkeeping ke liye dekho Energy methods in mechanics.)

Cell I — exam twist: coalescence

Common mistake Volume vs area — kya conserve karo?

Coalescence problems do flavours mein aati hain aur students dono ko mix up karte hain:

Physical conservation: jab bubbles merge hote hain toh air (volume, ya fixed temperature pe $P V$ ) truly conserved hoti hai — kuch bahar nahi jaata. Yeh honest model hai aur parent note ka cubic deta hai, $P_{0} r^{3} + 4 γ r^{2} = P_{0} (r_{1}^{3} + r_{2}^{3}) + 4 γ (r_{1}^{2} + r_{2}^{2})$ .
Simplified exam model: agar tum outside pressure $P_{0}$ ko poori tarah neglect karo, toh cubic ke $P_{0} r^{3}$ terms drop out ho jaate hain aur tum reh jaate ho $r^{2} = r_{1}^{2} + r_{2}^{2}$ ke saath — yeh ek area-conservation statement hai. Yeh idealisation hai, sirf small- $γ$ -dominates limit mein valid; yeh general law nahi hai. Neeche hum honest volume-conserving version solve karte hain aur phir dikhate hain ki neat 3–4–5 shortcut kahan se aata hai.

Worked example Ex 8 · Do soap bubbles merge — naya radius nikalo

Do soap bubbles ke radii $r_{1} = 3 cm$ aur $r_{2} = 4 cm$ hain, woh coalesce karte hain. Ise properly solve karo (air conserve karke) us limit mein jahan surface energy dominate kare, isliye $P_{0}$ terms negligible hain. Naya radius $r$ nikalo.

Forecast: naya radius $5 cm$ , $7 cm$ , ya exactly $r_{1} + r_{2} = 7 cm$ ke kareeb hoga?

Air ki honest conservation se shuru karo. Constant temperature pe, $\sum P_{i} V_{i}$ conserved hai. $P_{i} = P_{0} + \frac{4 γ}{r _{i}}$ aur $V_{i} = \frac{4}{3} π r_{i}^{3}$ ke saath, yeh parent cubic hai.
- Yeh step kyun? Air gayab nahi ho sakti; sirf surface rearrange hoti hai. Volume ( $P V$ ke through) physically conserved quantity hai, area nahi.
Surface-dominated limit $P_{0} \to 0$ lo. Tab $P_{i} \to \frac{4 γ}{r _{i}}$ , aur $\sum P_{i} V_{i}$ proportional ho jaata hai $\sum r_{i}^{2}$ se: $\frac{4 γ}{r _{i}} \cdot \frac{4}{3} π r_{i}^{3} = \frac{16}{3} π γ r_{i}^{2} \Rightarrow r^{2} = r_{1}^{2} + r_{2}^{2} .$
- Yeh step kyun? Volume se $r_{i}^{3}$ aur pressure se $1/ r_{i}$ milke $r_{i}^{2}$ chhod jaate hain — isliye neglect- $P_{0}$ model "area conservation" jaisa lagta hai. Yeh actually $P V$ law disguise mein hai.
Substitute karo: $r = r_{1}^{2} + r_{2}^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = 5 cm .$
- Yeh step kyun? Yahi neat 3–4–5 giveaway hai jo examiners love karte hain — radii "in quadrature" add hote hain, linearly nahi.

Verify: $r = 5 cm$ , aur $r < r_{1} + r_{2} = 7 cm$ jaisa hona chahiye (ek bubble do ke maqable mein zyada area-efficient hai) ✓. Yeh bhi $r > max (r_{1}, r_{2}) = 4 cm$ kyunki merged bubble dono ka air rakhti hai ✓.

Cell J — negative curvature: saddle jahan pressure cancel hota hai

Worked example Ex 9 · Soap-film neck (catenoid) —

$Δ P = 0$ kyun Ek soap film do coaxial rings ke beech span karti hai, ek waisted "neck" shape mein pull hoti hai (catenoid). Neck ke narrowest point pe film $R_{1} = 1 cm$ radius ke circular waist ke across inward curve karti hai, lekin axis ke saath (saddle) $R_{2} = 1 cm$ curvature ke saath doosri taraf outward curve karti hai. $γ = 0.030 N/m$ . Film ke across $Δ P$ nikalo.

Forecast: ek curved soap surface — surely har doosre example ki tarah pressure jump hogi? Signs dekho.

Hamare convention se signs assign karo. Waist ke across surface andar ki air ki taraf bulge karta hai, isliye $R_{1} = + 1 cm$ . Axis ke saath surface saddle hai — yeh ulti taraf curve karta hai (hugging circle ka centre bahar), isliye $R_{2} = - 1 cm$ negative hai.
- Yeh step kyun? Yeh woh case hai jiske liye sign convention banaya gaya tha. Saddle ek direction mein inside ki taraf aur perpendicular direction mein bahar ki taraf curve karta hai — opposite signs.
General formula apply karo, signs rakhte hue: $Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = γ (\frac{1}{+ 0.01} + \frac{1}{- 0.01}) = γ (100 - 100) = 0.$
- Yeh step kyun? Dono curvatures equal aur opposite hain, isliye unke reciprocals exactly cancel ho jaate hain. Ek direction ka inward pull doosre direction ke outward "unbending" se undo ho jaata hai.
Interpret karo. $Δ P = 0$ matlab film ke dono taraf air pressure same hai — film kuch bhi squeeze nahi kar rahi. Aisi equal-pressure film ek minimal surface hai (locally area-minimising); catenoid classic example hai (dekho Minimal surfaces & soap films).
- Yeh step kyun? Jab support karne ke liye koi pressure jump nahi hai, toh surface tension sirf area shrink kar sakti hai, isliye film sabse kum area wali shape mein settle ho jaati hai.

Verify: $\frac{1}{+ 0.01} + \frac{1}{- 0.01} = 100 - 100 = 0$ , isliye $Δ P = 0$ exactly ✓. Yahi promised edge case hai: ek radius negative hone se, terms cancel ho jaate hain aur genuinely curved surface bhi zero pressure jump rakh sakti hai — impossible agar sign bhool jao.

Figure neck dikhata hai: coral circle inward waist curvature $R_{1} > 0$ hai, jabki mint saddle-arc axis ke saath outward curvature $R_{2} < 0$ hai. Trace karo kaise ek "in" bend karta hai aur doosra "out" — yahi wajah hai ki woh cancel ho jaate hain.

Figure 5 — Saddle-shaped soap-film neck: $R_{1} > 0$ (waist) aur $R_{2} < 0$ (axis) cancel ho jaate hain, $Δ P = 0$ dete hain.

Recall

Recall Answers cover karo aur khud test karo

Do principal radii $R_{1}, R_{2}$ kya hote hain? ::: Most aur least curving ke perpendicular directions mein surface ko best hug karne wale do circles ke radii. Kaun se cases factor 2, 4, aur 1 dete hain? ::: Drop aur gas cavity ⇒ 2 (one skin); soap bubble ⇒ 4 (two skins); cylinder ⇒ 1 (ek radius infinite). Bulk water mein air bubble kaunsa factor use karta hai? ::: 2 — yeh ek skin hai, thin film nahi. $r \to \infty$ pe $Δ P$ kya hota hai? ::: Zero — flat surface ko koi pressure jump nahi chahiye. $r \to 0$ pe $Δ P$ kya hota hai? ::: Infinity — tiny curves infinitely hard squeeze karte hain. Principal radius negative kab count hota hai, aur kya ho sakta hai? ::: Jab surface doosri taraf curve kare (saddle/dimple), centre bahar ho — tab dono terms $Δ P = 0$ tak cancel ho sakte hain (minimal surface). Do connected bubbles: air kis taraf jaati hai? ::: Chote (high-pressure) bubble se bade wale ki taraf. Coalescence mein physically kya conserved hota hai, volume ya area? ::: Air (volume / $P V$ ). "Area conservation" sirf neglect- $P_{0}$ shortcut hai.