Surface tension — origin, Young-Laplace equation: Exercises

2.2.4 · D4 · Physics › Fluid Mechanics › Surface tension — origin, Young-Laplace equation

Poore note mein, parent note ke ye do headline results yaad rakho: $Δ P_{drop} = \frac{2 γ}{r}, Δ P_{soap bubble} = \frac{4 γ}{r}, Δ P = γ (\frac{1}{R _{1}} + \frac{1}{R _{2}}) .$ Yahan $γ$ (surface tension) force-per-length ya energy-per-area hai, $r$ sphere ka radius hai, aur $R_{1}, R_{2}$ do principal radii hain (surface do perpendicular directions mein kitni sharply curve karti hai). Jab tak aur na bataya jaye, $γ_{water} = 0.072 N/m$ aur atmospheric pressure $P_{0} = 1.013 \times 1 0^{5} Pa$ lo.

Level 1 — Recognition

Recall Solution — L1·Q1

KYA dhundhna hai: sirf ek hi fark hai — liquid–air interfaces ki ginti.

Ek drop mein one interface hota hai (andar liquid, bahar air) → $Δ P = \frac{2 γ}{r}$ .
Ek soap bubble ek thin film hai jisme dono taraf air hoti hai → two interfaces → $Δ P = \frac{4 γ}{r}$ .

$r = 2 \times 1 0^{- 3} m$ ke saath: $Δ P_{drop} = \frac{2 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 72 Pa, Δ P_{bubble} = \frac{4 ( 0.072 )}{2 \times 1 0 ^{- 3}} = 144 Pa .$ Bubble ka excess pressure drop se bilkul double hai — kyunki uske paas squeeze karne ke liye double skins hain.

Recall Solution — L1·Q2

Force per length: $\frac{N}{m}$ . Energy per area: $\frac{J}{m ^{2}} = \frac{N \cdot m}{m ^{2}} = \frac{N}{m}$ . Same unit $\Rightarrow$ dono definitions ek hi physical number describe karti hain. Isliye parent note mein $γ = F / L = energy / area$ likha ja saka.

Level 2 — Application

Recall Solution — L2·Q1

KYU energy, pressure nahi: question surface ke against kiya gaya kaam pooch raha hai, jo hai $γ \times (bana hua area)$ . Ek soap bubble ke two surfaces hote hain (inner + outer), har ek ka area $4 π R^{2}$ : $A_{total} = 2 \times 4 π R^{2} = 8 π R^{2} .$ $W = γ A_{total} = 0.030 \times 8 π (0.03)^{2} = 0.030 \times 8 π \times 9 \times 1 0^{- 4} .$ $W = 6.786 \times 1 0^{- 4} J \approx 6.79 \times 1 0^{- 4} J .$ Matlab: ek milli-joule se bhi kam — surfaces sasti hoti hain, isliye ek halki si saans ek bada bubble bana deti hai.

Recall Solution — L2·Q2

Ek film ke two faces hote hain, toh dono wire ko neeche kheenchte hain: $F = 2 γ L = 2 (0.025) (0.08) = 4.0 \times 1 0^{- 3} N = 4.0 mN .$ KYU 2 hai: ise bhool jaoge toh answer aadha ho jaayega — classic $F = 2 γ L$ frame result.

Level 3 — Analysis

Recall Solution — L3·Q1

KYU $1/ r$ compare karo: har bubble ka excess pressure $Δ P = \frac{4 γ}{r} \propto \frac{1}{r}$ hai. Chhota $r$ ⇒ zyada excess pressure.

Bubble B ( $r = 2 cm$ ) ka internal pressure zyada hai.
Air isliye B (chhote) se A (bade) ki taraf flow karegi — chhota bubble sickta hai, bada banta hai.

(b) Dono bubbles same atmosphere $P_{0}$ mein hain, toh driving difference excess pressures ka fark hai: $Δ P_{A} = \frac{4 ( 0.030 )}{0.04} = 3.0 Pa, Δ P_{B} = \frac{4 ( 0.030 )}{0.02} = 6.0 Pa .$ $Δ P_{A} - Δ P_{B} = 3.0 - 6.0 = - 3.0 Pa .$ Negative sign confirm karta hai ki B ka pressure A se $3.0 Pa$ zyada hai, air ko A ki taraf dhakelte hue. Figure dekho: tighter (redder) skin zyada squeeze karti hai.

Figure — Surface tension — origin, Young-Laplace equation

Recall Solution — L3·Q2

Middle film kya feel karta hai: ek taraf bubble B ka inside hai (zyada pressure $P_{B}$ ), doosri taraf bubble A ka inside hai (kam pressure $P_{A}$ ). Film low-pressure side ki taraf curve karti hai (A mein bulge karti hai), aur uska apna Young–Laplace jump uske across pressure difference ke barabar hona chahiye. Ek dividing film ke bhi two faces hote hain, toh: $mid film ke across jump \frac{4 γ}{r _{mid}} = P_{B} - P_{A} = \frac{4 γ}{r _{B}} - \frac{4 γ}{r _{A}} .$ $4 γ$ cancel karo: $\frac{1}{r _{mid}} = \frac{1}{r _{B}} - \frac{1}{r _{A}} = \frac{1}{0.02} - \frac{1}{0.04} = 50 - 25 = 25 m^{- 1} .$ $r_{mid} = \frac{1}{25} = 0.04 m = 4 cm = r_{A} .$ Neat result: $r_{mid} = \frac{r _{A} r _{B}}{r _{A} - r _{B}}$ ; yahan $r_{A}$ ke barabar hai kyunki $r_{A} = 2 r_{B}$ .

Level 4 — Synthesis

Recall Solution — L4·Q1

(a) Volume conserved hai (liquid almost incompressible hoti hai): $8 \times \frac{4}{3} π a^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = 8 a^{3} \Rightarrow R = 2 a = 2 mm .$ (b) Surface areas compare karo (drop ka ek interface hota hai): $\frac{A _{big}}{A _{small,total}} = \frac{4 π R ^{2}}{8 \times 4 π a ^{2}} = \frac{R ^{2}}{8 a ^{2}} = \frac{( 2 a ) ^{2}}{8 a ^{2}} = \frac{4}{8} = \frac{1}{2} .$ Surface energy $= γ A$ hai, toh energy half ho jaati hai — merged drop aath droplets ki surface energy ka aadha store karta hai. (c) Energy kam ho gayi, toh surface energy release hoti hai (heat / halki si warming ke roop mein). Isliye chhote droplets spontaneously milte hain: system ki energy kam hoti hai.

Recall Solution — L4·Q2

(a) Ek hemispherical meniscus ek interface hai jisme $R_{1} = R_{2} = r$ : $Δ P = γ (\frac{1}{r} + \frac{1}{r}) = \frac{2 γ}{r} = \frac{2 ( 0.072 )}{2.0 \times 1 0 ^{- 4}} = 720 Pa .$ Curved meniscus ke neeche pressure atmospheric se $720 Pa$ kam hai (surface liquid se door curve karti hai). (b) Ye deficit uthe hue column ke weight $ρ g h$ se balance hota hai: $ρ g h = \frac{2 γ}{r} \Rightarrow h = \frac{2 γ}{ρ g r} = \frac{2 ( 0.072 )}{1000 \times 9.8 \times 2.0 \times 1 0 ^{- 4}} .$ $h = \frac{0.144}{1.96} = 0.07347 m \approx 7.35 cm .$ Matlab: wahi Young–Laplace jump jo ek drop ko pressurise karta hai, paani ko ek thin tube mein upar kheenchta hai — physics same hai, bas curve doosri taraf hai.

Level 5 — Mastery

Recall Solution — L5·Q1

KYU $P V$ conservation: temperature fixed hai, toh trapped air ke liye Boyle's law $P V = const$ total enclosed air par lagta hai. Har bubble ka internal absolute pressure $P = P_{0} + \frac{4 γ}{r}$ hai (excess pressure atmosphere mein add hota hai), aur uska volume $\frac{4}{3} π r^{3}$ hai.

Total $P V$ conserve karo: $(P_{0} + \frac{4 γ}{r _{1}}) \frac{4}{3} π r_{1}^{3} + (P_{0} + \frac{4 γ}{r _{2}}) \frac{4}{3} π r_{2}^{3} = (P_{0} + \frac{4 γ}{r}) \frac{4}{3} π r^{3} .$ $\frac{4}{3} π$ cancel karo aur expand karo: $P_{0} r^{3} + 4 γ r^{2} = P_{0} (r_{1}^{3} + r_{2}^{3}) + 4 γ (r_{1}^{2} + r_{2}^{2}) .$ Numbers plug karo ( $r_{1} = 0.03, r_{2} = 0.04 m, 4 γ = 0.12$ ):

$P_{0} (r_{1}^{3} + r_{2}^{3}) = 1.013 \times 1 0^{5} (2.7 \times 1 0^{- 5} + 6.4 \times 1 0^{- 5}) = 1.013 \times 1 0^{5} \times 9.1 \times 1 0^{- 5} = 9.2183 J-ish (Pa\cdotm^{3}) .$
$4 γ (r_{1}^{2} + r_{2}^{2}) = 0.12 (9 \times 1 0^{- 4} + 1.6 \times 1 0^{- 3}) = 0.12 \times 2.5 \times 1 0^{- 3} = 3.0 \times 1 0^{- 4} .$

Toh equation hai: $1.013 \times 1 0^{5} r^{3} + 0.12 r^{2} = 9.2183 + 3.0 \times 1 0^{- 4} = 9.21860.$ Surface terms $P_{0} r^{3}$ ke mukable bahut chhoti hain, toh acchi accuracy ke liye $r^{3} \approx r_{1}^{3} + r_{2}^{3} = 9.1 \times 1 0^{- 5}$ : $r \approx (9.1 \times 1 0^{- 5})^{1/3} = 0.04497 m \approx 4.50 cm .$ Full cubic numerically solve karne par $r$ well under $0.01 mm$ shift hota hai, toh $r \approx 4.50 cm$ . Insight: ordinary atmospheric pressure par surface-energy terms almost matter nahi karti — merged volume essentially $r_{1}^{3} + r_{2}^{3}$ hai. Young–Laplace correction sirf vacuum mein dominate karta hai, jahan $P_{0} \to 0$ .

Recall Solution — L5·Q2

$P_{0} = 0$ hone par $P V$ equation ka $P_{0} r^{3}$ term khatam ho jaata hai. $P = \frac{4 γ}{r}$ use karo: $\frac{4 γ}{r _{1}} \cdot \frac{4}{3} π r_{1}^{3} + \frac{4 γ}{r _{2}} \cdot \frac{4}{3} π r_{2}^{3} = \frac{4 γ}{r} \cdot \frac{4}{3} π r^{3} .$ $\frac{4}{3} π \cdot 4 γ$ cancel karo: $r_{1}^{2} + r_{2}^{2} = r^{2} \Rightarrow r = r_{1}^{2} + r_{2}^{2} = 0.0 3^{2} + 0.0 4^{2} = 0.0025 = 0.05 m = 5 cm .$ KYU badla: koi external pressure nahi hai, toh sirf surface tension se pressure aata hai, isliye conserved quantity surface area ban jaati hai ( $4 γ / r \times r^{3} \propto r^{2}$ ) na ki volume. Merged bubble radii mein ek 3-4-5 right triangle hai — ek clean signature ki areas add hote hain, volumes nahi. Atmosphere mein, huge $P_{0}$ volume ko near-conserved quantity bana deta hai (Q1). Ye wahi "kaunsi quantity conserve hoti hai" wali theme hai jo coalescing drops mein thi, ab aur sharp ho gayi.

Recall wrap-up

Recall Cover-and-check

Drop vs soap-bubble excess pressure? ::: $2 γ / r$ vs $4 γ / r$ (ek surface vs do). Radius $R$ ka bubble phulane mein kaam? ::: $γ \times 8 π R^{2}$ (do surfaces). Radius $a$ ke aath drops milte hain — bada radius aur energy factor? ::: $R = 2 a$ ; surface energy half ho jaati hai. Air mein coalescing bubbles — kya nearly conserved hota hai? ::: total volume, $r^{3} \approx r_{1}^{3} + r_{2}^{3}$ . Vacuum mein coalescing bubbles — kya conserved hota hai? ::: total area, $r^{2} = r_{1}^{2} + r_{2}^{2}$ .