Question bank — Surface tension — origin, Young-Laplace equation

False — $\gamma$ skin ke saath tangentially act karta hai, use andar ki taraf kheenchta hai (inward-curving). Baahri push woh excess pressure hai jo tension gas ko supply karwata hai.

False — drop mein ek interface hai ($2\gamma /r$), soap bubble mein do hain ($4\gamma /r$), isliye bubble ka excess pressure double hota hai.

False — $\Delta P=4\gamma /r\propto 1/r$, isliye chhote bubble ka internal pressure zyada hota hai; zyada hawa ≠ zyada pressure.

True — $R_1=R_2=\infty$ ke saath, $\Delta P=\gamma (1/R_1+1/R_2)=0$; sirf curvature hi jump create karta hai.

False — ye dimensionally identical hain (${\text{J/m}}^2={\text{N⋅m/m}}^2=\text{N/m}$); force-per-length aur energy-per-area literally same number $\gamma$ hain.

False — bulk molecule ke har taraf neighbours hote hain, isliye cohesive pulls cancel hokar net force zero ho jaati hai; sirf ek surface molecule net inward pull feel karta hai.

True — $\Delta P=2\gamma /r$ inversely proportional hai $r$ se, isliye $r\to 2r$ karne par $\Delta P\to \Delta P/2$ ho jaata hai.

False — gas cavity mein ek liquid–gas interface hota hai, isliye $\Delta P=2\gamma /r$; "4" sirf thin soap film ke liye hai jiske do faces hote hain.

Error: ek film ke do faces hote hain, isliye $F=2\gamma L$ aur $\gamma =F/(2L)$; 2 chhod dene se jawab aadha aa jaata hai.

Error: $\frac{dA}{dr}=8\pi r$, isliye $dA=8\pi rdr$; $r^2$ ko differentiate karne wala factor bhool jaana poori $2\gamma /r$ ki derivation kharab kar deta hai.

Error: chhote bubble ka pressure zyada hota hai ($\propto 1/r$), isliye hawa usse bahar bade wale mein jaati hai — chhota sidhta hai, bada badhta hai.

Error: bilkul ulta hai — molecules attract karte hain (cohesion); surface molecules ke paas sirf upar se attract karne ke liye kam neighbours hote hain, jisse net inward pull aata hai.

Error: cylinder sirf ek direction mein curve karta hai ($R_1=r,R_2=\infty$), isliye $\Delta P=\gamma (1/r+0)=\gamma /r$ — sphere ki value ka aadha.

Error: $R_1$ aur $R_2$ independent principal radii hain; sphere sirf ek special case hai $R_1=R_2=r$, koi requirement nahi.

Sphere fixed volume ke liye minimum surface area hai, aur area minimize karna surface energy ($\gamma \times A$) minimize karta hai, jo nature hamesha favour karta hai.

Ek curved skin par tension ke forces ka ek chhota sa inward-pointing component hota hai; surface ko collapse hone se rokne ke liye andar ka pressure badhna chahiye, aur woh rise hi $\Delta P$ hai.

Soap bubble ek thin liquid shell hai jiske dono taraf hawa hai — do interfaces har ek $2\gamma /r$ deta hai — jabki raindrop ka ek hi liquid–air interface hota hai.

Ek molecule ko bulk se surface par le jaane se kuch attractive bonds toot jaate hain, uski energy badhti hai; woh energy cost per unit area exactly $\gamma$ hai.

Mechanical equilibrium par surface apne aap na badhta hai na sidhta hai, isliye baahri taraf push karne wala pressure work extra surface banane ki energy cost ke saath exactly balance hona chahiye.

Tighter curvature ka matlab hai skin ke saath tension vectors zyada steeply andar jhukti hain, isliye unka net inward component per unit area — hence $\Delta P\propto 1/r$ — zyada hota hai.

Zero — dono curvature terms vanish ho jaate hain, isliye koi pressure jump nahi hoti; isliye still open water ki top surface par koi excess pressure nahi hota.

$\Delta P=2\gamma /r\to \infty$ — bahut chhote droplets enormous internal pressure par hote hain, isliye fine mist aasaani se evaporate aur coalesce karta hai.

Radii opposite signs lete hain, isliye $\gamma (1/R_1+1/R_2)$ zero ya negative ho sakta hai; ek wire loop par soap film exactly aise minimal surface mein settle hoti hai jahan $\Delta P=0$ ho. Dekho Minimal surfaces & soap films.

Kisi taraf nahi — equal radii equal pressures dete hain, isliye system (unstable) balance mein hota hai; koi bhi chhota perturbation ek ko shrink aur doosre ko grow kar deta hai.

Cylinder $\gamma /r$ deta hai (sirf ek finite curvature) jabki sphere $2\gamma /r$ deta hai; same radius, lekin sphere ki double curvature jump ko double kar deti hai.

Uska radius of curvature effectively infinite hai, isliye $2\gamma /r\to 0$; surface-tension pressure tabhi significant hota hai jab $r$ chhota ho (millimetres ya usse kam).