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

False — $\gamma$ acts tangentially along the skin, pulling it together (inward-curving). The outward push is the excess pressure the tension forces the gas to supply.

False — the drop has one interface ($2\gamma /r$), the soap bubble has two ($4\gamma /r$), so the bubble's excess pressure is double.

False — $\Delta P=4\gamma /r\propto 1/r$, so the smaller bubble has the higher internal pressure; more air ≠ more pressure.

True — with $R_1=R_2=\infty$, $\Delta P=\gamma (1/R_1+1/R_2)=0$; only curvature creates the jump.

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

False — a bulk molecule has neighbours on all sides, so the cohesive pulls cancel to zero net force; only a surface molecule feels a net inward pull.

True — $\Delta P=2\gamma /r$ is inversely proportional to $r$, so $r\to 2r$ gives $\Delta P\to \Delta P/2$.

False — a gas cavity has a single liquid–gas interface, so $\Delta P=2\gamma /r$; the "4" is only for a thin soap film with two faces.

Error: a film has two faces, so $F=2\gamma L$ and $\gamma =F/(2L)$; dropping the 2 halves your answer.

Error: $\frac{dA}{dr}=8\pi r$, so $dA=8\pi rdr$; forgetting the factor from differentiating $r^2$ breaks the whole $2\gamma /r$ derivation.

Error: the smaller bubble has higher pressure ($\propto 1/r$), so air flows out of it into the larger one — the small shrinks, the big grows.

Error: it's the opposite — molecules attract (cohesion); surface molecules simply have fewer neighbours to attract them from above, giving a net inward pull.

Error: a cylinder curves in only one direction ($R_1=r,R_2=\infty$), so $\Delta P=\gamma (1/r+0)=\gamma /r$ — half the sphere's value.

Error: $R_1$ and $R_2$ are independent principal radii; a sphere is just the special case $R_1=R_2=r$, not a requirement.

A sphere is the minimum surface area for a fixed volume, and minimising area minimises surface energy ($\gamma \times A$), which nature always favours.

On a curved skin under tension the tension forces have a small inward-pointing component; to keep the surface from collapsing the inside pressure must rise, and that rise is $\Delta P$.

The soap bubble is a thin liquid shell with air on both sides — two interfaces each giving $2\gamma /r$ — while the raindrop has a single liquid–air interface.

Moving a molecule from the bulk to the surface breaks some attractive bonds, raising its energy; that energy cost per unit area is exactly $\gamma$.

At mechanical equilibrium the surface neither grows nor shrinks spontaneously, so the pressure work pushing out must exactly balance the energy cost of the extra surface created.

Tighter curvature means the tension vectors along the skin tilt more steeply inward, so their net inward component per unit area — hence $\Delta P\propto 1/r$ — is larger.

Zero — both curvature terms vanish, so no pressure jump exists; this is why still open water has no excess pressure at its top surface.

$\Delta P=2\gamma /r\to \infty$ — vanishingly small droplets are at enormous internal pressure, which is why fine mist evaporates and coalesces readily.

The radii take opposite signs, so $\gamma (1/R_1+1/R_2)$ can be zero or negative; a soap film on a wire loop settles into exactly such a minimal surface with $\Delta P=0$. See Minimal surfaces & soap films.

Neither — equal radii give equal pressures, so the system is in (unstable) balance; any tiny perturbation makes one shrink and the other grow.

The cylinder gives $\gamma /r$ (only one finite curvature) while the sphere gives $2\gamma /r$; same radius, but the sphere's double curvature doubles the jump.

Its radius of curvature is effectively infinite, so $2\gamma /r\to 0$; surface-tension pressure only becomes significant when $r$ is small (millimetres or less).