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irrotational flow has $\nabla \times \mathbf{v}=0$, so locally $\mathbf{v}=\nabla \Phi$; the simply connected fluid domain here permits a globally single-valued velocity potential. Incompressibility gives $\nabla \cdot \mathbf{v}=0$, hence ${\nabla }^2\Phi =0$. In a multiply connected domain, zero local vorticity alone would not rule out circulation and a multivalued potential.

Take $k>0$ so the disturbance decays at depth. Differentiating the given candidate potential yieldsThe $x$ and $z$ second derivatives of $\Phi$ cancel in the Laplacian. At $z=0$, $v_z=h_0\omega \cos (\omega t)\sin (kx)={\partial }_{t}Z$, so the linearized kinematic free-surface condition holds. This uses $k{h}_0\ll 1$; advection of the displaced surface is higher order.

Since ${\cos }^2(kx)+{\sin }^2(kx)=1$, the local squared speed is $h_0^2{\omega }^2{\cos }^2(\omega t)e^{2kz}$. The depth integral is thereforeIt is independent of horizontal position; no further spatial average is needed for this kinetic energy. Replacing the upper boundary $Z$ by zero changes it only beyond quadratic order in amplitude.

Relative to the flat surface, the gravitational energy at each horizontal position changes by $\rho g{\int }_0^{Z}zdz=\rho g{Z}^2/2$. Averaging over one wavelength givesThe energy of a deep-water standing gravity wave, averaged over wavelength, is $T+⟨U{⟩}_x$ and must be constant for a free inviscid standing wave, which requires equal coefficients of the complementary squared sine and cosine:As an independent check, the linearized dynamic boundary condition ${\partial }_t\Phi +gZ=0$ at $z=0$ gives exactly $g-{\omega }^2/k=0$. Infinite depth, negligible surface tension and no dissipative forcing are part of this gravity-wave model.

Archimedes' principle states that the resultant hydrostatic pressure force on a body equals the weight of the fluid displaced, directed opposite to gravity. Let $V_d$ be the displaced fluid region, with outward normal $\hat{\mathbf{n}}$ from the body into the fluid. Since the fluid pressure force is inward,Hydrostatic balance is $\nabla p={\rho }_f\mathbf{g}$, with downward gravitational acceleration $\mathbf{g}$. Thereforefor uniform fluid density and gravity. Uniform atmospheric pressure gives no net force on a closed surface. For a floating body, the same displaced-volume construction includes the free surface; any significant air buoyancy can be included as a second displaced fluid.

A freely floating body of mass $m$ satisfies $mg={\rho }_{f}g{V}_d$, so its displaced volume is $m/{\rho }_f$. A homogeneous body of density ${\rho }_b<{\rho }_f$ floats with submerged fraction ${\rho }_b/{\rho }_f$; if ${\rho }_b>{\rho }_f$, hydrostatic buoyancy alone cannot support it. The buoyant force acts through the centre of buoyancy, the centroid of the displaced volume for constant density, while weight acts through the centre of mass. Their relative positions determine rotational equilibrium and stability. For a fully submerged body, a centre of mass below the centre of buoyancy gives a restoring torque; a floating body's changing displaced shape additionally requires the metacentric height criterion. Surface tension, externally driven flow and contact forces can alter the simple force balance.

For the specified increment,Using the adiabat, or its first-order Taylor expansion,This is a very small cooling, about $61\mu \mathrm{K}$. There is no heat exchange, and the film is stretched against positive surface tension, so the internal energy increases, withThe cooling does not contradict this increase. In the linear model, $\gamma -T{\gamma }^{\prime }={\gamma }_0\simeq 0.065\mathrm{J}{\mathrm{m}}^{-2}$, soCreation of surface raises the internal energy more than the cooling lowers its temperature-dependent part. The change in surface tension during this tiny temperature fall changes the work only at a negligible higher order.

Treat the film at fixed material content, with area $A$ the coordinate conjugate to the surface tension in the stated work term. The Helmholtz free energy is a Legendre transform of the internal energy:Its natural independent variables are therefore $T,A$, soEquality of mixed second partial derivatives gives the Maxwell relation for the thermodynamics of a liquid film:The area convention matters: the supplied fundamental relation already specifies the coefficient of $dA$. There is no additional factor of two unless one changes from total interfacial area to a projected film area and changes the work term consistently.

Reading the original PDF graph gives approximatelyFor example, the points near $330\mathrm{K}$ and $450\mathrm{K}$ are about $22$ and $6.5\mathrm{mN}{\mathrm{m}}^{-1}$, respectively, giving this slope. The graph supports approximate values rather than exact fitted constants. The area increment is $\Delta A=10^{-4}{\mathrm{m}}^2$.

At fixed temperature, the surface tension is constant during this stretch. The work done on the film is positive:For a reversible isothermal process, the heat supplied is $Q=T\Delta S=-T{\gamma }^{\prime }\Delta A$. ThusThe film absorbs heat while its temperature stays constant. The first law gives the useful consistency check $\Delta U=Q+W_{\mathrm{on}}\simeq 6.5\mu \mathrm{J}$, agreeing with $(\gamma -T{\gamma }^{\prime })\Delta A$.

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.

The surface tension has dimensions $[\gamma ]=M{T}^{-2}$, while $[\rho ]=M{L}^{-3}$ and $[g]=L{T}^{-2}$. ConsequentlyMatching the dimension of height gives $2-n=1$, so $n=1$. Dimensional analysis fixes this exponent but cannot determine a dimensionless prefactor or a contact-angle dependence.

A rising parcel cools, lowering the saturation pressure. Once its water vapor supply exceeds that value, atmospheric supersaturation develops. Homogeneous nucleation of pure water is strongly inhibited at the small atmospheric supersaturations typical of clouds; existing aerosol particles provide cloud condensation nuclei. Their dry sizes and composition, together with the parcel's ascent rate, determine how many cloud droplets activate.

Köhler theory combines the Kelvin equation, which raises equilibrium vapor pressure over a curved surface, with Raoult's law, which lowers it for a chemical solution. In a dilute diameter convention, with dry diameter $D_d$ and wet diameter $D$,Here $s=e_v/E(T)-1$ is atmospheric supersaturation, ${\sigma }_w$ is surface tension, and ${\kappa }_h$ measures aerosol hygroscopicity, not the dry-adiabatic process exponent. The maximum of the Köhler curve is the critical supersaturation. Larger or more hygroscopic particles generally have lower thresholds. Cloud-droplet activation requires overcoming that maximum; total aerosol number is not the activated number.

Given an observed aerosol number size distribution $n_j(D_d)=d{N}_j/d\ln D_d$ for composition classes $j$, first assign each class a hygroscopicity and compute its critical supersaturation. If the parcel reaches maximum atmospheric supersaturation $s_{\max }$, the equilibrium-threshold estimate isFor each fitted lognormal distribution, this is its upper-tail integral; it can be evaluated analytically with an error function or numerically from measured bins.

The value $s_{\max }$ must be calculated, not guessed from the aerosol distribution alone. Prescribe parcel upward velocity $w$, initial pressure, temperature and humidity, then integrate expansion cooling, latent heat release, conservation of total water, and diffusional cloud droplet growth. Schematically,where $a_s,b_s>0$, $G$ includes water vapor diffusion and thermal conduction, and $q_l$ is liquid mass fraction. Rising motion produces atmospheric supersaturation; growing particles consume it. Their competition selects the peak and which particles cross their activation barriers. Assume a well-mixed closed parcel, liquid water, soluble spherical particles, specified mixing state, and negligible collision-coalescence and fallout during activation. Slow kinetic growth, weakly soluble particles or entrainment require a more detailed treatment than the threshold integral.

Oceanic magnetic anomalies. Newly formed oceanic crust contains magnetic minerals. As it cools through their magnetic blocking temperatures, it acquires thermoremanent magnetization recording the ambient field polarity. Continuing seafloor spreading carries that record away from the ridge. Geomagnetic reversals therefore generate approximately paired marine magnetic stripes parallel to the ridge, with older crust farther from its axis. A magnetometer measures the superposed field above this magnetized material; subtracting a reference field reveals magnetic anomalies. The anomaly sign and shape depend on magnetization direction, latitude, source depth and geometry, so a positive measured anomaly is not universally identical to normal rock polarity.^[ref]

Matching a sequence of stripes to an independently dated reversal chronology gives plate ages and spreading histories. A stripe of width $w$ formed over interval $\Delta t$ implies half-spreading rate $v=w/\Delta t$ under steady symmetric spreading; full separation speed is $2v$. Unequal widths on opposite flanks can reflect asymmetric spreading or later deformation. Alteration, remagnetization and overlapping magnetic sources complicate inversion. The sequence correlation is stronger evidence than an isolated stripe or a field-strength peak.

Rayleigh-Bénard convection. Consider a horizontal fluid layer of thickness $d$, hotter below by $\Delta T>0$. The conductive state is motionless with a linear temperature gradient. A rising warm parcel is buoyant, but viscosity resists its motion and thermal diffusion erases its temperature anomaly. In the Boussinesq approximation, density variations are retained in buoyancy while the flow is otherwise incompressible. The competing effects are measured by the Rayleigh number and Prandtl number,For stress-free impermeable, fixed-temperature boundaries, linear stationary disturbances with horizontal wavenumber $q/d$ giveMinimizing gives $q_c=\pi /\sqrt{2}$ and the critical Rayleigh number for an isothermal layer, ${\mathrm{Ra}}_c=27{\pi }^4/4\simeq 657.5$. Two rigid no-slip, fixed-temperature boundaries instead give approximately $1708$; the critical value is not independent of boundary conditions.^[ref]

Above onset, buoyancy can sustain overturning rolls and heat transfer exceeds pure conduction, with Nusselt number $\mathrm{Nu}>1$. Nonlinear advection eventually limits the disturbance; larger driving can produce time-dependent or turbulent states. The ideal Boussinesq problem does not require hexagonal cells: planform selection depends on boundaries and non-Boussinesq effects. Convection driven by surface tension at a free surface is a distinct mechanism. An adverse temperature gradient supplies driving, but overcoming diffusion and viscosity requires a finite threshold.

For potential flow, each velocity potential satisfies Laplace's equation in its fluid. The exact kinematic boundary condition on $y=\eta$ is ${\eta }_t+{\varphi }_{j,x}{\eta }_x={\varphi }_{j,y}$. Keeping only first-order terms givesThe linearized Bernoulli equation is $p_j=p_0-{\rho }_j({\varphi }_{j,t}+g\eta )$ at the interface, so pressure continuity givesFor a decaying disturbance away from the interface, put $\psi =kx-\omega t$ and takeThey satisfy Laplace's equation and the kinematic boundary conditions for $\eta =A\cos \psi$. Substituting in the dynamic condition gives the dispersion relationIf the heavier fluid is on top, ${\omega }^2<0$. Instead of oscillation, a Fourier disturbance has an exponentially growing component with growth rate $\sqrt{gk({\rho }_2-{\rho }_1)/({\rho }_1+{\rho }_2)}$. This is the Rayleigh-Taylor instability in the absence of surface tension.

The material surface is $F(x,y,t)=y-\eta (x,t)=0$. Its material derivative vanishes, giving the exact kinematic conditionFor irrotational flow $u=\nabla \varphi$, the Euler equations and $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$ integrate spatially to ${\varphi }_t+|\nabla \varphi {|}^2/2+gy+p/\rho =C(t)$. Adding a suitable function of time to $\varphi$ changes no velocity and makes the right side $p_a/\rho$. ThusTake $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product ${\varphi }_x{\eta }_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. ConsequentlyFor a harmonic wave $\eta =A\cos (kx-\omega t)$, the harmonic potential decaying with depth isIt satisfies the kinematic condition. Substituting into the dynamic condition givesThe phase speed is $c=\omega /k$ and the group velocity of a narrow wave packet is $c_g=d\omega /dk$. For the positive-frequency branch and $\sigma =T/\rho$,For positive surface tension this ratio exceeds one exactly when $\sigma k^2>g$, proving

For a bulk liquid and vapor at common fixed temperature and pressure, transfer $d{N}_l$ molecules from vapor to liquid while holding their sum fixed. The Gibbs free energy change is $dG=({\mu }_l-{\mu }_v)d{N}_l$. Thermodynamic equilibrium with both phases present requires stationarity under either sign of transfer, hence $\boxed{{\mu }_l={\mu }_v}$. This is the bulk coexistence condition; an interface's curvature adds a surface-energy cost.

In the capillarity approximation, take an incompressible drop with constant liquid number density $n_l$, bulk chemical potential ${\mu }_l$ at the specified reservoir conditions, and size-independent surface tension $\Gamma$. The liquid bulk is extensive, so its Gibbs thermodynamic free energy is $N_l{\mu }_l$. Creating interface area reversibly at fixed temperature adds $\Gamma A$. Equivalently, along growth at fixed bulk parameters the stated differential integrates as $d{G}_d={\mu }_{l}d{N}_l+\Gamma dA$, with the no-drop reference set to zero. For a sphere,This approximation uses bulk properties for the interior and neglects the molecular interface thickness, curvature dependence of surface tension and higher compressibility corrections. It does not apply quantitatively to drops only a few molecules across.

The vapor contains $N-N_l$ molecules. Treat its chemical potential as approximately unchanged by forming the drop, as for a large surrounding reservoir or negligible vapor depletion. ThenThe radius is limited by the available molecules in a finite sample; the constant-reservoir expression is a local nucleation model, not a description of arbitrarily complete vapor depletion. The excess over the vapor reference has derivativeIf ${\mu }_l>{\mu }_v$, it increases for every $R>0$, leaving only the stable boundary minimum $R=0$. If ${\mu }_l<{\mu }_v$, $R=0$ is a metastable local minimum and the additional stationary radius isAt that radius $\Delta G^{\prime \prime }=-8\pi \Gamma <0$, so it is an unstable maximum. The classical nucleation theory barrier is $16\pi {\Gamma }^3/[3n_l^2({\mu }_v-{\mu }_l{)}^2]$.

A fluctuation below $R_c$ tends to shrink because reducing its surface cost lowers thermodynamic free energy. A supercritical drop tends to grow, because its bulk free-energy gain outweighs the added surface cost. Growth continues while material supply and supersaturation allow it, eventually changing the vapor state or leading to macroscopic liquid. Curved-interface pressure and chemical thermodynamic equilibrium can also be described through the Young–Laplace equation and the Kelvin equation.

An amphiphile contains a solvent-compatible head and a solvent-avoiding tail. In water, aggregation can lower the Gibbs free energy by reducing exposed hydrocarbon area through the hydrophobic effect. This competes with loss of translational entropy, head-group repulsion and the cost of deforming the tails. Above the critical micelle concentration, added amphiphiles mainly enter aggregates rather than continuing to increase the isolated-molecule concentration proportionally.

The amphiphile packing parameter $P=v/(a_0{\ell }_c)$ compares tail volume $v$, preferred head area $a_0$ and tail length ${\ell }_c$. Rough geometric ranges favor spherical micelles for $P<1/3$, cylindrical aggregates for $1/3<P<1/2$, and lipid bilayers for $1/2<P<1$. These are organizing estimates rather than universal sharp phase boundaries. A bilayer can close into a vesicle to eliminate exposed edges; concentration, salt and temperature can change the preferred aggregate.

A simple experimental test varies concentration and measures surface tension: its slope against log concentration changes near the critical micelle concentration, while light scattering detects the appearance of larger aggregates. These observations distinguish the concentration threshold from an assumption that every molecule forms an aggregate at arbitrarily low concentration.

Creating a liquid surface changes the molecular environment and costs interfacial free energy. In the dispersion-only cleavage model, the attractive Van der Waals force per area has magnitude $A/(6\pi h^3)$. Separating two contacting halves from a molecular cutoff $h_0$ to infinity takes work per areaHere $A_H$ denotes the Hamaker constant, to distinguish it from an area. Cleavage creates two free surfaces, giving the Hamaker estimate of liquid surface tensionThe continuum attraction diverges at zero separation and neglects molecular impenetrability and short-range repulsion. The cutoff is therefore an effective contact separation of molecular size, commonly of order $0.1$–$0.3\mathrm{nm}$, not a limit that can be taken to zero.

An experimental limitation is the strong temperature dependence of surface tension. For water in equilibrium with its vapour, the recommended experimental values decrease from $71.98\mathrm{mN}{\mathrm{m}}^{-1}$ at $25{}^{\circ }\mathrm{C}$ to $58.92\mathrm{mN}{\mathrm{m}}^{-1}$ at $100{}^{\circ }\mathrm{C}$, and the fitted tension vanishes at the critical point. A fixed pairwise cleavage energy misses interfacial entropy, changing density and many-body correlations.^[ref] The dispersion-only model is unsuitable when cohesion is dominated by hydrogen bonds, ionic interactions or metallic bonding; liquid metals are a clear example of the latter failure.

The liquid does not wet the glass because $\cos 140^{\circ }<0$. The liquid meniscus is convex, with the capillary mercury level depressed below the reservoir. Balancing vertical surface-tension force against the hydrostatic column gives Jurin's law, using the inner radius $r=0.5\mathrm{mm}$:Thus the depression is approximately $10.1\mathrm{mm}$. The simple height law neglects small meniscus-volume corrections. Mercury's high surface tension reflects strong metallic bonding, not merely dispersion between weakly interacting molecular units.

A thin soap bubble has two interfaces, so the Young–Laplace equation gives $\Delta p=4\gamma /R=8\gamma /d$. HenceAt equal surface tension, the smaller bubble has the higher internal pressure. Air diffuses through their separating film from the smaller to the larger bubble, so the smaller shrinks and the larger grows. The pressure difference increases as the smaller shrinks. This is gas transfer driven by curvature, not necessarily immediate merging of the interfaces; coalescence additionally requires film rupture.

For complete wetting, the meniscus has curvature radius equal to the capillary radius $r=d/2$. The Young–Laplace equation gives the supporting pressure difference $\Delta p=2\gamma /r=4\gamma /d$. Balance it against the hydrostatic pressure $\rho gh$. Taking the sap's mass density to be water-like, $\rho \simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, givesThis is the height supported by the stipulated capillary-pressure model. The density assumption is needed because the surface tension and capillary diameter alone do not determine a height.

For a clean liquid interface at fixed temperature and composition, surface tension is the reversible free-energy cost per added area, $\gamma =(\partial F/\partial A{)}_{T,V,N}$, so $dF=\gamma dA$. Its units are both $\mathrm{J}{\mathrm{m}}^{-2}$ and $\mathrm{N}{\mathrm{m}}^{-1}$. Molecules at an interface have a different environment from those in the bulk; creating more interface changes their interactions and entropy. For constant $\gamma$, the interfacial free energy is $\gamma A$ up to a reference constant. This is not generally identical to the internal surface energy: the interfacial entropy density is $-d\gamma /dT$, giving internal-energy density $\gamma -Td\gamma /dT$.

A line of length $\ell$ in one interface transmits a tangential force $\gamma \ell$, perpendicular to the line. A soap film has two interfaces, giving force $2\gamma \ell$ on a sliding rim. Quasistatic extension by $dx$ creates total area $2\ell dx$, recovering that force from $dF=2\gamma \ell dx$. Uniform positive surface tension favors minimum area at fixed volume, hence an isolated spherical droplet when gravity is negligible. curvature gives the Young–Laplace equation, with outward normal chosen consistently:A spherical liquid drop has $\Delta p=2\gamma /R$; a thin soap bubble with two similar interfaces has approximately $4\gamma /R$. The capillary length $\sqrt{\gamma /(\Delta \rho g)}$ compares gravity with capillarity. surface tension also restores capillary waves, rather than providing a bulk static shear modulus.

Interfacial contamination and added surfactants can change $\gamma$ substantially; a nonuniform value produces tangential Marangoni stress. A rim-force measurement should therefore control temperature, cleanliness and the number of interfaces. Capillary-rise methods additionally need the contact angle, because the vertical component is proportional to $\gamma \cos \theta$. An independent frame-force measurement can distinguish a contact-angle change from a true change of surface tension.

In a regular solution, ideal combinatorial entropy competes with a nonzero unlike-contact energy. For equal molecular volumes and mole fraction $\varphi$ of species A, write mixing free energy per site aswhereHere $z$ is the coordination number and ${ϵ}_{ij}$ a pair-contact energy counted consistently in the lattice model. A positive $\chi$ penalizes unlike contacts; a negative value favors them. The logarithmic terms lower free energy on mixing. The sign of the enthalpy of mixing alone therefore does not determine miscibility.

At fixed temperature and composition, a homogeneous phase is globally stable only if no division into two compositions lowers its free energy. The common-tangent construction selects coexisting compositions ${\varphi }_a,{\varphi }_b$, equivalent to equality of both species' chemical potentials. Local stability to infinitesimal fluctuations instead requiresZero curvature gives the spinodal, while the coexistence boundary is the binodal. Between them a homogeneous state can be metastable and require nucleation; inside the spinodal its small composition fluctuations grow. The symmetric regular solution hasBelow this contact penalty, the free energy is convex at every composition. Above it a miscibility gap is possible around the central composition. If $\chi =A/T$ with $A>0$, heating promotes mixing, but specific interactions or temperature-dependent noncombinatorial terms can produce other trends.

For two polymers, let $N_A,N_B$ count equal-volume segments per chain, and now let $\varphi$ be the segment volume fraction of A. Connecting segments into chains reduces their independent translational entropy. The binary Flory-Huggins polymer blend hasThe energetic contact term remains extensive in the number of segments, while the favorable entropy per segment is reduced by $1/N_A$ and $1/N_B$. This is the key long-chain limit of polymer-blend miscibility: a small positive incompatibility can overcome the mixing entropy. The lattice treatment is a controlled model for this argument, rather than a claim that every real blend has composition-independent $\chi$.^[ref]

The local stability condition and critical interaction parameter becomeFor equal chain lengths, ${\chi }_c=2/N$. Similar segment chemistry and matched cohesive interactions can reduce the unlike-contact penalty; favorable specific interactions, such as complementary hydrogen bonding, can make mixing favorable even for long chains. Lower molar mass increases mixing entropy. temperature can help when it lowers the effective $\chi$, but heating is not universally sufficient. A suitable common solvent can facilitate processing or ternary mixing, and a block-copolymer compatibilizer can stabilize fine domains; neither guarantees that the final solvent-free binary material is one equilibrium molecular phase.

An oil–water emulsion illustrates emulsion stability versus molecular miscibility. A surfactant adsorbs with its water-affine head toward water and its oil-affine tail toward oil, lowering interfacial tension and the work $\gamma dA$ needed to make additional interface. Breaking a fixed oil volume $V_o$ into spherical droplets of radius $R$ gives total area $3V_o/R$, so creating small droplets still requires substantial area and usually mechanical work.

Adsorbed surfactants can oppose droplet coalescence through charged double-layer repulsion, hydrated-head interactions or steric repulsion from sufficiently extended groups. Adequate coverage, adsorption time, ionic conditions and film drainage affect the lifetime. A lower interfacial tension does not itself remove the bulk miscibility gap, and a protected emulsion is commonly kinetically stabilized. A thermodynamically stable microemulsion is a distinct situation requiring an appropriate composition and free-energy balance. The two liquids must not be called molecularly miscible simply because droplets disperse visibly.

For the capillary, balance upward contact-line force against hydrostatic column weight, neglecting meniscus volume and gas density. For inner radius $r$,The contact angle is measured through the liquid. complete wetting means ${\theta }_2=0$, so the contact-angle change of capillary rise givesThis follows from the stipulated wetting change, holding radius, bulk liquid surface tension and density fixed. One should not replace that premise with a general claim that grease always improves glass wetting; if contamination also changes the liquid surface tension, the original height and new angle alone would not determine the new height.

For two wide vertical plates separated by gap $d$, use unit width. There are two upward contact-line contributions and liquid cross-sectional area $d$, giving the capillary rise between parallel plates:Only one principal meniscus curvature contributes between wide plates, whereas a circular tube has two. For the same fluid and angle,If the plate gap equals the tube diameter, $d=2r$, the rise is half as large: the initial and completely wetting heights would be $2.50$ cm and $3.05$ cm. If instead $d=r$, the heights are equal. In either geometry, changing only the angle from $35^{\circ }$ to zero multiplies the height by $1/\cos 35^{\circ }$. Since the question specifies no plate spacing, the general gap-dependent relation is the unambiguous result; the half-height answer requires the equal-gap-to-diameter convention.