In a static meniscus, hydrostatic pressure variation $\rho g\ell$ balances capillary pressure $\gamma/\ell$, giving the <capillary length>
$$
\boxed{\ell=\sqrt{\frac\gamma{\rho g}}}.
$$
The small parameter satisfies
$$
\epsilon^3=\frac{h_0^2}{\ell^2},
\qquad
\boxed{\frac{h_0}{\ell}=\epsilon^{3/2}\ll1}.
$$
It is reasonable to match to a nearly static meniscus because the bulk pool has negligible thin-film viscous resistance, while its local capillary-hydrostatic shape approaches the vertical wall almost tangentially.

Since $h=h_0H$ and $x=h_0X/\epsilon$, dimensional curvature is
$$
h_{xx}=\frac{\epsilon^2}{h_0}H''.
$$
Matching it to $(2\rho g/\gamma)^{1/2}=\sqrt2/\ell$ gives
$$
\boxed{H''\sim\sqrt2\,\epsilon^{-1/2}=(\epsilon/2)^{-1/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 329 / 3 / c / Solution

{title2=$\ell_c=\sqrt{\gamma/(\rho g)}$}
{wiki}

The capillary length is the scale at which hydrostatic pressure and capillary pressure balance.


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 Incoming links: Capillary length