Question

a long capillary tube of radius r is put in contact with surface of a completely wtting liquid of density d and ver low viscosity. what is the range for the height of liquid at any time

Answer 1

The height of the liquid at any time lies in the range

$$\sqrt{\frac{2\gamma}{\rho\,r}}\,t \leq h(t) \leq \frac{2\gamma}{\rho\,g\,r}\,.$$

Answer 2

We start by recalling that when a capillary tube is brought in contact with a completely wetting liquid (contact angle θ = 0), the liquid rises until the capillary (surface‐tension) force is balanced by the weight of the liquid. In the static (equilibrium) case the balance is

$$2\pi r\,\gamma = \rho\,g\;\pi r^2\,h_{\rm eq}\,,$$

so that the equilibrium height is

$$h_{\rm eq}=\frac{2\gamma}{\rho\,g\,r}\,.$$

For a liquid with very low viscosity the rise is not monotonic but is governed by inertia. If we ignore gravity for very early times, the capillary force accelerates the mass of liquid; a scaling analysis yields

$$h(t)\sim \sqrt{\frac{2\gamma}{\rho\,r}}\,t\,.$$

Thus, during the inertial “startup” phase the liquid rises approximately linearly with time according to this relation. However, eventually gravity becomes important and the liquid does not rise indefinitely but rather “overshoots” and oscillates about the equilibrium height. Hence, the liquid height at any time will always lie between its initial inertial rise and the static (Jurin’s) height.

In summary, one acceptable answer is that the liquid height obeys

$$\sqrt{\frac{2\gamma}{\rho\,r}}\,t\quad\text{(inertial limit)} \quad\le h(t)\le \frac{2\gamma}{\rho\,g\,r}\quad\text{(equilibrium)}\,.$$