Question #42117

A capillary tube of radius $r$ is kept vertically in a large liquid bath. The liquid and the tube are initially at temperature $T_0$ and cool down to the ambient temperature $T_a$ according to Newton's law of cooling with a cooling constant $k$. The surface tension $\gamma$ and contact angle $\theta$ depend on temperature such that the quasi-static equilibrium height of the liquid column is $h_{eq}(T) = h_a - \beta(T - T_a)$, where $h_a$ is the equilibrium height at $T_a$ and $\beta$ is a positive constant. Assuming the mass of the liquid column remains approximately constant as $m = \pi r^2 \rho h_a$ (where $\rho$ is the liquid density) and neglecting viscous damping, the actual height of the liquid column can be expressed as $h(t) = h_{osc}(t) + h_{non-osc}(t)$, where $h_{osc}(t)$ is the oscillatory component. Which of the following represents $h_{non-osc}(t)$?