Question #43916

Two wide, parallel vertical plates are partially immersed in a liquid of density $\rho$ and surface tension $S$. The left plate is rigid and fixed. The right plate is a flexible bimetallic strip, rigidly clamped at the level of the free surface of the reservoir ($y=0$). Initially, the system is at temperature $T_0$, the plates are perfectly parallel with separation $d_0$, and the liquid rises to a height $h_0$ ($h_0 \gg d_0$). The true contact angle of the liquid with both plates is $\theta$ ($0 < \theta < \pi/2$).

The liquid in the reservoir is then heated to $T_0 + \Delta T$. The bimetallic strip reaches thermal equilibrium with the liquid only up to the new capillary height $h$. The portion above $h$ remains at $T_0$. Due to heating, the immersed portion of the strip develops a uniform curvature $\kappa$ (bending outward, away from the fixed plate). Assuming $\kappa$ is very small ($\kappa h_0 \ll 1$), the new equilibrium height is $h = h_0 + \Delta h$.

Which of the following gives the first-order change in height, $\Delta h$? (Assume $S$, $\rho$, and $\theta$ remain constant).