Three liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1 \rho_2 \rho_3$), having the same va

Question

Three liquids of densities $ho_1, ho_2$ and $ho_3$ (with $ho_1 > ho_2 > ho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $	heta_1, 	heta_2$ and $	heta_3$ obey:

Accepted Answer

1. **Capillary Rise Formula:** The height $h$ to which a liquid rises in a capillary tube of radius $r$ is given by the formula: $$ h = \frac{2T \cos heta}{r ho g} $$ where $T$ is surface tension, $ heta$ is the angle of contact, $ ho$ is the density of the liquid, and $g$ is acceleration due to gravity . 2. **Analyze Dependencies:** We are given that $h, T, r,$ and $g$ are the same for all three liquids. Rearranging the formula for the variable terms: $$ \cos heta = \frac{h r g}{2T} \cdot ho $$ Since $\frac{h r g}{2T}$ is a constant ($K$), we have: $$ \cos heta = K ho $$ This implies $\cos heta \propto ho$. 3. **Apply Conditions:** We are given densities $ ho_1 > ho_2 > ho_3$. Therefore: $$ \cos heta_1 > \cos heta_2 > \cos heta_3 $$ 4. **Behavior of Cosine:** For capillary rise ($h>0$), the angle of contact $ heta$ must be acute, i.e., $0 \le heta < \frac{\pi}{2}$. In this range, the cosine function is a decreasing function (as angle increases, cosine value decreases). Therefore, the inequality for the angles reverses: $$ heta_1 < heta_2 < heta_3 $$ 5. **Conclusion:** The correct relationship is $0 \le heta_1 < heta_2 < heta_3 < \frac{\pi}{2}$.

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