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(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos 	heta}{r ho g}$.Given that $h, T, r$,and $g$ are

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$	heta_1 < 	heta_2 < 	heta_3$

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(C) The height of the liquid rise in a capillary tube is given by the formula: $h = \frac{2T \cos heta}{r ho g}$.Given that $h, T, r$,and $g$ are constant for all three liquids.Therefore,$\frac{\cos heta}{ ho} = ext{constant}$.This implies: $\frac{\cos heta_1}{ ho_1} = \frac{\cos heta_2}{ ho_2} = \frac{\cos heta_3}{ ho_3}$.Since $ ho_1 > ho_2 > ho_3$,it follows that $\cos heta_1 > \cos heta_2 > \cos heta_3$.Because the cosine function is a decreasing function in the range $[0, \pi/2]$,a larger cosine value corresponds to a smaller angle.Therefore,$ heta_1 < heta_2 < heta_3$.

Three liquids have the same surface tension and densities $\rho_1, \rho_2$,and $\rho_3$ $(\rho_1 > \rho_2 > \rho_3)$. In three identical capillaries,the rise of the liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$,and $\theta_3$ are related as:

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