A student compares the equilibrium capillary rise of three straight-chain alcohols in identical narrow tubes coated with the same polar material. Measurements at \(25^{\circ}\text{C}\) are shown in the table. The contact angle \(\theta\) reflects the balance between adhesive attractions to the coating and cohesive attractions within the liquid.

Sample	Alcohol	Surface tension, \(\gamma\) (\(\text{mN/m}\))	Density, \(\rho\) (\(\text{g/mL}\))	\(\theta\)	\(\cos\theta\)
\(\mathrm{P}\)	\(\text{1-propanol}\), \(\text{CH}_3(\text{CH}_2)_2\text{OH}\)	\(24\)	\(0.80\)	\(0^{\circ}\)	\(1.00\)
\(\mathrm{H}\)	\(\text{1-hexanol}\), \(\text{CH}_3(\text{CH}_2)_5\text{OH}\)	\(26\)	\(0.81\)	\(30^{\circ}\)	\(0.87\)
\(\mathrm{N}\)	\(\text{1-nonanol}\), \(\text{CH}_3(\text{CH}_2)_8\text{OH}\)	\(29\)	\(0.83\)	\(60^{\circ}\)	\(0.50\)

For a liquid in a capillary tube of radius \(r\), the equilibrium rise \(h\) is given by

\[ h=\dfrac{2\gamma\cos\theta}{\rho g r} \]

Which choice correctly ranks, from greatest to least, both the surface tensions and the equilibrium capillary rises and gives the correct molecular-level explanation?