Sap has a density of 1050 kg/m^3, a contact angle of zero and the surface tension of water at 20.0 degrees Celsius, 0.0728 N/m. Capillary rise obeys h = 2 gamma cos(theta) / (rho g r). Taking g as 9.80 m/s^2, what tube radius would raise sap 100 m?

1. Rearrange the capillary relation for the radius: r = 2 gamma cos(theta) / (rho g h). 2. The contact angle is zero, so cos(theta) = 1 and the numerator is simply 2 gamma. 3. Numerator = 2(0.0728 N/m)(1) = 0.1456 N/m. 4. Denominator = (1050 kg/m^3)(9.80 m/s^2)(100 m) = 1.029 x 10^6 N/m^3. 5. r = (0.1456 N/m) / (1.029 x 10^6 N/m^3) = 1.41 x 10^-7 m. 6. Real xylem tubes have radii as small as 2.50 x 10^-5 m, about 180 times larger, so capillary action alone cannot lift sap to the top of a redwood. 7. The value 2.50 x 10^-5 m is that xylem radius offered as a trap, while 1.41 x 10^-5 m and 7.07 x 10^-7 m come from dropping the height or the factor of two. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action_