The sap in trees,which consists mainly of water in summer,rises in a system of capillaries of radius $r = 2.5 \times 10^{-5} \ m$. The surface tension of sap is $T = 7.28 \times 10^{-2} \ N/m$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?

(N/A) The height $h$ to which a liquid rises in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{r \rho g}$.
Given values are: $T = 7.28 \times 10^{-2} \ N/m$,$r = 2.5 \times 10^{-5} \ m$,$\theta = 0^{\circ}$,$\rho = 10^3 \ kg/m^3$ (density of water),and $g = 9.8 \ m/s^2$.
Substituting these values into the formula:
$h = \frac{2 \times (7.28 \times 10^{-2}) \times \cos 0^{\circ}}{(2.5 \times 10^{-5}) \times 10^3 \times 9.8}$
$h = \frac{14.56 \times 10^{-2}}{2.5 \times 10^{-2} \times 9.8} = \frac{14.56}{24.5} \approx 0.594 \ m \approx 0.6 \ m$.
Since the height reached by capillary action is only about $0.6 \ m$,it is insufficient to supply water to the tops of tall trees (which can be $10 \ m$ to $100 \ m$ high). Therefore,surface tension alone cannot account for the supply of water to the top of all trees.