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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 tress?

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r = radius of the capillary

r = 2.5 \times 10^{-5} m

S=T = 7.28 \times 10^{-2} N/m

g = 9.8 \frac{m}{s^{2}}

\theta =0, p = 1000 kg/m3

now, h =\frac{2 s \cos \theta}{rpg} = 2 \times 7.28 \times 10^{-2} \frac{\cos 0}{2.5 \times 10^{-5}\times 1000 \times 9.8}

 

h = \frac{104}{175}\times \frac{1000}{1000} = \frac{104}{175} = 0.594 \approx 0.6m

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