The sap in a tree rises in a system of capill | vedclass

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(B) Given: Radius of capillary $r = 2.5 	imes 10^{-5} \, m$,Surface tension $S = 7.28 	imes 10^{-2} \, N \, m^{-1}$,Angle of contact $	heta = 0^{\c

Vedclass · Accepted Answer

$0.59$

Vedclass · Answer

(B) Given: Radius of capillary $r = 2.5 	imes 10^{-5} \, m$,Surface tension $S = 7.28 	imes 10^{-2} \, N \, m^{-1}$,Angle of contact $	heta = 0^{\circ}$,Density $ho = 10^3 \, kg \, m^{-3}$,Acceleration due to gravity $g = 9.8 \, m \, s^{-2}$.The formula for the height of capillary rise is given by $h = \frac{2 S \cos 	heta}{r ho g}$.Substituting the values:$h = \frac{2 	imes (7.28 	imes 10^{-2}) 	imes \cos 0^{\circ}}{(2.5 	imes 10^{-5}) 	imes 10^3 	imes 9.8}$.Since $\cos 0^{\circ} = 1$:$h = \frac{14.56 	imes 10^{-2}}{2.5 	imes 10^{-2} 	imes 9.8} = \frac{14.56}{2.5 	imes 9.8} = \frac{14.56}{24.5} \approx 0.594 \, m$.Rounding to two decimal places,the height is $0.59 \, m$.

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