A capillary tube of radius 0.1 ,mm is dipped | vedclass

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(B) Given: Radius of capillary tube $r = 0.1 \,mm = 0.1 	imes 10^{-3} \,m$.Height of water rise $h = 2 \,cm = 2 	imes 10^{-2} \,m$.Surface tension $

Vedclass · Accepted Answer

$	heta = \cos^{-1}\left(\frac{1}{7.2}ight)$

Vedclass · Answer

(B) Given: Radius of capillary tube $r = 0.1 \,mm = 0.1 	imes 10^{-3} \,m$.Height of water rise $h = 2 \,cm = 2 	imes 10^{-2} \,m$.Surface tension $T = 0.072 \,N/m$.Density of water $ho = 10^3 \,kg/m^3$ and acceleration due to gravity $g = 10 \,m/s^2$.The formula for capillary rise is $h = \frac{2T \cos 	heta}{ho g r}$.Rearranging for $\cos 	heta$: $\cos 	heta = \frac{h ho g r}{2T}$.Substituting the values: $\cos 	heta = \frac{(2 	imes 10^{-2}) 	imes (10^3) 	imes (10) 	imes (0.1 	imes 10^{-3})}{2 	imes 0.072}$.$\cos 	heta = \frac{2 	imes 10^{-2} 	imes 10^4 	imes 0.1 	imes 10^{-3}}{0.144} = \frac{0.02}{0.144} = \frac{20}{144} = \frac{1}{7.2}$.Therefore, $	heta = \cos^{-1}\left(\frac{1}{7.2}ight)$.

$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

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