Water rises to a height of 10 cm in a capilla | vedclass

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(C) The height of liquid in a capillary tube is given by $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is surface tension,$r$ is the radius,$d$ is densit

Vedclass · Accepted Answer

$1:6$

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(C) The height of liquid in a capillary tube is given by $h = \frac{2T \cos 	heta}{rdg}$,where $T$ is surface tension,$r$ is the radius,$d$ is density,$g$ is acceleration due to gravity,and $	heta$ is the angle of contact.Rearranging for surface tension: $T = \frac{hrdg}{2 \cos 	heta}$.For water $(1)$ and mercury $(2)$: $\frac{T_1}{T_2} = \frac{h_1}{h_2} 	imes \frac{d_1}{d_2} 	imes \frac{\cos 	heta_2}{\cos 	heta_1}$.Given: $h_1 = 10 	ext{ cm}$,$h_2 = -3.112 	ext{ cm}$ (depression),$d_1 = 1 	ext{ g/cm}^3$,$d_2 = 13.6 	ext{ g/cm}^3$,$	heta_1 = 0^o$,$	heta_2 = 135^o$.Note: For capillary rise/fall,we consider the magnitude of height: $|h_2| = 3.112 	ext{ cm}$.$\frac{T_1}{T_2} = \frac{10}{3.112} 	imes \frac{1}{13.6} 	imes \frac{\cos(135^o)}{\cos(0^o)} = \frac{10}{3.112 	imes 13.6} 	imes \frac{1}{\sqrt{2}} \approx \frac{10}{42.32} 	imes 0.707 \approx \frac{1}{6}$.

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