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Question

(A) The height of the water column in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	h

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$\left( 1 - \frac{d}{R} \right)$

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(A) The height of the water column in a capillary tube is given by the formula $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$	heta$ is the contact angle,$r$ is the radius of the tube,$ho$ is the density of water,and $g$ is the acceleration due to gravity.Since $T, 	heta, r,$ and $ho$ are constants,the height $h$ is inversely proportional to $g$,i.e.,$h \propto \frac{1}{g}$.At the surface of the earth,the height is $x = \frac{k}{g}$,where $k = \frac{2T \cos 	heta}{r ho}$.At a depth $d$ inside a mine,the acceleration due to gravity becomes $g' = g \left( 1 - \frac{d}{R} ight)$.The new height is $y = \frac{k}{g'} = \frac{k}{g \left( 1 - \frac{d}{R} ight)}$.Therefore,the ratio $\frac{x}{y} = \frac{k/g}{k / [g(1 - d/R)]} = 1 - \frac{d}{R}$.

$A$ capillary tube is immersed vertically in water and the height of the water column is $x$. When this arrangement is taken into a mine of depth $d$,the height of the water column is $y$. If $R$ is the radius of earth,the ratio $\frac{x}{y}$ is

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