Water rises up to a height h in a capillary t | vedclass

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

Vedclass · Accepted Answer

$1-\frac{d}{R}$

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(B) The height of water in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$,where $T$ is surface tension,$r$ is the radius of the tube,$ho$ is the density of water,and $g$ is the acceleration due to gravity.On the surface of the earth,$h = \frac{2T \cos 	heta}{r ho g}$.At a depth $d$ inside a mine,the acceleration due to gravity becomes $g^{\prime} = g(1 - \frac{d}{R})$.The new height $h^{\prime}$ at depth $d$ is $h^{\prime} = \frac{2T \cos 	heta}{r ho g^{\prime}} = \frac{2T \cos 	heta}{r ho g(1 - \frac{d}{R})}$.Therefore,the ratio $\frac{h}{h^{\prime}}$ is given by:$\frac{h}{h^{\prime}} = \frac{\frac{2T \cos 	heta}{r ho g}}{\frac{2T \cos 	heta}{r ho g(1 - \frac{d}{R})}} = 1 - \frac{d}{R}$.

Water rises up to a height $h$ in a capillary tube immersed vertically in water. When this whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h^{\prime}$. If $R$ is the radius of the earth,then the ratio $\frac{h}{h^{\prime}}$ is

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