Water rises up to height x in a capillary tub | vedclass

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(B) The height of water 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

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$R:(R-d)$

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(B) The height of water 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 constant,we have $h \propto \frac{1}{g}$.At the surface of the earth,the acceleration due to gravity is $g_0$. Thus,$x = \frac{k}{g_0}$.At a depth $d$ inside a mine,the acceleration due to gravity is given by $g_d = g_0(1 - \frac{d}{R}) = g_0(\frac{R-d}{R})$.Thus,the new height is $Y = \frac{k}{g_d} = \frac{k}{g_0(\frac{R-d}{R})} = \frac{k}{g_0} \cdot \frac{R}{R-d}$.Substituting $x = \frac{k}{g_0}$,we get $Y = x \cdot \frac{R}{R-d}$.Therefore,the ratio $Y:x = \frac{R}{R-d}$,which is $R:(R-d)$.

Water rises up to height $x$ in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to height $Y$. If $R$ is the radius of the earth,then the ratio $Y:x$ is

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