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(B) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.From this expression,we can see that $h \pro

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

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(B) The height of the liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{r ho g}$.From this expression,we can see that $h \propto \frac{1}{g}$.Therefore,the ratio of the heights is $\frac{h_{1}}{h_{2}} = \frac{g_{2}}{g_{1}}$.At a depth $d$ below the surface of the Earth,the acceleration due to gravity is given by $g_{2} = g_{1} \left(1 - \frac{d}{R}ight)$,where $g_{1}$ is the acceleration due to gravity at the surface.Substituting this into the ratio,we get $\frac{h_{1}}{h_{2}} = \frac{g_{1}(1 - d/R)}{g_{1}} = 1 - \frac{d}{R}$.

$A$ capillary tube is vertically immersed in water,and water rises up to a height $h_{1}$. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h_{2}$. The ratio $h_{1} / h_{2}$ is ($R =$ radius of earth).

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