$A$ 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).

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

In a capillary tube of area of cross-section '$a$',water rises to a height '$h$'. To what height will water rise in a capillary tube of area of cross-section $4a$?

Water rises in a vertical capillary tube up to a height of $2.0 \, cm$. If the tube is inclined at an angle of $60^{\circ}$ with the vertical,then up to what length will the water rise in the tube?

Water rises up to $10 \, cm$ height in a long capillary tube. If this tube is immersed in water so that the height above the water surface is only $8 \, cm$,then:

$A$ narrow glass capillary tube of radius $1.00 \ mm$ is dipped in a trough containing mercury. The amount by which the level of mercury dips down relative to the normal level outside the tube will be approximately $..... \ mm$.
$[$Surface tension of mercury at room temperature $= 0.465 \ N/m$,density of mercury $= 13.6 \times 10^3 \ kg/m^3$,angle of contact with glass $= 135^{\circ}$,$g = 9.8 \ m/s^2]$