The action of a nib split at the top is explained by

$A$ 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

$A$ capillary tube of radius $0.1 \ mm$ is partly dipped in water (surface tension $70 \ dyn/cm$ and glass-water contact angle $\simeq 0^{\circ}$) inclined at $30^{\circ}$ with the vertical. The length of water risen in the capillary is . . . . . . $cm$. (Take $g = 980 \ cm/s^2$)

$A$ $U$-tube is such that the diameter of one limb is $0.4\,mm$ and that of the other is $d\,mm$. If the surface tension of water contained in the tube is $0.07\,N/m$ and the difference in the levels of liquid in the limbs is $3.6\,cm$,then the value of $d$ is:

In a capillary tube having area of cross-section $A$,the water rises to a height $h$. If the cross-sectional area is reduced to $\frac{A}{9}$,the rise of water in the capillary tube is

Liquid $A$ rises to a height of $10 \ cm$ in a capillary tube and liquid $B$ falls to a depth of $2 \ cm$ in the same tube. The densities of $A$ and $B$ are $1 \ g/cm^3$ and $10 \ g/cm^3$ respectively. The contact angles of $A$ and $B$ with the tube are $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tensions of $A$ and $B$ are $S_A$ and $S_B$,then the ratio $\frac{S_B}{S_A}$ is: