Two capillary tubes $P$ and $Q$ are dipped in water. The height of the water level in capillary $P$ is $2/3$ of the height in capillary $Q$. The ratio of their diameters is:

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

$A$ $U$-tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2} \, N/m$. The angle of contact is zero and the density of water is $10^3 \, kg/m^3$. If $g = 10 \, m/s^2$,then the difference in the liquid levels of the two limbs is:

When a capillary tube is dipped vertically in water,the water rises to a height of $2.0 \, cm$. If the capillary tube is tilted at an angle of $60^\circ$ with the vertical,what will be the length of the water column in the capillary tube in $cm$?

Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. The rise of liquid in two capillaries is $h_1$ and $h_2$ respectively. If the surface tensions of liquids are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ is (Assume that their angles of contact are same).

$A$ glass capillary tube is in the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically,water rises in it to a height $h$,where the radius of its cross section is $b$. If the surface tension of water is $S$,its density is $\rho$,and its contact angle with glass is $\theta$,the value of $h$ will be ($g$ is the acceleration due to gravity).