Three liquids have the same surface tension and densities $\rho_1, \rho_2$ and $\rho_3$ $(\rho_1 < \rho_2 < \rho_3)$. In three identical capillaries, the rise of liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$ and $\theta_3$ are related as:

$A$ glass capillary tube of internal radius $r = 0.25 \, mm$ is immersed in water. The top end of the tube projects $2 \, cm$ above the surface of the water. At what angle does the liquid meet the tube (in $^{\circ}$)? (Surface tension of water $T = 0.07 \, N/m$,density $\rho = 1000 \, kg/m^3$,$g = 9.8 \, m/s^2$)

$A$ 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).

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio $4:3$. If their surface tensions are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ of liquids in the two capillary tubes is (Their angle of contacts are same).

$A$ capillary tube made of glass with a radius of $0.15\, mm$ is dipped vertically in a beaker filled with methylene iodide (surface tension $= 0.05\, N m^{-1}$,density $= 667\, kg m^{-3}$),which rises to a height $h$ in the tube. It is observed that the two tangents drawn from the liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^{\circ}$ with one another. Then $h$ is close to $...... m$ $(g = 10\, m s^{-2})$