The correct relation for the capillary rise $h$ in a tube of radius $r$ is:

Two tubes of same length and diameters $4 \ mm$ and $8 \ mm$ are joined together to form a $U$-shaped tube open at both the ends. If the $U$-tube contains water,then the difference between the levels of water in the two limbs of the tube is (Surface tension of water at the temperature of experiment is $7.3 \times 10^{-2} \ N \ m^{-1}$,angle of contact $= 0^{\circ}$,density of water $= 1.0 \times 10^3 \ kg \ m^{-3}$ and acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $mm$)

If a capillary tube of radius $1 \,mm$ is immersed in water, the mass of water rising in the capillary tube is $m$. If the radius of the capillary tube is doubled, then the mass of water that rises in the same capillary tube will be

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

Three liquids have the same surface tension and densities $\varrho_{1}, \varrho_{2}$,and $\varrho_{3}$ $(\varrho_{1} > \varrho_{2} > \varrho_{3})$. 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:

$A$ $20\, cm$ long capillary tube is dipped in water. The water rises up to $8\, cm$. If the entire arrangement is put in a freely falling elevator,the length of water column in the capillary tube will be ...... $cm$.