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

$A$ uniform capillary tube of length $l$ and inner radius $r$ with its upper end sealed is submerged vertically into water. The outside pressure is $p_{0}$ and surface tension of water is $\gamma$. When a length $x$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of $x$ is

The radius of a capillary tube is $2 \times 10^{-3} \ m$. If a liquid of weight $6.28 \times 10^{-4} \ N$ can be supported in the capillary,what is the surface tension of the liquid?

$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

$A$ capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5 \, g$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is $........ \, g$.

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 $mm$)