Water rises in a capillary tube to a height of $3\, cm$ when one end is dipped vertically in it. If the surface tension of water is $75 \times 10^{-3}\, N/m$,then the diameter of the capillary tube will be....... $mm$. (Assume $g = 10\, m/s^2$)

Why does the water rise in a capillary tube? Explain.

In a capillary tube of radius $R$,a straight thin metal wire of radius $r$ $(R > r)$ is inserted symmetrically,and one end of the combination is dipped vertically in water such that the lower end of the combination is at the same level. The rise of water in the capillary tube is $[T =$ surface tension of water,$\rho =$ density of water,$g =$ gravitational acceleration$]$.

$A$ glass rod of radius $1.0\,mm$ is inserted symmetrically into a vertical capillary tube of radius $2.0\,mm$ such that their lower ends are at the same level. This arrangement is now dipped in water. The height to which water will rise into the tube will be ...... $mm$ (Surface Tension of water $T = 75 \times 10^{-3}\,N/m$,density $\rho = 10^3\,kg/m^3$,$g = 10\,m/s^2$).

Two narrow bores of diameter $5.0 \, mm$ and $8.0 \, mm$ are joined together to form a $U$-shaped tube open at both ends. If this $U$-tube contains water,what is the difference in the level of two limbs of the tube? [Take surface tension of water $T = 7.3 \times 10^{-2} \, Nm^{-1}$,angle of contact $= 0$,$g = 10 \, ms^{-2}$ and density of water $\rho = 1.0 \times 10^{3} \, kg \, m^{-3}$] (in $mm$)

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).