The maximum length of a water column that can stay without falling in a vertically held capillary tube of diameter $1 \ mm$ and open at both ends is (Acceleration due to gravity $= 10 \ ms^{-2}$ and surface tension of water $= 0.07 \ Nm^{-1}$) (in $cm$)

Radius of a capillary is $2 \times 10^{-3} \,m$. $A$ liquid of weight $6.2 \times 10^{-4} \,N$ may remain in the capillary. Then the surface tension of the liquid will be $:-$

The angle of contact between glass and water is $0^\circ$ and it rises in a capillary up to $6 \text{ cm}$ when its surface tension is $70 \text{ dynes/cm}$. Another liquid of surface tension $140 \text{ dynes/cm}$,angle of contact $60^\circ$ and relative density $2$ will rise in the same capillary by ........ $\text{cm}$.

What is capillary action? Derive the formula for the rise of liquid in a capillary tube immersed vertically in a liquid.

$A$ 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 $\rho = 1000 \, kg/m^3$,$g = 9.8 \, m/s^2$)

Two narrow bores of diameters $3.0 \;mm$ and $6.0 \;mm$ are joined together to form a $U$-tube open at both ends. If the $U$-tube contains water,what is the difference (in $mm$) in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3 \times 10^{-2} \;N m^{-1}$. Take the angle of contact to be zero and density of water to be $1.0 \times 10^{3} \;kg m^{-3}$ $(g = 9.8 \;m s^{-2})$.