The lower end of a glass capillary tube is dipped in water. Water rises to a height of $8 \ cm$. The tube is then broken at a height of $6 \ cm$. The new height of water column and new angle of contact will be $:-$

$A$ capillary tube of radius '$r$' is immersed in water and water rises to a height of '$h$'. The mass of water in the capillary tube is $5 \times 10^{-3} \ kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ}$. The mass of liquid which rises into the capillary tube now is (in $kg$):

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

Ploughing helps to retain water by soil by:

$A$ capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$. The mass of water in the capillary tube is $M$. If the radius of the tube is halved,the mass of water that will rise in the capillary tube will now 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