$A$ $20\, cm$ 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 ...... $cm$.

$A$ capillary tube is vertically immersed in water,and water rises up to a height $h_{1}$. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h_{2}$. The ratio $h_{1} / h_{2}$ is ($R =$ radius of earth).

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 $mm$)

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

$A$ 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 $kg$):

Water rises up to a height $h$ in a capillary tube of a certain diameter. This capillary tube is replaced by a similar tube of half the diameter. Now,the water will rise to a height of: