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

Three liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$),having the same value of surface tension $T$,rise to the same height in three identical capillaries. The angles of contact $\theta_1, \theta_2$ and $\theta_3$ obey:

During the capillary rise of a liquid in a capillary tube,the surface of contact that remains constant is of

One end of a capillary tube is dipped in water,the rise of water column is $h$. The upward force of $98 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of the water column. The inner circumference of the capillary is (surface tension of water $= 7 \times 10^{-2} \text{ Nm}^{-1}$) (in $\text{ cm}$)

The action of a nib split at the top is explained by

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 $mm$)