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

According to Poiseuille's law,the pressure drop per unit length required to overcome viscous forces is $\Delta P = \frac{8 \eta v}{r^2}$,where $r$ is the radius of the cross-section,$v$ is the fluid velocity,and $\eta$ is the coefficient of viscosity. $A$ capillary tube of radius $a$ is dipped in a liquid of density $\rho$,surface tension $T$,and coefficient of viscosity $\eta$. The liquid starts rising in it so that its height $h(t)$ is a function of time $t$. The resulting rate of change of the momentum of the liquid column in the capillary (taking vertically up to be the positive direction and the contact angle to be close to $0^{\circ}$) is $-\pi a^2 \rho gh + F$. Then $F$ is ($g$ is the acceleration due to gravity):

Water rises to a height of $2.2 \text{ cm}$ in a glass capillary tube. The height to which the same water rises in another capillary having $\frac{1}{4}$ of the cross-sectional area is: (in $\text{ cm}$)

Water rises to a height of $20 \ mm$ in a capillary tube. If the radius of the capillary tube is made $\frac{1}{3}rd$ of its previous value,what is the new height of the capillary rise (in $mm$)?

Water rises up to $10 \, cm$ height in a long capillary tube. If this tube is immersed in water so that the height above the water surface is only $8 \, cm$,then:

Two tubes of same length and diameters $4 \ mm$ and $8 \ mm$ are joined together to form a $U$-shaped tube open at both the ends. If the $U$-tube contains water,then the difference between the levels of water in the two limbs of the tube is (Surface tension of water at the temperature of experiment is $7.3 \times 10^{-2} \ N \ m^{-1}$,angle of contact $= 0^{\circ}$,density of water $= 1.0 \times 10^3 \ kg \ m^{-3}$ and acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $mm$)