Three liquids have the same surface tension and densities $\varrho_{1}, \varrho_{2}$,and $\varrho_{3}$ $(\varrho_{1} > \varrho_{2} > \varrho_{3})$. In three identical capillaries,the rise of liquid is the same. The corresponding angles of contact $\theta_{1}, \theta_{2}$,and $\theta_{3}$ are related as:

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):

In a capillary,the pressure below the curved surface of water will be:

The sap in trees,which consists mainly of water in summer,rises in a system of capillaries of radius $r = 2.5 \times 10^{-5} \ m$. The surface tension of sap is $T = 7.28 \times 10^{-2} \ N/m$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?

In a capillary tube having area of cross-section $A$,the water rises to a height $h$. If the cross-sectional area is reduced to $\frac{A}{9}$,the rise of water in the capillary tube is

If a capillary tube is tilted to $45^o$ and $60^o$ from the vertical,then the ratio of the lengths $l_1$ and $l_2$ of the liquid columns in it will be: