By inserting a capillary tube up to a depth $l$ in water,the water rises to a height $h$. If the lower end of the capillary is closed inside water,the capillary is taken out,and then the closed end is opened,to what height will the water remain in the tube?

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

Water rises in a capillary tube up to a height of $4 \, cm$. If it is tilted to $30^{\circ}$ from the vertical,then the length of the water column in it will be

In a capillary tube experiment, a vertical $30 \text{ cm}$ long capillary tube is dipped in water. Water rises up to a height of $10 \text{ cm}$ due to capillarity. If this experiment is conducted in a freely falling elevator, then the length of the water column becomes

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. What will be the mass of water that rises in a capillary tube of radius $\frac{r}{4}$?

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):