Two vertical parallel plates are partially submerged in water. The distance between the plates is equal to $d$. Water rises due to surface tension $T$,the width of the plates is $l$,and the contact angle of water with glass is $0^o$. Find the force of attraction between the plates.

If water rises in a capillary tube up to $3 \, cm$,what is the diameter of the capillary tube? (Surface tension of water $T = 7.2 \times 10^{-2} \, N/m$,density $\rho = 10^3 \, kg/m^3$,$g = 10 \, m/s^2$)

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):

$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

When one end of a capillary tube is dipped in water,the height of the water column is $h$. The upward force of $108 \ dyne$ due to surface tension is balanced by the force due to the weight of the water column. What is the inner circumference of the capillary (in $cm$)? (Surface tension of water $T = 7.2 \times 10^{-2} \ N/m$)

$A$ $20 cm$ long capillary tube is dipped vertically in water and the liquid rises up to $10 cm$. If the entire system is kept in a freely falling platform, the length of the water column in the tube will be (in $cm$)