$A$ $U$-tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2} \, N/m$. The angle of contact is zero and the density of water is $10^3 \, kg/m^3$. If $g = 10 \, m/s^2$,then the difference in the liquid levels of the two limbs is:

$A$ capillary tube is taken from the Earth's surface to the Moon's surface. What happens to the rise of the liquid column on the Moon's surface? (Acceleration due to gravity on the Earth's surface is six times that of the Moon's surface.)

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):

$A$ $20 \ cm$ 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 the water column in the capillary tube will be (in $cm$)?

Water rises up to height $x$ in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to height $Y$. If $R$ is the radius of the earth,then the ratio $Y:x$ is

$A$ glass capillary tube of internal radius $r = 0.25 \, mm$ is immersed in water. The top end of the tube projects $2 \, cm$ above the surface of the water. At what angle does the liquid meet the tube (in $^{\circ}$)? (Surface tension of water $T = 0.07 \, N/m$,density $\rho = 1000 \, kg/m^3$,$g = 9.8 \, m/s^2$)