$A$ glass capillary tube of inner diameter $0.28 \,mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that the water level in the tube is the same as that in the vessel is ............ $\times 10^3 \,N/m^2$ (surface tension of water $= 0.07 \,N/m$ and atmospheric pressure $= 10^5 \,N/m^2$).

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):

$A$ glass capillary of radius $0.35 \ mm$ is inclined at $60^{\circ}$ with the vertical in water. The length of the water column in the capillary is (surface tension of water $= 7 \times 10^{-2} \ N/m$,acceleration due to gravity $g = 10 \ m/s^2$,$\cos 0^{\circ} = 1$,$\cos 60^{\circ} = 0.5$,density of water $\rho = 10^3 \ kg/m^3$) (in $cm$)

Water rises to a height $h$ in a capillary tube at the surface of the Earth. On the surface of the moon,the height of the water column in the same capillary tube will be:

Radius of a capillary is $2 \times 10^{-3} \,m$. $A$ liquid of weight $6.2 \times 10^{-4} \,N$ may remain in the capillary. Then the surface tension of the liquid will be $:-$

In a capillary tube of area of cross-section $a$,water rises to a height $h$. To what height will water rise in a capillary tube of area of cross-section $4a$?