Liquid rises to a height $2 \ cm$ in a capillary tube; in that case,the angle of contact between the solid and the liquid is $0^{\circ}$. The tube is lowered more now,so that the capillary is only $1 \ cm$ above the liquid. In this case,the angle of contact between the solid and liquid is $......^{\circ}$.

Water rises to a height of $3 \,cm$ in a capillary tube. If the cross-sectional area of the capillary tube is reduced to $1/9$th of its initial area, then the water will rise to a height of: (in $\,cm$)

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):

The kerosene oil rises up in the wick of a lamp ...........

When a capillary tube is dipped vertically in water,the rise of water in the capillary is $h$. The angle of contact is $0^{\circ}$. Now,the tube is depressed so that its length above the water surface is $\frac{h}{3}$. The new apparent angle of contact is $(\cos 0^{\circ} = 1)$.

Kerosene oil rises up the wick in a lantern due to: