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 $4a$?

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):

Given below are two statements:
Statement $I$: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well.
Statement $II$: The rise of a liquid in a capillary tube does not depend on the inner radius of the tube.
In the light of the above statements,choose the correct answer from the options given below:

In a capillary tube,water rises to $3\, mm$. The height of water that will rise in another capillary tube having one-third radius of the first is ........ $mm$.

In a capillary tube of radius $R$,a straight thin metal wire of radius $r$ $(R > r)$ is inserted symmetrically,and one end of the combination is dipped vertically in water such that the lower end of the combination is at the same level. The rise of water in the capillary tube is $[T =$ surface tension of water,$\rho =$ density of water,$g =$ gravitational acceleration$]$.

In a capillary tube,water rises by $1.2 \ mm$. The height of water that will rise in another capillary tube having half the radius of the first is ........ $mm$.