In a capillary tube of radius '$R$',a straight thin metal wire of radius '$r$' is inserted symmetrically and one end of the combination is dipped vertically in water such that the lower end of the capillary and the thin wire are at the same level. If '$T$' is the surface tension of water and '$\rho$' is the density of water, then the rise of water in the capillary is:

$A$ uniform capillary tube of length $l$ and inner radius $r$ with its upper end sealed is submerged vertically into water. The outside pressure is $p_{0}$ and surface tension of water is $\gamma$. When a length $x$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of $x$ is

$A$ capillary tube of radius $R$ is dipped in water,and water rises to a height $H$. The mass of water in the capillary tube is $M$. If a capillary tube of half the radius is dipped in water,what will be the mass of the water in the capillary tube?

$A$ $20 cm$ 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 $cm$)

The surface tension for pure water in a capillary tube experiment is

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):