$A$ capillary tube is placed at angles of $30^o$ and $60^o$ with the vertical. What is the ratio of the length of the liquid column in the capillary tube?

The lower end of a capillary tube is dipped into water and it is observed that the water in the capillary tube rises by $7.5 \ cm$. Find the radius of the capillary tube used,if the surface tension of water is $7.5 \times 10^{-2} \ N \ m^{-1}$. The angle of contact between water and glass is $0^{\circ}$ and the acceleration due to gravity is $10 \ m \ s^{-2}$.

When one end of a capillary tube is dipped in water,the height of water column is $h$. The upward force of $105 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of water column. The inner circumference of the capillary tube is (Surface tension of water $= 7 \times 10^{-2} \text{ N/m}$) (in $\text{ cm}$)

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 $\,cm$)

In a $U$-shaped tube,the radius of one limb is $2 \ mm$ and that of the other limb is $4 \ mm$. $A$ liquid of surface tension $0.03 \ Nm^{-1}$,density $1500 \ kgm^{-3}$,and angle of contact zero is taken in the tube. The difference in the heights of the levels of the liquid in the two limbs is (Acceleration due to gravity $= 10 \ ms^{-2}$) (in $mm$)

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):