Liquid $A$ rises to a height of $10 \ cm$ in a capillary tube and liquid $B$ falls to a depth of $2 \ cm$ in the same tube. The densities of $A$ and $B$ are $1 \ g/cm^3$ and $10 \ g/cm^3$ respectively. The contact angles of $A$ and $B$ with the tube are $0^{\circ}$ and $135^{\circ}$ respectively. If the surface tensions of $A$ and $B$ are $S_A$ and $S_B$, then the ratio $\frac{S_B}{S_A}$ is:

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

$A$ capillary tube $(A)$ is dipped in water. Another identical tube $(B)$ is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

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):

Water rises up to a height of $4 \,cm$ in a capillary tube. The lower end of the capillary tube is at a depth of $8 \,cm$ below the water level. The mouth pressure required to blow an air bubble at the lower end of the capillary will be '$X$' $cm$ of water,where $X$ is equal to

Water rises to a height of $2\, cm$ in a capillary tube. If the tube is tilted $60^{\circ}$ from the vertical,water will rise in the tube to a length of ...... $cm$.