Two capillary tubes of length $L$ and $2L$ and of radius $R$ and $2R$ are connected in series. The net rate of flow of fluid through them will be (given rate of flow through a single capillary,$X = \frac{\pi P R^4}{8 \eta L}$)

Two capillary tubes of different diameters are dipped in water. The rise of water is

Water rises up to $10 \, cm$ height in a long capillary tube. If this tube is immersed in water so that the height above the water surface is only $8 \, cm$,then:

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 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 $10 \ m \ s^{-2}$.

Water rises to a height of $20 \,mm$ in a capillary tube of cross-sectional area $A$. If the area of cross-section of the tube is made $\frac{A}{4}$, then water will rise to a height of (in $\,cm$)