Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. What will be the mass of water that rises in a capillary of radius $\frac{r}{4}$?

When a capillary tube is dipped vertically in water,the rise of water in the capillary is $h$. The angle of contact is $0^{\circ}$. Now,the tube is depressed so that its length above the water surface is $\frac{h}{3}$. The new apparent angle of contact is $(\cos 0^{\circ} = 1)$.

$A$ capillary tube of radius '$r$' is immersed in water and water rises to a height of '$h$'. The mass of water in the capillary tube is $5 \times 10^{-3} \ kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ}$. The mass of liquid which rises into the capillary tube now is (in $kg$):

When a capillary tube is dipped in water,it rises up to $8 \ cm$ in the tube. What happens when the tube is pushed down such that its end is only $5 \ cm$ above the outside water level?

Surface tension of two liquids (having same densities), $T_1$ and $T_2$, are measured using the capillary rise method utilizing two tubes with inner radii of $r_1$ and $r_2$ where $r_1 > r_2$. The measured liquid heights in these tubes are $h_1$ and $h_2$ respectively. [Ignore the weight of the liquid above the lowest point of the meniscus]. If $T_1 = T_2$, which of the following relations is satisfied?

$A$ $20 \ cm$ long capillary tube is dipped in water. The water rises up to $8 \ cm$. If the entire arrangement is put in a freely falling elevator,the length of the water column in the capillary tube will be (in $cm$)?