$A$ $20 \ cm$ 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 will be ....... $cm$.

Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio $4:3$. If their surface tensions are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ of liquids in the two capillary tubes is (Their angle of contacts are same).

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. The mass of water that will rise in a capillary tube of radius $\frac{r}{3}$ will be:

Two capillary tubes $P$ and $Q$ are dipped in water. The height of the water level in capillary $P$ is $2/3$ of the height in capillary $Q$. The ratio of their diameters is:

$A$ capillary tube of radius $0.1 \ mm$ is partly dipped in water (surface tension $70 \ dyn/cm$ and glass-water contact angle $\simeq 0^{\circ}$) inclined at $30^{\circ}$ with the vertical. The length of water risen in the capillary is . . . . . . $cm$. (Take $g = 980 \ cm/s^2$)

$A$ $20 \ cm$ 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 $cm$)?