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 tube of radius $\frac{r}{4}$?

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:

Water rises to a height of $10$ cm in a capillary tube and mercury falls to a depth of $3.112$ cm in the same capillary tube. If the density of mercury is $13.6 \text{ g/cm}^3$ and the angle of contact for mercury is $135^o$,the ratio of surface tension of water and mercury is (Assume density of water = $1 \text{ g/cm}^3$ and angle of contact for water = $0^o$)

$A$ capillary tube is vertically immersed in water,and water rises up to a height $h_{1}$. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h_{2}$. The ratio $h_{1} / h_{2}$ is ($R =$ radius of earth).

If a capillary tube of radius $1 \,mm$ is immersed in water, the mass of water rising in the capillary tube is $m$. If the radius of the capillary tube is doubled, then the mass of water that rises in the same capillary tube will be

In a capillary tube of radius $R$,a straight thin metal wire of radius $r$ $(R > r)$ is inserted symmetrically,and one end of the combination is dipped vertically in water such that the lower end of the combination is at the same level. The rise of water in the capillary tube is $[T =$ surface tension of water,$\rho =$ density of water,$g =$ gravitational acceleration$]$.