Water rises to a height of $3 \,cm$ in a capillary tube. If the cross-sectional area of the capillary tube is reduced to $1/9$th of its initial area, then the water will rise to a height of: (in $\,cm$)

Two capillary tubes $A$ and $B$ of the same internal diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. If the surface tensions of these two liquids are in the ratio $6:5$,then the ratio of rise of liquid in capillary $A$ to that in $B$ is (assume their angles of contact are nearly equal).

$A$ 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 $g = 980 \ cm/s^2$)

Two capillary tubes of the same diameter are kept vertically in two different liquids whose densities are in the ratio $4:3$. The rise of liquid in two capillaries is $h_1$ and $h_2$ respectively. If the surface tensions of liquids are in the ratio $6:5$,the ratio of heights $\left(\frac{h_1}{h_2}\right)$ is (Assume that their angles of contact are same).

Water rises to a height of $15 \,mm$ in a capillary tube having cross-sectional area $A$. If the cross-sectional area of the tube is made $A / 3$, then the water will rise to a height of:

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 = $0^o$)