Water rises up to a height $h$ in a capillary tube of a certain diameter. This capillary tube is replaced by a similar tube of half the diameter. Now,the water will rise to a height of:

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

Liquid rises to a height $2 \ cm$ in a capillary tube; in that case,the angle of contact between the solid and the liquid is $0^{\circ}$. The tube is lowered more now,so that the capillary is only $1 \ cm$ above the liquid. In this case,the angle of contact between the solid and liquid is $......^{\circ}$.

When one end of a capillary tube is dipped in water,the height of water column is $h$. The upward force of $105 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of water column. The inner circumference of the capillary tube is (Surface tension of water $= 7 \times 10^{-2} \text{ N/m}$) (in $\text{ cm}$)

$A$ liquid (density $= 10^3 \ kg/m^3$) rises to a height of $10 \ cm$ in a capillary tube. If the angle of contact of the liquid-glass pair is $0^{\circ}$ and the radius of the tube is $2 \ mm$,then the surface tension of the liquid is:

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?