Two capillary tubes of same diameter are put vertically one each in two liquids whose relative densities are $0.8$ and $0.6$ and surface tensions are $60$ and $50 \text{ dyne/cm}$ respectively. The ratio of the heights of the liquids in the two tubes $\frac{h_1}{h_2}$ is:

The lower end of a glass capillary tube is dipped in water. Water rises to a height of $8 \ cm$. The tube is then broken at a height of $6 \ cm$. The new height of water column and new angle of contact will be $:-$

$A$ uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $\sigma$. The angle of contact between water and the wall of the capillary tube is $\theta$. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?
$(A)$ For a given material of the capillary tube,$h$ decreases with increase in $r$.
$(B)$ For a given material of the capillary tube,$h$ is independent of $\sigma$.
$(C)$ If this experiment is performed in a lift going up with a constant acceleration,then $h$ decreases.
$(D)$ $h$ is proportional to contact angle $\theta$.

The angle of contact between glass and water is $0^{\circ}$ and water rises in a glass capillary up to $6 \ cm$ (Surface tension of water is $T$). Another liquid of surface tension $2T$,angle of contact $60^{\circ}$ and relative density $2$ will rise in the same capillary up to (Given: $\cos 0^{\circ}=1, \cos 60^{\circ}=0.5$) (in $cm$)

When a part of a straight capillary tube is placed vertically in a liquid, the liquid rises up to a certain height $h$. If the inner radius of the capillary tube, density of the liquid, and surface tension of the liquid decrease by $1\%$ each, then the height of the liquid in the tube will change by . . . . . . $\%$.

Water rises to a height of $2.2 \text{ cm}$ in a glass capillary tube. The height to which the same water rises in another capillary having $\frac{1}{4}$ of the cross-sectional area is: (in $\text{ cm}$)