The radius of a capillary tube is $2 \times 10^{-3} \ m$. If a liquid of weight $6.28 \times 10^{-4} \ N$ can be supported in the capillary,what is the surface tension of the liquid?

Three liquids have the same surface tension and have densities $\rho_1, \rho_2$, and $\rho_3$ $(\rho_1 > \rho_2 > \rho_3)$. In three identical capillaries, the rise of liquid is the same. The corresponding angles of contact $\theta_1, \theta_2$, and $\theta_3$ are related as:

Why does the water rise in a capillary tube? Explain.

$A$ capillary tube of radius '$r$' is immersed in water and water rises to a height of '$h$'. The mass of water in the capillary tube is $5 \times 10^{-3} \ kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^{\circ}$. The mass of liquid which rises into the capillary tube now is (in $kg$):

$A$ 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 heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)