When a capillary tube is dipped vertically in water,the rise of water in the capillary is $h$. The angle of contact is $0^{\circ}$. Now,the tube is depressed so that its length above the water surface is $\frac{h}{3}$. The new apparent angle of contact is $(\cos 0^{\circ} = 1)$.

The correct relation for the capillary rise $h$ in a tube of radius $r$ is:

Two capillary tubes of radii $0.2 \ cm$ and $0.4 \ cm$ are dipped in the same liquid. The ratio of the heights to which the liquid will rise in the tubes is:

Water rises to a height of $16.3 \, cm$ in a capillary tube of height $18 \, cm$ above the water level. If the tube is cut at a height of $12 \, cm$ above the water level,then:

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. The mass of water that will rise in a capillary of radius $\frac{r}{5}$ will be:

$A$ 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 $kg$):