Water rises in a vertical capillary tube up to a height of $2.0 \, cm$. If the tube is inclined at an angle of $60^{\circ}$ with the vertical,then up to what length will the water rise in the tube?

Water rises in a capillary tube of radius $r$ up to a height $h$. The mass of water in the capillary is $m$. What will be the mass of water that rises in a capillary tube of radius $\frac{r}{4}$?

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 = $0^o$)

$A$ capillary tube of radius $R$ is immersed in water and water rises in it to a height $H$. The mass of water in the capillary tube is $M$. If the radius of the tube is halved,the mass of water that will rise in the capillary tube will now be

If the surface tension of water is $0.06 \, N/m$,then the capillary rise in a tube of diameter $1 \, mm$ is ....... $cm$ (assume $\theta = 0^\circ$,$g = 10 \, m/s^2$,and density of water $\rho = 1000 \, kg/m^3$).

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $75 \times 10^{-4} \, N$ force due to the weight of the liquid. If the surface tension of water is $6 \times 10^{-2} \, N m^{-1}$,the inner circumference of the capillary must be