$A$ capillary tube is immersed vertically in water,and the water rises to a height $h_1$ on the Earth's surface. When this arrangement is taken into a mine of depth $d$ below the Earth's surface,the height of the water column is $h_2$. If $R$ is the radius of the Earth,the ratio $\frac{h_2}{h_1}$ is:

Two narrow bores of diameter $5.0 \, mm$ and $8.0 \, mm$ are joined together to form a $U$-shaped tube open at both ends. If this $U$-tube contains water,what is the difference in the level of two limbs of the tube? [Take surface tension of water $T = 7.3 \times 10^{-2} \, Nm^{-1}$,angle of contact $= 0$,$g = 10 \, ms^{-2}$ and density of water $\rho = 1.0 \times 10^{3} \, kg \, m^{-3}$] (in $mm$)

$A$ capillary tube is placed at angles of $30^o$ and $60^o$ with the vertical. What is the ratio of the length of the liquid column in the capillary tube?

Water rises to a height $3$ cm in a capillary tube. If the cross-sectional area of the capillary tube is reduced to $1/3^{rd}$ of its initial area, then water will rise to a height of:

Water rises to a height of $10 \, cm$ in a capillary tube and mercury falls to a depth of $3.1 \, cm$ in the same capillary tube. If the density of mercury is $13.6 \, g/cm^3$ and the angle of contact for mercury is $135^{\circ}$,the approximate ratio of surface tensions of water and mercury is

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 $\rho = 1000 \, kg/m^3$).