Two capillary tubes of radii $R$ and $2R$ are dipped in water. The water rises to heights $h_1$ and $h_2$ respectively in the two tubes. If $m_1$ and $m_2$ be the masses of water ascended in the tubes,then $m_1: m_2=$

$A$ 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:

If water rises in a capillary tube up to $3 \, cm$,what is the diameter of the capillary tube? (Surface tension of water $T = 7.2 \times 10^{-2} \, N/m$,density $\rho = 10^3 \, kg/m^3$,$g = 10 \, m/s^2$)

$A$ 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 $kg$):

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 = $0^o$)

Water rises up to a height $h$ in a capillary tube on the surface of the earth. The value of $h$ will increase if the experimental setup is kept in: