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 $m_1: m_2=$

Water rises to a height $h$ in a capillary tube at the surface of the Earth. On the surface of the moon,the height of the water column in the same capillary tube will be:

$A$ capillary tube of radius $0.1 \,mm$ is dipped in water. The water rises to a height of $2 \,cm$ in the tube. If the surface tension of water is $0.072 \,N/m$, the contact angle between water and the wall of the tube is:

$A$ capillary tube of radius $0.1 \ mm$ is partly dipped in water (surface tension $70 \ dyn/cm$ and glass-water contact angle $\simeq 0^{\circ}$) inclined at $30^{\circ}$ with the vertical. The length of water risen in the capillary is . . . . . . $cm$. (Take $g = 980 \ cm/s^2$)

$A$ capillary tube is vertically immersed in water,and water rises up to a height $h_{1}$. When the whole arrangement is taken to a depth $d$ in a mine,the water level rises up to a height $h_{2}$. The ratio $h_{1} / h_{2}$ is ($R =$ radius of earth).

$A$ liquid (density $= 10^3 \ kg/m^3$) rises to a height of $10 \ cm$ in a capillary tube. If the angle of contact of the liquid-glass pair is $0^{\circ}$ and the radius of the tube is $2 \ mm$,then the surface tension of the liquid is: