$A$ 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

When a long glass capillary tube of radius $0.015 \; cm$ is dipped in a liquid,the liquid rises to a height of $15 \; cm$ within it. If the contact angle between the liquid and glass is close to $0^{\circ}$,the surface tension of the liquid,in $milliNewton \; m^{-1}$ is $.....$
$[\rho_{\text{liquid}} = 900 \; kg \; m^{-3}, g = 10 \; ms^{-2}]$ (Give answer in closest integer)

Water rises up to a height of $4 \,cm$ in a capillary tube. The lower end of the capillary tube is at a depth of $8 \,cm$ below the water level. The mouth pressure required to blow an air bubble at the lower end of the capillary will be '$X$' $cm$ of water,where $X$ is equal to

$A$ $U$-tube is such that the diameter of one limb is $0.4\,mm$ and that of the other is $d\,mm$. If the surface tension of water contained in the tube is $0.07\,N/m$ and the difference in the levels of liquid in the limbs is $3.6\,cm$,then the value of $d$ is:

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

$A$ $U$-tube with limbs of diameters $5\, mm$ and $2\, mm$ contains water of surface tension $7 \times 10^{-2} \, N/m$. The angle of contact is zero and the density of water is $10^3 \, kg/m^3$. If $g = 10 \, m/s^2$,then the difference in the liquid levels of the two limbs is: