If $M$ is the mass of water that rises in a capillary tube of radius $r,$ then the mass of water which will rise in a capillary tube of radius $2r$ is

$A$ 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:

In a $U$-shaped tube,the radius of one limb is $2 \ mm$ and that of the other limb is $4 \ mm$. $A$ liquid of surface tension $0.03 \ Nm^{-1}$,density $1500 \ kgm^{-3}$,and angle of contact zero is taken in the tube. The difference in the heights of the levels of the liquid in the two limbs is (Acceleration due to gravity $= 10 \ ms^{-2}$) (in $mm$)

When a capillary is dipped in water,water rises to a height $h$. If the length of the capillary is made less than $h$,then

$A$ capillary tube $(A)$ is dipped in water. Another identical tube $(B)$ is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

When a capillary tube is dipped in water,it rises up to $8 \ cm$ in the tube. What happens when the tube is pushed down such that its end is only $5 \ cm$ above the outside water level?