Radius of a capillary is $2 \times 10^{-3} \,m$. $A$ liquid of weight $6.2 \times 10^{-4} \,N$ may remain in the capillary. Then the surface tension of the liquid will be $:-$

The radius of the bore of a capillary tube is $r$ and the angle of contact of the liquid is $\theta$. When the tube is dipped in the liquid,the radius of curvature of the meniscus of liquid rising in the tube 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:

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

Why does the water rise in a capillary tube? Explain.

How does an oil lamp produce light through a wick that is lit using a match?