Two capillary tubes $P$ and $Q$ are dipped vertically in water. The height of water level in capillary tube $P$ is $\frac{2}{3}$ of the height in capillary tube $Q$. The ratio of their diameters is

One end of a capillary tube is dipped in water,the rise of water column is $h$. The upward force of $98 \text{ dyne}$ due to surface tension is balanced by the force due to the weight of the water column. The inner circumference of the capillary is (surface tension of water $= 7 \times 10^{-2} \text{ Nm}^{-1}$) (in $\text{ cm}$)

Water rises against gravity in a capillary tube when its one end is dipped into water because

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?

$A$ capillary tube is immersed vertically in water and the height of the water column is $x$. When this arrangement is taken into a mine of depth $d$,the height of the water column is $y$. If $R$ is the radius of earth,the ratio $\frac{x}{y}$ is

What is capillary action? Derive the formula for the rise of liquid in a capillary tube immersed vertically in a liquid.