The correct curve between the height or depression $h$ of liquid in a capillary tube and its radius $r$ is

$A$ 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).

Two capillary tubes of length $L$ and $2L$ and of radius $R$ and $2R$ are connected in series. The net rate of flow of fluid through them will be (given rate of flow through a single capillary,$X = \frac{\pi P R^4}{8 \eta L}$)

$A$ capillary tube is taken from the Earth's surface to the Moon's surface. What happens to the rise of the liquid column on the Moon's surface? (Acceleration due to gravity on the Earth's surface is six times that of the Moon's surface.)

$A$ 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:

When two capillary tubes of different diameters are dipped vertically,the rise of the liquid is