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:

The height of the water column in a capillary tube is given by the formula:

$$ h = \frac{2S \cos \theta}{\rho g r} $$

where $$ S $$ is the surface tension, $$ \theta $$ is the contact angle, $$ \rho $$ is the density of water, $$ g $$ is the acceleration due to gravity, and $$ r $$ is the radius of the capillary tube.

Since the capillary tube and the liquid (water) are the same in both scenarios, $$ S $$, $$ \theta $$, $$ \rho $$, and $$ r $$ are constant. Therefore, the height $$ h $$ is inversely proportional to $$ g $$:

$$ h \propto \frac{1}{g} $$

At the Earth's surface, the height is $$ x $$, so:

$$ x \propto \frac{1}{g_{\text{surface}}} $$

In the mine at depth $$ d $$, the height is $$ y $$, so:

$$ y \propto \frac{1}{g_{\text{mine}}} $$

The ratio $$ \frac{x}{y} $$ is:

$$ \frac{x}{y} = \frac{g_{\text{mine}}}{g_{\text{surface}}} $$

Now, we need to find how $$ g $$ changes with depth. The acceleration due to gravity at a depth $$ d $$ below the Earth's surface is given by:

$$ g_{\text{mine}} = g_{\text{surface}} \left(1 - \frac{d}{R}\right) $$

where $$ R $$ is the radius of the Earth. This is because, at depth $$ d $$, only the mass of the Earth within the sphere of radius $$ R - d $$ contributes to gravity, and under the assumption of uniform density, $$ g $$ decreases linearly with depth.

Therefore, substituting this into the ratio:

$$ \frac{g_{\text{mine}}}{g_{\text{surface}}} = 1 - \frac{d}{R} $$

So, the ratio $$ \frac{x}{y} $$ is:

$$ \frac{x}{y} = 1 - \frac{d}{R} $$

Comparing with the options:

A. $$ \left(\frac{R-d}{R+d}\right) $$

B. $$ \left(\frac{R+d}{R-d}\right) $$

C. $$ \left(1 - \frac{2d}{R}\right) $$

D. $$ \left(1 - \frac{d}{R}\right) $$

Option D matches the derived expression.

Hence, the correct answer is Option D.