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 the Earth, the ratio $\dfrac{x}{y}$ is:
A. $\left( 1-\dfrac{d}{R} \right)$
B. $\left( 1-\dfrac{2d}{R} \right)$
C. $\left( \dfrac{R-d}{R+d} \right)$
D. $\left( \dfrac{R+d}{R-d} \right)$

Hint: The given problem is based on the acceleration due to gravity. So, we should calculate the acceleration due to gravity for both Earth’s surface and in the mine of depth ‘d’. Then we shall equate the ratio to the gained equation and find the answer.

Complete step by step answer:
Gravity is the force with which the earth attracts every object towards the centre. It is denoted by ‘g’ and its value at the surface of earth is of about -9.8 ms$^{-2}$.
Acceleration due to gravity is the acceleration possessed by an object because of the gravity. It has both magnitude and direction, which means it is a vector quantity. The formula is given by the combination of Newton’s second law and gravitational law of forces.
When a capillary tube is placed on the surface of earth, it will possess an acceleration due to gravity. Let us consider ‘g’ to be the acceleration due to gravity on Earth’s surface.
As the rise in the capillary tube is directly proportional to the acceleration due to gravity, the whole apparatus is taken into a mine of depth ‘d’. The expression or value for acceleration due to gravity in mine can be given as
$\dfrac{g}{1-\dfrac{d}{R}}$ where. ‘R’ is the radius of Earth.
‘x’, ‘y’ will be the rise of height of the water column in both cases i.e., on Earth’s surface and in mine of depth ‘d’. The ratio can be written as
 $\dfrac{x}{y}=\dfrac{g}{\dfrac{g}{1-\dfrac{d}{R}}}$
On solving, we get
$\dfrac{x}{y}=g\times \dfrac{1-\dfrac{d}{R}}{g}$
$\dfrac{x}{y}=\left( 1-\dfrac{d}{R} \right)$
Therefore, the correct answer for the given question is option (A).

Note: The SI unit of acceleration due to gravity is measured in metres per second square ($m/s^2$).
Acceleration due to gravity is least at the equator and maximum at the poles. The value is always negative because when going down, it is moving in a negative direction.