Question

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

Answer 1

Hint – To solve this problem, firstly learn about Gravitation, the concept behind it and how will it affect the values asked in our question. By using this knowledge and then applying the formulas required, we will easily find the required values and hence approach the solution to our question.

Step-By-Step answer:
Let us take in consideration when the capillary is placed in water on the surface of the earth, so that the acceleration due to gravity is g .When the capillary is placed in the mine of depth d, the value of g will become, $\dfrac{g}{{1 - \dfrac{d}{R}}}$. Rise in capillary tube $\propto$ to acceleration due to gravity so, the rise in capillary tube when it is taken to a mine of depth $d\,\,is\,\,y$, by using this information we can easily measure the ratio between $x\,\,and\,\,y$ .
Gravitation: It states that any two bodies having mass attract each other with force $\propto$ to the product of their mass and $\dfrac{1}{ \propto }$ to the square of distance between them.
Acceleration because of gravity is the acceleration that is gained by an object due to gravitational force. Its SI unit is $m/{s^2}$ .It has both magnitude and direction; hence it is a vector quantity.
When capillary is kept in water on the surface of earth, the acceleration because of gravity is $g$. The rise in the capillary tube is directly proportional to the acceleration because of gravity.
Therefore, when the entire apparatus is taken into a mine of depth $d$ .
The value of the acceleration because of gravity will become,$\dfrac{g}{{1 - \dfrac{d}{R}}}$
Thus, the rise in the capillary tube will be $x,y$ in the 2 cases.
So,
$\Rightarrow \,\,\dfrac{x}{y} = \dfrac{g}{{\dfrac{g}{{1 - \dfrac{d}{R}}}}} \\\ \Rightarrow \,\,1 - \dfrac{d}{R} \\\$

Hence, the correct answer is option A - $\left( {1 - \dfrac{d}{R}} \right)$

Note - Acceleration because of gravity is the acceleration gained by the object due to gravitational force. Its S.I unit is $m/{s^{ - 1}}$. The standard value of $g$ on the surface of the earth at sea level is $9.8m/{s^2}$. Acceleration due to gravity is represented by $g$.