Question

A capillary tube when immersed vertically in water, the rise of water column is upto height $h_1$ on earth's surface. When this arrangement is taken into a mine of depth 'd', below earth's surface, the height of the water column is $h_2$. If R is the radius of the earth, the ratio $h_2/h_1$ is ______.

• A. $\frac{R+d}{R}$
• B. $\frac{R-d}{R}$
• C. $\frac{R}{R+d}$
• D. $\frac{R}{R-d}$

Hint:

Gravity is maximum at the Earth's surface and decreases whether you go up into space or down into a mine. Since capillary rise is inversely proportional to gravity, the water will ALWAYS rise higher ($h_2 > h_1$) in a mine or in space!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are dealing with capillary rise in two different gravitational environments (surface vs. deep mine). We must link the capillary rise formula with the variation of gravity with depth to find the ratio of heights.

Step 2: Detailed Explanation:
The height ($h$) to which a liquid rises in a capillary tube is given by Jurin's Law:
$h = \frac{2T \cos \theta}{r \rho g}$
where:
$T$ = surface tension
$\theta$ = angle of contact
$r$ = radius of capillary
$\rho$ = density of liquid
$g$ = acceleration due to gravity
Since the tube and liquid remain physically identical in both locations, all parameters ($T, \theta, r, \rho$) are constant. Thus, the capillary rise is strictly inversely proportional to local gravity:
$h \propto \frac{1}{g}$
Therefore, the ratio of heights is the inverse ratio of gravities:
$\frac{h_2}{h_1} = \frac{g_{\text{surface}}}{g_{\text{mine}}}$ --- (Equation 1)
The acceleration due to gravity at a depth $d$ below the Earth's surface ($g_{\text{mine}}$) is given by:
$g_{\text{mine}} = g_{\text{surface}} \left( 1 - \frac{d}{R} \right)$
$g_{\text{mine}} = g_{\text{surface}} \left( \frac{R - d}{R} \right)$
Substitute this expression for $g_{\text{mine}}$ back into Equation 1:
$\frac{h_2}{h_1} = \frac{g_{\text{surface}}}{ g_{\text{surface}} \left( \frac{R - d}{R} \right) }$
The $g_{\text{surface}}$ terms cancel out perfectly:
$\frac{h_2}{h_1} = \frac{1}{ \frac{R - d}{R} }$
$\frac{h_2}{h_1} = \frac{R}{R - d}$
Because gravity is weaker down in the mine, the water column feels "lighter" and gets pulled up higher by the constant surface tension!

Step 3: Final Answer:
The ratio is $\frac{R}{R-d}$, matching option (d).