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 Concept:
Capillary rise $h$ is given by $h = \frac{2T \cos \theta}{r \rho g}$. Thus, $h \propto \frac{1}{g}$.

Step 2: Formula Application:
At depth $d$, gravity is $g_d = g(1 - \frac{d}{R}) = g(\frac{R-d}{R})$. Since $h \propto 1/g$, we have $\frac{h_2}{h_1} = \frac{g}{g_d}$.

Step 3: Explanation:
$\frac{h_2}{h_1} = \frac{g}{g(\frac{R-d}{R})} = \frac{R}{R-d}$.

Step 4: Final Answer:
The ratio $h_2/h_1$ is $\frac{R}{R-d}$.