Question

If a capillary tube is immersed vertically in water, rise of water in capillary is $h_1$. When the whole arrangement is taken to a depth $d$ in a mine, the water level rises to $h_2$. The ratio $\dfrac{h_2}{h_1}$ is $\left(R = \text{radius of earth}\right)$

• A. $\left(1 + \dfrac{d^2}{R^2}\right)$
• B. $\left(1 + \dfrac{d}{R}\right)$
• C. $\left(1 - \dfrac{d^2}{R^2}\right)$
• D. $\left(1 - \dfrac{d}{R}\right)$

Hint:

Capillary rise increases when acceleration due to gravity decreases.

Answer 1

The Correct Option is D

Step 1: Expression for capillary rise.
Capillary rise is given by: \[ h = \dfrac{2T\cos\theta}{\rho g r} \] Hence, $h \propto \dfrac{1}{g}$.
Step 2: Value of acceleration due to gravity at depth $d$.
Acceleration due to gravity at depth $d$ is: \[ g_d = g\left(1 - \dfrac{d}{R}\right) \] Step 3: Relation between $h_1$ and $h_2$.
\[ \dfrac{h_2}{h_1} = \dfrac{g}{g_d} \] Step 4: Substitute value of $g_d$.
\[ \dfrac{h_2}{h_1} = \dfrac{g}{g\left(1 - \dfrac{d}{R}\right)} = \left(1 - \dfrac{d}{R}\right) \]