Question

A capillary tube of glass is taken from the Earth's surface to the surface of the Moon. The rise of the water column on the Moon (given contact angle between glass and water is $k$$=0$ degree and acceleration due to gravity on the Earth is $6times$ that of the Moon) is :

Answer 1

Step 1. Given Data:

The angle between glass and water, $\theta =$$0°$.

Acceleration due to gravity on the Earth is $6times$ that of the Moon, $g_{earth=}6g_{moon}$

Step 2. Formula used:

$h=\frac{2T\cos \theta }{r\rho g}$

where $h$ is the capillary rise,

$T$ is the surface tension of the liquid,

$\theta$ is the angle between glass and water i.e $0$$°$,

$\rho$ being the density of liquid,

$r$ bring the inner radius of the capillary,

and $g$ is the acceleration due to gravity.

Step 3. Solution:

The rise in the capillary is as follows,

Capillary rise, $h=\frac{2T\cos \theta }{r\rho g}$

where $T$ is the surface tension of the liquid.

This can be written as $h=\frac{k}{g}$ where $k$ is constant

so,

, $$\begin{gathered} h_{moon}=\frac{k}{\left({\displaystyle \frac{g}{6}}\right)} \\ {h}_{moon}=\frac{6k}{g}=6h_{earth} \end{gathered}$$

Hence , the rise of the water column on the Moon is six times that on the Earth Surface.