# Question #3c446

Apr 18, 2017

Density

#### Explanation:

The density of mercury $\left(13593 \frac{k g}{m} ^ 3\right)$ is much more than that of ethanol$\left(789 \frac{k g}{m} ^ 3\right)$ and water $\left(1000 \frac{k g}{m} ^ 3\right)$. What this means is that height of mercury required to produce a given pressure is much less than the height of a column of water or ethanol.

Let's do a simple calculation to illustrate the point.

The pressure a manometer normally measures is of the order of atmospheres. $1 a t m o s p h e r e = {10}^{5} P a$ in SI units.

Now, $\mathrm{dg} h = P$ $\implies h = \frac{P}{\mathrm{dg}}$ where
$P$ = pressure
$d$ = density of material used
$g$ = acceleration due to gravity. Let us take it as $10 \frac{m}{s} ^ 2$ for simplicity.
$h$ = height of column

$\therefore$ height of mercury column required to produce/measure a pressure of $1 a t m = {10}^{5} P a$ is:-

$\frac{100000}{13593 \cdot 10} \approx 0.735 m = \textcolor{red}{73.5 c m}$

Similarly height of water column for $1 a t m = {10}^{5} P a$ is:-

$\frac{100000}{1000 \cdot 10} = 10 m = \textcolor{red}{1000 c m}$

and for ethanol is:-

$\frac{100000}{789 \cdot 10} \approx 12.67 m = \textcolor{red}{1267 c m}$

Now you can easily see that the height of mercury column required i.e. $73.5 c m$ is much more feasible to use than heights of $1000 c m$ or $1267 c m$.