Chapter 20, Problem 36A

If there were a liquid twice as dense as mercury, and if it were used to make a barometer, how tall would the column be?

To determine

The height of the column.

Answer to Problem 36A

The height of the column of liquid is half the height of the mercury column.

Explanation of Solution

Given:

The density of the column is $\rho =2{\rho }_{\text{mercury}}$ .

Formula used:

The expression for pressure is,

  $P=\rho gh$

Here, $\rho$ is the density, $g$ is the acceleration due to gravity, and $h$ is the height.

Calculation:

Consider the density of the mercury is ${\rho }_{\text{mercury}}$ . The height $\left(h_{\text{mc}}\right)$ of the column of mercury is,

  $\begin{array}{l}P={\rho }_{\text{mercury}}g{h}_{\text{mc}}\\ h_{\text{mc}}=\frac{P}{{\rho }_{\text{mercury}}g}\end{array}$

Consider the mercury in the column is replaced by a liquid of density $\rho$ .

Consider a liquid used is twice dense as compared to mercury.

  $\rho =2{\rho }_{\text{mercury}}$

The height $\left(h_{\text{lc}}\right)$ of the column of liquid is,

  $\begin{array}{l}P=\rho g{h}_{\text{lc}}\\ h_{\text{lc}}=\frac{P}{\rho g}\\ h_{\text{lc}}=\frac{P}{\left(2{\rho }_{\text{mercury}}\right)g}\\ h_{\text{lc}}=\frac{1}{2}\left(\frac{P}{{\rho }_{\text{mercury}}g}\right)\\ h_{\text{lc}}=\frac{1}{2}{h}_{\text{mc}}\end{array}$

Conclusion:

Thus, the height of the column of liquid is half the height of the mercury column.