To find the height of the column of gallium that would be sustained in a barometer at 1 atm pressure, we can follow these steps:
1. Understanding the problem:
- We know the density of mercury ([tex]\(\rho_{Hg}\)[/tex]) is [tex]\(13.6 \text{ gm/cm}^3\)[/tex].
- The typical height of a mercury column in a barometer at 1 atm pressure is approximately 76 cm.
- We are given that the density of gallium ([tex]\(\rho_{Ga}\)[/tex]) is [tex]\(6.09 \text{ gm/cm}^3\)[/tex].
2. Setting up the relationship:
- The pressure exerted by a column of fluid in a barometer is given by the product [tex]\( P = h \cdot \rho \cdot g \)[/tex], where:
- [tex]\(h\)[/tex] is the height of the fluid column,
- [tex]\(\rho\)[/tex] is the density of the fluid,
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- Since both columns are at the same pressure in a barometer (1 atm), we can set up the equation:
[tex]\[
h_{\text{Hg}} \cdot \rho_{\text{Hg}} \cdot g = h_{\text{Ga}} \cdot \rho_{\text{Ga}} \cdot g
\][/tex]
3. Simplifying the equation:
- Notice that [tex]\(g\)[/tex] (acceleration due to gravity) cancels out from both sides of the equation since it is a common factor.
- This leaves us with:
[tex]\[
h_{\text{Hg}} \cdot \rho_{\text{Hg}} = h_{\text{Ga}} \cdot \rho_{\text{Ga}}
\][/tex]
4. Solving for the unknown height ( [tex]\(h_{\text{Ga}}\)[/tex] ):
- Rearrange the equation to solve for [tex]\(h_{\text{Ga}}\)[/tex]:
[tex]\[
h_{\text{Ga}} = \frac{h_{\text{Hg}} \cdot \rho_{\text{Hg}}}{\rho_{\text{Ga}}}
\][/tex]
- Substitute the known values into the equation:
[tex]\[
h_{\text{Ga}} = \frac{76 \text{ cm} \cdot 13.6 \text{ gm/cm}^3}{6.09 \text{ gm/cm}^3}
\][/tex]
5. Calculate the height:
- Plugging in the numbers:
[tex]\[
h_{\text{Ga}} = \frac{1033.6}{6.09} \approx 169.72 \text{ cm}
\][/tex]
Therefore, the height of the column of gallium that would be sustained in a barometer at 1 atm pressure is approximately [tex]\(169.72 \text{ cm}\)[/tex].
