1 atmospheric pressure is taken as the pressure exerted by a mercury column of height 76 cm . Wh at is the approximate height of an air column that will produce one atmospheric pressure on its bottom ? [Assume that the temperature and density of a ir and th e value ofg remains constant throughout the a ir column .]
`[rho_("mercury")=13.6xx10^3kg//m^3` and `rho_("air")=1.3kg//m^3`]

Options

A. 8 km
B. 6 km
C. 5 km
D. 4 km

Explanation

To find the approximate height of an air column that will produce one atmospheric pressure at its bottom, we can use the relationship between pressure, density, gravitational acceleration, and height.

### Step-by-Step Solution:

1. **Understand the relationship of pressure**:
The pressure exerted by a fluid column can be expressed as:
\[ P = \rho g h \]
where \( P \) is the pressure, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid column.

2. **Set up the equation for mercury**:
For mercury, we know that 1 atmospheric pressure (P) is equivalent to the pressure exerted by a mercury column of height 76 cm. Therefore, we can write:
\[ P = \rho_{\text{mercury}} g h_1 \]
where \( h_1 = 76 \, \text{cm} = 0.76 \, \text{m} \) and \( \rho_{\text{mercury}} = 13.6 \times 10^3 \, \text{kg/m}^3 \).

3. **Set up the equation for air**:
For the air column, we want to find the height \( h_2 \) that produces the same pressure:
\[ P = \rho_{\text{air}} g h_2 \]
where \( \rho_{\text{air}} = 1.3 \, \text{kg/m}^3 \).

4. **Equate the pressures**:
Since both expressions represent the same pressure (1 atm), we can equate them:
\[ \rho_{\text{mercury}} g h_1 = \rho_{\text{air}} g h_2 \]

5. **Cancel out \( g \)**:
Since \( g \) is constant and appears on both sides, we can cancel it out:
\[ \rho_{\text{mercury}} h_1 = \rho_{\text{air}} h_2 \]

6. **Rearrange to find \( h_2 \)**:
Rearranging the equation gives:
\[ h_2 = \frac{\rho_{\text{mercury}} h_1}{\rho_{\text{air}}} \]

7. **Substitute known values**:
Substitute the known values into the equation:
\[ h_2 = \frac{(13.6 \times 10^3 \, \text{kg/m}^3)(0.76 \, \text{m})}{1.3 \, \text{kg/m}^3} \]

8. **Calculate \( h_2 \)**:
Performing the calculation:
\[ h_2 = \frac{(13.6 \times 10^3)(0.76)}{1.3} \approx 7.95 \times 10^3 \, \text{m} \]

9. **Convert to kilometers**:
To express the height in kilometers:
\[ h_2 \approx 7.95 \times 10^3 \, \text{m} \approx 8 \, \text{km} \]

### Final Answer:
The approximate height of the air column that will produce one atmospheric pressure at its bottom is **8 km**.