What is the barometric height of a liquid of density 3.4 g `cm^(-3)` at a place, where that for mercury barometer is 70 cm?

Options

A. 70 cm
B. 140 cm
C. 280 cm
D. 340 cm

Answer

To find the barometric height of a liquid with a density of 3.4 g/cm³, given that the height of mercury in a barometer is 70 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**: The height of a liquid column in a barometer is determined by the atmospheric pressure acting on it. The pressure exerted by a column of liquid is given by the formula:
\[ P = \rho \cdot g \cdot h \]
where \( P \) is the pressure, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the liquid column.

2. **Set Up the Equation**: For mercury, we have:
\[ P_a = \rho_{Hg} \cdot g \cdot h_{Hg} \]
For the liquid with density \( \rho_{liquid} \):
\[ P_a = \rho_{liquid} \cdot g \cdot h_{liquid} \]
Since both expressions equal the atmospheric pressure \( P_a \), we can equate them:
\[ \rho_{Hg} \cdot g \cdot h_{Hg} = \rho_{liquid} \cdot g \cdot h_{liquid} \]

3. **Cancel Out \( g \)**: The acceleration due to gravity \( g \) is common in both equations, so we can cancel it out:
\[ \rho_{Hg} \cdot h_{Hg} = \rho_{liquid} \cdot h_{liquid} \]

4. **Rearrange to Find \( h_{liquid} \)**: We can rearrange the equation to solve for the height of the liquid \( h_{liquid} \):
\[ h_{liquid} = \frac{\rho_{Hg} \cdot h_{Hg}}{\rho_{liquid}} \]

5. **Substitute Known Values**: We know:
- \( \rho_{Hg} = 13.6 \, \text{g/cm}^3 \)
- \( h_{Hg} = 70 \, \text{cm} \)
- \( \rho_{liquid} = 3.4 \, \text{g/cm}^3 \)

Substituting these values into the equation:
\[ h_{liquid} = \frac{13.6 \, \text{g/cm}^3 \cdot 70 \, \text{cm}}{3.4 \, \text{g/cm}^3} \]

6. **Calculate**: Performing the calculation:
\[ h_{liquid} = \frac{952 \, \text{g/cm}^2}{3.4 \, \text{g/cm}^3} = 280 \, \text{cm} \]

### Final Answer:
The barometric height of the liquid is **280 cm**.

---