At room temperature mercury has a density of 13.6 g/cc while liquid bromoform `(CHBr_3)` has a density of 3.02 g cc. How high a column of bromoform will be supported by a pressure that supports a column of mercury 200 mm high?

Options

A. 98.67 cm
B. 40.99 cm
C. 90.07 cm
D. 100 cm

Answer

To solve the problem, we need to compare the pressures exerted by two different liquids: mercury and bromoform. The pressure exerted by a liquid column can be calculated using the formula:

\[ P = \rho g h \]

where:
- \( P \) is the pressure,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity (which remains constant for both liquids),
- \( h \) is the height of the liquid column.

### Step-by-Step Solution:

1. **Identify the Variables:**
- For mercury (Hg):
- Density \( \rho_1 = 13.6 \, \text{g/cm}^3 \)
- Height \( h_1 = 200 \, \text{mm} = 20 \, \text{cm} \) (conversion from mm to cm)
- For bromoform (CHBr₃):
- Density \( \rho_2 = 3.02 \, \text{g/cm}^3 \)
- Height \( h_2 \) is what we need to find.

2. **Set Up the Pressure Equations:**
Since the pressures exerted by both columns are equal, we can set up the equation:
\[ \rho_1 g h_1 = \rho_2 g h_2 \]

3. **Cancel Out the Common Terms:**
The acceleration due to gravity \( g \) appears on both sides of the equation, so we can cancel it out:
\[ \rho_1 h_1 = \rho_2 h_2 \]

4. **Rearrange to Solve for \( h_2 \):**
We can rearrange the equation to solve for \( h_2 \):
\[ h_2 = \frac{\rho_1 h_1}{\rho_2} \]

5. **Substitute the Known Values:**
Substitute the values for \( \rho_1 \), \( h_1 \), and \( \rho_2 \):
\[ h_2 = \frac{13.6 \, \text{g/cm}^3 \times 20 \, \text{cm}}{3.02 \, \text{g/cm}^3} \]

6. **Calculate \( h_2 \):**
Perform the calculation:
\[ h_2 = \frac{272 \, \text{g/cm}^2}{3.02 \, \text{g/cm}^3} \approx 90.07 \, \text{cm} \]

### Final Result:
The height of the bromoform column that will be supported by the same pressure as a 200 mm column of mercury is approximately **90.07 cm**.

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