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 of how high a column of bromoform will be supported by the same pressure that supports a column of mercury 200 mm high, we can use the concept of hydrostatic pressure. The pressure exerted by a liquid column is given by 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 we can cancel out since it is constant for both liquids),
- \( h \) is the height of the liquid column.

### Step-by-Step Solution:

1. **Identify the known values:**
- Density of mercury (\( \rho_1 \)) = 13.6 g/cc
- Height of mercury column (\( h_1 \)) = 200 mm
- Density of bromoform (\( \rho_2 \)) = 3.02 g/cc
- We need to find the height of the bromoform column (\( h_2 \)).

2. **Set up the equation using the hydrostatic pressure formula:**
Since the pressure exerted by both columns is equal, we can write:
\[ \rho_1 g h_1 = \rho_2 g h_2 \]
Here, \( g \) cancels out from both sides:
\[ \rho_1 h_1 = \rho_2 h_2 \]

3. **Rearrange the equation to solve for \( h_2 \):**
\[ h_2 = \frac{\rho_1 h_1}{\rho_2} \]

4. **Substitute the known values into the equation:**
- Convert the height of mercury from mm to cm for consistency (200 mm = 20 cm).
\[ h_2 = \frac{13.6 \, \text{g/cc} \times 20 \, \text{cm}}{3.02 \, \text{g/cc}} \]

5. **Calculate \( h_2 \):**
\[ h_2 = \frac{272}{3.02} \approx 90.07 \, \text{cm} \]

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

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