A 1.00 m tall container is filled to the brim , partway with mercury and the rest of the way with water . The container is open to the atmosphere . What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure ?

Options

A. `0.74 ` m
B. `0.88` m
C. `0.43 ` m
D. `0.52` m

Answer

To solve the problem, we need to find the depth of mercury (h) in a 1.00 m tall container filled with mercury and water, such that the absolute pressure at the bottom of the container is twice the atmospheric pressure.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The container is 1.00 m tall and is filled with mercury and water. The total height of the liquid column (mercury + water) is 1.00 m.

2. **Define Variables**:
- Let \( h \) be the depth of the mercury.
- Therefore, the depth of the water will be \( 1.00 - h \).

3. **Pressure at the Bottom**:
- The absolute pressure at the bottom of the container can be expressed as:
\[ P_{bottom} = P_{atm} + P_{mercury} + P_{water} \]
- Where:
- \( P_{atm} \) = atmospheric pressure (approximately \( 1.01 \times 10^5 \) Pa)
- \( P_{mercury} = \rho_{mercury} \cdot g \cdot h \)
- \( P_{water} = \rho_{water} \cdot g \cdot (1 - h) \)

4. **Substituting Known Values**:
- The density of mercury (\( \rho_{mercury} \)) is approximately \( 13600 \, \text{kg/m}^3 \).
- The density of water (\( \rho_{water} \)) is approximately \( 1000 \, \text{kg/m}^3 \).
- The acceleration due to gravity (\( g \)) is approximately \( 9.81 \, \text{m/s}^2 \).

5. **Setting Up the Equation**:
- We want the absolute pressure at the bottom to be twice the atmospheric pressure:
\[ P_{bottom} = 2 \cdot P_{atm} \]
- Therefore:
\[ P_{atm} + \rho_{mercury} \cdot g \cdot h + \rho_{water} \cdot g \cdot (1 - h) = 2 \cdot P_{atm} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \rho_{mercury} \cdot g \cdot h + \rho_{water} \cdot g \cdot (1 - h) = P_{atm} \]

7. **Substituting Values**:
- Plugging in the values:
\[ 13600 \cdot 9.81 \cdot h + 1000 \cdot 9.81 \cdot (1 - h) = 1.01 \times 10^5 \]

8. **Simplifying the Equation**:
- This simplifies to:
\[ 133416h + 9810(1 - h) = 101000 \]
- Expanding gives:
\[ 133416h + 9810 - 9810h = 101000 \]
- Combining like terms:
\[ (133416 - 9810)h = 101000 - 9810 \]
\[ 123606h = 91990 \]

9. **Solving for h**:
- Now, solving for \( h \):
\[ h = \frac{91990}{123606} \approx 0.743 \, \text{m} \]

### Final Answer:
The depth of the mercury must be approximately **0.743 m**.