A column of mercury of 10cm length is contained in the middle of a narrow horizontal 1m long tube which is closed at both the ends. Both the halves of the tube contain air at a pressure of 76 cm of mercury. By what distance will the column of mercury be displaced if the tube is held vertically?

Explanation

To solve the problem of how much the column of mercury will be displaced when the tube is held vertically, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial setup consists of a horizontal tube that is 1 meter long with a 10 cm long column of mercury in the middle. The pressure in both halves of the tube is 76 cm of mercury.

**Hint:** Visualize the tube and the mercury column's position before it is turned vertically.

### Step 2: Analyze the Forces Acting on the Mercury Column
When the tube is held vertically, the weight of the mercury column will act downward, while the air pressure on both sides will act upward. The mercury will be displaced due to the change in pressure distribution.

**Hint:** Remember that the pressure exerted by a column of fluid is given by \( P = \rho g h \).

### Step 3: Set Up the Equilibrium Condition
In the horizontal position, the pressure at the top and bottom of the mercury column is equal. When the tube is turned vertically, we need to consider the new pressures at the top and bottom of the mercury column.

Let \( P_1 \) be the pressure at the top and \( P_2 \) be the pressure at the bottom after displacement \( x \).

**Hint:** Use the hydrostatic pressure formula to express \( P_1 \) and \( P_2 \).

### Step 4: Write the Pressure Equations
1. The initial pressure \( P \) in the tube is 76 cm of mercury, which can be converted to pressure units:
\[ P = \rho g h = 76 \text{ cmHg} \]
2. After displacement, the pressures can be expressed as:
\[ P_1 = P + \rho g (45 + x) \quad \text{(for the upper side)} \]
\[ P_2 = P + \rho g (45 - x) \quad \text{(for the lower side)} \]

**Hint:** Remember that the density of mercury (\( \rho \)) and gravitational acceleration (\( g \)) are constants.

### Step 5: Set Up the Equation for Equilibrium
At equilibrium, the pressures must balance:
\[ P_1 = P_2 \]
Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ P + \rho g (45 + x) = P + \rho g (45 - x) \]

**Hint:** Simplify the equation by canceling out the common terms.

### Step 6: Solve for Displacement \( x \)
Rearranging the equation gives:
\[ \rho g (45 + x) = \rho g (45 - x) \]
This simplifies to:
\[ x + x = 0 \quad \Rightarrow \quad 2x = 0 \]
This indicates that we need to account for the weight of the mercury column as well.

### Step 7: Calculate the Final Displacement
Using the balance of forces and the fact that the total pressure difference must equal the weight of the mercury column:
\[ \frac{45 \times 76}{45 + x} + 10 = \frac{45 \times 76}{45 - x} \]
Solving this equation will yield the value of \( x \).

After solving, we find:
\[ x \approx 2.95 \text{ cm} \]

### Conclusion
The column of mercury will be displaced by approximately **2.95 cm** when the tube is held vertically.

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