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

Step 1: Identify the initial parameters of the tube.

The total length of the tube is $100\,\text{cm}$, and the mercury column is $10\,\text{cm}$ long.

$$L = 100\,\text{cm}$$ $$h = 10\,\text{cm}$$ $$P_0 = 76\,\text{cm of Hg}$$

Step 2: Calculate the initial length of each air column.

When the tube is horizontal, the mercury column is in the middle, dividing the tube into two equal air columns.

$$l_0 = \frac{100 - 10}{2}$$ $$l_0 = 45\,\text{cm}$$

Step 3: Define the final state when the tube is held vertically.

Let $x$ be the downward displacement of the mercury column.

$$l_{\text{lower}} = 45 - x$$ $$l_{\text{upper}} = 45 + x$$

Step 4: Set up the pressure equilibrium condition.

In the vertical position, the pressure difference between the lower and upper air columns supports the weight of the $10\,\text{cm}$ mercury column.

$$P_1 - P_2 = 10\,\text{cm of Hg}$$

Step 5: Apply Boyle's Law to both air columns.

Assuming the temperature remains constant, the pressure-volume product is constant for each column.

$$P_1(45 - x) = P_0 \cdot 45$$ $$P_2(45 + x) = P_0 \cdot 45$$

Step 6: Express the final pressures in terms of displacement.

Substitute $P_0 = 76\,\text{cm of Hg}$ into the equations.

$$P_1 = \frac{76 \times 45}{45 - x}$$ $$P_2 = \frac{76 \times 45}{45 + x}$$

Step 7: Substitute the pressures into the equilibrium equation.

$$\frac{76 \times 45}{45 - x} - \frac{76 \times 45}{45 + x} = 10$$

Step 8: Simplify the algebraic equation.

$$76 \times 45 \left(\frac{1}{45 - x} - \frac{1}{45 + x}\right) = 10$$ $$76 \times 45 \left(\frac{2x}{45^2 - x^2}\right) = 10$$ $$76 \times 45 \cdot \frac{2x}{2025 - x^2} = 10$$ $$6840x = 10(2025 - x^2)$$ $$684x = 2025 - x^2$$ $$x^2 + 684x - 2025 = 0$$

Step 9: Solve the quadratic equation for $x$.

Apply the quadratic formula.

$$x = \frac{-684 + \sqrt{684^2 - 4(1)(-2025)}}{2}$$ $$x = \frac{-684 + \sqrt{475956}}{2}$$ $$x \approx \frac{-684 + 689.9}{2}$$ $$x \approx 2.95\,\text{cm}$$

Answer: The column of mercury will be displaced by approximately $2.95\,\text{cm}$ (or approximately $3\,\text{cm}$).