An open glass tube is immersed in mercury in such a way that a length of $8\,\text{cm}$ extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional $46\,\text{cm}$. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = $76\,\text{cm of Hg}$)

Step 1: Identify the initial state of the trapped air.

Initially, the tube is open to the atmosphere, so the pressure of the trapped air equals the atmospheric pressure, and its length is the portion of the tube extending above the mercury.

$$L_1 = 8\,\text{cm}$$ $$P_1 = P_{\text{atm}} = 76\,\text{cm of Hg}$$

Step 2: Determine the new position of the top of the tube.

When the tube is raised vertically by an additional $46\,\text{cm}$, the total height of the top of the tube above the external mercury level increases.

$$H = 8 + 46 = 54\,\text{cm}$$

Step 3: Establish the geometric and pressure relationships in the final state.

Let $L_2$ be the final length of the air column, and $h$ be the height of the mercury column inside the tube above the external level.

$$L_2 + h = 54$$ $$h = 54 - L_2$$

From the pressure balance at the mercury interface, the final pressure $P_2$ of the trapped air is related to the atmospheric pressure.

$$P_2 + h = P_{\text{atm}}$$ $$P_2 = 76 - h$$

Step 4: Apply Boyle's Law.

Since the temperature is constant and the tube has a uniform cross-sectional area, the pressure-volume product remains constant.

$$P_1 L_1 = P_2 L_2$$ $$76 \times 8 = (76 - h) L_2$$

Step 5: Solve the quadratic equation for the final air column length.

Substitute $h = 54 - L_2$ into the equation.

$$608 = (76 - (54 - L_2)) L_2$$ $$608 = (22 + L_2) L_2$$ $$L_2^2 + 22L_2 - 608 = 0$$

Apply the quadratic formula to find the roots.

$$\Delta = 22^2 - 4(1)(-608) = 2916 = 54^2$$ $$L_2 = \frac{-22 \pm 54}{2}$$

Since the length must be positive, we discard the negative root.

$$L_2 = \frac{-22 + 54}{2} = 16\,\text{cm}$$

Step 6: Verify the physical consistency of the solution.

Calculate the final height of the mercury column and the final pressure to verify Boyle's Law.

$$h = 54 - 16 = 38\,\text{cm}$$ $$P_2 = 76 - 38 = 38\,\text{cm of Hg}$$ $$P_2 L_2 = 38 \times 16 = 608 = P_1 L_1$$

Answer: The length of the air column above the mercury in the tube is $16\,\text{cm}$.