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Step-by-Step Solution
Step 1: Identify the initial state of the air in the tube
Initially, the tube is open to the atmosphere and a column of air of length
$8\,\text{cm}$ is trapped above the mercury level. Because it is open to atmospheric
pressure, the initial pressure of the air column is
$76\,\text{cm of Hg}$. Denote this initial pressure as
$P_1 = 76\,\text{cm of Hg}$ and the initial volume (proportional to the length) of the air
column as $V_1 = 8\,\text{cm}$.
Step 2: Note the process of sealing and lifting the tube
Next, the open end is sealed, so no air can enter or escape. The tube is then
raised vertically by $46\,\text{cm}$ above the mercury reservoir. The total
vertical distance from the free surface of mercury to the top of the tube
now becomes $8 + 46 = 54\,\text{cm}$.
Step 3: Express the final state of the air column
Let the final length of the air column (above the mercury inside the tube)
be $L\,\text{cm}$. The remainder of the tube length, namely
$54 - L\,\text{cm}$, will be occupied by the mercury inside the tube.
Because the air inside the tube expands when the tube is raised (its pressure
is now lower than atmospheric), the mercury level inside the tube is pushed
upward above the outside free surface of mercury. Consequently, if
$H = 54 - L$ is the height of the mercury column inside the tube above the
free surface, the final air pressure $P_2$ is given by:
$P_2 = 76 - H = 76 - (54 - L) = 76 - 54 + L = 22 + L$ (in cm of Hg).
Step 4: Apply Boyle's Law
Since the temperature and the amount of air remain constant, Boyle's law
applies:
$P_1 \times V_1 = P_2 \times V_2$.
Substituting the known values:
$$
76 \times 8 = (22 + L) \times L
$$
Hence,
$$
608 = L^2 + 22\,L.
$$
Step 5: Solve the quadratic equation
Rearrange the equation to standard form:
$$
L^2 + 22\,L - 608 = 0.
$$
By checking integer factors or using the quadratic formula, one finds
$L = 16\,\text{cm}$ as the physically valid root.
Step 6: Conclude the final length of air column
Therefore, the final length of the air column above the mercury in the tube is
$16\,\text{cm}$.
Illustration
