1. Initial situation

A glass tube is immersed vertically in mercury such that $8\,\text{cm}$ of its length is above the mercury surface.

Then the top end is closed while in this position.

So the trapped air column initially has:

• length $l_1 = 8\,\text{cm}$
• pressure equal to atmospheric pressure, since the mercury level inside and outside are at the same level before raising.

Hence, $P_1 = 76\,\text{cm of Hg}$

2. After raising the closed tube by $46\,\text{cm}$

Now the top end rises by $46\,\text{cm}$ more, so the top of the tube is at height $8 + 46 = 54\,\text{cm}$ abovethe outside mercury surface.

Let the final length of trapped air column be $l$ cm.

Then the mercury level inside the tube will be at a depth $54 - l$ above the outside mercury surface.

Therefore the pressure of trapped air inside becomes less than atmospheric by $(54-l)$ cm of Hg: $P_2 = 76 - (54-l)$ $P_2 = 22 + l$

3. Apply Boyle's law

Assuming constant temperature, $P_1V_1 = P_2V_2$ Since tube has uniform cross-section, volume is proportional to length: $P_1 l_1 = P_2 l$

So, $76 \times 8 = (22 + l)l$

$608 = l^2 + 22l$

$l^2 + 22l - 608 = 0$

4. Solve the quadratic

$l^2 + 22l - 608 = 0$

Discriminant: $D = 22^2 + 4 \times 608 = 484 + 2432 = 2916$ $\sqrt{2916} = 54$

Thus, $l = \frac{-22 \pm 54}{2}$

Possible values: $l = \frac{32}{2} = 16$ or $l = \frac{-76}{2} = -38$

Negative length is not physical, so $l = 16\,\text{cm}$

5. Check with options

The correct option is: A: $16\,\text{cm}$