An open glass tube is immersed in mercury in such a way that a length of 8 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 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

We have a straight glass tube whose bore (cross-sectional area) is uniform. Initially both ends are open and the tube is dipped into a mercury bath in such a way that $$8\ \text{cm}$$ of its length projects above the outside mercury level. Hence the column inside the tube that lies above the mercury in the tube is an $$8\ \text{cm}$$ long column of air which is at the atmospheric pressure $$P_{\text{atm}} = 76\ \text{cm of Hg}$$, because the upper end is still open to the atmosphere.

Now the upper (projecting) end is closed and instantly sealed. At that instant the trapped air has

$$P_1 = 76\ \text{cm Hg}, \qquad V_1 \propto 8\ \text{cm}.$$

Next the whole tube is lifted vertically upward by an additional $$46\ \text{cm}$$ while its lower end remains immersed in the same mercury bath. Consequently, the top of the tube now stands

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

above the outside mercury surface.

Let the new length of the trapped air column be $$L\ \text{cm}$$ (this is what we have to find). Then the mercury inside the tube must have risen above the outside level by

$$H = 54 - L\ \text{cm}.$$

To relate the new air pressure to this rise, we use the hydrostatic pressure relation:

Moving upward through a column of mercury of height $$H$$ reduces the pressure by $$H\ \text{cm of Hg}$$. Hence the pressure just beneath the mercury-air interface inside the tube is

$$P_{\text{interface}} = P_{\text{atm}} - H = 76 - (54 - L) = 76 - 54 + L = 22 + L\ \text{cm Hg}.$$

This pressure is the same as the pressure $$P_2$$ of the trapped air above the mercury, because the air is in direct contact with the mercury surface. Therefore,

$$P_2 = 22 + L\ \text{cm Hg}.$$

The process occurs slowly enough for us to treat the trapped air as undergoing an isothermal change (temperature is effectively constant). Hence we can apply Boyle’s law, which states

$$P_1 V_1 = P_2 V_2.$$

Because the cross-sectional area of the tube is constant, the volume is proportional to the length of the air column, so we have

$$P_1 \times (\text{length }8) = P_2 \times (\text{length }L).$$

Substituting the known values,

$$76 \times 8 = (22 + L)\, L.$$

That gives the quadratic equation

$$L^2 + 22L - 608 = 0.$$

Solving it by the quadratic formula,

$$L = \frac{-22 \pm \sqrt{22^2 + 4 \times 608}}{2} = \frac{-22 \pm \sqrt{484 + 2432}}{2} = \frac{-22 \pm \sqrt{2916}}{2} = \frac{-22 \pm 54}{2}.$$

The negative root is not physically meaningful (it would give a negative length). Taking the positive root,

$$L = \frac{32}{2} = 16\ \text{cm}.$$

Hence the length of the trapped air column after the tube has been raised is $$16\ \text{cm}$$.

Hence, the correct answer is Option A.