An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the merc

Q: Why is the initial pressure exactly 76 cm of Hg?

Before the tube was sealed, it was open to the atmosphere. Therefore, the air inside was in equilibrium with the outside air, meaning its pressure was exactly equal to the atmospheric pressure, which is given as $76\text{ cm of Hg}$.

Q: Why does the mercury rise inside the tube when we pull it up?

When you pull the sealed tube up, you are increasing the volume available for the trapped air. According to Boyle's Law, an increase in volume leads to a decrease in pressure. Because the pressure inside the tube is now less than the atmospheric pressure outside, the outside atmosphere pushes the mercury up into the tube until the pressures balance out.

Q: How did we get the equation $p_2 = 76 - (54 - x)$?

We used the principle that pressure at the same horizontal level is equal. At the level of the outside mercury surface, the pressure is $76\text{ cm of Hg}$. Inside the tube, at this same level, the pressure is the sum of the trapped air's pressure ($p_2$) and the weight of the mercury column above it, which has a height of $(54 - x)$. So, $p_2 + (54 - x) = 76$.

Q: Why can we ignore the cross-sectional area of the tube?

The volume of a cylinder is Area $\times$ Height. When we set up Boyle's Law ($p_1V_1 = p_2V_2$), the Area ($A$) appears on both sides of the equation ($p_1 \cdot h_1 \cdot A = p_2 \cdot h_2 \cdot A$). Because it's a non-zero constant, it mathematically cancels out, meaning the final height is independent of how wide the tube is.

Q: What if the quadratic equation gave two positive roots?

In physical problems like this, if a quadratic yields two positive roots, you must check the physical constraints. For instance, the length of the air column $x$ cannot be greater than the total length of the tube above the mercury ($54\text{ cm}$). One of the roots would inevitably violate such a boundary condition.

Question

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)

Options:

(A) 16 cm

(B) 22 cm

(C) 38 cm

(D) 6 cm

Sigma Percentile · Accepted Answer

### The Physics of Trapped Air: A Journey Through Boyle's Law and Fluid Statics

Imagine you are in a laboratory, holding a glass tube partially submerged in a pool of shimmering liquid mercury. This classic setup is not just a visual treat; it is a profound demonstration of the invisible forces that govern our atmosphere and the gases around us. Let's dive into the mechanics of this fascinating problem.

### Analyzing the Initial State

We start with an open glass tube immersed in mercury, with exactly $8	ext{ cm}$ extending above the surface. When we seal the top, we trap a specific amount of air inside. Because the tube was open to the atmosphere just before sealing, the pressure of this trapped air, $p_1$, is perfectly balanced with the outside world.

Therefore, $p_1 = p_{atm} = 76	ext{ cm of Hg}$.

If we assume the cross-sectional area of the tube is $A$, the initial volume of this trapped air is simply the area multiplied by the height: $V_1 = 8A$.

### The Upward Pull and Fluid Statics

Now, the plot thickens. We raise the tube vertically by an additional $46	ext{ cm}$. The total length of the tube extending above the outside mercury level is now $8 + 46 = 54	ext{ cm}$.

As we pull the tube up, the pressure inside drops, and the atmospheric pressure pushes mercury up into the tube. Let's call the new length of the air column $x$. This means the mercury has risen inside the tube to a height of $(54 - x)	ext{ cm}$.

How do we find the new pressure, $p_2$, of the trapped air? We rely on the fundamental principle of fluid statics: **the pressure at the same horizontal level in a continuous static fluid must be equal**.

If we look at the horizontal level of the outside mercury surface, the pressure is $p_{atm}$. Inside the tube at this exact same level, the pressure is the sum of the trapped air's pressure ($p_2$) and the pressure exerted by the column of mercury above it.

$$p_2 + (54 - x) = 76$$

Solving for $p_2$, we get a beautiful expression dependent entirely on our unknown $x$:

$$p_2 = 76 - (54 - x) = 22 + x$$

### Boyle's Law in Action

Because this process happens slowly, the temperature of the trapped air remains constant. This is the perfect scenario to deploy **Boyle's Law**, which states that for a fixed mass of gas at a constant temperature, the product of pressure and volume is constant.

$$p_1V_1 = p_2V_2$$

Let's substitute our knowns and expressions into this master equation:

$$76 	imes (8A) = (22 + x) 	imes (xA)$$

### The Final Calculation

The cross-sectional area $A$ elegantly cancels out from both sides, leaving us with a pure algebraic equation:

$$608 = 22x + x^2$$

Rearranging this into a standard quadratic form, we get:

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

To solve this, we look for two numbers that multiply to $-608$ and add to $22$. Those numbers are $38$ and $-16$. Factoring the quadratic gives:

$$(x + 38)(x - 16) = 0$$

This yields two mathematical solutions: $x = -38$ and $x = 16$. Since a physical length cannot be negative, we discard $-38$.

Thus, the new length of the air column is exactly **$16	ext{ cm}$**. By carefully balancing fluid pressures and applying the laws of thermodynamics, we've successfully decoded the hidden mechanics of the trapped air!

Select Answer:

Visualized Solution

Initial State Analysis

Final State Geometry

Pressure in Final State

Applying Boyle's Law

Substituting Values

Algebraic Simplification

Solving the Quadratic Equation

Final Conclusion

The Sigma Insight: Kinetic Theory of Gases

The Physics of Trapped Air

Analyzing the Initial State

The Upward Pull and Fluid Statics

Boyle's Law in Action

The Final Calculation

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