A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm c

Q: Why do we use $5 \cos 60^\circ$ for the pressure difference?

Because pressure in this context is often measured in 'cm of Hg', which directly corresponds to the vertical height of the mercury column. When the tube is tilted at $60^\circ$ to the vertical, the effective vertical height of the 5 cm mercury column is $5 \cos 60^\circ$.

Q: Why is the process considered isothermal?

The problem states that the temperature of the system is kept at $30^\circ\text{C}$. Since the temperature is constant and the trapped air behaves like an ideal gas, the process is isothermal, allowing us to use Boyle's Law.

Q: Why does the total length of the tube remain constant?

The tube is a rigid, sealed container. Tilting it doesn't stretch or compress the glass. Therefore, the sum of the lengths of the two air columns and the mercury column must be the same in both the horizontal and inclined states.

Q: What if the tube was open at one end?

If one end were open, the pressure at that end would simply be the atmospheric pressure. You would only need to apply Boyle's Law to the sealed end, and the force balance on the mercury would involve atmospheric pressure.

Q: Can we solve this using standard SI units (Pascals)?

Absolutely! You could convert the pressures to Pascals using $p = \rho g h$, where $\rho$ is the density of mercury. However, working directly in 'cm of Hg' is a massive shortcut that saves time and reduces calculation errors, which is crucial for JEE.

Question

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle $5	ext{ cm}$ containing mercury and the two equal ends containing air at the same pressure $p$. When the tube is held at an angle of $60^{\circ}$ with the vertical direction, the length of the air column above and below the mercury column are $46	ext{ cm}$ and $44.5	ext{ cm}$ respectively. Calculate the pressure $p$ in centimetre of mercury. (The temperature of the system is kept at $30^{\circ}	ext{C}$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a sealed glass tube lying flat on a table. Inside, a $5	ext{ cm}$ plug of mercury sits perfectly in the middle, trapping two equal columns of air on either side. Because the tube is horizontal, the mercury doesn't exert any net force on either air column, meaning the pressure $p$ is identical on both sides.

Since the tube is a rigid container, its total length is a strict constant. When we eventually tilt the tube, the lengths of the air columns will change to $46	ext{ cm}$ and $44.5	ext{ cm}$, while the mercury remains $5	ext{ cm}$ long.

We can use this to find the initial length $x$ of each air column:
$$2x + 5 = 46 + 5 + 44.5$$
$$2x = 90.5 \implies x = 45.25	ext{ cm}$$

### The Master Equation

Now, we tilt the tube so it makes an angle of $60^\circ$ with the vertical. This is equivalent to a $30^\circ$ angle with the horizontal. Gravity now pulls the mercury plug downwards along the incline, compressing the bottom air column and allowing the top one to expand.

Let the new pressure in the top column be $p_1$ and the bottom column be $p_2$. The mercury plug is in equilibrium, so the forces acting on it must balance. The pressure from below pushes up, the pressure from above pushes down, and the component of the mercury's weight acts down the incline.

Balancing these forces gives us the pressure difference. If we express pressure directly in "cm of Hg", the pressure difference is simply the effective vertical height of the mercury column:
$$p_2 - p_1 = l_{	ext{Hg}} \cos 60^\circ = 5 \cos 60^\circ = 2.5	ext{ cm of Hg}$$

Since the temperature is held constant at $30^\circ	ext{C}$, the trapped air undergoes an **isothermal process**. We can apply Boyle's Law ($pV = 	ext{constant}$) to both air columns to express $p_1$ and $p_2$ in terms of the initial pressure $p$.

For the top column:
$$p \cdot (45.25 \cdot A) = p_1 \cdot (46 \cdot A) \implies p_1 = \frac{45.25}{46} p$$

For the bottom column:
$$p \cdot (45.25 \cdot A) = p_2 \cdot (44.5 \cdot A) \implies p_2 = \frac{45.25}{44.5} p$$

### Final Calculation

We now substitute these expressions back into our master force balance equation:
$$\frac{45.25}{44.5} p - \frac{45.25}{46} p = 2.5$$

Factoring out $45.25 p$, we get:
$$45.25 p \left( \frac{46 - 44.5}{44.5 	imes 46} ight) = 2.5$$
$$45.25 p \left( \frac{1.5}{2047} ight) = 2.5$$

Finally, isolating $p$:
$$p = \frac{2.5 	imes 2047}{45.25 	imes 1.5} = \frac{5117.5}{67.875} \approx 75.4	ext{ cm of Hg}$$

The initial pressure of the air in the tube was $75.4	ext{ cm of Hg}$. A beautiful synthesis of fluid mechanics and thermodynamics!

Enter Numerical Value:

Visualized Solution

$x = 45.25 cm$

$Inclined State$

$p_{2} - p_{1} = 2.5 cm of Hg$

$p_{1} = \frac{45.25}{46} p$

$p_{2} = \frac{45.25}{44.5} p$

$45.25 p (\frac{1.5}{2047}) = 2.5$

$p \approx 75.4 cm of Hg$

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=45.25 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inclined State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2​−p1​=2.5 cm of Hg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p1​=4645.25​p

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p2​=44.545.25​p

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 45.25p(20471.5​)=2.5

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p≈75.4 cm of Hg