The apparatus shown in figure consists of four glass columns connected by horizontal sections. The h

Q: Why do we equate the pressure at the top horizontal tube?

Because the liquid is in static equilibrium. If the pressure at the top of column B was different from the top of column C, liquid would flow through the connecting tube until the pressures equalized.

Q: Why does the pressure decrease when we go up column B?

As you move upwards in a fluid, there is less fluid above you pushing down, so the hydrostatic pressure decreases by $\rho g h$.

Q: Why are the liquid levels in A and D different?

The columns are at different temperatures. Column A is hotter ($95^\circ\text{C}$), so its liquid is less dense and expands, requiring a higher column to balance the pressure compared to the cooler, denser liquid in column D.

Q: Can we use $\rho_t = \rho_0 (1 - \gamma t)$ instead of $\rho_t = \frac{\rho_0}{1 + \gamma t}$?

Yes, for small temperature changes, the binomial approximation $\rho_0 (1 + \gamma t)^{-1} \approx \rho_0 (1 - \gamma t)$ is valid. However, using the exact fractional form is safer and often leads to cleaner cancellations in ratio problems like this one.

Q: Why did we divide $\gamma$ by 3 at the end?

The question specifically asks for the *linear* coefficient of thermal expansion ($\alpha$). The formula for density variation gives us the *volume* coefficient of expansion ($\gamma$). For isotropic materials, $\gamma \approx 3\alpha$.

Question

The apparatus shown in figure consists of four glass columns connected by horizontal sections. The height of two central columns $B$ and $C$ are $49	ext{ cm}$ each. The two outer columns $A$ and $D$ are open to the atmosphere. $A$ and $C$ are maintained at a temperature of $95^\circ	ext{C}$ while the columns $B$ and $D$ are maintained at $5^\circ	ext{C}$. The height of the liquid in $A$ and $D$ measured from the base line are $52.8	ext{ cm}$ and $51	ext{ cm}$ respectively. Determine the linear coefficient of thermal expansion of the liquid.

Sigma Percentile · Accepted Answer

The problem presents a fascinating scenario where thermal expansion and hydrostatics intertwine. We have a complex U-tube apparatus with four columns maintained at different temperatures. Our goal is to find the linear coefficient of thermal expansion of the liquid inside.

### Analyzing the Setup

Let's break down the geometry and temperatures. We have two outer columns, $A$ and $D$, which are open to the atmosphere. The two inner columns, $B$ and $C$, are connected by a horizontal tube at the top.

The temperatures are alternating: column $A$ is at $95^\circ	ext{C}$, $B$ is at $5^\circ	ext{C}$, $C$ is at $95^\circ	ext{C}$, and $D$ is at $5^\circ	ext{C}$. Because the liquid in the hotter columns expands and becomes less dense, it requires a taller column to exert the same hydrostatic pressure. This is why the liquid level in $A$ ($h_1 = 52.8	ext{ cm}$) is higher than in $D$ ($h_2 = 51	ext{ cm}$). The central columns $B$ and $C$ have a height of $h = 49	ext{ cm}$.

### The Master Equation

Since the liquid is in static equilibrium, the pressure at any continuous horizontal level must be consistent. The most strategic place to equate pressures is the top horizontal tube connecting columns $B$ and $C$. Let's calculate this pressure from both the left and the right sides.

Starting from the left (column $A$), the pressure at the open surface is atmospheric pressure, $p_0$. As we go down column $A$ to the bottom, the pressure increases by $ho_{95^\circ	ext{C}} g h_1$. Then, as we go up column $B$ to the top horizontal tube, the pressure decreases by $ho_{5^\circ	ext{C}} g h$.

$$p_{	ext{top}} = p_0 + ho_{95^\circ	ext{C}} g h_1 - ho_{5^\circ	ext{C}} g h$$

Now, let's trace from the right (column $D$). Starting at $p_0$, we go down column $D$, adding $ho_{5^\circ	ext{C}} g h_2$, and then up column $C$, subtracting $ho_{95^\circ	ext{C}} g h$.

$$p_{	ext{top}} = p_0 + ho_{5^\circ	ext{C}} g h_2 - ho_{95^\circ	ext{C}} g h$$

Equating these two expressions for $p_{	ext{top}}$, the $p_0$ and $g$ terms cancel out beautifully:

$$ho_{95^\circ	ext{C}} h_1 - ho_{5^\circ	ext{C}} h = ho_{5^\circ	ext{C}} h_2 - ho_{95^\circ	ext{C}} h$$

Rearranging to group the densities, we get:

$$ho_{95^\circ	ext{C}} (h_1 + h) = ho_{5^\circ	ext{C}} (h_2 + h)$$

$$\frac{ho_{95^\circ	ext{C}}}{ho_{5^\circ	ext{C}}} = \frac{h_2 + h}{h_1 + h}$$

### Density Variation with Temperature

We know that the density of a liquid varies with temperature according to the relation:

$$ho_t = \frac{ho_0}{1 + \gamma t}$$

where $\gamma$ is the coefficient of volume expansion. Substituting this into our ratio:

$$\frac{\frac{ho_0}{1 + 95\gamma}}{\frac{ho_0}{1 + 5\gamma}} = \frac{1 + 5\gamma}{1 + 95\gamma}$$

Equating this to our height ratio:

$$\frac{1 + 5\gamma}{1 + 95\gamma} = \frac{51 + 49}{52.8 + 49} = \frac{100}{101.8}$$

### Final Calculation

Now, we just need to solve this linear equation for $\gamma$:

$$101.8 (1 + 5\gamma) = 100 (1 + 95\gamma)$$

$$101.8 + 509\gamma = 100 + 9500\gamma$$

$$1.8 = 8991\gamma$$

$$\gamma = \frac{1.8}{8991} \approx 2 	imes 10^{-4} / ^\circ	ext{C}$$

The question specifically asks for the **linear** coefficient of thermal expansion, $\alpha$. For isotropic liquids, the volume coefficient is three times the linear coefficient ($\gamma = 3\alpha$).

$$\alpha = \frac{\gamma}{3} = \frac{2 	imes 10^{-4}}{3} \approx 6.7 	imes 10^{-5} / ^\circ	ext{C}$$

This elegant problem demonstrates how macroscopic measurements of liquid columns can precisely determine microscopic thermal properties!

Visualized Solution

\text{Understanding the Apparatus}

\text{Hydrostatic Equilibrium}

\text{Pressure from Left Side}

\text{Pressure from Right Side}

\text{Equating Pressures}

\text{Density Variation with Temperature}

\text{Calculating } \gamma

\text{Linear Coefficient of Expansion}

\text{Food for Thought}

The Sigma Insight: Thermal Expansion

Analyzing the Setup

The Master Equation

Density Variation with Temperature

Final Calculation

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