Two vessel connected by a pipe with a sliding plug contain mercury. In one vessel, the height of murcury column is `39.2 cm` and its temperature is `0^@ C`, while in the other, the height of mercury column is `40 cm` and its temperature is `100^@ C`. Find the coefficient if cubical expansion for mercury. The volume of the connecting pipe should be neglected.

Explanation

To find the coefficient of cubical expansion (γ) for mercury, we can follow these steps:

### Step 1: Understand the relationship between pressure and height of the mercury column
Since the two vessels are connected by a pipe, the pressure at the bottom of each mercury column must be equal. Therefore, we can write:
\[ \rho_0 g h_0 = \rho_{100} g h_{100} \]
where:
- \( \rho_0 \) = density of mercury at 0°C
- \( h_0 \) = height of mercury at 0°C = 39.2 cm
- \( \rho_{100} \) = density of mercury at 100°C
- \( h_{100} \) = height of mercury at 100°C = 40 cm

### Step 2: Express the densities in terms of the initial density and the coefficient of cubical expansion
The density of mercury at a temperature \( T \) can be expressed as:
\[ \rho_T = \frac{\rho_0}{1 + \gamma \Delta T} \]
where \( \Delta T \) is the change in temperature. For our case:
- At 0°C: \( \Delta T = 0 \)
- At 100°C: \( \Delta T = 100 \)

Thus, we have:
\[ \rho_{100} = \frac{\rho_0}{1 + 100\gamma} \]

### Step 3: Substitute the density expressions into the pressure equation
Substituting the expressions for \( \rho_0 \) and \( \rho_{100} \) into the pressure equation gives:
\[ \rho_0 g h_0 = \frac{\rho_0}{1 + 100\gamma} g h_{100} \]
We can cancel \( \rho_0 g \) from both sides:
\[ h_0 = \frac{h_{100}}{1 + 100\gamma} \]

### Step 4: Rearranging the equation to solve for γ
Rearranging the equation gives:
\[ h_0 (1 + 100\gamma) = h_{100} \]
\[ h_0 + 100h_0\gamma = h_{100} \]
\[ 100h_0\gamma = h_{100} - h_0 \]
\[ \gamma = \frac{h_{100} - h_0}{100h_0} \]

### Step 5: Substitute the known values
Now substituting the values:
- \( h_0 = 39.2 \) cm
- \( h_{100} = 40 \) cm

We get:
\[ \gamma = \frac{40 - 39.2}{100 \times 39.2} \]
\[ \gamma = \frac{0.8}{3920} \]
\[ \gamma = 0.00020408 \, \text{per °C} \]

### Step 6: Final answer
Thus, the coefficient of cubical expansion for mercury is:
\[ \gamma \approx 2.04 \times 10^{-4} \, \text{per °C} \]