A glass flask of volume `1000 cm^(3)` is completely filled with mercury at `0^(@)C`. The coefficient of cubical epansioni of mercury is `182xx10^(-6)//.^(@)C` and that of glas ils `30xx10^(-6)//.^(@)C`, how much mercury will over flow? When heated to `100^(@)C`

Options

A. `30 cm^(3)`
B. `18.2 cm^(3)`
C. `15.2 cm^(2)`
D. `3 cm^(2)`

Answer

To solve the problem of how much mercury will overflow when heated from \(0^\circ C\) to \(100^\circ C\), we will follow these steps:

### Step 1: Calculate the change in volume of mercury
The formula for the change in volume due to thermal expansion is given by:

\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]

Where:
- \(V_0\) = initial volume of mercury = \(1000 \, \text{cm}^3\)
- \(\gamma\) = coefficient of cubical expansion of mercury = \(182 \times 10^{-6} \, \text{°C}^{-1}\)
- \(\Delta T\) = change in temperature = \(100 - 0 = 100 \, \text{°C}\)

Substituting the values:

\[ \Delta V_m = 1000 \cdot (182 \times 10^{-6}) \cdot 100 \]

Calculating this gives:

\[ \Delta V_m = 1000 \cdot 182 \cdot 10^{-6} \cdot 100 = 18.2 \, \text{cm}^3 \]

### Step 2: Calculate the change in volume of the glass flask
Using the same formula for the glass flask:

\[ \Delta V_g = V_0 \cdot \gamma_g \cdot \Delta T \]

Where:
- \(\gamma_g\) = coefficient of cubical expansion of glass = \(30 \times 10^{-6} \, \text{°C}^{-1}\)

Substituting the values:

\[ \Delta V_g = 1000 \cdot (30 \times 10^{-6}) \cdot 100 \]

Calculating this gives:

\[ \Delta V_g = 1000 \cdot 30 \cdot 10^{-6} \cdot 100 = 3 \, \text{cm}^3 \]

### Step 3: Determine the net change in volume of mercury
The net change in volume of mercury that will cause it to overflow is given by:

\[ \text{Overflow} = \Delta V_m - \Delta V_g \]

Substituting the values:

\[ \text{Overflow} = 18.2 \, \text{cm}^3 - 3 \, \text{cm}^3 = 15.2 \, \text{cm}^3 \]

### Conclusion
Thus, the amount of mercury that will overflow when heated to \(100^\circ C\) is:

\[ \boxed{15.2 \, \text{cm}^3} \]

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