A glass flask of volume 1 litre is fully filled with mercury at `0^(@)C`. Both the flask and mercury are now heated to `100^(@)C`. If the coefficient of volume expansion of mercury is `1.82xx10^(-4)//""^(@)C`, volume coefficient of linear expansion of glass is `10xx10^(-6)//""^(@)C`, the amount of mercury which overflows is

Options

A. 15.2 ml
B. 17.2 ml
C. 19.2 ml
D. 21.2 ml

Explanation

To find the amount of mercury that overflows from a glass flask when both the flask and mercury are heated from \(0^\circ C\) to \(100^\circ C\), we can follow these steps:

### Step 1: Understand the Problem
We have a glass flask of volume 1 liter filled with mercury. When heated, both the mercury and the flask expand, but they do so at different rates due to their different coefficients of expansion. The goal is to find out how much mercury overflows when the temperature is raised to \(100^\circ C\).

### Step 2: Calculate the Volume Expansion of Mercury
The volume expansion of mercury can be calculated using the formula:
\[ \Delta V_m = V_0 \cdot \gamma_m \cdot \Delta T \]
where:
- \(V_0 = 1 \, \text{liter} = 1000 \, \text{cm}^3\)
- \(\gamma_m = 1.82 \times 10^{-4} \, \text{°C}^{-1}\) (coefficient of volume expansion of mercury)
- \(\Delta T = 100^\circ C - 0^\circ C = 100^\circ C\)

Substituting the values:
\[ \Delta V_m = 1000 \, \text{cm}^3 \cdot (1.82 \times 10^{-4}) \cdot 100 \]
\[ \Delta V_m = 1000 \cdot 1.82 \times 10^{-2} = 18.2 \, \text{cm}^3 \]

### Step 3: Calculate the Volume Expansion of Glass
The volume expansion of the glass flask can be calculated using the formula:
\[ \Delta V_g = V_0 \cdot (3\alpha_g) \cdot \Delta T \]
where:
- \(\alpha_g = 10 \times 10^{-6} \, \text{°C}^{-1}\) (coefficient of linear expansion of glass)

Substituting the values:
\[ \Delta V_g = 1000 \, \text{cm}^3 \cdot (3 \cdot 10 \times 10^{-6}) \cdot 100 \]
\[ \Delta V_g = 1000 \cdot 3 \times 10^{-4} = 0.3 \, \text{cm}^3 \]

### Step 4: Calculate the Overflow of Mercury
The amount of mercury that overflows can be found by taking the difference between the volume expansion of mercury and the volume expansion of glass:
\[ \Delta V_{\text{overflow}} = \Delta V_m - \Delta V_g \]
Substituting the values:
\[ \Delta V_{\text{overflow}} = 18.2 \, \text{cm}^3 - 0.3 \, \text{cm}^3 = 17.9 \, \text{cm}^3 \]

### Step 5: Convert to Milliliters
Since \(1 \, \text{cm}^3 = 1 \, \text{mL}\), the amount of mercury that overflows is:
\[ \Delta V_{\text{overflow}} = 17.9 \, \text{mL} \]

### Final Answer
The amount of mercury that overflows is approximately **17.9 mL**.

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