A one litre flask contains certain quantity of mercury. If the volume of air inside the flask remains the same at all temperatures then the volume of mercury in the flask is (volume expansion coefficient of mercury is 20 times that of flask)

Options

A. 100 cc
B. 50 cc
C. 200 cc
D. 150 cc

Explanation

To solve the problem, we need to analyze the thermal expansion of mercury and the flask. Given that the volume of air inside the flask remains constant at all temperatures, we can derive the volume of mercury in the flask.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a 1-liter flask containing mercury.
- The volume of air inside the flask remains constant at all temperatures.
- The volume expansion coefficient of mercury (γ_Hg) is 20 times that of the flask (γ_F).

2. **Initial Volume of the Flask**:
- The initial volume of the flask (V_flask) is given as 1 liter, which can be converted to cubic meters:
\[ V_{\text{flask}} = 1 \, \text{liter} = 10^{-3} \, \text{m}^3 \]

3. **Volume Expansion Relationship**:
- The volume expansion of a substance can be expressed as:
\[ \Delta V = \gamma \cdot V_0 \cdot \Delta T \]
- Where:
- \(\Delta V\) is the change in volume,
- \(\gamma\) is the volume expansion coefficient,
- \(V_0\) is the initial volume,
- \(\Delta T\) is the change in temperature.

4. **Setting Up the Equation**:
- Let the initial volume of mercury be \(V_{Hg}\).
- The volume of air in the flask is \(V_{\text{air}} = V_{\text{flask}} - V_{Hg}\).
- Since the volume of air remains constant, any increase in the volume of the flask must be compensated by a decrease in the volume of mercury.

5. **Volume Expansion of Flask and Mercury**:
- The volume expansion of the flask can be expressed as:
\[ \Delta V_{\text{flask}} = \gamma_F \cdot V_{\text{flask}} \cdot \Delta T \]
- The volume expansion of mercury can be expressed as:
\[ \Delta V_{\text{Hg}} = \gamma_{Hg} \cdot V_{Hg} \cdot \Delta T \]

6. **Using the Given Relationship**:
- Since \(\gamma_{Hg} = 20 \cdot \gamma_F\), we can substitute this into the equation:
\[ \Delta V_{\text{Hg}} = 20 \cdot \gamma_F \cdot V_{Hg} \cdot \Delta T \]

7. **Equating the Volume Changes**:
- For the volume of air to remain constant:
\[ \Delta V_{\text{flask}} + \Delta V_{\text{Hg}} = 0 \]
- This leads to:
\[ \gamma_F \cdot V_{\text{flask}} \cdot \Delta T + 20 \cdot \gamma_F \cdot V_{Hg} \cdot \Delta T = 0 \]
- Dividing through by \(\Delta T\) (assuming \(\Delta T \neq 0\)):
\[ \gamma_F \cdot V_{\text{flask}} + 20 \cdot \gamma_F \cdot V_{Hg} = 0 \]

8. **Solving for \(V_{Hg}\)**:
- Rearranging gives:
\[ V_{Hg} = -\frac{V_{\text{flask}}}{20} \]
- Since volumes cannot be negative, we take the absolute value:
\[ V_{Hg} = \frac{V_{\text{flask}}}{20} = \frac{10^{-3}}{20} = 5 \times 10^{-5} \, \text{m}^3 \]

9. **Converting to Centimeters Cubed**:
- To convert to cubic centimeters (cc):
\[ V_{Hg} = 5 \times 10^{-5} \, \text{m}^3 \times 10^6 \, \text{cc/m}^3 = 50 \, \text{cc} \]

### Final Answer:
The volume of mercury in the flask is **50 cc**.