A two litre glass flask contains some mercury. It is found that at all temperatures the volume of the air inside the flask remains the same. The volume of the mercury inside the flask is (`alpha`for glass `=9xx10^-6//^@C,gamma` for mercury `1.8xx10^-4//^@C)`

Options

A. 1500 cc
B. 150 cc
C. 3000 cc
D. 300 cc

Answer

To solve the problem step by step, we will analyze the situation involving the glass flask containing mercury and the properties of thermal expansion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a 2-liter glass flask containing mercury.
- The volume of air inside the flask remains constant at all temperatures.
- The coefficient of linear expansion for glass (α) is \(9 \times 10^{-6} \, ^\circ C^{-1}\).
- The coefficient of volumetric expansion for mercury (γ) is \(1.8 \times 10^{-4} \, ^\circ C^{-1}\).

2. **Identify the Relationship**:
- The volume of the air inside the flask remains constant, which implies that any expansion of the glass flask must be compensated by the contraction of mercury, or vice versa.
- The volumetric expansion coefficient for glass (γ_f) can be calculated from the linear expansion coefficient (α) using the formula:
\[ \gamma_f = 3\alpha \]

3. **Calculate the Volumetric Expansion Coefficient for Glass**:
- Substitute the value of α into the equation:
\[ \gamma_f = 3 \times (9 \times 10^{-6}) = 27 \times 10^{-6} \, ^\circ C^{-1} \]

4. **Setting Up the Equation**:
- Since the volume of air remains constant, the expansion of the glass flask must equal the expansion of the mercury:
\[ \Delta V_f = \Delta V_m \]
- The change in volume can be expressed as:
\[ \Delta V = V_0 \cdot \gamma \cdot \Delta T \]
- Therefore, we can write:
\[ V_f \cdot \gamma_f \cdot \Delta T = V_m \cdot \gamma_m \cdot \Delta T \]

5. **Simplifying the Equation**:
- Since \(\Delta T\) is common on both sides, we can cancel it out:
\[ V_f \cdot \gamma_f = V_m \cdot \gamma_m \]

6. **Substituting Known Values**:
- The volume of the flask \(V_f = 2 \, \text{liters} = 2000 \, \text{cc}\).
- Substitute the values of γ_f and γ_m:
\[ 2000 \cdot (27 \times 10^{-6}) = V_m \cdot (1.8 \times 10^{-4}) \]

7. **Solving for \(V_m\)**:
- Rearranging the equation gives:
\[ V_m = \frac{2000 \cdot (27 \times 10^{-6})}{1.8 \times 10^{-4}} \]
- Calculate \(V_m\):
\[ V_m = \frac{2000 \cdot 27}{1.8} \times \frac{10^{-6}}{10^{-4}} = \frac{54000}{1.8} \times 10^{-2} \]
\[ V_m = 3000 \times 10^{-2} = 300 \, \text{cc} \]

8. **Final Result**:
- The volume of mercury in the flask is \(300 \, \text{cc}\).