A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass`=9xx10^(-6)//^(@)C` and the coefficient of volume expansion of `Hg=1.8xx10^(-4)//^(@)C` ?

Options

A. a. 50 `cm^(3)`
B. b. 100 `cm^(3)`
C. c. 150 `cm^(3)`
D. d. 200 `cm^(3)`

Answer

To solve the problem, we need to determine the volume of mercury in a 1-L flask, given that the volume of air inside the flask remains constant at different temperatures. The coefficients of linear and volumetric expansion for the glass and mercury are provided.

### Step-by-Step Solution

1. **Identify the Given Data:**
- Volume of the flask (V_flask) = 1 L = 1000 cm³
- Coefficient of linear expansion of glass (α_glass) = 9 × 10^(-6) /°C
- Coefficient of volumetric expansion of mercury (γ_mercury) = 1.8 × 10^(-4) /°C

2. **Calculate the Coefficient of Volume Expansion for Glass:**
- The volumetric expansion coefficient (γ) for glass is related to the linear expansion coefficient (α) by the formula:
\[ \gamma_{glass} = 3 \times \alpha_{glass} \]
- Substituting the value:
\[ \gamma_{glass} = 3 \times (9 \times 10^{-6}) = 27 \times 10^{-6} \, /°C \]

3. **Change in Volume of Glass:**
- The change in volume of the glass (ΔV_glass) can be expressed as:
\[ \Delta V_{glass} = V_{glass} \times \gamma_{glass} \times \Delta T \]
- Since the volume of the flask is 1 L (1000 cm³), we have:
\[ \Delta V_{glass} = 1000 \times (27 \times 10^{-6}) \times \Delta T \]

4. **Change in Volume of Mercury:**
- The change in volume of mercury (ΔV_mercury) is given by:
\[ \Delta V_{mercury} = V_{mercury} \times \gamma_{mercury} \times \Delta T \]
- Substituting the value of γ_mercury:
\[ \Delta V_{mercury} = V_{mercury} \times (1.8 \times 10^{-4}) \times \Delta T \]

5. **Setting the Change in Volumes Equal:**
- Since the volume of air inside the flask remains constant, the change in volume of glass must equal the change in volume of mercury:
\[ \Delta V_{glass} = \Delta V_{mercury} \]
- Therefore:
\[ 1000 \times (27 \times 10^{-6}) \times \Delta T = V_{mercury} \times (1.8 \times 10^{-4}) \times \Delta T \]

6. **Canceling ΔT:**
- Since ΔT is common in both sides, we can cancel it out:
\[ 1000 \times (27 \times 10^{-6}) = V_{mercury} \times (1.8 \times 10^{-4}) \]

7. **Solving for V_mercury:**
- Rearranging the equation to find V_mercury:
\[ V_{mercury} = \frac{1000 \times (27 \times 10^{-6})}{1.8 \times 10^{-4}} \]
- Calculating the right-hand side:
\[ V_{mercury} = \frac{1000 \times 27}{1.8} \times \frac{10^{-6}}{10^{-4}} = \frac{27000}{1.8} \times 10^{-2} \]
- Performing the division:
\[ V_{mercury} = 1500 \times 10^{-2} = 150 \, \text{cm}^3 \]

### Final Answer:
The volume of mercury in the flask is **150 cm³**.