A brass rod and a lead rod each 80 cm long at `0^@C` are clamped together at one end with their free ends coinciding. The separatioin of free ends of the rods if the system is placed in a steam bath is (`alpha_("brass")=18xx10^(-6)//^(@)C` and `alpha_("lead")=28xx10^(-6)//^(@)C`)

Options

A. 0.2 mm
B. 0.8 mm
C. 1.4 mm
D. 1.6 mm

Explanation

To solve the problem, we need to calculate the separation between the free ends of a brass rod and a lead rod when they are heated from 0°C to the temperature of steam (100°C).

### Step-by-step Solution:

1. **Identify the Initial Length and Temperature Change**:
- Both rods have an initial length \( L_0 = 80 \, \text{cm} = 800 \, \text{mm} \).
- The temperature change \( \Delta T = 100^\circ C - 0^\circ C = 100^\circ C \).

2. **Identify the Coefficients of Linear Expansion**:
- Coefficient of linear expansion for brass, \( \alpha_{\text{brass}} = 18 \times 10^{-6} \, /^\circ C \).
- Coefficient of linear expansion for lead, \( \alpha_{\text{lead}} = 28 \times 10^{-6} \, /^\circ C \).

3. **Calculate the Final Length of Each Rod**:
- The final length of the brass rod after heating:
\[ L_{\text{brass}} = L_0 (1 + \alpha_{\text{brass}} \Delta T) = 800 \, \text{mm} \left(1 + 18 \times 10^{-6} \times 100\right) \]
\[ = 800 \, \text{mm} \left(1 + 0.0018\right) = 800 \, \text{mm} \times 1.0018 = 801.44 \, \text{mm} \]

- The final length of the lead rod after heating:
\[ L_{\text{lead}} = L_0 (1 + \alpha_{\text{lead}} \Delta T) = 800 \, \text{mm} \left(1 + 28 \times 10^{-6} \times 100\right) \]
\[ = 800 \, \text{mm} \left(1 + 0.0028\right) = 800 \, \text{mm} \times 1.0028 = 802.24 \, \text{mm} \]

4. **Calculate the Separation Between the Free Ends**:
- The separation between the free ends of the rods is given by:
\[ \text{Separation} = L_{\text{lead}} - L_{\text{brass}} = 802.24 \, \text{mm} - 801.44 \, \text{mm} = 0.8 \, \text{mm} \]

### Final Answer:
The separation between the free ends of the rods when placed in a steam bath is \( 0.8 \, \text{mm} \).

---