A brass rod and a lead rod each 80.00 cm long at `0^@C` are clamped together at one end with their free ends coinciding. The separation of the free ends of the rods it the system is placed in steam bath is (coefficients of linear expansions of brass and lead are `2xx10^(-5)//^@C and 1.0 xx 10^(-5)//^@C` respectively)

Options

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

Answer

To solve the problem of finding the separation between the free ends of a brass rod and a lead rod when placed in a steam bath, we will follow these steps:

### Step 1: Understand the problem
We have two rods, one made of brass and the other made of lead, each with an initial length of 80.00 cm at 0°C. They are clamped together at one end, and we need to find the separation of their free ends when the temperature is raised to 100°C.

### Step 2: Identify the coefficients of linear expansion
The coefficients of linear expansion for the materials are given as:
- Brass: \( \alpha_b = 2 \times 10^{-5} \, \text{°C}^{-1} \)
- Lead: \( \alpha_l = 1.0 \times 10^{-5} \, \text{°C}^{-1} \)

### Step 3: Calculate the change in length for each rod
The formula for the change in length due to thermal expansion is:
\[ L' = L_0 (1 + \alpha \Delta T) \]
Where:
- \( L' \) is the final length
- \( L_0 \) is the initial length
- \( \alpha \) is the coefficient of linear expansion
- \( \Delta T \) is the change in temperature

For both rods:
- Initial length \( L_0 = 80.00 \, \text{cm} = 0.80 \, \text{m} \)
- Change in temperature \( \Delta T = 100°C - 0°C = 100°C \)

**For Brass:**
\[ L_b' = 0.80 \, \text{m} \times (1 + 2 \times 10^{-5} \times 100) \]
\[ L_b' = 0.80 \, \text{m} \times (1 + 0.002) \]
\[ L_b' = 0.80 \, \text{m} \times 1.002 = 0.8016 \, \text{m} \]

**For Lead:**
\[ L_l' = 0.80 \, \text{m} \times (1 + 1.0 \times 10^{-5} \times 100) \]
\[ L_l' = 0.80 \, \text{m} \times (1 + 0.001) \]
\[ L_l' = 0.80 \, \text{m} \times 1.001 = 0.8008 \, \text{m} \]

### Step 4: Calculate the separation between the free ends
The separation between the free ends of the rods is given by the difference in their lengths:
\[ \text{Separation} = L_b' - L_l' \]
Substituting the values we calculated:
\[ \text{Separation} = 0.8016 \, \text{m} - 0.8008 \, \text{m} = 0.0008 \, \text{m} \]
Converting to millimeters:
\[ \text{Separation} = 0.0008 \, \text{m} \times 1000 \, \text{mm/m} = 0.8 \, \text{mm} \]

### Final Answer
The separation of the free ends of the rods when placed in the steam bath is **0.8 mm**.

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