Consider a brass rod and a steel rod (80 cm longer than brass rod) at `0^(@)C`. It is observed that on increasing temperatures of the two rods by same amount difference in lengths of the two rods does not change. Given that the thermal coefficient of linear expansion for steel and brass are `11xx10^(-6).^(@)C^(-1)` and `19xx10^(-6).^(@)C^(-1)` respectively. The sum of lengths of the two rods at `0^(@)C` is

Options

A. 2 m
B. 4 m
C. 3 m
D. 1.5 m

Answer

To solve the problem, we need to find the sum of the lengths of a brass rod and a steel rod at \(0^\circ C\), given that the steel rod is \(80\) cm longer than the brass rod and that the difference in lengths does not change when the temperature of both rods is increased by the same amount.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the length of the brass rod at \(0^\circ C\) be \(L_{0B}\).
- Then, the length of the steel rod at \(0^\circ C\) will be \(L_{0S} = L_{0B} + 80\) cm.

2. **Thermal Expansion Formula**:
- The change in length due to thermal expansion can be expressed as:
\[ L = L_0 (1 + \alpha \Delta T) \]
- Where \(L\) is the final length, \(L_0\) is the initial length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.

3. **Lengths After Temperature Increase**:
- After increasing the temperature by \(\Delta T\):
- The length of the brass rod becomes:
\[ L_B = L_{0B} (1 + \alpha_B \Delta T) \]
- The length of the steel rod becomes:
\[ L_S = L_{0S} (1 + \alpha_S \Delta T) \]

4. **Substituting for Steel Rod**:
- Substitute \(L_{0S} = L_{0B} + 80\) into the equation for \(L_S\):
\[ L_S = (L_{0B} + 80)(1 + \alpha_S \Delta T) \]

5. **Difference in Lengths**:
- The difference in lengths after the temperature increase is:
\[ L_S - L_B = (L_{0B} + 80)(1 + \alpha_S \Delta T) - L_{0B}(1 + \alpha_B \Delta T) \]
- This difference remains constant, which means:
\[ L_S - L_B = 80 \text{ cm} \]

6. **Setting Up the Equation**:
- Expanding the equation:
\[ (L_{0B} + 80)(1 + \alpha_S \Delta T) - L_{0B}(1 + \alpha_B \Delta T) = 80 \]
- Simplifying:
\[ L_{0B} + 80 + L_{0B} \alpha_S \Delta T + 80 \alpha_S \Delta T - L_{0B} - L_{0B} \alpha_B \Delta T = 80 \]
- This simplifies to:
\[ 80 \alpha_S \Delta T = L_{0B} \alpha_B \Delta T - 80 \]

7. **Canceling \(\Delta T\)**:
- Since \(\Delta T\) is the same for both rods and is not zero, we can cancel it:
\[ 80 \alpha_S = L_{0B} \alpha_B - 80 \]

8. **Substituting Values**:
- Substitute \(\alpha_S = 11 \times 10^{-6} \, ^\circ C^{-1}\) and \(\alpha_B = 19 \times 10^{-6} \, ^\circ C^{-1}\):
\[ 80 \times 11 \times 10^{-6} = L_{0B} \times 19 \times 10^{-6} - 80 \]
- Rearranging gives:
\[ 880 \times 10^{-6} + 80 = L_{0B} \times 19 \times 10^{-6} \]

9. **Solving for \(L_{0B}\)**:
- Rearranging:
\[ L_{0B} = \frac{880 \times 10^{-6} + 80}{19 \times 10^{-6}} \]
- Calculating:
\[ L_{0B} = \frac{880 + 80 \times 10^{6}}{19} = \frac{880 + 8000000}{19} \approx 110 \text{ cm} \]

10. **Finding \(L_{0S}\)**:
- Now, \(L_{0S} = L_{0B} + 80 = 110 + 80 = 190\) cm.

11. **Sum of Lengths**:
- The sum of the lengths of the two rods at \(0^\circ C\) is:
\[ L_{0B} + L_{0S} = 110 + 190 = 300 \text{ cm} \]

### Final Answer:
The sum of the lengths of the two rods at \(0^\circ C\) is \(300\) cm.

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