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 the brass rod and the steel rod at \(0^\circ C\). Let's denote the length of the brass rod as \(L_b\) and the length of the steel rod as \(L_s\). According to the problem, the steel rod is 80 cm (or 0.8 m) longer than the brass rod. Therefore, we can express the lengths as follows:

1. **Define the lengths:**
\[ L_s = L_b + 0.8 \quad \text{(1)} \]

2. **Thermal expansion formula:**
The change in length due to thermal expansion is given by:
\[ \Delta L = L_0 \alpha \Delta T \]
where \(L_0\) is the original length, \(\alpha\) is the coefficient of linear expansion, and \(\Delta T\) is the change in temperature.

3. **Change in lengths for both rods:**
For the brass rod:
\[ \Delta L_b = L_b \cdot \alpha_b \cdot \Delta T \quad \text{(2)} \]
For the steel rod:
\[ \Delta L_s = L_s \cdot \alpha_s \cdot \Delta T \quad \text{(3)} \]

4. **Setting up the equation:**
According to the problem, the difference in lengths after heating does not change:
\[ \Delta L_s - \Delta L_b = 0 \quad \Rightarrow \quad \Delta L_s = \Delta L_b \]
Substituting equations (2) and (3):
\[ L_s \cdot \alpha_s \cdot \Delta T = L_b \cdot \alpha_b \cdot \Delta T \]
Since \(\Delta T\) is the same for both rods, we can cancel it out:
\[ L_s \cdot \alpha_s = L_b \cdot \alpha_b \quad \text{(4)} \]

5. **Substituting \(L_s\) from equation (1) into equation (4):**
\[ (L_b + 0.8) \cdot \alpha_s = L_b \cdot \alpha_b \]
Expanding this gives:
\[ L_b \cdot \alpha_s + 0.8 \cdot \alpha_s = L_b \cdot \alpha_b \]

6. **Rearranging the equation:**
\[ L_b \cdot \alpha_s - L_b \cdot \alpha_b = -0.8 \cdot \alpha_s \]
Factoring out \(L_b\):
\[ L_b (\alpha_s - \alpha_b) = -0.8 \cdot \alpha_s \]
Thus,
\[ L_b = \frac{-0.8 \cdot \alpha_s}{\alpha_s - \alpha_b} \quad \text{(5)} \]

7. **Substituting the values of \(\alpha_b\) and \(\alpha_s\):**
Given:
\(\alpha_b = 19 \times 10^{-6} \, ^\circ C^{-1}\) and \(\alpha_s = 11 \times 10^{-6} \, ^\circ C^{-1}\),
substituting these values into equation (5):
\[ L_b = \frac{-0.8 \cdot (11 \times 10^{-6})}{(11 \times 10^{-6}) - (19 \times 10^{-6})} \]

8. **Calculating \(L_b\):**
\[ L_b = \frac{-0.8 \cdot 11 \times 10^{-6}}{-8 \times 10^{-6}} = \frac{0.8 \cdot 11}{8} = 1.1 \, \text{m} \]

9. **Finding \(L_s\):**
Using equation (1):
\[ L_s = L_b + 0.8 = 1.1 + 0.8 = 1.9 \, \text{m} \]

10. **Calculating the total length:**
The total length of the two rods at \(0^\circ C\) is:
\[ L_{total} = L_b + L_s = 1.1 + 1.9 = 3.0 \, \text{m} \]

Thus, the sum of the lengths of the two rods at \(0^\circ C\) is **3.0 m**.