Two rods , one of aluminium and the other made of steel, having initial length `l_(1)` and `l_(2)` are connected together to from a sinlge rod of length `l_(1)+l_(2)` . The coefficient of linear expansion for aluminium and steel are `alpha_(a)` and `alpha_(s)` for `AC` and `BC` . If the distance `DC` remains constant for small changes in temperature,

Options

A. `(alpha_(s))/(alpha_(a))`
B. `(alpha_(a))/(alpha_(s))`
C. `(alpha_(s))/((alpha_(a)+alpha_(s)))`
D. `(alpha_(a))/((alpha_(a)+alpha_(s)))`

Explanation

To solve the problem, we need to analyze the situation involving two rods made of aluminum and steel, which are connected together. We'll denote the initial lengths of the aluminum and steel rods as \( l_1 \) and \( l_2 \), respectively. The coefficients of linear expansion for aluminum and steel are \( \alpha_a \) and \( \alpha_s \).

### Step-by-Step Solution

1. **Understand the Problem**: We have two rods, one made of aluminum and the other made of steel. When the temperature changes by \( \Delta T \), both rods expand. We need to find the ratio \( \frac{l_1}{l_1 + l_2} \) under the condition that the change in length of both rods is the same.

2. **Change in Length Formula**: The change in length \( \Delta L \) of a rod due to temperature change is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where \( L \) is the initial length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

3. **Set Up the Equations**: For the aluminum rod, the change in length \( \Delta L_1 \) is:
\[ \Delta L_1 = l_1 \cdot \alpha_a \cdot \Delta T \]
For the steel rod, the change in length \( \Delta L_2 \) is:
\[ \Delta L_2 = l_2 \cdot \alpha_s \cdot \Delta T \]

4. **Equate the Changes in Length**: Since the problem states that the distance \( DC \) remains constant, we have:
\[ \Delta L_1 = \Delta L_2 \]
Thus, we can write:
\[ l_1 \cdot \alpha_a \cdot \Delta T = l_2 \cdot \alpha_s \cdot \Delta T \]
We can cancel \( \Delta T \) from both sides (assuming \( \Delta T \neq 0 \)):
\[ l_1 \cdot \alpha_a = l_2 \cdot \alpha_s \]

5. **Express \( l_2 \) in terms of \( l_1 \)**: Rearranging the equation gives:
\[ l_2 = \frac{l_1 \cdot \alpha_a}{\alpha_s} \]

6. **Find \( l_1 + l_2 \)**: Now, we can express the total length of the combined rod:
\[ l_1 + l_2 = l_1 + \frac{l_1 \cdot \alpha_a}{\alpha_s} = l_1 \left(1 + \frac{\alpha_a}{\alpha_s}\right) \]

7. **Calculate the Ratio**: Now we can find the ratio \( \frac{l_1}{l_1 + l_2} \):
\[ \frac{l_1}{l_1 + l_2} = \frac{l_1}{l_1 \left(1 + \frac{\alpha_a}{\alpha_s}\right)} = \frac{1}{1 + \frac{\alpha_a}{\alpha_s}} = \frac{\alpha_s}{\alpha_a + \alpha_s} \]

8. **Final Answer**: Thus, the ratio \( \frac{l_1}{l_1 + l_2} \) is:
\[ \frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s} \]