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 single rod of length `l_1+l_2.` The coefficients of linear expansion for aluminium and steel are `alpha_a and alpha_s` and respectively. If the length of each rod increases by the same amount when their temperature are raised by `t^0C,` then find the ratio `l_1//(l_1+l_2)`

Options

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

Explanation

To solve the problem, we need to analyze the behavior of the two rods made of aluminum and steel when they are heated. We will use the concept of linear expansion.

### Step-by-Step Solution:

1. **Understanding Linear Expansion**:
The change in length (\( \Delta L \)) of a material due to temperature change can be expressed as:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]
where \( \alpha \) is the coefficient of linear expansion, \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.

2. **Setting Up the Problem**:
Let:
- \( l_1 \) = initial length of the aluminum rod
- \( l_2 \) = initial length of the steel rod
- \( \alpha_a \) = coefficient of linear expansion for aluminum
- \( \alpha_s \) = coefficient of linear expansion for steel
- \( t \) = temperature change in °C

The lengths of the rods after heating will be:
- For aluminum: \( l_1' = l_1 + \Delta L_a = l_1 + \alpha_a l_1 t \)
- For steel: \( l_2' = l_2 + \Delta L_s = l_2 + \alpha_s l_2 t \)

3. **Equal Increase in Length**:
According to the problem, both rods increase in length by the same amount:
\[ \Delta L_a = \Delta L_s \]
Thus, we have:
\[ \alpha_a l_1 t = \alpha_s l_2 t \]

4. **Canceling \( t \)**:
Since \( t \) is a common factor and is not zero, we can cancel it from both sides:
\[ \alpha_a l_1 = \alpha_s l_2 \]

5. **Finding the Ratio**:
Rearranging the equation gives:
\[ \frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a} \]

6. **Expressing the Ratio in Terms of Total Length**:
We need to find the ratio \( \frac{l_1}{l_1 + l_2} \):
Let \( l_2 = k l_1 \) where \( k = \frac{\alpha_s}{\alpha_a} \). Then:
\[ l_1 + l_2 = l_1 + k l_1 = l_1(1 + k) \]
Therefore:
\[ \frac{l_1}{l_1 + l_2} = \frac{l_1}{l_1(1 + k)} = \frac{1}{1 + k} \]

7. **Substituting \( k \)**:
Substituting \( k \) back in gives:
\[ \frac{l_1}{l_1 + l_2} = \frac{1}{1 + \frac{\alpha_s}{\alpha_a}} = \frac{\alpha_a}{\alpha_a + \alpha_s} \]

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