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. `alpha_s//alpha_a`
B. `alpha_a//alpha_s`
C. `alpha_s//(alpha_a+alpha_s)`
D. `alpha_a//(alpha_s+alpha_s)`

Explanation

To solve the problem, we need to find the ratio \( \frac{l_1}{l_1 + l_2} \) given that two rods (one made of aluminum and the other made of steel) expand by the same amount when their temperature is raised by \( t \) degrees Celsius.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have two rods:
- Rod 1 (Aluminum) with initial length \( l_1 \) and coefficient of linear expansion \( \alpha_a \).
- Rod 2 (Steel) with initial length \( l_2 \) and coefficient of linear expansion \( \alpha_s \).

2. **Write the Change in Length Formulas**:
The change in length for each rod when the temperature is increased by \( t \) degrees Celsius can be expressed as:
- For aluminum rod:
\[ \Delta L_1 = \alpha_a \cdot t \cdot l_1 \]
- For steel rod:
\[ \Delta L_2 = \alpha_s \cdot t \cdot l_2 \]

3. **Set the Changes in Length Equal**:
Since it is given that both rods increase by the same amount, we can set the two expressions equal to each other:
\[ \Delta L_1 = \Delta L_2 \]
Therefore:
\[ \alpha_a \cdot t \cdot l_1 = \alpha_s \cdot t \cdot l_2 \]

4. **Cancel \( t \) from Both Sides**:
Assuming \( t \neq 0 \), we can divide both sides by \( t \):
\[ \alpha_a \cdot l_1 = \alpha_s \cdot l_2 \]

5. **Rearrange the Equation**:
Rearranging gives us:
\[ \frac{l_1}{l_2} = \frac{\alpha_s}{\alpha_a} \]

6. **Express \( l_1 + l_2 \)**:
From the ratio, we can express \( l_2 \) in terms of \( l_1 \):
\[ l_2 = \frac{\alpha_a}{\alpha_s} l_1 \]

7. **Substitute \( l_2 \) into \( l_1 + l_2 \)**:
Now, substituting \( l_2 \) back into the total length:
\[ l_1 + l_2 = l_1 + \frac{\alpha_a}{\alpha_s} l_1 = l_1 \left(1 + \frac{\alpha_a}{\alpha_s}\right) \]

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

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