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_a)//(alpha_s))`

Explanation

To solve the problem, we need to find the ratio \( \frac{l_1}{l_1 + l_2} \) given that the increase in length of both rods is the same when the temperature is raised by \( t \) degrees Celsius.

### Step-by-step Solution:

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

2. **Setting Up the Equations**:
For the aluminium rod:
\[ \Delta L_a = l_1 \cdot \alpha_a \cdot t \]
For the steel rod:
\[ \Delta L_s = l_2 \cdot \alpha_s \cdot t \]

3. **Equating the Changes in Length**:
Since both rods increase in length by the same amount:
\[ \Delta L_a = \Delta L_s \]
Therefore:
\[ l_1 \cdot \alpha_a \cdot t = l_2 \cdot \alpha_s \cdot t \]

4. **Canceling \( t \)**:
Since \( t \) is common in both terms and is not zero, we can cancel it out:
\[ l_1 \cdot \alpha_a = l_2 \cdot \alpha_s \]

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

6. **Expressing \( l_1 \) in Terms of \( l_2 \)**:
We can express \( l_1 \) in terms of \( l_2 \):
\[ l_1 = \frac{l_2 \cdot \alpha_s}{\alpha_a} \]

7. **Finding \( l_1 + l_2 \)**:
Now, substituting \( l_1 \) into \( l_1 + l_2 \):
\[ l_1 + l_2 = \frac{l_2 \cdot \alpha_s}{\alpha_a} + l_2 = l_2 \left( \frac{\alpha_s}{\alpha_a} + 1 \right) \]

8. **Finding the Ratio \( \frac{l_1}{l_1 + l_2} \)**:
Now we can find the desired ratio:
\[ \frac{l_1}{l_1 + l_2} = \frac{\frac{l_2 \cdot \alpha_s}{\alpha_a}}{l_2 \left( \frac{\alpha_s}{\alpha_a} + 1 \right)} \]
Simplifying this gives:
\[ \frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_s + \alpha_a} \]

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