Two rods, one of aluminium and other made of steel, having initial lengths `l_(1) and l_(2)` are connected together to form a single rod of length `(l_(1)+l_(2))`. The coefficient of linear expansions for aluminium and steel are `alpha_(a)` and `alpha_(s)` respectively. If length of each rod increases by same amount when their tempertures are raised by `t^(@)C`, 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 analyze the linear expansion of both rods and derive the required ratio. Here’s a step-by-step solution:

### Step 1: Understand Linear Expansion
The change in length due to thermal expansion can be expressed using the formula:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]
where \(\Delta L\) is the change in length, \(\alpha\) is the coefficient of linear expansion, \(L_0\) is the original length, and \(\Delta T\) is the change in temperature.

### Step 2: Write the Change in Length for Each Rod
For the aluminium rod (length \(L_1\)):
\[ \Delta L_a = \alpha_a \cdot L_1 \cdot t \]
For the steel rod (length \(L_2\)):
\[ \Delta L_s = \alpha_s \cdot L_2 \cdot t \]

### Step 3: Set the Changes in Length Equal
According to the problem, the change in length for both rods is the same:
\[ \Delta L_a = \Delta L_s \]
This leads to:
\[ \alpha_a \cdot L_1 \cdot t = \alpha_s \cdot L_2 \cdot t \]

### Step 4: Cancel Out the Common Factor
Since \(t\) is common in both terms, we can cancel it out (assuming \(t \neq 0\)):
\[ \alpha_a \cdot L_1 = \alpha_s \cdot L_2 \]

### Step 5: Rearrange the Equation
Rearranging gives us the ratio of the lengths:
\[ \frac{L_2}{L_1} = \frac{\alpha_a}{\alpha_s} \]

### Step 6: Express \(L_1\) in Terms of \(L_2\)
From the ratio, we can express \(L_2\) in terms of \(L_1\):
\[ L_2 = L_1 \cdot \frac{\alpha_a}{\alpha_s} \]

### Step 7: Find the Total Length
The total length of the combined rod is:
\[ L = L_1 + L_2 = L_1 + L_1 \cdot \frac{\alpha_a}{\alpha_s} = L_1 \left(1 + \frac{\alpha_a}{\alpha_s}\right) \]

### Step 8: Find the Ratio \( \frac{L_1}{L} \)
Now we can find the ratio of \(L_1\) to the total length \(L\):
\[ \frac{L_1}{L} = \frac{L_1}{L_1 \left(1 + \frac{\alpha_a}{\alpha_s}\right)} = \frac{1}{1 + \frac{\alpha_a}{\alpha_s}} \]

### Step 9: Simplify the Ratio
This can be simplified further:
\[ \frac{L_1}{L} = \frac{1}{\frac{\alpha_s + \alpha_a}{\alpha_s}} = \frac{\alpha_s}{\alpha_a + \alpha_s} \]

### Final Result
Thus, the required ratio is:
\[ \frac{L_1}{L_1 + L_2} = \frac{\alpha_s}{\alpha_a + \alpha_s} \]