To solve the problem step-by-step, we will analyze the situation involving three rods connected in series, each with different Young's moduli and coefficients of linear expansion.
### Step 1: Understand the Setup
We have three rods of lengths \( L_1, L_2, \) and \( L_3 \) connected in series. The Young's moduli are \( Y, 2Y, \) and \( 3Y \), and the coefficients of linear expansion are \( \alpha, 2\alpha, \) and \( 3\alpha \) respectively. The rods are fixed between two rigid walls.
### Step 2: Change in Length Due to Temperature
When the temperature increases, each rod will try to expand. The change in length due to temperature for each rod can be expressed as:
- For Rod 1:
\[
\Delta L_1 = L_1 \cdot \alpha \cdot \Delta T
\]
- For Rod 2:
\[
\Delta L_2 = L_2 \cdot 2\alpha \cdot \Delta T
\]
- For Rod 3:
\[
\Delta L_3 = L_3 \cdot 3\alpha \cdot \Delta T
\]
### Step 3: Change in Length Due to Stress
Since the rods are fixed, they will also experience compressive stress due to the expansion of adjacent rods. The change in length due to stress for the middle rod (Rod 2) can be expressed using Young's modulus:
\[
\Delta L_{2, stress} = \frac{\sigma L_2}{2Y}
\]
where \( \sigma \) is the stress applied on the rod.
### Step 4: Condition for Rod 2
According to the problem, the length of the middle rod (Rod 2) does not change with temperature rise. Therefore, the total change in length for Rod 2 must be zero:
\[
\Delta L_2 + \Delta L_{2, stress} = 0
\]
Substituting the expressions we derived:
\[
L_2 \cdot 2\alpha \cdot \Delta T - \frac{\sigma L_2}{2Y} = 0
\]
This simplifies to:
\[
\sigma = 4\alpha Y \Delta T
\]
### Step 5: Change in Length for Rod 1 and Rod 3
For Rod 1, the total change in length due to temperature and stress is:
\[
\Delta L_1 + \Delta L_{1, stress} = 0
\]
This gives:
\[
L_1 \cdot \alpha \cdot \Delta T - \frac{\sigma L_1}{Y} = 0
\]
Substituting \( \sigma \):
\[
L_1 \cdot \alpha \cdot \Delta T - \frac{4\alpha Y \Delta T L_1}{Y} = 0
\]
This simplifies to:
\[
\alpha \Delta T = 4\alpha \Delta T
\]
For Rod 3, we have:
\[
\Delta L_3 + \Delta L_{3, stress} = 0
\]
This gives:
\[
L_3 \cdot 3\alpha \cdot \Delta T - \frac{\sigma L_3}{3Y} = 0
\]
Substituting \( \sigma \):
\[
L_3 \cdot 3\alpha \cdot \Delta T - \frac{4\alpha Y \Delta T L_3}{3Y} = 0
\]
This simplifies to:
\[
3\alpha \Delta T = \frac{4\alpha \Delta T}{3}
\]
### Step 6: Relating \( L_1 \) and \( L_3 \)
From the equations derived, we can relate \( L_1 \) and \( L_3 \):
\[
\frac{L_1}{L_3} = \frac{5}{9}
\]
Thus, we need to find \( \frac{9L_1}{L_3} \):
\[
\frac{9L_1}{L_3} = 9 \cdot \frac{5}{9} = 5
\]
### Final Answer
The value of \( \frac{9L_1}{L_3} \) is \( 5 \).
