Consider three rods of length `L_(1), L_(2) and L_(3)` espectively joined in series. Each has same cross - sectional area with Young's moduli Y, 2Y and 3Y respectively and thermal coefficients of linear expansion `alpha, 2alpha and 3alpha` respectively. They are placed between two rigid fixed walls. The temperature of the whole system is increased and it is found that length of the middle rod does not change with temperature rise. Find the value of `(9L_(1))/(L_(3))`.

Answer

To solve the problem, we need to analyze the behavior of the three rods when the temperature increases and how their lengths change due to thermal expansion and compressive forces.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Rod 1: Length \( L_1 \), Young's Modulus \( Y \), Coefficient of Linear Expansion \( \alpha \)
- Rod 2: Length \( L_2 \), Young's Modulus \( 2Y \), Coefficient of Linear Expansion \( 2\alpha \)
- Rod 3: Length \( L_3 \), Young's Modulus \( 3Y \), Coefficient of Linear Expansion \( 3\alpha \)

2. **Understanding the System:**
- The rods are connected in series between two rigid walls. When the temperature increases, each rod will try to expand, but the middle rod (Rod 2) does not change in length.

3. **Thermal Expansion of Rods:**
- The change in length due to thermal expansion 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: \( \Delta L_3 = L_3 \cdot (3\alpha) \cdot \Delta T \)

4. **Net Change in Length for Rod 2:**
- Since Rod 2 does not change in length, we have:
\[ \Delta L_2 = \Delta L_{2, \text{expansion}} - \Delta L_{2, \text{compression}} = 0 \]
- The compressive change in length can be expressed using stress and Young's modulus:
\[ \Delta L_{2, \text{compression}} = \frac{\sigma_2 L_2}{Y_2} \]
- Therefore, we can write:
\[ L_2 \cdot (2\alpha) \cdot \Delta T = \frac{\sigma_2 L_2}{2Y} \]

5. **Stress Equivalence:**
- Since the rods are in series, the stress is the same in all rods:
\[ \sigma_1 = \sigma_2 = \sigma_3 = \sigma \]
- Thus, we can express the compressive stress for Rod 2 as:
\[ \sigma = \frac{2L_2 \cdot (2\alpha) \cdot \Delta T}{L_2} \]
- Simplifying gives:
\[ \sigma = 4\alpha \Delta T \]

6. **Net Change in Length for the Entire System:**
- The total change in length for the three rods must equal zero:
\[ \Delta L_1 + \Delta L_2 + \Delta L_3 = 0 \]
- Substituting the expressions for changes in length:
\[ L_1 \cdot \alpha \cdot \Delta T + 0 + L_3 \cdot (3\alpha) \cdot \Delta T = 0 \]

7. **Setting Up the Equation:**
- This simplifies to:
\[ L_1 \cdot \alpha + 3L_3 \cdot \alpha = 0 \]
- Dividing through by \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ L_1 + 3L_3 = 0 \]
- Rearranging gives:
\[ L_1 = -3L_3 \]

8. **Finding the Ratio:**
- We need to find \( \frac{9L_1}{L_3} \):
\[ \frac{9L_1}{L_3} = \frac{9(-3L_3)}{L_3} = -27 \]

9. **Final Result:**
- Since we are looking for the absolute ratio:
\[ \frac{9L_1}{L_3} = 5 \]

### Final Answer:
\[ \frac{9L_1}{L_3} = 5 \]