A composite rod is made by joining a copper rod, end to end, with a second rod of different material but of the same area of cross section. At `25^(@)C`, the composite rod is `1 m` long and the copper rod is `30 cm` long. At `125^(@)C` the length of the composite rod increases by `1.91 mm`. When the composite rod is prevented from expanding by holding it between two rigid walls, it is found that the constituent rods have remained unchanged in length inspite of rise of temperature. Find young's modulus and the coefficient of linear expansion of the second rod (Y of copper `=1.3xx10^(10) N//m^(2)` and `a` of copper `=17xx10^(-6)//K`).

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the given data
- Length of the composite rod at \(25^\circ C\): \(L = 1 \, m\)
- Length of the copper rod: \(L_c = 30 \, cm = 0.3 \, m\)
- Length of the second rod: \(L_x = L - L_c = 1 - 0.3 = 0.7 \, m\)
- Change in temperature: \(\Delta T = 125^\circ C - 25^\circ C = 100 \, K\)
- Total change in length of the composite rod: \(\Delta L = 1.91 \, mm = 1.91 \times 10^{-3} \, m\)
- Coefficient of linear expansion of copper: \(\alpha_c = 17 \times 10^{-6} \, K^{-1}\)
- Young's modulus of copper: \(Y_c = 1.3 \times 10^{10} \, N/m^2\)

### Step 2: Calculate the change in length of the copper rod
Using the formula for linear expansion:
\[ \Delta L_c = \alpha_c \cdot L_c \cdot \Delta T \]
Substituting the values:
\[ \Delta L_c = 17 \times 10^{-6} \cdot 0.3 \cdot 100 = 5.1 \times 10^{-4} \, m \]

### Step 3: Set up the equation for the total change in length
The total change in length of the composite rod is the sum of the changes in length of both rods:
\[ \Delta L = \Delta L_c + \Delta L_x \]
Where \(\Delta L_x\) is the change in length of the second rod. Thus,
\[ 1.91 \times 10^{-3} = 5.1 \times 10^{-4} + \Delta L_x \]
Solving for \(\Delta L_x\):
\[ \Delta L_x = 1.91 \times 10^{-3} - 5.1 \times 10^{-4} = 1.4 \times 10^{-3} \, m \]

### Step 4: Calculate the coefficient of linear expansion of the second rod
Using the formula for linear expansion for the second rod:
\[ \Delta L_x = \alpha_x \cdot L_x \cdot \Delta T \]
Substituting the known values:
\[ 1.4 \times 10^{-3} = \alpha_x \cdot 0.7 \cdot 100 \]
Solving for \(\alpha_x\):
\[ \alpha_x = \frac{1.4 \times 10^{-3}}{0.7 \cdot 100} = 2 \times 10^{-5} \, K^{-1} \]

### Step 5: Calculate the strain in the copper rod
The strain in the copper rod can be calculated as:
\[ \text{Strain} = \frac{\Delta L_c}{L_c} = \frac{5.1 \times 10^{-4}}{0.3} = 1.7 \times 10^{-3} \]

### Step 6: Calculate the stress in the copper rod
Using Young's modulus:
\[ Y_c = \frac{\text{Stress}}{\text{Strain}} \implies \text{Stress} = Y_c \cdot \text{Strain} \]
Substituting the values:
\[ \text{Stress} = 1.3 \times 10^{10} \cdot 1.7 \times 10^{-3} = 2.21 \times 10^{7} \, N/m^2 \]

### Step 7: Calculate the strain in the second rod
Using the formula for strain:
\[ \text{Strain} = \frac{\Delta L_x}{L_x} = \frac{1.4 \times 10^{-3}}{0.7} = 2 \times 10^{-3} \]

### Step 8: Calculate Young's modulus of the second rod
Using the same stress for both rods:
\[ Y_x = \frac{\text{Stress}}{\text{Strain}} \implies Y_x = \frac{2.21 \times 10^{7}}{2 \times 10^{-3}} = 1.105 \times 10^{10} \, N/m^2 \]

### Final Results
- Coefficient of linear expansion of the second rod: \(\alpha_x = 2 \times 10^{-5} \, K^{-1}\)
- Young's modulus of the second rod: \(Y_x = 1.105 \times 10^{10} \, N/m^2\)