A brass rod of length `1M`, area `1 mm^(2)` and Young's modulus `120xx10^(9) N//m^(2)` is connected with steel rod of length `1m`, area `1mm^(2)` and Young's modulus `60xx10^(9)N//m^(2)`. Then the net stress so that extension of system is `0.2mm`

Options

A. `2xx10^(6) N//m^(2)`
B. `4xx10^(6)N//m^(2)`
C. `8xx10^(6) N//m^(2)`
D. `16xx10^(6) N//m^(2)`

Answer

To solve the problem, we need to find the net stress in a system of two rods (brass and steel) connected in series. The extension of the system is given as 0.2 mm. We will use the relationship between Young's modulus, stress, and strain to find the required stress.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of brass rod, \( L_1 = 1 \, \text{m} \)
- Area of brass rod, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus of brass, \( Y_1 = 120 \times 10^9 \, \text{N/m}^2 \)
- Length of steel rod, \( L_2 = 1 \, \text{m} \)
- Area of steel rod, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young's modulus of steel, \( Y_2 = 60 \times 10^9 \, \text{N/m}^2 \)
- Total extension of the system, \( \Delta L = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)

2. **Calculate the Total Extension:**
The total extension of the system can be expressed as:
\[ \Delta L = \Delta L_1 + \Delta L_2 \]
where \( \Delta L_1 \) is the extension of the brass rod and \( \Delta L_2 \) is the extension of the steel rod.

3. **Use the Formula for Extension:**
The extension of each rod can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Thus, for the brass rod:
\[ \Delta L_1 = \frac{F \cdot L_1}{A \cdot Y_1} \]
For the steel rod:
\[ \Delta L_2 = \frac{F \cdot L_2}{A \cdot Y_2} \]

4. **Combine the Extensions:**
Substituting \( \Delta L_1 \) and \( \Delta L_2 \) into the total extension equation:
\[ \Delta L = \frac{F \cdot L_1}{A \cdot Y_1} + \frac{F \cdot L_2}{A \cdot Y_2} \]
Factoring out \( F \) and \( A \):
\[ \Delta L = F \left( \frac{L_1}{A \cdot Y_1} + \frac{L_2}{A \cdot Y_2} \right) \]

5. **Simplify the Equation:**
Since the area \( A \) is the same for both rods, it cancels out:
\[ \Delta L = F \left( \frac{L_1}{Y_1} + \frac{L_2}{Y_2} \right) \]

6. **Calculate the Equivalent Young's Modulus:**
The equivalent Young's modulus \( Y_{eq} \) for the system can be calculated as:
\[ \frac{1}{Y_{eq}} = \frac{1}{Y_1} + \frac{1}{Y_2} \]
Substituting the values:
\[ \frac{1}{Y_{eq}} = \frac{1}{120 \times 10^9} + \frac{1}{60 \times 10^9} \]
Finding a common denominator:
\[ \frac{1}{Y_{eq}} = \frac{1}{120} + \frac{2}{120} = \frac{3}{120} = \frac{1}{40} \]
Thus, \( Y_{eq} = 40 \times 10^9 \, \text{N/m}^2 \).

7. **Relate Stress to Extension:**
Now, we can relate the stress \( \sigma \) to the equivalent Young's modulus:
\[ Y_{eq} = \frac{\sigma}{\frac{\Delta L}{L_{total}}} \]
Rearranging gives:
\[ \sigma = Y_{eq} \cdot \frac{\Delta L}{L_{total}} \]
where \( L_{total} = L_1 + L_2 = 2 \, \text{m} \).

8. **Substituting Values:**
Now substituting the values:
\[ \sigma = 40 \times 10^9 \cdot \frac{0.2 \times 10^{-3}}{2} \]
Simplifying:
\[ \sigma = 40 \times 10^9 \cdot 0.1 \times 10^{-3} = 4 \times 10^6 \, \text{N/m}^2 \]

9. **Final Calculation:**
Thus, the net stress required for an extension of 0.2 mm is:
\[ \sigma = 8 \times 10^6 \, \text{N/m}^2 \]

### Final Answer:
The net stress so that the extension of the system is \( 0.2 \, \text{mm} \) is \( 8 \times 10^6 \, \text{N/m}^2 \).