In an experiment, brass and steel wires of length 1 m each with areas of cross section `1mm^(2)` are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is, [Given, the Young’s Modulus for steel and brass are, respectively, `120 xx 10^(9) N//m^(2) and 60 xx 10^(9) N//m^(2)`

Options

A. `8.0 xx 10^(6) N//m^(2)`
B. `1.2 xx 10^(6) N//m^(2)`
C. `4.0 xx 10^(6) N//m^(2)`
D. `0.2 xx 10^(6) N//m^(2)`

Explanation

To solve the problem, we need to find the stress required to produce a net elongation of 0.2 mm in a system of brass and steel wires connected in series. We will use the given Young's moduli for brass and steel and the formula for stress.

### Step-by-Step Solution:

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

2. **Calculate the Individual Elongations:**
The total elongation in the series system is the sum of the elongations in each wire:
\[ \Delta L = \Delta L_1 + \Delta L_2 \]
Using the formula for elongation:
\[ \Delta L = \frac{F L}{A Y} \]
For steel wire:
\[ \Delta L_1 = \frac{F L_1}{A Y_1} = \frac{F \cdot 1}{1 \times 10^{-6} \cdot 120 \times 10^9} \]
For brass wire:
\[ \Delta L_2 = \frac{F L_2}{A Y_2} = \frac{F \cdot 1}{1 \times 10^{-6} \cdot 60 \times 10^9} \]

3. **Set Up the Equation:**
Substitute the elongations into the total elongation equation:
\[ 0.2 \times 10^{-3} = \frac{F}{1 \times 10^{-6} \cdot 120 \times 10^9} + \frac{F}{1 \times 10^{-6} \cdot 60 \times 10^9} \]

4. **Combine the Terms:**
Factor out \( F \):
\[ 0.2 \times 10^{-3} = F \left( \frac{1}{1 \times 10^{-6} \cdot 120 \times 10^9} + \frac{1}{1 \times 10^{-6} \cdot 60 \times 10^9} \right) \]
Simplifying the fractions:
\[ 0.2 \times 10^{-3} = F \left( \frac{1}{1.2 \times 10^3} + \frac{1}{0.6 \times 10^3} \right) \]
\[ = F \left( \frac{5 + 10}{6 \times 10^3} \right) = F \left( \frac{15}{6 \times 10^3} \right) = F \left( \frac{5}{2 \times 10^3} \right) \]

5. **Solve for Force \( F \):**
Rearranging gives:
\[ F = \frac{0.2 \times 10^{-3} \cdot 2 \times 10^3}{5} = \frac{0.4}{5} = 0.08 \, \text{N} \]

6. **Calculate Stress:**
Stress \( \sigma \) is given by:
\[ \sigma = \frac{F}{A} = \frac{0.08}{1 \times 10^{-6}} = 8 \times 10^6 \, \text{N/m}^2 \]

### Final Answer:
The stress required to produce a net elongation of 0.2 mm is \( \sigma = 8 \times 10^6 \, \text{N/m}^2 \).