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. `1.8xx10^(6)N//m^(2)`
B. `1.2xx10^(6)N//m^(2)`
C. `8.0xx10^(6)N//m^(2)`
D. `0.2xx10^(6)N//m^(2)`

Explanation

To solve the problem, we need to determine the stress required to produce a net elongation of 0.2 mm in the combined brass and steel wires connected in series.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of each wire, \( L = 1 \, \text{m} \)
- Cross-sectional area, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young's Modulus for steel, \( Y_s = 120 \times 10^9 \, \text{N/m}^2 \)
- Young's Modulus for brass, \( Y_b = 60 \times 10^9 \, \text{N/m}^2 \)
- Total elongation, \( \Delta L = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)

2. **Calculate the Stiffness (k) of Each Wire:**
The stiffness \( k \) of a wire is given by the formula:
\[ k = \frac{Y \cdot A}{L} \]
For the steel wire:
\[ k_s = \frac{Y_s \cdot A}{L} = \frac{120 \times 10^9 \cdot 1 \times 10^{-6}}{1} = 120 \times 10^3 \, \text{N/m} \]
For the brass wire:
\[ k_b = \frac{Y_b \cdot A}{L} = \frac{60 \times 10^9 \cdot 1 \times 10^{-6}}{1} = 60 \times 10^3 \, \text{N/m} \]

3. **Calculate the Equivalent Stiffness (k_eq) for Series Connection:**
For wires in series, the equivalent stiffness is given by:
\[ \frac{1}{k_{eq}} = \frac{1}{k_s} + \frac{1}{k_b} \]
Substituting the values:
\[ \frac{1}{k_{eq}} = \frac{1}{120 \times 10^3} + \frac{1}{60 \times 10^3} \]
Finding a common denominator:
\[ \frac{1}{k_{eq}} = \frac{1}{120} + \frac{2}{120} = \frac{3}{120} = \frac{1}{40} \]
Therefore:
\[ k_{eq} = 40 \times 10^3 \, \text{N/m} \]

4. **Calculate the Force Required for the Given Elongation:**
Using Hooke's Law:
\[ F = k_{eq} \cdot \Delta L \]
Substituting the values:
\[ F = 40 \times 10^3 \cdot (0.2 \times 10^{-3}) = 8 \, \text{N} \]

5. **Calculate the Stress:**
Stress is defined as:
\[ \text{Stress} = \frac{F}{A} \]
Substituting the values:
\[ \text{Stress} = \frac{8}{1 \times 10^{-6}} = 8 \times 10^6 \, \text{N/m}^2 \]

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

---