A steel wire and a copper wire of equal length and equal cross- sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of
copper/steel
(a) the stresses developed . In the two wires,
(b) the strains developed. (Y of steel `= 2xx10^(11) N//m^(2)`)

Explanation

To solve the problem, we need to find the ratio of stresses and strains developed in a steel wire and a copper wire that are joined end to end and subjected to the same tension.

### Given Data:
- Young's modulus of steel (Y_steel) = \(2 \times 10^{11} \, \text{N/m}^2\)
- Young's modulus of copper (Y_copper) = \(1.3 \times 10^{11} \, \text{N/m}^2\)
- Both wires have equal length (L) and equal cross-sectional area (A).

### Step-by-Step Solution:

#### (a) Finding the Ratio of Stresses Developed

1. **Understanding Stress**:
Stress (σ) is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]

2. **Applying Tension**:
Since both wires are subjected to the same tension (T) and have the same cross-sectional area (A), the stress in both wires will be the same:
\[ \sigma_{steel} = \frac{T}{A} \quad \text{and} \quad \sigma_{copper} = \frac{T}{A} \]

3. **Calculating the Ratio of Stresses**:
\[ \frac{\sigma_{copper}}{\sigma_{steel}} = \frac{T/A}{T/A} = 1 \]
Therefore, the ratio of stresses developed in copper to steel is:
\[ \text{Ratio of stresses} = 1:1 \]

#### (b) Finding the Ratio of Strains Developed

1. **Understanding Strain**:
Strain (ε) is defined as the change in length (ΔL) divided by the original length (L):
\[ \epsilon = \frac{\Delta L}{L} \]

2. **Using Young's Modulus**:
Young's modulus (Y) relates stress and strain:
\[ Y = \frac{\sigma}{\epsilon} \]
Rearranging gives:
\[ \epsilon = \frac{\sigma}{Y} \]

3. **Calculating Strains for Both Wires**:
For steel:
\[ \epsilon_{steel} = \frac{\sigma_{steel}}{Y_{steel}} = \frac{T/A}{Y_{steel}} \]
For copper:
\[ \epsilon_{copper} = \frac{\sigma_{copper}}{Y_{copper}} = \frac{T/A}{Y_{copper}} \]

4. **Calculating the Ratio of Strains**:
\[ \frac{\epsilon_{copper}}{\epsilon_{steel}} = \frac{T/A}{Y_{copper}} \div \frac{T/A}{Y_{steel}} = \frac{Y_{steel}}{Y_{copper}} \]
Substituting the values:
\[ \frac{\epsilon_{copper}}{\epsilon_{steel}} = \frac{2 \times 10^{11}}{1.3 \times 10^{11}} = \frac{2}{1.3} = \frac{20}{13} \]

5. **Final Ratio of Strains**:
Therefore, the ratio of strains developed in copper to steel is:
\[ \text{Ratio of strains} = 20:13 \]

### Final Answers:
- (a) The ratio of stresses developed: **1:1**
- (b) The ratio of strains developed: **20:13**