To solve the problem, we need to find the ratio of the diameter of the copper wire to the diameter of the steel wire given that the tension in each wire is the same.
### Step-by-Step Solution:
1. **Identify the Given Data:**
- Mass of the rigid bar, \( m = 15 \, \text{kg} \)
- Length of each wire, \( L = 2 \, \text{m} \)
- Young's modulus of copper, \( Y_{\text{copper}} = 1.1 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus of steel, \( Y_{\text{steel}} = 1.9 \times 10^{11} \, \text{N/m}^2 \)
2. **Calculate the Weight of the Bar:**
\[
W = mg = 15 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 147.15 \, \text{N} \approx 150 \, \text{N}
\]
3. **Determine the Tension in Each Wire:**
Since the bar is supported symmetrically by three wires, the total tension (T) in the three wires must equal the weight of the bar:
\[
3T = W \implies T = \frac{W}{3} = \frac{150 \, \text{N}}{3} = 50 \, \text{N}
\]
4. **Understand the Relationship Between Stress, Strain, and Young's Modulus:**
The stress (\( \sigma \)) in each wire is given by:
\[
\sigma = \frac{T}{A}
\]
where \( A \) is the cross-sectional area of the wire. The strain (\( \epsilon \)) is given by:
\[
\epsilon = \frac{\Delta L}{L}
\]
Young's modulus (\( Y \)) is defined as:
\[
Y = \frac{\sigma}{\epsilon}
\]
5. **Since the Tension and Strain are the Same:**
For copper and steel, we have:
\[
\frac{T}{A_{\text{copper}}} = Y_{\text{copper}} \cdot \epsilon
\]
\[
\frac{T}{A_{\text{steel}}} = Y_{\text{steel}} \cdot \epsilon
\]
Since the tension and strain are the same in both wires, we can equate the two expressions:
\[
\frac{A_{\text{copper}}}{A_{\text{steel}}} = \frac{Y_{\text{steel}}}{Y_{\text{copper}}}
\]
6. **Cross-Sectional Area of Wires:**
The cross-sectional area of a wire is given by:
\[
A = \frac{\pi d^2}{4}
\]
Therefore, the ratio of the areas becomes:
\[
\frac{A_{\text{copper}}}{A_{\text{steel}}} = \frac{\frac{\pi d_{\text{copper}}^2}{4}}{\frac{\pi d_{\text{steel}}^2}{4}} = \frac{d_{\text{copper}}^2}{d_{\text{steel}}^2}
\]
7. **Substituting into the Area Ratio:**
Thus, we have:
\[
\frac{d_{\text{copper}}^2}{d_{\text{steel}}^2} = \frac{Y_{\text{steel}}}{Y_{\text{copper}}}
\]
Taking the square root gives:
\[
\frac{d_{\text{copper}}}{d_{\text{steel}}} = \sqrt{\frac{Y_{\text{steel}}}{Y_{\text{copper}}}} = \sqrt{\frac{1.9 \times 10^{11}}{1.1 \times 10^{11}}} = \sqrt{\frac{19}{11}}
\]
8. **Final Result:**
Therefore, the ratio of the diameter of the copper wire to the diameter of the steel wire is:
\[
\frac{d_{\text{copper}}}{d_{\text{steel}}} = \sqrt{\frac{19}{11}}
\]
