To solve the problem, we will use the relationship between Young's modulus, stress, strain, and the dimensions of the wires.
### Step-by-Step Solution:
1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress (force per unit area) to strain (relative change in length). Mathematically, it can be expressed as:
\[
Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}
\]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length.
2. **Given Information**:
- Young's moduli ratio: \( \frac{Y_A}{Y_B} = \frac{7}{4} \)
- Length of wire A, \( L_A = 2 \, m \)
- Length of wire B, \( L_B = 1.5 \, m \)
- Radius of wire B, \( r_B = 2 \, mm = 2 \times 10^{-3} \, m \)
- Radius of wire A, \( r_A = R \)
3. **Area Calculation**:
The cross-sectional area \( A \) of a wire is given by:
\[
A = \pi r^2
\]
Therefore, the areas for wires A and B are:
\[
A_A = \pi R^2 \quad \text{and} \quad A_B = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) \, m^2
\]
4. **Setting Up the Equation**:
Since both wires stretch by the same length \( \Delta L \) under the same load \( F \), we can equate the expressions for \( \Delta L \):
\[
\Delta L_A = \frac{F L_A}{A_A Y_A} \quad \text{and} \quad \Delta L_B = \frac{F L_B}{A_B Y_B}
\]
Setting these equal gives:
\[
\frac{F L_A}{\pi R^2 Y_A} = \frac{F L_B}{\pi (4 \times 10^{-6}) Y_B}
\]
The \( F \) and \( \pi \) terms cancel out:
\[
\frac{L_A}{R^2 Y_A} = \frac{L_B}{(4 \times 10^{-6}) Y_B}
\]
5. **Substituting Values**:
Substitute \( L_A = 2 \, m \), \( L_B = 1.5 \, m \), and the ratio of Young's moduli:
\[
\frac{2}{R^2 Y_A} = \frac{1.5}{(4 \times 10^{-6}) Y_B}
\]
Using \( \frac{Y_A}{Y_B} = \frac{7}{4} \) implies \( Y_B = \frac{4}{7} Y_A \):
\[
\frac{2}{R^2 Y_A} = \frac{1.5}{(4 \times 10^{-6}) \left(\frac{4}{7} Y_A\right)}
\]
Simplifying gives:
\[
\frac{2}{R^2} = \frac{1.5 \cdot 7}{(4 \times 10^{-6}) \cdot 4}
\]
6. **Solving for \( R^2 \)**:
Rearranging gives:
\[
R^2 = \frac{2 \cdot (4 \times 10^{-6}) \cdot 4}{1.5 \cdot 7}
\]
Calculating the right-hand side:
\[
R^2 = \frac{32 \times 10^{-6}}{10.5} \approx 3.047619 \times 10^{-6}
\]
Taking the square root:
\[
R \approx \sqrt{3.047619 \times 10^{-6}} \approx 1.74 \times 10^{-3} \, m \approx 1.74 \, mm
\]
### Final Answer:
The value of \( R \) is approximately **1.74 mm**.
