Young's moduli of two wires A and B are in the ratio `7:4` wire A is `2m` long and has radius R. . wire B is `1.5 m `long has radius `2mm` for a given load , then the value of `R` is close to:

Options

A. `1.85mm`
B. `1.9mm`
C. `1.7mm`
D. `1.3mm`

Answer

To solve the problem, we will use the relationship between Young's modulus, length, area, and the given ratio of Young's moduli for the two wires.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
The Young's modulus (Y) is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
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 Data**:
- 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: \( r_B = 2 \, mm = 0.002 \, m \)

3. **Cross-sectional Areas**:
- Area of wire A: \( A_A = \pi R^2 \)
- Area of wire B: \( A_B = \pi (0.002)^2 = \pi \times 0.000004 \, m^2 \)

4. **Equating Young's Modulus for Both Wires**:
Since the load (force) is the same for both wires, we can set up the equation:
\[ \frac{Y_A \cdot A_A}{L_A} = \frac{Y_B \cdot A_B}{L_B} \]

5. **Substituting the Known Values**:
Substitute the areas and the ratio of Young's moduli:
\[ \frac{7 \cdot \pi R^2}{2} = \frac{4 \cdot \pi \times 0.000004}{1.5} \]

6. **Canceling \(\pi\)**:
The \(\pi\) cancels out from both sides:
\[ \frac{7 R^2}{2} = \frac{4 \times 0.000004}{1.5} \]

7. **Calculating the Right Side**:
Calculate the right side:
\[ \frac{4 \times 0.000004}{1.5} = \frac{0.000016}{1.5} \approx 0.00001067 \]

8. **Setting Up the Equation**:
Now we have:
\[ 7 R^2 = 2 \times 0.00001067 \]
\[ 7 R^2 = 0.00002134 \]

9. **Solving for \(R^2\)**:
\[ R^2 = \frac{0.00002134}{7} \approx 0.000003052 \]

10. **Finding \(R\)**:
\[ R \approx \sqrt{0.000003052} \approx 0.00174 \, m = 1.74 \, mm \]

### Final Answer:
The value of \( R \) is approximately **1.74 mm**.