Two elastic wire `A & B` having length `l_(A)=2m` and `l_(B)=1.5m` have young's modules ratio `(Y_(A))/(Y_(B))=(7)/(4)`. If `r_(B)=2mm` then the radius of `A` given that due to application of the same force change in length in both `A & B` is same

Options

A. `1.7mm`
B. `1.9mm`
C. `2.7mm`
D. `2mm`

Answer

To solve the problem, we will use the relationship between Young's modulus, force, length, area, and change in length. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
The formula for Young's modulus \( Y \) is given by:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
where:
- \( F \) is the applied force,
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the change in length.

2. **Setting Up the Equations**:
For wire A:
\[ Y_A = \frac{F \cdot L_A}{\pi r_A^2 \cdot \Delta L} \]
For wire B:
\[ Y_B = \frac{F \cdot L_B}{\pi r_B^2 \cdot \Delta L} \]

3. **Taking the Ratio of Young's Moduli**:
Given the ratio of Young's moduli:
\[ \frac{Y_A}{Y_B} = \frac{7}{4} \]
Substituting the expressions for \( Y_A \) and \( Y_B \):
\[ \frac{\frac{F \cdot L_A}{\pi r_A^2 \cdot \Delta L}}{\frac{F \cdot L_B}{\pi r_B^2 \cdot \Delta L}} = \frac{7}{4} \]

4. **Simplifying the Ratio**:
Canceling out common terms \( F \), \( \pi \), and \( \Delta L \):
\[ \frac{L_A \cdot r_B^2}{L_B \cdot r_A^2} = \frac{7}{4} \]

5. **Substituting Known Values**:
We know:
- \( L_A = 2 \, \text{m} \)
- \( L_B = 1.5 \, \text{m} \)
- \( r_B = 2 \, \text{mm} = 0.002 \, \text{m} \)

Substitute these values into the equation:
\[ \frac{2 \cdot (0.002)^2}{1.5 \cdot r_A^2} = \frac{7}{4} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 8 \cdot (0.002)^2 = 7 \cdot 1.5 \cdot r_A^2 \]

7. **Calculating**:
Calculate \( 8 \cdot (0.002)^2 \):
\[ 8 \cdot (0.000004) = 0.000032 \]
And \( 7 \cdot 1.5 = 10.5 \):
\[ 0.000032 = 10.5 \cdot r_A^2 \]

8. **Solving for \( r_A^2 \)**:
Rearranging gives:
\[ r_A^2 = \frac{0.000032}{10.5} \]
Calculating \( r_A^2 \):
\[ r_A^2 = 0.00000304761905 \]

9. **Finding \( r_A \)**:
Taking the square root:
\[ r_A = \sqrt{0.00000304761905} \approx 0.001743 \, \text{m} = 1.743 \, \text{mm} \]

10. **Final Answer**:
Rounding to two decimal places, the radius of wire A is approximately:
\[ r_A \approx 1.74 \, \text{mm} \]