Two parallel wires A and B of same meterial are fixed to rigid support at the upper ends and subjected to same load at the lower ends. The subjeted to same load at the lower ends. The lengths of the wire are in the ration 4 : 5 and their radii are in the ratio 4 : 3 the increase in the length of wire A is 1 mm. Calculate the increase in the length of wire A is 1mm. Calculate the increase in the length of the wire B.

Explanation

To solve the problem, we will use the concept of Young's modulus and the relationship between stress, strain, and the dimensions of the wires.

### Step-by-Step Solution:

1. **Identify Given Values:**
- For wire A:
- Length \( L_A = 4L \)
- Radius \( r_A = 4r \)
- Increase in length \( \Delta L_A = 1 \text{ mm} \)
- For wire B:
- Length \( L_B = 5L \)
- Radius \( r_B = 3r \)
- Increase in length \( \Delta L_B = ? \)

2. **Calculate the Cross-Sectional Areas:**
- The cross-sectional area \( A \) of a wire is given by the formula:
\[ A = \pi r^2 \]
- For wire A:
\[ A_A = \pi (4r)^2 = 16\pi r^2 \]
- For wire B:
\[ A_B = \pi (3r)^2 = 9\pi r^2 \]

3. **Use Young's Modulus Formula:**
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ Y = \frac{F L}{A \Delta L} \]

4. **Set Up the Equations for Both Wires:**
- For wire A:
\[ Y = \frac{F}{A_A} \cdot \frac{L_A}{\Delta L_A} = \frac{F}{16\pi r^2} \cdot \frac{4L}{1 \text{ mm}} \]
- For wire B:
\[ Y = \frac{F}{A_B} \cdot \frac{L_B}{\Delta L_B} = \frac{F}{9\pi r^2} \cdot \frac{5L}{\Delta L_B} \]

5. **Equate the Young's Modulus for Both Wires:**
- Since both wires are made of the same material, their Young's modulus values are equal:
\[ \frac{F}{16\pi r^2} \cdot \frac{4L}{1 \text{ mm}} = \frac{F}{9\pi r^2} \cdot \frac{5L}{\Delta L_B} \]

6. **Cancel Common Terms:**
- Cancel \( F \), \( \pi \), and \( r^2 \) from both sides:
\[ \frac{4L}{16} = \frac{5L}{\Delta L_B} \]

7. **Solve for \( \Delta L_B \):**
- Rearranging gives:
\[ \Delta L_B = \frac{5L \cdot 16}{4L} = 20 \text{ mm} \]

8. **Final Result:**
- The increase in length of wire B is:
\[ \Delta L_B = 2.5 \text{ mm} \]

### Summary:
The increase in the length of wire B is **2.5 mm**.