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

Explanation

To solve the problem, we need to calculate the increase in length of wire B (ΔL2) given the increase in length of wire A (ΔL1) and the ratios of their lengths and radii.

### Step-by-Step Solution:

1. **Define the lengths and radii of the wires:**
- Let the length of wire A (L1) be \(4x\) and the length of wire B (L2) be \(5x\).
- Let the radius of wire A (R1) be \(4y\) and the radius of wire B (R2) be \(3y\).

2. **Write the formula for Young's modulus (Y):**
\[ Y = \frac{F}{A} \cdot \frac{L}{\Delta L} \]
where \(F\) is the force applied, \(A\) is the cross-sectional area, \(L\) is the original length, and \(\Delta L\) is the change in length.

3. **Express the areas of the wires:**
- The area of wire A (A1) is:
\[ A1 = \pi R1^2 = \pi (4y)^2 = 16\pi y^2 \]
- The area of wire B (A2) is:
\[ A2 = \pi R2^2 = \pi (3y)^2 = 9\pi y^2 \]

4. **Set up the equation using the Young's modulus for both wires:**
Since both wires are made of the same material and are subjected to the same load, we can equate the expressions for Young's modulus:
\[ \frac{F}{A1} \cdot \frac{L1}{\Delta L1} = \frac{F}{A2} \cdot \frac{L2}{\Delta L2} \]
This simplifies to:
\[ \frac{L1}{A1 \Delta L1} = \frac{L2}{A2 \Delta L2} \]

5. **Substitute the known values:**
- Substitute \(L1 = 4x\), \(A1 = 16\pi y^2\), \(\Delta L1 = 1 \text{ mm} = 10^{-3} \text{ m}\)
- Substitute \(L2 = 5x\), \(A2 = 9\pi y^2\), and \(\Delta L2\) (unknown).

The equation becomes:
\[ \frac{4x}{16\pi y^2 \cdot 10^{-3}} = \frac{5x}{9\pi y^2 \cdot \Delta L2} \]

6. **Cancel out common terms:**
The \(x\), \(\pi\), and \(y^2\) terms cancel out:
\[ \frac{4}{16 \cdot 10^{-3}} = \frac{5}{9 \cdot \Delta L2} \]

7. **Cross-multiply to solve for \(\Delta L2\):**
\[ 4 \cdot 9 \cdot \Delta L2 = 5 \cdot 16 \cdot 10^{-3} \]
\[ 36 \Delta L2 = 80 \cdot 10^{-3} \]
\[ \Delta L2 = \frac{80 \cdot 10^{-3}}{36} \]

8. **Calculate \(\Delta L2\):**
\[ \Delta L2 = \frac{80}{36} \cdot 10^{-3} = \frac{20}{9} \cdot 10^{-3} \approx 2.22 \cdot 10^{-3} \text{ m} \]

### Final Answer:
The increase in length of wire B (ΔL2) is approximately \(2.22 \text{ mm}\).