The radii and Young's moduli of two uniform wires A and B are in the ratio `2:1` and `1:2` respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire A is one percent, the percentage increase in length of the wire B is

Options

A. `1.0`
B. `1.5`
C. `2.0`
D. `3.0`

Explanation

To solve the problem, we need to find the percentage increase in length of wire B given the percentage increase in length of wire A and the ratios of their radii and Young's moduli.

### Step-by-Step Solution:

1. **Identify Given Ratios**:
- The ratio of the radii of wires A and B is given as \( r_A : r_B = 2 : 1 \).
- The ratio of Young's moduli of wires A and B is given as \( Y_A : Y_B = 1 : 2 \).

2. **Express Radii and Young's Moduli**:
- Let \( r_B = r \). Then \( r_A = 2r \).
- Let \( Y_A = Y \). Then \( Y_B = 2Y \).

3. **Understand the Relationship of Length Increase**:
- The increase in length \( \Delta L \) of a wire under a longitudinal force \( F \) is given by the formula:
\[ \frac{\Delta L}{L} = \frac{F}{A \cdot Y} \]
- Where \( A \) is the cross-sectional area of the wire. For a circular wire, \( A = \pi r^2 \).

4. **Relate the Increase in Length for Both Wires**:
- For wire A:
\[ \frac{\Delta L_A}{L_A} = \frac{F}{\pi (r_A^2) Y_A} = \frac{F}{\pi (4r^2) Y} = \frac{F}{4\pi r^2 Y} \]
- For wire B:
\[ \frac{\Delta L_B}{L_B} = \frac{F}{\pi (r_B^2) Y_B} = \frac{F}{\pi (r^2) (2Y)} = \frac{F}{2\pi r^2 Y} \]

5. **Calculate the Ratio of Length Increases**:
- The ratio of the percentage increases in lengths of wires A and B can be expressed as:
\[ \frac{\Delta L_A / L_A}{\Delta L_B / L_B} = \frac{(F / (4\pi r^2 Y))}{(F / (2\pi r^2 Y))} = \frac{2}{4} = \frac{1}{2} \]

6. **Use the Given Percentage Increase in Length of Wire A**:
- We are given that the percentage increase in length of wire A is 1%. Thus:
\[ \frac{\Delta L_A}{L_A} = 0.01 \]
- Therefore, using the ratio we found:
\[ \frac{\Delta L_B}{L_B} = 2 \times \frac{\Delta L_A}{L_A} = 2 \times 0.01 = 0.02 \]

7. **Convert to Percentage**:
- The percentage increase in length of wire B is:
\[ \Delta L_B = 0.02 \times 100\% = 2\% \]

### Final Answer:
The percentage increase in length of wire B is **2%**.