A piece of copper wire has twice the radius of a piece of steel wire. Young's modulus for steel is twice that of the copper. One end of the copper wire is joined to one end of the steel wire so that both can be subjected to the same longitudinal force. By what fraction of its length will the steel have stretched when the length of the copper has increased by `1%`?

Options

A. `1%`
B. `2%`
C. `2.5%`
D. `3%`

Explanation

To solve the problem, we need to analyze the relationship between the elongation of the copper and steel wires when subjected to the same longitudinal force.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Let the radius of the steel wire be \( r \).
- The radius of the copper wire is \( 2r \) (twice that of the steel wire).
- Young's modulus for steel \( Y_s = 2Y_c \) (twice that of copper).
- The elongation of the copper wire \( \Delta L_c = 1\% \) of its original length \( L_c \).

2. **Understanding Young's Modulus:**
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ \Delta L = \frac{F L}{A Y} \]
- The area \( A \) of a wire is given by \( A = \pi r^2 \).

3. **Calculate the Elongation for Copper Wire:**
- For the copper wire:
\[ \Delta L_c = \frac{F L_c}{\pi (2r)^2 Y_c} = \frac{F L_c}{4 \pi r^2 Y_c} \]

4. **Calculate the Elongation for Steel Wire:**
- For the steel wire:
\[ \Delta L_s = \frac{F L_s}{\pi r^2 Y_s} = \frac{F L_s}{\pi r^2 (2Y_c)} = \frac{F L_s}{2 \pi r^2 Y_c} \]

5. **Relate the Elongations:**
- Since both wires are subjected to the same force, we can set up a ratio of their elongations:
\[ \frac{\Delta L_s}{\Delta L_c} = \frac{L_s}{L_c} \cdot \frac{Y_c}{2Y_c} \cdot \frac{4}{1} = \frac{4 L_s}{2 L_c} = 2 \frac{L_s}{L_c} \]

6. **Substituting the Given Elongation of Copper:**
- Given that \( \Delta L_c = 0.01 L_c \) (1% elongation):
\[ \Delta L_s = 2 \cdot \frac{L_s}{L_c} \cdot 0.01 L_c = 0.02 L_s \]

7. **Finding the Fractional Increase in Length of Steel Wire:**
- The fractional increase in length of the steel wire is:
\[ \frac{\Delta L_s}{L_s} = \frac{0.02 L_s}{L_s} = 0.02 = 2\% \]

### Final Answer:
The steel wire will stretch by **2%** of its original length when the length of the copper wire has increased by 1%.