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 wire and the steel wire when subjected to the same force. Here’s the step-by-step solution:

### Step 1: Understand the given data
- Let the radius of the steel wire be \( r_s \).
- The radius of the copper wire is \( r_c = 2r_s \) (twice the radius of steel).
- Let the Young's modulus of copper be \( Y_c \).
- The Young's modulus of steel is \( Y_s = 2Y_c \) (twice that of copper).
- The increase in length of the copper wire, \( \Delta L_C \), is given as \( 1\% \) of its original length \( L_C \).

### Step 2: Write the formula for elongation
The elongation (change in length) of a wire under a tensile force can be expressed using Young's modulus:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F \) is the applied force,
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( Y \) is Young's modulus.

### Step 3: Calculate the cross-sectional areas
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
- For the steel wire:
\[ A_s = \pi r_s^2 \]
- For the copper wire:
\[ A_c = \pi (2r_s)^2 = 4\pi r_s^2 \]

### Step 4: Set up the equations for elongation
Using the formula for elongation for both wires:
- For copper:
\[ \Delta L_C = \frac{F \cdot L_C}{A_c \cdot Y_c} = \frac{F \cdot L_C}{4\pi r_s^2 \cdot Y_c} \]
- For steel:
\[ \Delta L_S = \frac{F \cdot L_S}{A_s \cdot Y_s} = \frac{F \cdot L_S}{\pi r_s^2 \cdot 2Y_c} \]

### Step 5: Relate the elongations
Since both wires are subjected to the same force \( F \), we can relate their elongations:
\[ \frac{\Delta L_C}{L_C} = \frac{F}{4\pi r_s^2 Y_c} \quad \text{and} \quad \frac{\Delta L_S}{L_S} = \frac{F}{\pi r_s^2 (2Y_c)} \]

### Step 6: Divide the two equations
Now, we can divide the two elongation equations:
\[ \frac{\Delta L_C / L_C}{\Delta L_S / L_S} = \frac{(F / (4\pi r_s^2 Y_c))}{(F / (\pi r_s^2 (2Y_c)))} \]
This simplifies to:
\[ \frac{\Delta L_C / L_C}{\Delta L_S / L_S} = \frac{1}{2} \]

### Step 7: Substitute known values
Given that \( \Delta L_C = 0.01 L_C \) (1% increase), we can substitute this into the equation:
\[ \frac{0.01}{\Delta L_S / L_S} = \frac{1}{2} \]
This implies:
\[ \Delta L_S / L_S = 0.02 \]

### Step 8: Conclusion
Thus, the steel wire will have stretched by \( 2\% \) of its length when the length of the copper wire has increased by \( 1\% \).

### Final Answer:
The steel wire will have stretched by **2%** of its length.

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