To determine which wire experiences the maximum extension when the same force is applied to each wire, we can use the formula derived from Young's modulus of elasticity. The extension (l) in the wire can be expressed as:
\[
l = \frac{F \cdot L}{Y \cdot A}
\]
Where:
- \( l \) = extension in the wire
- \( F \) = force applied
- \( L \) = original length of the wire
- \( Y \) = Young's modulus of the material
- \( A \) = cross-sectional area of the wire
The cross-sectional area \( A \) can be expressed in terms of the diameter \( d \) of the wire as:
\[
A = \frac{\pi d^2}{4}
\]
Substituting this into the extension formula gives:
\[
l = \frac{F \cdot L}{Y \cdot \left(\frac{\pi d^2}{4}\right)} = \frac{4F \cdot L}{Y \cdot \pi d^2}
\]
From this equation, we can see that the extension \( l \) is directly proportional to the ratio \( \frac{L}{d^2} \):
\[
l \propto \frac{L}{d^2}
\]
Now, we need to evaluate the ratio \( \frac{L}{d^2} \) for each option provided:
1. **Option A**: \( L = 400 \, \text{cm}, d = 0.01 \, \text{mm} = 0.001 \, \text{cm} \)
\[
\frac{L}{d^2} = \frac{400}{(0.001)^2} = \frac{400}{0.000001} = 400000000
\]
2. **Option B**: \( L = 300 \, \text{cm}, d = 0.03 \, \text{mm} = 0.003 \, \text{cm} \)
\[
\frac{L}{d^2} = \frac{300}{(0.003)^2} = \frac{300}{0.000009} = 33333333.33
\]
3. **Option C**: \( L = 200 \, \text{cm}, d = 0.02 \, \text{mm} = 0.002 \, \text{cm} \)
\[
\frac{L}{d^2} = \frac{200}{(0.002)^2} = \frac{200}{0.000004} = 50000000
\]
4. **Option D**: \( L = 500 \, \text{cm}, d = 0.05 \, \text{mm} = 0.005 \, \text{cm} \)
\[
\frac{L}{d^2} = \frac{500}{(0.005)^2} = \frac{500}{0.000025} = 20000000
\]
Now, we can compare the values of \( \frac{L}{d^2} \):
- Option A: \( 400000000 \)
- Option B: \( 33333333.33 \)
- Option C: \( 50000000 \)
- Option D: \( 20000000 \)
The maximum value of \( \frac{L}{d^2} \) is for **Option A**, which means that **Option A** will have the maximum extension in the wire.
### Final Answer:
**Option A has the maximum extension in the wire.**
