To determine in which case the extension of the wire will be maximum, we can use the relationship derived from Young's modulus. The extension (ΔL) of a wire under tension is given by the formula:
\[
\Delta L = \frac{F \cdot L}{Y \cdot A}
\]
Where:
- \(F\) is the force applied (tension in the wire),
- \(L\) is the original length of the wire,
- \(Y\) is Young's modulus of the material (constant for the same material),
- \(A\) is the cross-sectional area of the wire.
The cross-sectional area \(A\) of a wire with diameter \(d\) is given by:
\[
A = \frac{\pi d^2}{4}
\]
Substituting this into the extension formula gives:
\[
\Delta L = \frac{F \cdot L}{Y \cdot \left(\frac{\pi d^2}{4}\right)} = \frac{4F \cdot L}{Y \cdot \pi d^2}
\]
From this expression, we can see that the extension ΔL is directly proportional to the length \(L\) and inversely proportional to the square of the diameter \(d^2\):
\[
\Delta L \propto \frac{L}{d^2}
\]
Now, we can analyze the given options by calculating \( \frac{L}{d^2} \) for each case.
1. **Option A**: \(L = 200 \, \text{mm}, d = 0.5 \, \text{mm}\)
\[
\frac{L}{d^2} = \frac{200}{(0.5)^2} = \frac{200}{0.25} = 800
\]
2. **Option B**: \(L = 300 \, \text{cm} = 3000 \, \text{mm}, d = 1 \, \text{mm}\)
\[
\frac{L}{d^2} = \frac{3000}{(1)^2} = 3000
\]
3. **Option C**: \(L = 50 \, \text{cm} = 500 \, \text{mm}, d = 0.05 \, \text{mm}\)
\[
\frac{L}{d^2} = \frac{500}{(0.05)^2} = \frac{500}{0.0025} = 200000
\]
4. **Option D**: \(L = 100 \, \text{cm} = 1000 \, \text{mm}, d = 0.2 \, \text{mm}\)
\[
\frac{L}{d^2} = \frac{1000}{(0.2)^2} = \frac{1000}{0.04} = 25000
\]
Now, we can compare the values calculated:
- Option A: 800
- Option B: 3000
- Option C: 200000
- Option D: 25000
From these calculations, we can see that **Option C** has the maximum value of \( \frac{L}{d^2} \), which means it will have the maximum extension.
Thus, the case with the maximum extension of the wire is **Option C**.
