The length L and diameter d of four wires of the same material are given below. Which of these will elongate the most, when the same tension is applied ?

Options

A. L=50 cm, d=0.5 mm
B. L=100 cm, d=1.0 mm
C. L=200 cm, d=2.0 mm
D. L=300 cm, L=3.0 mm

Answer

To determine which wire will elongate the most when the same tension is applied, we will use the formula for elongation based on Young's modulus. The elongation (ΔL) of a wire can be expressed as:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Where:
- \( \Delta L \) = elongation
- \( F \) = force (tension) applied
- \( L \) = original length of the wire
- \( A \) = cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

Since all wires are made of the same material, Young's modulus \( Y \) will be constant for all wires. We can also express the cross-sectional area \( A \) in terms of the diameter \( d \):

\[ A = \frac{\pi d^2}{4} \]

Substituting this into the elongation formula gives:

\[ \Delta L = \frac{F \cdot L}{\frac{\pi d^2}{4} \cdot Y} = \frac{4F \cdot L}{\pi d^2 \cdot Y} \]

Since \( Y \) and \( F \) are constant for all wires, we can compare the elongation based on the term \( \frac{L}{d^2} \).

### Steps to Solve:

1. **Identify the Length and Diameter of Each Wire**:
- Wire 1: \( L_1 = 50 \, \text{m}, \, d_1 = 0.5 \times 10^{-3} \, \text{m} \)
- Wire 2: \( L_2 = 100 \, \text{m}, \, d_2 = 1 \times 10^{-3} \, \text{m} \)
- Wire 3: \( L_3 = 200 \, \text{m}, \, d_3 = 2 \times 10^{-3} \, \text{m} \)
- Wire 4: \( L_4 = 300 \, \text{m}, \, d_4 = 3 \times 10^{-3} \, \text{m} \)

2. **Calculate \( \frac{L}{d^2} \) for Each Wire**:
- For Wire 1:
\[ \frac{L_1}{d_1^2} = \frac{50}{(0.5 \times 10^{-3})^2} = \frac{50}{0.25 \times 10^{-6}} = 200 \times 10^{6} = 2 \times 10^{8} \]
- For Wire 2:
\[ \frac{L_2}{d_2^2} = \frac{100}{(1 \times 10^{-3})^2} = \frac{100}{1 \times 10^{-6}} = 100 \times 10^{6} = 1 \times 10^{8} \]
- For Wire 3:
\[ \frac{L_3}{d_3^2} = \frac{200}{(2 \times 10^{-3})^2} = \frac{200}{4 \times 10^{-6}} = 50 \times 10^{6} = 5 \times 10^{7} \]
- For Wire 4:
\[ \frac{L_4}{d_4^2} = \frac{300}{(3 \times 10^{-3})^2} = \frac{300}{9 \times 10^{-6}} = \frac{100}{3} \times 10^{6} \approx 33.33 \times 10^{6} \]

3. **Compare the Values**:
- Wire 1: \( 2 \times 10^{8} \)
- Wire 2: \( 1 \times 10^{8} \)
- Wire 3: \( 5 \times 10^{7} \)
- Wire 4: \( 33.33 \times 10^{6} \)

4. **Conclusion**:
The wire with the highest value of \( \frac{L}{d^2} \) will elongate the most. From the calculations, Wire 1 has the highest value, so it will elongate the most.

### Final Answer:
**Wire 1 will elongate the most when the same tension is applied.**