Question

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

• A. Length 50 cm, diameter = 0.5 mm
• B. Length 100 cm, diameter - 1 mm
• C. Length 200 cm, diameter = 2 mm
• D. Length = 300 cm, diameter = 3 mm

Answer 1

The Correct Option is A

To determine which wire will have the largest extension when the same tension is applied, we must consider the formula for the extension of a wire under tension, known as the formula of elongation for materials:

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

Where:

• \Delta L is the extension (elongation) of the wire.
• F is the force (tension) applied.
• L is the original length of the wire.
• A is the cross-sectional area of the wire.
• Y is the Young's modulus of the material (same for all wires as they are made of the same material).

The formula shows that the extension is directly proportional to the original length and inversely proportional to the cross-sectional area. Since Young's modulus and the force applied are the same for all wires, we need to compare the ratio \frac{L}{A}.

The cross-sectional area A is calculated using:

A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}

Let's calculate \frac{L}{A} for each wire:

1. Wire 1: Length L_1 = 50 cm, Diameter d_1 = 0.5 mm = 0.05 cm
   A_1 = \frac{\pi (0.05)^2}{4}
   \frac{L_1}{A_1} = \frac{50}{\frac{\pi (0.05)^2}{4}}
2. Wire 2: Length L_2 = 100 cm, Diameter d_2 = 1 mm = 0.1 cm
   A_2 = \frac{\pi (0.1)^2}{4}
   \frac{L_2}{A_2} = \frac{100}{\frac{\pi (0.1)^2}{4}}
3. Wire 3: Length L_3 = 200 cm, Diameter d_3 = 2 mm = 0.2 cm
   A_3 = \frac{\pi (0.2)^2}{4}
   \frac{L_3}{A_3} = \frac{200}{\frac{\pi (0.2)^2}{4}}
4. Wire 4: Length L_4 = 300 cm, Diameter d_4 = 3 mm = 0.3 cm
   A_4 = \frac{\pi (0.3)^2}{4}
   \frac{L_4}{A_4} = \frac{300}{\frac{\pi (0.3)^2}{4}}

We need to find which wire has the highest \frac{L}{A} value, because that will give the largest extension:

• Wire 1: \frac{L_1}{A_1} = \frac{50}{\frac{\pi (0.05)^2}{4}} = \frac{50 \times 4}{\pi \times 0.0025} \approx \frac{200}{0.00785} \approx 25477
• Wire 2: \frac{L_2}{A_2} = \frac{100}{\frac{\pi (0.1)^2}{4}} = \frac{100 \times 4}{\pi \times 0.01} \approx \frac{400}{0.0314} \approx 12739
• Wire 3: \frac{L_3}{A_3} = \frac{200}{\frac{\pi (0.2)^2}{4}} = \frac{200 \times 4}{\pi \times 0.04} \approx \frac{800}{0.1256} \approx 6370
• Wire 4: \frac{L_4}{A_4} = \frac{300}{\frac{\pi (0.3)^2}{4}} = \frac{1200}{\pi \times 0.09} \approx \frac{1200}{0.2826} \approx 4244

The value \frac{L}{A} is highest for Wire 1, which means Wire 1 will have the largest extension.

Thus, the correct answer is Length 50 cm, diameter = 0.5 mm.