To find the increase in length of the wire when a force is applied, we can use the formula for tensile stress and strain:

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

where:

• $F$ is the applied force,
• $L$ is the original length of the wire,
• $A$ is the cross-sectional area of the wire,
• $Y$ is the tensile modulus of the material.

Given Values

1. Original length $L = 12.00 \, \text{m}$
2. Diameter of the wire $d = 1.50 \, \text{mm} = 0.0015 \, \text{m}$
3. Tensile modulus $Y = 7.00 \times 10^{10} \, \text{N/m}^2$
4. Force applied $F = 500 \, \text{N}$

Step 1: Calculate the Cross-sectional Area $A$

Since the wire is circular, the area $A$ can be found using: $A = \pi \left(\frac{d}{2}\right)^2$ $A = \pi \left(\frac{0.0015}{2}\right)^2 = \pi \times (0.00075)^2$ $A \approx 1.767 \times 10^{-6} \, \text{m}^2$

Step 2: Apply the Formula to Find $\Delta L$

Substitute the values into the formula: $\Delta L = \frac{(500 \, \text{N}) \times (12.00 \, \text{m})}{(1.767 \times 10^{-6} \, \text{m}^2) \times (7.00 \times 10^{10} \, \text{N/m}^2)}$

Calculating the numerator: $500 \times 12 = 6000 \, \text{N m}$

Calculating the denominator: $(1.767 \times 10^{-6}) \times (7.00 \times 10^{10}) \approx 1236.9$

Now divide: $\Delta L = \frac{6000}{1236.9} \approx 4.85 \, \text{mm}$

Answer

The increase in length of the wire is approximately $48.5 \, \text{mm}$.

Correct choice: 48.5 mm

Would you like more details on any part of this solution?

Here are 5 additional practice questions to consider:

1. What would the increase in length be if the tensile modulus doubled?
2. How would the length change if the diameter was 2 mm?
3. Calculate the increase in length for a 1000 N force instead.
4. How would the length change if the wire was only 6 m long?
5. What tensile modulus would result in an increase of 60 mm?

Tip: Remember, tensile modulus $Y$ measures a material's resistance to being deformed elastically, so a higher $Y$ means less stretching under the same force.