A wire is 12.00 m long and has a diameter of 1.50 mm. The tensile modulus of the wire is 7.00 × 1010 N/m2. If a force of 500 N is applied to end of the wire, then the increase in length of the wire is

Group of answer choices

38.5 mm.

40.2 mm.

44.1 mm.

48.5 mm.

51.50 mm.

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.

A wire is 12.00 m long and has a diameter of 1.50 mm. The tensile modulus of the wire is 7.00 × 10^10 N/m^2. If a force of 500 N is applied to end of the wire, then the increase in length of the wire is:

Group of answer choices:

38.5 mm.

40.2 mm.

44.1 mm.

48.5 mm.

51.50 mm.

Solve for the increase in length of a 12.00 m wire with a 1.50 mm diameter and tensile modulus of 7.00 × 10^10 N/m^2 under a 500 N force. Detailed solution using elasticity and tensile modulus formulas. Ideal for high school physics students (Grades 10-12).

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