A truck is pulling a car out of a ditch by means fo a steel cable that is `9.1m` long and has a radius of `5mm`. When the car just beings to move, the tension in the cable is `800 N`. If Young's modulus for steel is `2xx10^(11) N//m^(2)` then the strecth in the cable is (neraly)

Options

A. `5xx10^(-3)m`
B. `0.5xx10^(-3)m`
C. `3xx10^(-3)m`
D. `0.3xx10^(-3)m`

Explanation

To find the stretch in the steel cable when the truck pulls the car, we can use the relationship defined by Young's modulus. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Length of the cable, \( L = 9.1 \, m \)
- Radius of the cable, \( r = 5 \, mm = 5 \times 10^{-3} \, m \)
- Tension in the cable, \( F = 800 \, N \)
- Young's modulus for steel, \( Y = 2 \times 10^{11} \, N/m^2 \)

### Step 2: Calculate the cross-sectional area of the cable
The area \( A \) of the cable can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (5 \times 10^{-3})^2 = \pi (25 \times 10^{-6}) \, m^2 \]
\[ A \approx 3.14 \times 25 \times 10^{-6} = 7.85 \times 10^{-5} \, m^2 \]

### Step 3: Use Young's modulus to find the stretch \( \Delta L \)
Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} \]
Rearranging this gives:
\[ \Delta L = \frac{F L}{A Y} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ \Delta L = \frac{800 \, N \times 9.1 \, m}{7.85 \times 10^{-5} \, m^2 \times 2 \times 10^{11} \, N/m^2} \]

### Step 5: Calculate the numerator and denominator
Calculating the numerator:
\[ 800 \times 9.1 = 7280 \, N \cdot m \]

Calculating the denominator:
\[ 7.85 \times 10^{-5} \times 2 \times 10^{11} = 1.57 \times 10^{7} \, N \]

### Step 6: Calculate \( \Delta L \)
Now, substituting these values back into the equation:
\[ \Delta L = \frac{7280}{1.57 \times 10^{7}} \approx 4.63 \times 10^{-4} \, m \]

### Step 7: Convert to millimeters
To express the stretch in millimeters:
\[ \Delta L \approx 4.63 \times 10^{-4} \, m = 0.463 \, mm \]

### Final Answer
The stretch in the cable is approximately \( 0.463 \, mm \).

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