A tow truck is pulling a car out of a ditch by means of steel cable that is 9.1 m long and has a radius of 0.50 cm. When the car just begins to move, the tension in the cable is 890 N. How much has the cable stretched ?

Options

A. `2.6xx10^(-6)m`
B. `2.3xx10^(-2)m`
C. `1.7xx10^(-4)m`
D. `5.2xx10^(-4)m`

Answer

To solve the problem of how much the steel cable has stretched when the tow truck pulls the car out of the ditch, we can use the formula derived from Young's modulus. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Length of the cable (L) = 9.1 m
- Radius of the cable (r) = 0.50 cm = 0.005 m (conversion from cm to m)
- Tension in the cable (F) = 890 N
- Young's modulus for steel (Y) = \(2 \times 10^{11} \, \text{N/m}^2\)

### Step 2: Calculate the cross-sectional area (A) of the cable
The cross-sectional area of a circular cable is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (0.005)^2 = \pi (0.000025) \approx 7.85 \times 10^{-5} \, \text{m}^2 \]

### Step 3: Use Young's modulus to find the extension (ΔL)
Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging this gives:
\[ \Delta L = \frac{F L}{A Y} \]
Substituting the known values:
\[ \Delta L = \frac{890 \times 9.1}{7.85 \times 10^{-5} \times 2 \times 10^{11}} \]

### Step 4: Calculate ΔL
Calculating the numerator:
\[ 890 \times 9.1 = 8099 \]
Calculating the denominator:
\[ 7.85 \times 10^{-5} \times 2 \times 10^{11} = 1.57 \times 10^{7} \]
Now substituting these values into the equation:
\[ \Delta L = \frac{8099}{1.57 \times 10^{7}} \approx 0.000515 \, \text{m} \]

### Step 5: Convert ΔL to a more convenient unit
\[ \Delta L \approx 5.15 \times 10^{-4} \, \text{m} = 0.515 \, \text{mm} \]

### Step 6: Round off the answer
Rounding off gives:
\[ \Delta L \approx 5.2 \times 10^{-4} \, \text{m} \]

### Final Answer
The cable has stretched approximately \(5.2 \times 10^{-4} \, \text{m}\) or \(0.52 \, \text{mm}\).

---