A `5.0xx10^(2)N` object is hung from the end of a cross - sectional area `0.010 cm^(2)`. The wire stretches from its original length of 200.00 cm to 200.50 cm.
What is the stress on the wire ?

Options

A. `5.0xx10^(2)N//m^(2)`
B. `1.0xx10^(6)N//m^(2)`
C. `1.0xx10^(8)N//m^(2)`
D. `5.0xx10^(8)N//m^(2)`

Answer

To find the stress on the wire, we can use the formula for stress, which is defined as the force applied per unit area. The formula is given by:

\[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \]

### Step 1: Identify the given values
- Force (F) = \( 5.0 \times 10^2 \, \text{N} = 500 \, \text{N} \)
- Cross-sectional area (A) = \( 0.010 \, \text{cm}^2 \)

### Step 2: Convert the area from cm² to m²
To convert the area from cm² to m², we use the conversion factor \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \).

\[ A = 0.010 \, \text{cm}^2 \times 10^{-4} \, \text{m}^2/\text{cm}^2 = 0.010 \times 10^{-4} \, \text{m}^2 = 1.0 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Substitute the values into the stress formula
Now, we can substitute the values of force and area into the stress formula:

\[ \text{Stress} = \frac{F}{A} = \frac{500 \, \text{N}}{1.0 \times 10^{-6} \, \text{m}^2} \]

### Step 4: Calculate the stress
Perform the calculation:

\[ \text{Stress} = 500 \times 10^6 \, \text{N/m}^2 = 5.0 \times 10^8 \, \text{N/m}^2 \]

### Final Answer
The stress on the wire is:

\[ \text{Stress} = 5.0 \times 10^8 \, \text{N/m}^2 \]

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