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 elongation strain on the wire ?

Options

A. `1.0xx10^(2)`
B. `1.0xx10^(-2)`
C. `5.0xx10^(2)`
D. `2.5xx10^(-3)`

Answer

To find the elongation strain on the wire, we can follow these steps:

### Step 1: Identify the given values
- Original length of the wire (L₀) = 200.00 cm
- Final length of the wire (L) = 200.50 cm
- Force applied (not needed for strain calculation)

### Step 2: Calculate the change in length (ΔL)
The change in length (ΔL) can be calculated as:
\[ \Delta L = L - L₀ \]
Substituting the values:
\[ \Delta L = 200.50 \, \text{cm} - 200.00 \, \text{cm} = 0.50 \, \text{cm} \]

### Step 3: Convert the original length to the same units as the change in length
Since the change in length is in centimeters, we can keep the original length in centimeters:
\[ L₀ = 200.00 \, \text{cm} \]

### Step 4: Calculate the elongational strain (ε)
The elongational strain (ε) is defined as the change in length divided by the original length:
\[ \epsilon = \frac{\Delta L}{L₀} \]
Substituting the values:
\[ \epsilon = \frac{0.50 \, \text{cm}}{200.00 \, \text{cm}} = \frac{0.50}{200} = 0.0025 \]

### Step 5: Express the strain in scientific notation
To express 0.0025 in scientific notation:
\[ 0.0025 = 2.5 \times 10^{-3} \]

### Final Answer
The elongation strain on the wire is:
\[ \epsilon = 2.5 \times 10^{-3} \]

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