Length of wire at room temperature is `4.55` m , when the temperature increases up to `100^(@) C ` then its length becomes `4.57` m . The coefficient of linear expansion `(alpha)` of the given wire is

Options

A. `5.021 xx 10^(-5) K^(-1)`
B. `6.021 xx 10^(-5) K^(-1)`
C. `7.021 xx 10^(-5) K^(-1)`
D. `8.021 xx 10^(-5) K^(-1)`

Explanation

To find the coefficient of linear expansion (α) of the wire, we can use the formula:

\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]

Where:
- \( \Delta L \) is the change in length of the wire,
- \( L_0 \) is the original length of the wire,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

### Step 1: Identify the given values
- Original length of the wire, \( L_0 = 4.55 \, \text{m} \)
- New length of the wire, \( L = 4.57 \, \text{m} \)
- Change in length, \( \Delta L = L - L_0 = 4.57 \, \text{m} - 4.55 \, \text{m} = 0.02 \, \text{m} \)
- Initial temperature, \( T_1 = 27^\circ C \) (room temperature)
- Final temperature, \( T_2 = 100^\circ C \)
- Change in temperature, \( \Delta T = T_2 - T_1 = 100^\circ C - 27^\circ C = 73^\circ C \)

### Step 2: Convert temperature change to Kelvin
Since the change in temperature in Celsius is the same as in Kelvin, we can directly use:
\[ \Delta T = 73 \, \text{K} \]

### Step 3: Substitute the values into the formula
Now, substituting the values into the formula for \( \Delta L \):
\[ 0.02 = 4.55 \cdot \alpha \cdot 73 \]

### Step 4: Solve for α
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{0.02}{4.55 \cdot 73} \]

Calculating the denominator:
\[ 4.55 \cdot 73 = 332.15 \]

Now substituting this back into the equation for \( \alpha \):
\[ \alpha = \frac{0.02}{332.15} \approx 6.021 \times 10^{-5} \, \text{K}^{-1} \]

### Final Answer
The coefficient of linear expansion \( \alpha \) of the given wire is approximately:
\[ \alpha \approx 6.021 \times 10^{-5} \, \text{K}^{-1} \]

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