Length of wire at room temperature is `4.55` m , when the temperature increases upto `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 follow these steps:

### Step 1: Understand the formula for linear expansion
The formula for linear expansion is given by:
\[ L_2 = L_1 (1 + \alpha (T_2 - T_1)) \]
Where:
- \(L_1\) = original length of the wire at initial temperature \(T_1\)
- \(L_2\) = new length of the wire at temperature \(T_2\)
- \(\alpha\) = coefficient of linear expansion
- \(T_1\) = initial temperature
- \(T_2\) = final temperature

### Step 2: Identify the given values
From the problem:
- \(L_1 = 4.55 \, \text{m}\)
- \(L_2 = 4.57 \, \text{m}\)
- \(T_1 = 27^\circ C\) (room temperature)
- \(T_2 = 100^\circ C\)

### Step 3: Convert temperatures to Kelvin
To use the formula, we can convert the temperatures to Kelvin:
- \(T_1 = 27^\circ C + 273 = 300 \, \text{K}\)
- \(T_2 = 100^\circ C + 273 = 373 \, \text{K}\)

### Step 4: Calculate the change in temperature
Now, calculate the change in temperature:
\[ \Delta T = T_2 - T_1 = 373 \, \text{K} - 300 \, \text{K} = 73 \, \text{K} \]

### Step 5: Substitute values into the linear expansion formula
Now, substitute the values into the linear expansion formula:
\[ 4.57 = 4.55 (1 + \alpha \cdot 73) \]

### Step 6: Solve for α
First, divide both sides by \(4.55\):
\[ \frac{4.57}{4.55} = 1 + 73\alpha \]
Now, calculate \(\frac{4.57}{4.55}\):
\[ \frac{4.57}{4.55} \approx 1.0044 \]
So we have:
\[ 1.0044 = 1 + 73\alpha \]
Now, subtract 1 from both sides:
\[ 0.0044 = 73\alpha \]
Now, divide both sides by 73 to solve for \(\alpha\):
\[ \alpha = \frac{0.0044}{73} \approx 6.021 \times 10^{-5} \, \text{K}^{-1} \]

### 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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