The diameter of iron wheel is `1cm`. If its temprature is increased by `700^(0)C` What is the increase in circumference of the wheel?
`(alpha = 12 xx 10^(-6 //0)C)`

Options

A. `0.0264cm`
B. `0.264 cm`
C. `2.64cm`
D. `26.4cm`

Explanation

To find the increase in the circumference of an iron wheel when its temperature is increased, we can follow these steps:

### Step 1: Understand the relationship between circumference and diameter
The circumference \( C \) of a circle is given by the formula:
\[ C = \pi \times d \]
where \( d \) is the diameter of the circle.

### Step 2: Calculate the initial circumference
Given that the diameter of the iron wheel is \( 1 \, \text{cm} \):
\[ C_0 = \pi \times 1 \, \text{cm} = \pi \, \text{cm} \approx 3.14 \, \text{cm} \]

### Step 3: Identify the change in temperature
The temperature is increased by \( 700^{\circ}C \), so:
\[ \Delta T = 700^{\circ}C \]

### Step 4: Use the coefficient of linear expansion
The coefficient of linear expansion \( \alpha \) for iron is given as:
\[ \alpha = 12 \times 10^{-6} \, \text{per} \, ^{\circ}C \]

### Step 5: Calculate the increase in circumference
The increase in circumference \( \Delta C \) can be calculated using the formula:
\[ \Delta C = C_0 \times \alpha \times \Delta T \]
Substituting the known values:
\[ \Delta C = \pi \, \text{cm} \times (12 \times 10^{-6} \, \text{per} \, ^{\circ}C) \times (700^{\circ}C) \]
Calculating this:
\[ \Delta C = 3.14 \, \text{cm} \times 12 \times 10^{-6} \times 700 \]
\[ \Delta C = 3.14 \times 12 \times 700 \times 10^{-6} \]
\[ \Delta C = 3.14 \times 8400 \times 10^{-6} \]
\[ \Delta C = 26376 \times 10^{-6} \, \text{cm} \]
\[ \Delta C = 0.026376 \, \text{cm} \]

### Step 6: Round the result
Rounding \( 0.026376 \, \text{cm} \) gives approximately:
\[ \Delta C \approx 0.0264 \, \text{cm} \]

### Final Result
The increase in circumference of the wheel is approximately \( 0.0264 \, \text{cm} \).

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