An iron tyre is to be fitted onto a wooden wheel 1.0 m in diameter. The diameter of the tyre is 6 mm smaller than that of wheel the tyre should be heated so that its temperature increases by a minimum of (coefficient of volume expansion of iron is `3.6xx10^-5//^@C`)

Options

A. `167^@C`
B. `334^@C`
C. `500^@C`
D. `1000^@C`

Explanation

To solve the problem of fitting an iron tyre onto a wooden wheel, we need to determine the minimum temperature increase required for the tyre to fit properly. Here’s a step-by-step solution:

### Step 1: Identify the diameters
The diameter of the wooden wheel is given as 1.0 m. To convert this to millimeters:
\[ \text{Diameter of the wheel} = 1.0 \, \text{m} = 1000 \, \text{mm} \]

The diameter of the tyre is 6 mm smaller than that of the wheel:
\[ \text{Diameter of the tyre} = 1000 \, \text{mm} - 6 \, \text{mm} = 994 \, \text{mm} \]

### Step 2: Calculate the radius of the tyre
The radius of the tyre can be calculated as:
\[ \text{Radius of the tyre} = \frac{\text{Diameter of the tyre}}{2} = \frac{994 \, \text{mm}}{2} = 497 \, \text{mm} \]

### Step 3: Determine the change in diameter required
To fit the tyre onto the wheel, we need to increase the diameter of the tyre by 6 mm. Therefore, the change in diameter (\(\Delta D\)) is:
\[ \Delta D = 6 \, \text{mm} \]

### Step 4: Calculate the change in radius
The change in radius (\(\Delta r\)) is half of the change in diameter:
\[ \Delta r = \frac{\Delta D}{2} = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} \]

### Step 5: Relate the change in radius to temperature change
The change in radius due to thermal expansion can be expressed as:
\[ \Delta r = r \cdot \alpha \cdot \Delta T \]
where:
- \(r\) is the original radius (497 mm),
- \(\alpha\) is the coefficient of linear expansion (which is \(\frac{\gamma}{3}\) for volume expansion, where \(\gamma = 3.6 \times 10^{-5} \, \text{°C}^{-1}\)),
- \(\Delta T\) is the change in temperature.

### Step 6: Substitute the values
First, calculate \(\alpha\):
\[ \alpha = \frac{\gamma}{3} = \frac{3.6 \times 10^{-5}}{3} = 1.2 \times 10^{-5} \, \text{°C}^{-1} \]

Now substitute into the equation:
\[ 3 \, \text{mm} = 497 \, \text{mm} \cdot (1.2 \times 10^{-5} \, \text{°C}^{-1}) \cdot \Delta T \]

### Step 7: Solve for \(\Delta T\)
Rearranging the equation gives:
\[ \Delta T = \frac{3 \, \text{mm}}{497 \, \text{mm} \cdot 1.2 \times 10^{-5} \, \text{°C}^{-1}} \]

Calculating this:
\[ \Delta T = \frac{3}{497 \cdot 1.2 \times 10^{-5}} \approx \frac{3}{5.964 \times 10^{-3}} \approx 502.5 \, \text{°C} \]

### Final Answer
The minimum temperature increase required for the iron tyre to fit onto the wooden wheel is approximately:
\[ \Delta T \approx 500 \, \text{°C} \]