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. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the dimensions
The diameter of the wooden wheel is given as 1.0 m, which is equivalent to 1000 mm. The diameter of the tyre is 6 mm smaller than that of the wheel.

- **Diameter of the wheel (D_w)** = 1000 mm
- **Diameter of the tyre (D_t)** = D_w - 6 mm = 1000 mm - 6 mm = 994 mm

### Step 2: Calculate the radii
Next, we calculate the radii of both the wheel and the tyre.

- **Radius of the wheel (R_w)** = D_w / 2 = 1000 mm / 2 = 500 mm
- **Radius of the tyre (R_t)** = D_t / 2 = 994 mm / 2 = 497 mm

### Step 3: Determine the required expansion
To fit the tyre onto the wheel, the radius of the tyre must expand by the difference between the radius of the wheel and the radius of the tyre.

- **Required change in radius (ΔR)** = R_w - R_t = 500 mm - 497 mm = 3 mm

### Step 4: Use the formula for linear expansion
The formula for linear expansion is given by:

\[ \Delta R = R \cdot \alpha \cdot \Delta T \]

Where:
- \(\Delta R\) = change in radius (3 mm)
- \(R\) = initial radius of the tyre (497 mm)
- \(\alpha\) = coefficient of linear expansion

### Step 5: Calculate the coefficient of linear expansion
The coefficient of volume expansion (\(\gamma\)) for iron is given as \(3.6 \times 10^{-5} \, / ^\circ C\). The relationship between the coefficient of volume expansion and the coefficient of linear expansion is:

\[ \alpha = \frac{\gamma}{3} \]

Calculating \(\alpha\):

\[ \alpha = \frac{3.6 \times 10^{-5}}{3} = 1.2 \times 10^{-5} \, / ^\circ C \]

### Step 6: Substitute values into the expansion formula
Now we can substitute the values into the expansion formula to find \(\Delta T\):

\[ 3 \, \text{mm} = 497 \, \text{mm} \cdot (1.2 \times 10^{-5}) \cdot \Delta T \]

### Step 7: Solve for \(\Delta T\)
Rearranging the equation to solve for \(\Delta T\):

\[ \Delta T = \frac{3 \, \text{mm}}{497 \, \text{mm} \cdot (1.2 \times 10^{-5})} \]

Calculating \(\Delta T\):

\[ \Delta T = \frac{3}{497 \cdot 1.2 \times 10^{-5}} \approx \frac{3}{5.964 \times 10^{-3}} \approx 503.05 \, ^\circ C \]

### Conclusion
The minimum temperature increase required to fit the iron tyre onto the wooden wheel is approximately **503 °C**.