A hole is drilled in a copper sheet. The diameter of the hole is `4.24 cm` at `27.0^(@)C`. What is the change in the diameter of the hole when the sheet is heated to `227^(0)C`? `alpha` for copper `= 1.70 xx 10^(-5)K^(-1)`

Options

A. `1.44xx10^(-2) cm`
B. `2.44xx10^(-3) cm`
C. `1.44xx10^(-2) mm`
D. `2.44xx10^(-3) mm`

Answer

To solve the problem of finding the change in the diameter of a hole in a copper sheet when heated, we can follow these steps:

### Step 1: Identify the given values
- Initial diameter of the hole, \( D_0 = 4.24 \, \text{cm} \)
- Initial temperature, \( T_1 = 27.0 \, ^\circ C \)
- Final temperature, \( T_2 = 227.0 \, ^\circ C \)
- Coefficient of linear expansion for copper, \( \alpha = 1.70 \times 10^{-5} \, \text{K}^{-1} \)

### Step 2: Calculate the change in temperature
The change in temperature (\( \Delta T \)) can be calculated as:
\[ \Delta T = T_2 - T_1 = 227.0 \, ^\circ C - 27.0 \, ^\circ C = 200.0 \, ^\circ C \]

### Step 3: Use the formula for change in diameter
The change in diameter (\( \Delta D \)) due to thermal expansion can be calculated using the formula:
\[ \Delta D = D_0 \cdot \alpha \cdot \Delta T \]

### Step 4: Substitute the values into the formula
Substituting the known values:
\[ \Delta D = 4.24 \, \text{cm} \cdot (1.70 \times 10^{-5} \, \text{K}^{-1}) \cdot (200.0 \, \text{K}) \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ \Delta D = 4.24 \cdot 1.70 \times 10^{-5} \cdot 200.0 \]
\[ \Delta D = 4.24 \cdot 1.70 \cdot 200.0 \times 10^{-5} \]
\[ \Delta D = 4.24 \cdot 340 \times 10^{-5} \]
\[ \Delta D = 1441.6 \times 10^{-5} \, \text{cm} \]
\[ \Delta D = 1.4416 \times 10^{-2} \, \text{cm} \]

### Step 6: Final answer
The change in the diameter of the hole when the sheet is heated to \( 227^\circ C \) is approximately:
\[ \Delta D \approx 1.44 \times 10^{-2} \, \text{cm} \]