The coefficient of linear expansion of aluminium is `23xx10^(-6)//""^@C` . A circular hole in an aluminium plate 2.725 is cm in diameter at `0^@C` .What is the diameter of the hole if the temperature of the plate is raised to `100^@C` ?

Options

A. 0.0063 cm
B. 2.728 cm
C. 2.731 cm
D. 2.757 cm

Answer

To find the new diameter of the circular hole in the aluminum plate when the temperature is raised from \(0^\circ C\) to \(100^\circ C\), we will use the formula for linear expansion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Coefficient of linear expansion (\(\alpha\)) of aluminum: \(23 \times 10^{-6} \, ^\circ C^{-1}\)
- Initial diameter (\(D_0\)) of the hole at \(0^\circ C\): \(2.725 \, \text{cm}\)
- Initial temperature (\(T_0\)): \(0^\circ C\)
- Final temperature (\(T_f\)): \(100^\circ C\)

2. **Calculate the Change in Temperature (\(\Delta T\)):**
\[ \Delta T = T_f - T_0 = 100^\circ C - 0^\circ C = 100^\circ C \]

3. **Use the Linear Expansion Formula:**
The formula for linear expansion is given by:
\[ D_f = D_0 \times (1 + \alpha \Delta T) \]
where:
- \(D_f\) is the final diameter,
- \(D_0\) is the initial diameter,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

4. **Substitute the Values into the Formula:**
\[ D_f = 2.725 \, \text{cm} \times \left(1 + (23 \times 10^{-6}) \times 100\right) \]

5. **Calculate the Expansion Factor:**
First, calculate \((23 \times 10^{-6}) \times 100\):
\[ 23 \times 10^{-6} \times 100 = 0.0023 \]
Therefore, the expansion factor becomes:
\[ 1 + 0.0023 = 1.0023 \]

6. **Calculate the Final Diameter:**
Now substitute back into the equation for \(D_f\):
\[ D_f = 2.725 \, \text{cm} \times 1.0023 \]
\[ D_f \approx 2.725 \times 1.0023 \approx 2.731 \, \text{cm} \]

### Final Answer:
The diameter of the hole at \(100^\circ C\) is approximately **2.731 cm**.

---