An aluminium flagpole is 30 m high . By how much does its length increase as the temperature increases by `15^@C` ?

Answer

To solve the problem of how much the length of an aluminum flagpole increases when the temperature increases by \(15^\circ C\), we can use the formula for linear thermal expansion:

\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]

Where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length of the object,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Original length of the flagpole, \(L_0 = 30 \, m\)
- Coefficient of linear expansion for aluminum, \(\alpha = 2.5 \times 10^{-5} \, \text{per Kelvin}\)
- Change in temperature, \(\Delta T = 15 \, \text{degrees Celsius} = 15 \, K\) (since a change in degrees Celsius is equivalent to a change in Kelvin).

2. **Substitute the Values into the Formula:**
\[ \Delta L = 30 \, m \cdot (2.5 \times 10^{-5} \, \text{per K}) \cdot 15 \, K \]

3. **Calculate the Change in Length:**
\[ \Delta L = 30 \cdot 2.5 \times 10^{-5} \cdot 15 \]
\[ \Delta L = 30 \cdot 3.75 \times 10^{-4} \]
\[ \Delta L = 1.125 \times 10^{-2} \, m \]
\[ \Delta L = 0.01125 \, m \]

4. **Convert to Millimeters (if needed):**
\[ \Delta L = 0.01125 \, m \times 1000 \, \text{mm/m} = 11.25 \, mm \]

5. **Final Result:**
The length of the aluminum flagpole increases by approximately \(0.01125 \, m\) or \(11.25 \, mm\).