The scale on a steel meter stick is calibrated at `15^@ C`. What is the error in the reading of `60 cm` at `27^@C` ? `alpha_(steel) =1.2 xx 10^-5 (.^@ C)^-1`.

Explanation

To solve the problem, we need to calculate the error in the reading of a steel meter stick that is calibrated at 15°C when the actual temperature is 27°C. We will use the formula for linear thermal expansion to find the change in length, which will give us the error in the reading.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length (L) = 60 cm
- Initial temperature (T₁) = 15°C
- Final temperature (T₂) = 27°C
- Coefficient of linear expansion for steel (α) = 1.2 × 10⁻⁵ (°C)⁻¹

2. **Calculate the Change in Temperature (ΔT):**
\[ \Delta T = T_2 - T_1 = 27°C - 15°C = 12°C \]

3. **Use the Formula for Change in Length (ΔL):**
The formula for the change in length due to thermal expansion is:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

4. **Substitute the Values into the Formula:**
\[ \Delta L = 60 \, \text{cm} \cdot (1.2 \times 10^{-5} \, \text{°C}^{-1}) \cdot (12 \, \text{°C}) \]

5. **Calculate ΔL:**
\[ \Delta L = 60 \cdot 1.2 \times 10^{-5} \cdot 12 \]
\[ = 60 \cdot 1.2 \cdot 12 \times 10^{-5} \]
\[ = 864 \times 10^{-5} \, \text{cm} \]
\[ = 0.00864 \, \text{cm} \]

6. **Determine the Error in the Reading:**
The error in the reading of the meter stick at 27°C is:
\[ \Delta L = 0.00864 \, \text{cm} \]

### Final Answer:
The error in the reading of 60 cm at 27°C is **0.00864 cm**.