A steel tape is correctly calibrated at `20^(@)C` and is used to measure the length of a table at `30^(@)C`. Find the percentage error in the measurement of length. `[alpha_(steel) = 11 xx 10^(-5)//.^(@)C]`

Explanation

To solve the problem, we need to find the percentage error in the measurement of length when using a steel tape that is calibrated at 20°C to measure a length at 30°C.

### Step-by-Step Solution:

1. **Identify the Coefficient of Linear Expansion**:
The coefficient of linear expansion for steel is given as:
\[ \alpha_{\text{steel}} = 11 \times 10^{-5} \, \text{°C}^{-1} \]

2. **Determine the Temperature Change**:
The temperature change (ΔT) when measuring from 20°C to 30°C is:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 30°C - 20°C = 10°C \]

3. **Calculate the Change in Length (ΔL)**:
The change in length (ΔL) due to thermal expansion can be calculated using the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where \(L\) is the original length measured at 20°C. Since we are looking for the percentage error, we will express ΔL in terms of L.

Therefore:
\[ \Delta L = L \cdot (11 \times 10^{-5}) \cdot 10 \]
\[ \Delta L = L \cdot 1.1 \times 10^{-4} \]

4. **Calculate the Percentage Error**:
The percentage error in the measurement of length is given by:
\[ \text{Percentage Error} = \left( \frac{\Delta L}{L} \right) \times 100 \]
Substituting ΔL from the previous step:
\[ \text{Percentage Error} = \left( \frac{L \cdot 1.1 \times 10^{-4}}{L} \right) \times 100 \]
\[ \text{Percentage Error} = 1.1 \times 10^{-4} \times 100 \]
\[ \text{Percentage Error} = 0.011 \times 100 = 0.11\% \]

### Final Answer:
The percentage error in the measurement of length is **0.11%**.