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 of finding the percentage error in the measurement of length using a steel tape calibrated at 20°C and used at 30°C, we can follow these steps:

### Step 1: Understand the concept of linear expansion
The linear expansion of a material is given by the formula:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]
where:
- \(\Delta L\) is the change in length,
- \(\alpha\) is the coefficient of linear expansion,
- \(L_0\) is the original length,
- \(\Delta T\) is the change in temperature.

### Step 2: Identify the values given in the problem
- Coefficient of linear expansion of steel, \(\alpha = 11 \times 10^{-5} \, ^\circ C^{-1}\)
- Original calibration temperature, \(T_0 = 20^\circ C\)
- Measurement temperature, \(T = 30^\circ C\)

### Step 3: Calculate the change in temperature
\[ \Delta T = T - T_0 = 30^\circ C - 20^\circ C = 10^\circ C \]

### Step 4: Substitute the values into the linear expansion formula
Since we are interested in the percentage error, we can express it as:
\[ \text{Percentage Error} = \left( \frac{\Delta L}{L_0} \right) \times 100 \]
Substituting \(\Delta L\):
\[ \text{Percentage Error} = \left( \frac{\alpha \cdot L_0 \cdot \Delta T}{L_0} \right) \times 100 = \alpha \cdot \Delta T \times 100 \]

### Step 5: Substitute the known values
Now substituting the values of \(\alpha\) and \(\Delta T\):
\[ \text{Percentage Error} = (11 \times 10^{-5}) \cdot (10) \cdot 100 \]

### Step 6: Calculate the percentage error
\[ \text{Percentage Error} = (11 \times 10^{-5} \times 10 \times 100) = 0.11\% \]

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

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