A steel tape is callibrated at `20^@ C`. On a cold day when the temperature is `-15^@ C`, what will be the percentage error in the tape ?

Explanation

To find the percentage error in the steel tape calibrated at \(20^\circ C\) when the temperature drops to \(-15^\circ C\), we can follow these steps:

### Step 1: Understand the problem
The steel tape is calibrated at \(20^\circ C\) and we need to find out how much it will expand or contract when the temperature changes to \(-15^\circ C\). The change in length due to temperature change can be calculated using the formula for linear expansion.

### Step 2: Identify the formula for linear expansion
The change in length (\(\Delta L\)) of a material due to temperature change can be expressed as:
\[ \Delta L = L \cdot \alpha \cdot \Delta \theta \]
where:
- \(L\) is the original length of the tape,
- \(\alpha\) is the coefficient of linear expansion for steel (approximately \(1.2 \times 10^{-5} \, \text{°C}^{-1}\)),
- \(\Delta \theta\) is the change in temperature.

### Step 3: Calculate the change in temperature
The change in temperature (\(\Delta \theta\)) is calculated as:
\[ \Delta \theta = T_{\text{final}} - T_{\text{initial}} = -15^\circ C - 20^\circ C = -35^\circ C \]

### Step 4: Substitute values into the formula
Now, substituting the values into the formula for \(\Delta L\):
\[ \Delta L = L \cdot (1.2 \times 10^{-5}) \cdot (-35) \]

### Step 5: Calculate the percentage error
The percentage error in the tape can be calculated using the formula:
\[ \text{Percentage Error} = \frac{\Delta L}{L} \times 100 \]
Substituting \(\Delta L\) into the percentage error formula, we have:
\[ \text{Percentage Error} = \frac{L \cdot (1.2 \times 10^{-5}) \cdot (-35)}{L} \times 100 \]
This simplifies to:
\[ \text{Percentage Error} = (1.2 \times 10^{-5}) \cdot (-35) \times 100 \]

### Step 6: Calculate the final value
Calculating the value:
\[ \text{Percentage Error} = -1.2 \times 35 \times 10^{-3} = -0.042 \% \]

### Conclusion
The percentage error in the tape when the temperature is \(-15^\circ C\) is approximately \(-0.042\%\).

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