Question:
A 30 m long tape is standardized at \(25^{\circ}\text{C}\). It was used to measure the length of a line which came out to be 200 m. The temperature during the measurement was \(35^{\circ}\text{C}\). The coefficient of expansion of the tape was \(11 \times 10^{-6}\) per \(^{\circ}\text{C}\). The correction in the measured length (in mm) due to change in temperature is (in integer).
A 30 m long tape is standardized at \(25^{\circ}\text{C}\). It was used to measure the length of a line which came out to be 200 m. The temperature during the measurement was \(35^{\circ}\text{C}\). The coefficient of expansion of the tape was \(11 \times 10^{-6}\) per \(^{\circ}\text{C}\). The correction in the measured length (in mm) due to change in temperature is (in integer).
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Temperature correction on a tape is alpha x (Tm - Ts) x measured length; add it since the tape ran hot and long.
Updated On: Jul 17, 2026
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Correct Answer: 22
Solution and Explanation
Step 1: Recall the temperature correction formula for a survey tape.
A steel tape is manufactured to be exactly correct in length only at its standardization temperature. If it is used at a different temperature, it expands or contracts, so every reading taken with it is slightly wrong. The correction to be applied to the measured length is
\[ C_t = \alpha (T_m - T_s) \, L \]
where \(\alpha\) is the coefficient of thermal expansion of the tape, \(T_m\) is the temperature during measurement, \(T_s\) is the standardization temperature, and \(L\) is the measured length.
Step 2: Substitute the given values.
\[ \alpha = 11 \times 10^{-6} \ \text{per} \ ^{\circ}\text{C}, \quad T_m = 35^{\circ}\text{C}, \quad T_s = 25^{\circ}\text{C}, \quad L = 200 \ \text{m} \]
\[ C_t = 11 \times 10^{-6} \times (35 - 25) \times 200 \]
Step 3: Work out the correction.
\[ C_t = 11 \times 10^{-6} \times 10 \times 200 = 11 \times 10^{-6} \times 2000 = 0.022 \ \text{m} \]
Converting to millimetres,
\[ C_t = 0.022 \times 1000 = 22 \ \text{mm} \]
Step 4: Interpret the sign of the correction.
Since the field temperature (\(35^{\circ}\text{C}\)) is higher than the standardization temperature (\(25^{\circ}\text{C}\)), the tape has expanded and become longer than its nominal length. A longer tape lays off a longer true distance for the same number of tape lengths read, so the true measured length is greater than the recorded 200 m: the correction is added.
Final Answer:
The temperature correction to be added to the measured length is 22 mm.
\[ \boxed{C_t = 22 \ \text{mm}} \]
A steel tape is manufactured to be exactly correct in length only at its standardization temperature. If it is used at a different temperature, it expands or contracts, so every reading taken with it is slightly wrong. The correction to be applied to the measured length is
\[ C_t = \alpha (T_m - T_s) \, L \]
where \(\alpha\) is the coefficient of thermal expansion of the tape, \(T_m\) is the temperature during measurement, \(T_s\) is the standardization temperature, and \(L\) is the measured length.
Step 2: Substitute the given values.
\[ \alpha = 11 \times 10^{-6} \ \text{per} \ ^{\circ}\text{C}, \quad T_m = 35^{\circ}\text{C}, \quad T_s = 25^{\circ}\text{C}, \quad L = 200 \ \text{m} \]
\[ C_t = 11 \times 10^{-6} \times (35 - 25) \times 200 \]
Step 3: Work out the correction.
\[ C_t = 11 \times 10^{-6} \times 10 \times 200 = 11 \times 10^{-6} \times 2000 = 0.022 \ \text{m} \]
Converting to millimetres,
\[ C_t = 0.022 \times 1000 = 22 \ \text{mm} \]
Step 4: Interpret the sign of the correction.
Since the field temperature (\(35^{\circ}\text{C}\)) is higher than the standardization temperature (\(25^{\circ}\text{C}\)), the tape has expanded and become longer than its nominal length. A longer tape lays off a longer true distance for the same number of tape lengths read, so the true measured length is greater than the recorded 200 m: the correction is added.
Final Answer:
The temperature correction to be added to the measured length is 22 mm.
\[ \boxed{C_t = 22 \ \text{mm}} \]
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