Question

A distance of 1000 m was measured in the field at a mean temperature of 60°F using a tape of coefficient of thermal expansion \(5 \times 10^{-6}\) /°F. If the standardization temperature is 80°F, the correction value for temperature is

• A. −0.1 m
• B. 0.1 m
• C. −0.12 m
• D. +0.12 m

Hint:

When the field temperature differs from the temperature at which the tape was standardized, the tape changes length and a temperature correction must be applied.

Answer 1

The Correct Option is A

Concept:
When the field temperature differs from the temperature at which the tape was standardized, the tape changes length and a temperature correction must be applied. The correction is given by \[ C_t = \alpha (T_m - T_0) L \] where \(\alpha\) is the coefficient of thermal expansion, \(T_m\) is the mean field temperature, \(T_0\) is the standardization temperature, and \(L\) is the measured length.

Step 1:
Substitute the given values: \(\alpha = 5 \times 10^{-6}\) /°F, \(T_m = 60\)°F, \(T_0 = 80\)°F, \(L = 1000\) m. \[ C_t = 5 \times 10^{-6} \times (60 - 80) \times 1000. \]

Step 2:
Evaluate: \[ C_t = 5 \times 10^{-6} \times (-20) \times 1000 = -0.1 \text{ m}. \] The negative sign means the field temperature is lower than the standard, so the tape is shorter and the correction is subtractive.

Answer: Option (1) — −0.1 m.