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

Step 1: Write down the values.
Length \(L = 1000\) m, field temperature \(T_m = 60^\circ\)F, standard temperature \(T_0 = 80^\circ\)F, coefficient \(\alpha = 5 \times 10^{-6}\) per \(^\circ\)F.

Step 2: Recall the temperature correction formula.
\[ C_t = \alpha (T_m - T_0) L \]

Step 3: Find the temperature difference.
\[ T_m - T_0 = 60 - 80 = -20^\circ \text{F} \]

Step 4: Put the numbers in.
\[ C_t = (5 \times 10^{-6}) \times (-20) \times 1000 \]

Step 5: Simplify.
\[ C_t = 5 \times 10^{-6} \times (-20000) = -0.1 \text{ m} \]
The minus sign shows the tape was colder, so it shrank and the true length is a bit less.

Step 6: State the answer.
The temperature correction is \(-0.1\) m.
\[ \boxed{-0.1 \text{ m}} \]