Question

A steel tape is calibrated at ${20}^{\circ }C$. On a cold day when the temperature is $-{15}^{\circ }C$ percentage error in the tape will be $[{\alpha }_{\text{steel}}=1.2\times {10}^{-5\cdot 0}{C}^{-1}]$

• A. $0.038\%$
• B. $0.042\%$
• C. $0.012\%$
• D. $0.018\%$

Answer 1

Answer: (A)

$0.038\%$

$L=L_0(1+\alpha \Delta T)$

$\Delta L=|L-L_0|$

$\begin{array}{c}\Delta L=L_0\alpha \Delta T\\ \frac{\Delta L}{L_0}=\alpha \Delta T\end{array}$

$\begin{array}{c}=1.2\times {10}^{-5}\times (20-(-15\left)\right)\\ =1.2\times {10}^5\times 35\\ =38.4\times {10}^{-5}\end{array}$

$\%\text{error}=0.038\%$