Question

A surveyor has a steel measuring tape that is calibrated to be 100.000 m long (i.e., accurate to$\pm 1mm$) at $20°C$. If she measures the distance between two stakes to be 65.175 m on a $3°C$day, does she need to add or subtract a correction factor to get the true distance? How large, in mm, is the correction factor?

Short answer

She needs to subtract a correction factor to get the true distance and the correction factor is mm is 12.18 mm

Step-by-step solution

Description 

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through $∆T$, as well as length changes through $∆L$and the length's fractional changes $∆L/L$are effective to the temperature's change $∆T$

Result on findings 

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through $∆T$, as well as length changes through $∆L$and the length's fractional changes $∆L/L$are effective to the temperature's change $∆T$.

The equation (18.8) in the form is:

$\frac{∆L}{L}=\alpha ∆T$ (1)

In this case, $\alpha$is the coefficient. Table 18.4 effectively gives the coefficient expression of linear for common materials, for instance

$\alpha =1.2\times {10}^{-5\circ }{C}^{-1}$

The equation for $∆L$is

$∆L=\alpha L∆T$ (2)

In this case, the subject of L is the initial length and the temperature effectively changes from $T_1=20°Cto{T}_2=3°C.$

Therefore, the change is temperature is

$$\begin{gathered} ∆T=T_2-T_1 \\ =3°C-20°C \\ =-17°C \end{gathered}$$

Thus, plug the value for $∆L,Tand\alpha$into equation 2 in order to get $∆L$

$$\begin{gathered} ∆L=\alpha L∆T \\ =(1.1\times {10}^{-5°}{C}^{-1})(65.175m)(-17°C) \\ =-12.18\times {10}^{-3}m \\ =-12.18mm \end{gathered}$$

There is a requirement for subtracting the correction factor in terms of getting true distance and the correction factor is same for change in length.

Therefore, correction factor $=12.18mm$