Question

A surveyor’s 30.0 m steel tape is correct at $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$. The distance between two points, as measured by this tape on a day when its temperature is $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$, is $\mathbf{25}\mathbf{.}\mathbf{970}\mathbf{}\mathit{m}$ . What is the true distance between the points?

Short answer

The true distance between the points is 25.965 m .

Step-by-step solution

Given data

Length of tape: 30.0 m

measured length is: 25.970

room temperature: $5°C$

the temperature at which tape is correct: $20°C$

 Concept of Linear Expansion in length.

The linear change in length, when the temperature change is given, is given by the linear expansion expression,

$\mathbf{L}\mathbf{=}{\mathbf{L}}_{\mathbf{0}}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{\alpha \Delta T}\mathbf{)}$ ...(i)

Here, L is the expanded length, L₀ is the original length, α is the coefficient of expansion of the material and T is the absolute temperature.

In a steel scale, when the temperature fluctuates (increases), the marking distances also increases which gives error in measurements.

 Step 2:  Determination of true distance between the markings on the tape.

For steel scale, $\alpha =(1.2\times {10}^{-5}\hspace{0.33em}{\left({}^{o}C\right)}^{-1})$

Given in question, $L_o=25.970m$and $\Delta T=(5.00{\hspace{0.33em}}^{o}C-20.0\hspace{0.33em}{}^{o}C)=\hspace{0.33em}-15\hspace{0.33em}{}^{o}C$

Substitute all the values in equation (i),

$$\begin{gathered} L=\left(25.970m\right)\left(1+1.2\times {10}^{-5}{\left({}^{o}C\right)}^{-1}\left(15{}^{o}C\right)\right) \\ =25.965m \end{gathered}$$

Thus, the true distance between the tape markings is 25.965 m .