Chapter 13, Problem 73GP

(a)

To determine

Change in reading of precise steel tape at ${34}^{\circ }\text{C}$ .

Explanation of Solution

Introduction:

When any object has been heated to a certain temperature, the dimension of the objects increases due to the thermal expansion.

As the temperature is increased from ${20}^{\circ }\text{C}$ to ${34}^{\circ }\text{C}$ , this will cause increase in the length of the steel tape due to thermal expansion. Thus, the steel tape made will measure more than the true length of every object. Thus, it will read low as compared to the correct length of the object.

Conclusion:

A precise steel tape when it has been used at ${34}^{\circ }\text{C}$ , it will read low.

(b)

To determine

To Find: The percentage error.

Answer to Problem 73GP

The percentage error in the measurement of the length is $0.02\mathrm{\%}$ .

Explanation of Solution

Given:

Coefficient of expansion, $\mathrm{\alpha }=12\times {10}^{-6}°{\text{C}}^{\text{-1}}$

Initial temperature, ${\mathrm{T}}_1={20}^{\circ }\text{C}$

Final temperature, ${\mathrm{T}}_2={34}^{\circ }\text{C}$

Formula used:

Percentage error is calculated by the following formula,

  $\begin{array}{l}\mathrm{\%}\text{ error }=\frac{{\mathrm{l}}_0}{\mathrm{l}}\\ \mathrm{l}={\mathrm{l}}_0+\mathrm{\Delta }\mathrm{l}\\ \mathrm{\Delta }\mathrm{l}=\mathrm{\alpha }\times {\mathrm{l}}_0\times \mathrm{\Delta }\mathrm{T}\end{array}$

Here, $\mathrm{\alpha }$ is coefficient of expansion.

  $\mathrm{\Delta }\mathrm{T}$ is the change in temperature.

  $\mathrm{\Delta }\mathrm{l}$ is the change in length.

Calculation:

Substitute the given values,

  $\begin{array}{l}\mathrm{\%}\text{ error }=\frac{{\mathrm{l}}_0}{\mathrm{l}}=\frac{{\mathrm{l}}_0}{{\mathrm{l}}_0+\mathrm{\Delta }\mathrm{l}}=\frac{{\mathrm{l}}_0}{{\mathrm{l}}_0+\left(\mathrm{\alpha }\times {\mathrm{l}}_0\times \mathrm{\Delta }\mathrm{T}\right)}\\ \frac{{\mathrm{l}}_0}{\mathrm{l}}=\frac{{\mathrm{l}}_0}{{\mathrm{l}}_0\left(1+\mathrm{\alpha }\times \mathrm{\Delta }\mathrm{T}\right)}=\frac{1}{1+\mathrm{\alpha }\times \mathrm{\Delta }\mathrm{T}}=\frac{1}{1+\left(12\times {10}^{-6}\times 14\right)}\\ \frac{{\mathrm{l}}_0}{\mathrm{l}}=0.9998\\ {\mathrm{l}}_0=0.9998\times \mathrm{l}=99.98\mathrm{\%}(\mathrm{l})\\ \text{Error in }\mathrm{\%}=(100-99.98)\mathrm{\%}=0.02\text{ \%}\end{array}$

Conclusion:

Thus, the percentage error in the measurement of the length is $0.02\text{ \%}$ .