Question

The diameter of a wire, as measured by a screw gauge, was found to be 2.620, 2.625, 2.628, and 2.626 centimeters. Calculate the percentage error.

Answer 1

Screw gauge is an instrument for measuring the thickness of the small object (here wire). Generally, it is used where the thickness of the objects is very less. As it is a measuring instrument and so some sort of error is always present while measuring or calculating.

Here, in the question, the same wire’s diameter has been measured by the same screw gauge four times with different readings, and we need to determine the percentage error in the measurement compared with the true measurement. For this, we need to first evaluate the mean of the given measurements and then the mean of the absolute error of the measurements. The percentage error is the ratio of the mean absolute error and the mean of the measurements.

Complete step by step answer:
When a screw gauge is used to measure the diameters of the wires, then the average value of the measurements will be considered as the true value of the diameter of the wire.

Let us consider the mean of the diameter of the wire (true value) to be ${d_m}$.

Mean (or the average value) is the ratio of the sum of the measurements and the total number of the trials. So, here substitute all the given values in the formula ${d_m} = \dfrac{{{d_1} + {d_2} + {d_3} + {d_4}}}{4}$, we get

${d_m} = \dfrac{{{d_1} + {d_2} + {d_3} + {d_4}}}{4} \\\ = \dfrac{{2.620 + 2.625 + 2.628 + 2.626}}{4} \\\ = \dfrac{{10.499}}{4} \\\ = 2.62475 - - - - (i) \\\$

Now, we have to evaluate the absolute errors of each measurement by following the formula $\vartriangle {d_i} = {d_m} - {d_i}$ as:

$\vartriangle {d_1} = {d_m} - {d_1} \\\ = 2.62475 - 2.620 \\\ = 0.00475 \\\$

Similarly, for the second, third, and fourth measurements, we get:
$\vartriangle {d_2} = {d_m} - {d_2} \\\ = 2.62475 - 2.625 \\\ = - 0.00025 \\\ \vartriangle {d_3} = {d_m} - {d_3} \\\ = 2.62475 - 2.628 \\\ = - 0.00325 \\\ \vartriangle {d_1} = {d_m} - {d_1} \\\ = 2.62475 - 2.626 \\\ = - 0.00125 \\\$

Now, the mean of the absolute error can be calculated as:
$\Delta {d_m} = \dfrac{{\left| {\Delta {d_1}} \right| + \left| {\Delta {d_1}} \right| + \left| {\Delta {d_2}} \right| + \left| {\Delta {d_4}} \right|}}{4} \\\ = \dfrac{{\left| {0.00475} \right| + \left| { - 0.00025} \right| + \left| { - 0.00325} \right| + \left| { - 0.00125} \right|}}{4} \\\ = \dfrac{{0.00475 + 0.00025 + 0.00325 + 0.00125}}{4} \\\ = \dfrac{{0.0095}}{4} \\\ = 0.002375 - - - - (ii) \\\$

Substitute ${d_m} = 2.62475$ and $\Delta {d_m} = 0.002375$ in the formula $\% e = \dfrac{{\vartriangle {d_m}}}{{{d_m}}} \times 100$ to determine the percentage error in the measurement of the diameter of the gauge.

$\% e = \dfrac{{\vartriangle {d_m}}}{{{d_m}}} \times 100 \\\ = \dfrac{{0.002375}}{{2.62475}} \times 100 \\\ = 0.0009048 \times 100 \\\ = 0.090\% \\\$

Hence, the percentage error while calculating the diameter of a wire as measured by the screw gauge is 0.09%.

Note:
Students should note here that while calculating the mean of the absolute error, sign convention has to be neglected, i.e., absolute error signifies the deviation from the true value regardless of the sign either positive or negative.