Question

The diameter of a wire as measured by a screw gauge was found to be 1.328,1.330,1.325,1.334,1.336cm.calculate (a) percentage error.​

Answer 1

First we have to find the mean value, i.e., the true value.

We have $\displaystyle\overline{a}=\dfrac{1}{n}\sum_{i=1}^{n}a_i.$ So,

$\overline{a}=\dfrac{1.328+1.330+1.325+1.334+1.336}{5}=1.331\ cm$

Now we find the absolute errors of each measurements.

We have $\Delta a=|\overline{a}-a|.$ So,

$|1.331-1.328|=0.003\ cm\\\\|1.331-1.330|=0.001\ cm\\\\|1.331-1.325|=0.006\ cm\\\\|1.331-1.334|=0.003\ cm\\\\|1.331-1.336|=0.005\ cm$

Now the mean absolute error has to be taken.

We have $\displaystyle\overline{\Delta a}=\dfrac{1}{n}\sum_{i=1}^{n}\Delta a_i.$ So,

$\overline{\Delta a}=\dfrac{0.003+0.001+0.006+0.003+0.005}{5}=0.004\ cm$

Now we can find the percentage error.

We have percentage error, $\delta a=\dfrac{\overline{\Delta a}}{\overline{a}}\times100.$ So,

$\delta a=\dfrac{0.004}{1.331}\times100=\mathbf{0.3\%}$