The diameter of a wire as measured by screw gauge was found to be $2.620, 2.625, 2.628 a n d 2. 626 c m$ . Calculate percentage error.

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Solution

Mean value of the diameter, $a_{m} = \frac{2.620 + 2.625 + 2.628 + 2.626}{4} = 2.625$

Thus, $Δ a_{1} = a_{m} - a_{1} = 2.625 - 2.620 = 0.005$
$Δ a_{2} = a_{m} - a_{2} = 2.625 - 2.625 = 0.000$
$Δ a_{3} = a_{m} - a_{3} = 2.625 - 2.628 = - 0.003$
$Δ a_{4} = a_{m} - a_{4} = 2.625 - 2.626 = - 0.001$

Mean absolute error , $Δ a_{m e a n} = \frac{| Δ a_{1} | + | Δ a_{2} | + | Δ a_{3} | + | Δ a_{4} |}{4} = \frac{0.005 + 0.000 + 0.003 + 0.001}{4} = 0.002$

Fractional error $= \pm \frac{Δ a_{m e a n}}{a_{m}} = \pm \frac{0.002}{2.625} = \pm 0.0008$

Percentage error $=$ fractional error $\times 100 = \pm 0.0008 \times 100 = 0.08$ %

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