InterviewSolution
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InterviewSolution
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The diameter of a ball was measured five times with the aid of a micrometer whose absolute error `(Deltad_("inst"))= pm 0.01 mm`. The results of measuring the diameter of the ball are `d_(1)=5.27 mm, d_(2)=5.30 mm, d_(3)=5.28 mm, d_(4)=5.32 mm` and `d_(5)=5.28 mm`. Find (a) mean value of ball diameter (b) mean absolute error (c) result of measurement (d) relative error (e) persentage error. |
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Answer» (a) The mean value of the ball diameter, i.e., `d_(m)=(5.27+5.30+5.28+5.32+5.28)/(5)=5.29 mm` (b) The absolute error in the measurements are : `Deltad_(1)=|d_(m)-d_(1)|=0.02 mm` `Deltad_(2)=|d_(m)-d_(2)|=0.01 mm` `Deltad_(3)=|d_(m)-d_(3)|=0.01 mm` `Deltad_(4)=|d_(m)-d_(4)|=0.03 mm` `Deltad_(5)=|d_(m)-d_(5)|=0.01 mm` Mean absolute error, `Deltad_("mean")=(0.02+0.01+0.01+0.03+0.01)/(5)` `cong 0.02 mm`. (c) Since the mean absolute error `(Deltad_("mean"))` is greater than the instrumental error `(Delta d_("inst"))`, the result of measurement is `d=d_(m) pm Deltad_("mean")=(5.29 pm 0.02) mm` (As a rule, we take either `Deltad_("mean")` or `Deltad_("inst")`, depending upon which of these errors is greater) (d) Relative error, `(Deltad_("mean"))/d_(m)=0.02/5.29=0.004` (e) Percentage error, `(Deltad_("mean"))/d_(m)xx100=0.004xx100=0.4 %` |
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