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.

Explanation

(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 %`