The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(a) mean value of diameter (b) absolute error in each measurement
(c) mean absolute error (d) fractional error
(e) percentage error (f) Express the result in terms of percentage error

Explanation

(i) Mean value of diameter.
`a_(m)=(2.620+2.625+2.630+2.628+2.62)/(5)`
`=2.6258 cm=2.626 cm` (rounding off to three decimal places)
(ii) Taking `a_(m)` as the true value, the absolute errors in different observations are
`Deltaa_(1)=2.626-2.620=+0.006cm`
`Deltaa_(2)=2.626-2.625=+0.001cm`
`Deltaa_(3)=2.626-2.630=-0.004cm`
`Deltaa_(4)=2.626-2.628=-0.002cm`
`Deltaa_(5)=2.626-2.626=-0.000cm`
(iii) Mean absolute error,
`Deltaa_(mean)=(|Delata_(1)|+|Deltaa_(2)|+|Deltaa_(3)|+|Deltaa_(4)|+|Deltaa_(5)|)/(5)`
`(0.006+0.001+0.004+0.002+0.000)/(5)`
0.0026=0.003 (rounding off to three decimal places)
(iv) Fractional error =`pm(Deltaa_(mean))/(a_(mean))=pm(0.003)/(2.626)=pm0.001`
(v) Percentage error `=pm0.001xx100=pm0.1%`
(vi) Diameter of wire can be written as, ltbr. `d=2.626cmpm0.1%`