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 (d) (f) Express the result in terms of percentage error

Answer

A, B, C, D

Explanation

(a) Mean value of diameter
`a_m = (2.620 + 2.625 + 2.630 + 2.628 + 2.626)/5`
= 2.6258 cm
= 2.626 cm.
(b) 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 = +0.001 cm `
` Deltaa_3 = 2.623-2.630 = -0.004 cm`
`Deltaa_4 = 2.626-2.628 = -0.002 cm`
`Deltaa_5 = 2.626 - 2.626 = 0.000 cm. `
(c) Mean absolute error,
`Deltaa_(mean) = (|Deltaa_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`
(d)Fractional error = `+- (Deltaa_(mean))/(a_(m)) = +-(0.003)/(2.626) = +-0.001`
(e) Percentage error = ` +-0.001 xx 100 = +-0.1%`
(f) Diameter of wire can be written as ,
`d= 2.626 +-0.1%` .