Question

The diameter ofthe diameter of a wire as measured by a screw gauge was found to be 0.026cm,0.028cm,0.029cm,0.027cm 0.024cm,0.027cm. calculate i)mean value of the diameter .ii)mean absolute error.iii) relative error. iv) % error. aslo express the result in term of absolute errors and percentage error????
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Answer 1

Answer:

Explanation:

a(mean) = (0.026+0.028+0.029+0.027+0.024+0.027)/6

=0.0268 cm

da1=0.0008

da2=0.0012

da3=0.0022

da4=0.0002

da5=0.0028

da6=0.0002

da(mean) = 0.0012

R.E.= 0.0012/0.0268 =0.045

%error= 4.5