Question

The diameter of a thin wire is measured with screw gauge.In successive meaurements : The values of the diameter of the wire are 2.04mm, 2.06mm,2.08mm,2.07mmand 2.05mm.Find (i) thediameter of the wire (ii) absolute errors in each measurement (iii) Mean absolute error (iv) Relative error (ii) Percentage error

Answer 1

Answer:

2.04+2.06+2.08+2.072.05=10.30/5=2.06

So absolute error is 2.06

Now mean absolute error is

2.04_2.06= 0.02

2.06_2.06=0.00

2.08_2.06=0.02

2.07_2.06=0.01

2.05_2.06=0.01

Mean absolute error is 0.02+0.00+0.02+0.01+0.01=0.06/5=0.01

Real value is 2.06+_0.01

Realative error=mean absolute error /mean

0.01/2.06=0.004854

Percentage error is=realative error ×100

0.004854×100=0.4854%

Percentage error is=0.485

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Explanation: