(1) Diameter of a wire is measured with micrometer screw gauge for five times as
0.421 cm, 0.422 cm, 0.421 cm, 0.423 cm, 0.423 cm. Calculate (i) average corrected
reading, (ii
) average absolute error and (ii) percentage error in measuring the
diameter of the wire if zero error of micrometer is +0.002 cm
Ans. (i) Average corrected reading Xm = 0.420 cm
(ii) Average absolute error oxavg = 0.0012 cm
(iii) Percentage error = 0.2857%
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mayushinde027:
plz give explanation
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Given : Diameter of a wire is measured with micrometer screw gauge for five times as :
0.421 cm, 0.422 cm, 0.421 cm, 0.423 cm, 0.423 cm.
zero error of micrometer is +0.002 cm
To Find : (i) average corrected reading,
Solution:
0.421 cm, 0.422 cm, 0.421 cm, 0.423 cm, 0.423 cm.
Total Readings = 5
Sum of Readings = 0.421 + 0.422 + 0.421 + 0.423 + 0.423
= 2.110
Average = ( 2.110 / 5) = 0.422
zero error of micrometer is +0.002 cm
Hence average corrected reading = 0.422 - 0.002 = 0.420 cm
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