Math, asked by dhirajmhatre15112004, 5 months ago

(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%​

Answers

Answered by jayathulungjayathulu
3

Answer:

hlo good evening.......


mayushinde027: plz give explanation
jayathulungjayathulu: Don't know
Answered by amitnrw
5

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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