Question

[Ans : -15]
vi) In a workshop a worker measures the length
of a steel plate with a Vernier callipers
having a least count 0.01 cm. Four such
measurements of the length yielded the
following values: 3.11 cm, 3.13 cm, 3.14
cm, 3.14 cm. Find the mean length, the
mean absolute error and the percentage
error in the measured value of the
length.
[Ans: 3.13 cm,​

Answer 1

Given:

a1=3.11; a2=3.13; a3=3.14; a4=3.14

Mean length (am) = a1+a2+q3+a4/4

=3.11+3.13+3.14+3.14/4

= 12.52/4

(am) = 3.13cm.

Absolute error:

∆a1=|am - a1 | = | 3.13-3.11 | = 0.02

∆a2=|am -a2 | = | 3.13-3.13 | =0

∆a3=|am -a3 | = | 3.13-3.14 | =0.01

∆a4=|am -a4 | = | 3.13-3.14 | =0.01

Mean absolute error:

∆am=∆a1+∆a2+∆a3+∆a4/4

∆am=0.02+0+0.01+0.01/4

∆am=0.01cm

% error = ∆am/am

=0.01/3.13*100

=0.31%

Answer 2

Answer:

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