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Question
In a workshop, a worker measures the length of a steel plate with Vernier calipers 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.
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Solution
Given:
a1 = 3.11 cm, a2 = 3.13 cm,
a3 = 3.14 cm, a4 = 3.14 cm
Least count L.C. = 0.01 cm.
To find:
i. Mean length `("a"_"mean")`
ii. Mean absolute error `(triangle "a"_"mean")`
iii. Percentage error.
Formulae:
1. `"a"_"mean" = ("a"_1 + "a"_2 + "a"_3 + "a"_4)/4`
2. `triangle "a"_"n" = |"a"_"mean" - "a"_"n"|`
3. `triangle "a"_"mean" = (triangle"a"_1 + triangle"a"_2 + triangle"a"_3 + triangle"a"_4)/4`
4. Percentage error = `(triangle "a"_"mean")/("a"_"mean") xx 100`
Calculation:
From formula (i),
`"a"_"mean" = (3.11 + 3.13 + 3.14 + 3.14)/4`
= 3.13 cm
From formula (ii),
`triangle"a"_1 = |3.13 - 3.11| = 0.02` cm
`triangle"a"_2 = |3.13 - 3.13| = 0`
`triangle"a"_3 = |3.13 - 3.14| = 0.01` cm
`triangle"a"_4 = |3.13 - 3.14| = 0.01` cm
From formula (iii),
`triangle"a"_"mean" = (0.02 + 0 + 0.01 + 0.01)/4` = 0.01 cm
From formula (iii),
% error = `0.01/3.13 xx 100`
`= 1/3.13`
= 0.3196 .....(using reciprocal table)
= 0.32 %
Ans:
- Mean length is 3.13 cm.
- Mean absolute error is 0.01 cm.
- Percentage error is 0.32 %.
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