Question

Following observations were taken with a vernier callipers while measuring the length of a cylinder :
`3.29 cm,3.28 cm,3.31 cm,3.28 cm,3.27 cm,3.29 cm,3.20 cm`. Then find :
(a) Most accurate length of the cylinder
(b) Absolute error in each observation
(c) Mean absolute error
(d) Relative error
(e) Percentage error
Express the result in terms absolute error and percentage error.

Answer 1

(a) Most accurate length of the cylinder will be the mean length `(barl)`
`barl = (3.19+ 3.28+ 3.29+ 3.31+ 3.28+ 3.27+ 3.29+ 3.30)/(8) = 3.288765 cm or barl = 3.29` cm
(b) Absolute error in the first reading = 3.29-3.29 =0.00 cm
Absolute error in the second reading = 3.29 - 3.28 = 0.01 cm
Absolute error in the third reading = 3.29 -3.29 = 0.00 cm
Absolute error in the forth reading = 3.29 -3.31=-0.02 cm
Absolute error in the fifth reading = 3.29 - 3.28 k= 0.01 cm
Absolute error in the sixth reading = 3.29 - 3.27 = 0.02 cm
Absolute error in the seventh reading = 3.29 - 3.209=0.00 cm
Absolute error in the last reading = 3.29 -3.30 = -0.01
(c) Mean absolute error =` bar(Deltal) = (0.00+ 0.01+0.00+0.02+0.01+0.02+0.00+0.01)/(8) = 0.01 cm `
(d) Relative error in length = `(bar(Deltal))/(barl) = (0.01)/(3.29)= 0.0030395= 0.003`
(e) Percentage in length = `(bar(Delta l))/(barl)xx 100 = 0.003 xx 100 = 0.3 %`
So length `l = 3.29 cm pm 0.01 cm " "` (in terms of absolute error)
or `l= 3.29 cm pm 0.30 %` (in terms percentage error)