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.

Explanation

(a) Most accurate length of the cylinder will be the mean length `(vecl)`
`vecl=(3.29+3.28+3.29+3.31+3.28+3.27+3.29+3.30)/(8)=3.28875 cm "or" vecl=3.29cm`
(b) Absolute error in the first reading `=3.29-3.29=0.00cm`
Absolute error in the second reading =3.29-3.28=0.01cm
Absolute error in the thrid reading =2.29-3.29=0.00cm
Absolute error in the forth reading =3.39-3.31=-0.02cm
Absolute error in the fifith reading =3.29-3.28=0.01cm
Absolute error in the sixth reading =3.29-3.27=0.02cm
Absolute error in the seventh reading =3.29-3.29=0.00cm
Absolute error in the last reading =3.29-3.30=-0.01cm
(c) Mean absolute error `=vecDeltal=(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 `=(vecDeltal)/(l)=(0.01)/(3.29)=0.0030395=0.003`
(e) Percentage error `=(vecDeltal)/(vecl)xx100=0.003xx100=0.3%`
so length `l=3.29cm+-0.01cm` (in terms of absolute error)
or `t=3.29cm +- 0.30%` (in terms percentage error)