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 `(bar(l))= 3.28875 cm= 3.29 cm`
(b) Absolute error in the first reading `= 3.29 - 3.29 = 0.00 cm`
Absolute error 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 fourth reading `=3.39- 3.31 =-0.02 cm`
Absolute error in the fifth reading `=3.29 - 3.28 = 0.01 cm`
Absolute error in the sixth reading `= 3.29 - 3.27 = 0.02 cm`
Absolute in the seventh reading `=3.29 - 3.29 = 0.00 cm`
Absolute error in the last reading `=3.29 - 3.30 = -0.01 cm`
(c) Mean absolute error `=bar(Deltal)=(0.00+0.01+0.00+0.02+0.01+0.02+0.00+0.01)/(8)=0.01cm`
(d) Relative error in length `= (bar(Deltal))/(l)=(0.01)/(3.29)=0.0030395=0.003`
(e) Percentage error `= (bar(Deltal))/(l)xx100=0.003 xx100 =0.3%`
So length `l = 3.29 cm pm 0.01 cm` (in terms of absolute error)
`rArr l=3.29cm pm 0.30%` (in terms of percentage error)