(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)