Using a Vernier Caliper, the length of a cylinder in different measurements is found to be 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm and 2.29 cm. Find the mean value, absolute error the relative error and the percentage error of the cylinder.

Explanation

The given readings are 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm, and 2.29 cm
The mass value vec(l) = (2.36 + 2.27 + 2.26 + 2.28 + 2.31 + 2.28 + 2.29)/(7) = (16.05)/(7) = 2.29` cm
Abolute errors in the measurement are:
`|Deltal_(1)| = |2.29 - 2.36| = 0.07`
`|Deltal_(2)| = |2.29 - 2.27| = 0.02`
`|Deltal_(3)| = |2.29 - 2.26| = 0.03`
`|Deltal_(4)| = |2.29 - 2.28| = 0.01`
`|Deltal_(5)| = |2.29 - 2.31| = 0.02`
`|Deltal_(6)| = |2.29 - 2.28| = 0.01`
`|Deltal_(7)| = |2.29 - 2.29| = 0.00`
Mean A bsolute error
`Deltal_("mean") = (0.07+0.02+0.03+0.01+0.02+0.01+0.00)/(7) = (0.16)/(7) = 0.02`
Relative error = `(Deltal_("mean"))/(overset(-)(l)) = pm (0.02)/(2.29) = pm 8.7 xx 10^(-3)`
Percentage error `= pm 8.7 xx 10^(-3) xx 100 = 0.87% xx 100 = pm (8.7 xx 10^(-1)) = 0.9%`