Errors In Physical Quantity

 In an experiment the mass of an iron ball in five successive measures are given by 3.51 gm,3.52 gm,3.49 gm, 3.54 gm,3.53 gm calculate a) the mean value of the mass of balls b) the absolute error in each measurement c) mean absolute error d) relative error e) percentage error

a) M_mean=(3.51+3.52+3.49+3.54+3.53) / 5

M_mean= 3.518=3.52 gm

b) the absolute error

∆m₁=M_mean-m₁ and so on up to 5 measurements

. ^. . errors in each measurements are

3.52-3.51=0.01

3.52-3.52=0.00

3.52-3.49=0.03

3.52-3.54=0.02

3.52-3.53=0.01

c) Mean absoute error=(0.01+0.00+0.03+0.02+0.01) / 5

Mean absoute error= ∆m_mean=0.014

d) relative error= ∆m_mean  / M_mean=0.014 / 3.52 =0.003977

relative error= 0.003977

e) percentage error=relative error x 100=0.003977× 100=0.3977%