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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 ∆m 1 = M mean - m 1  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%

Errors in Physical Quantities

Notes on this topic is available on the home page link The radius of a sphere is expressed as 110.4 ± 0.2 cm Find the percentage error in    a) radius   b) surface area    c) volume a) Percentage error = ( Δx/x )×100 • • • percentage error in radius= (0.2/110.4) x 100 = 0.18% b) Surface area = 4 πr 2 • • • percentage error in Surface area = 2 × Percentage error in radius=2×0.18= 0.36% ( derivation for the expression will be posting   using calculus soon) c) Volume = 4/3(πr 3 ) • • • percentage error in volume = 3 × Percentage error in radius=3×0.18= 0.54% Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761