Question

A metal block has length breadth and thickness is equal to (5 +- 0.01) (3+- 0.02) (1+- 0.01) calculate mass abosolute error or the total area of metal block

Answer 1

given, Length , L = (5 ± 0.01) m

breadth , B = (3 ± 0.02) m

and thickness , H = (1 ± 0.01) m

we know,

total area of metal block (cuboid), A = 2(LB + BH + HL)

so, A = 2(5 × 3 + 3 × 1 + 1 × 5) m²

= 2(15 + 3 + 5 ) m²

= 46 m²

now finding error in total area.i.e., ∆A

formula of total area , A = 2(LB + BH + HL)

taking log both sides,

logA = log2 + 2(logL + logB + logH)

now differentiating both sides,

dA/A = 0 + 2 × dL/L + 2dB/B + 2dH/H

if dA , dL,.. are comparable then, dA → ∆L, dL →∆L...

so, formula is..

∆A/A = 2∆L/L + 2∆B/B + 2∆H/H

putting, A = 46, ∆L = 0.01 , L = 5 , ∆B = 0.02, B = 3 , ∆H = 0.01 and H = 1

so, ∆A/46 =2( 0.01/5 + 0.02/3 + 0.01/1)

or, ∆A = 92(0.03 + 0.10 + 0.15)/15

= 92 × 0.28/15

≈ 1.72 m²

so, absolute error in total area is 1.72 m²