The length breadth and thickness of a strip are

  1. If the length, breadth and thickness of a strip are 10.0± 0.1 cm,1.00 ± 0.01 cm and 0.100± 0.001 cm, respectively, then the most probable percentage error in its volume will be
  2. The length, breadth and thickness of a block are measured with the help of a meter scale. given ℓ = (5.12 + 0.01)cm, b = (10.15 + 0.01)cm, t = (5.28 + 0.01)cm. The percentage error in volume is:


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If the length, breadth and thickness of a strip are 10.0± 0.1 cm,1.00 ± 0.01 cm and 0.100± 0.001 cm, respectively, then the most probable percentage error in its volume will be

Given l ± Δ l = ( 10.0 ± 0.1 ) c m b ± Δ b = ( 1.00 ± 0.01 ) c m t ± Δ t = ( 0.100 ± 0.001 ) c m As we know, volume of strip ( V ) = Length × Breadth × Thickness Therefore, percentage error in V is given by, Δ V V × 100 = ± Δ l l × 100 ± Δ b b × 100 ± Δ t t × 100 By putting in the given equation we get, Δ V V × 100 = ± 0.1 l 0 × 100 ± 0.01 1 × 100 ± 0.001 0.1 × 100 Δ V V × 100 = ( ± 0.01 ± 0.01 ) % Δ V V % = ± 0.03 % Final Answer : Δ V V % = ± 0.03 %

The length, breadth and thickness of a block are measured with the help of a meter scale. given ℓ = (5.12 + 0.01)cm, b = (10.15 + 0.01)cm, t = (5.28 + 0.01)cm. The percentage error in volume is:

Given that, Length of the block l = ( 5 . 1 2 + 0 . 0 1 )cm Breadth of the block b = ( 1 0 . 1 5 + 0 . 0 1 ) cm Thickness of the block t = ( 5 . 2 8 + 0 . 0 1 ) cm We know that, Relative error in volume of cuboid V d V ​ = l d l ​ + b d b ​ + t d t ​ V d V ​ = 5 . 1 2 0 . 0 1 ​ + 1 0 . 1 5 0 . 0 1 ​ + 5 . 2 8 0 . 0 1 ​ V d V ​ = 5 1 2 1 ​ + 1 0 1 5 1 ​ + 5 2 8 1 ​ Now, the percentage error in volume V Δ V ​ × 1 0 0 = 0 . 4 8 % Δ V = 0 . 4 8 % Hence, the percentage error in volume is 0.48.

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