Question

The length, breadth and height of a rectangular block are found to be 15.12 ± 0.02 cm, 7.86 ± 0.01 cm and 4.16 ± 0.02 cm, respectively. Compute the percentage error in the volume of the block.

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Answer 1

Given : The length , breadth and height of a rectangular block are found to be (15.12 ± 0.02) cm , (7.86 ± 0.01) cm and (4.16 ± 0.02) cm respectively.

To find : The percentage error in the volume of the block.

solution : volume of cuboid , V = l × b × h

Taking log base e both sides we get,

logV = logl + logb + logh

differentiating both sides we get,

dV/V = dl/l + db/b + dh/h

so fractional error in volume , ∆V/V = ∆l/l +∆b/b + ∆h/h

and hence percentage error in volume, ∆V/V × 100 = (∆l/l + ∆b/b + ∆h/h) × 100

here l = 15.12 , ∆l = 0.02, b = 7.86, ∆b = 0.01 , h = 4.16 and ∆h = 0.02

% error in the volume = (0.02/15.12 + 0.01/7.86 + 0.02/4.16) × 100

= 0.74 %

Therefore the percentage error in volume of the block is 0.74 %