If the length of a cylinder is l = (4.00+- 0.001)cm radius r =(0.0250+-0.001)cm, and
mass m= (6.25+-0.01)gm
Calculate the percentage error in the determination of density
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length of cylinder , l = (4.00 ± 0.001) cm
radius of base of cylinder , r = (0.025 ± 0.001) cm
mass of cylinder, m = (6.25 ± 0.01) gm
we know, density , d = m/v
and volume of cylinder , v = πr²l
so, density , d = m/(πr²l)
then fractional error can be written as,
∆d/d = ∆m/m + 2 × ∆r/r + ∆l/l
given, ∆m = 0.01g , m = 6.25 g , ∆r = 0.001, r = 0.025 , ∆l = 0.001 and l = 4.00
so, ∆d/d = (0.01)/(6.25) + 2 × (0.001)/(0.025) + (0.001)/(4.00)
= 0.0016 + 0.08 + 0.00025
= 0.08185
percentage error of density = 100 × fractional error of density
= 100 × 0.08185
= 8.185 %
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