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Question
Solve the numerical example.
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) g. Calculate the percentage error in the determination of density.
Solution
Given:
l = (4.00 ± 0.001) cm,
In order to have same precision, we use,
(4.000 ± 0.001), r = (0.0250 ± 0.001) cm,
In order to have same precision, we use, (0.025 ± 0.001) m = (6.25 ± 0.01) g
To find: percentage error in density
Formulae:
1. Relative error in volume, `(Delta"V")/"V" = (2Delta"r")/"r" + (Deltal)/l` ....(∵ Volume of cylinder, V = πr2l)
2. Relative error `(Deltarho)/rho = (Delta"m")/"m" + (Delta"V")/"V"` ....[∵ Density (ρ) = `("mass"("m"))/("volume"("v"))`]
3. Percentage error = Relative error × 100 %
Calculation:
From formulae (i) and (ii),
∴ `(Deltarho)/rho = (Delta"m")/"m" + (2Delta"r")/"r" + (Deltal)/l`
`= 0.01/6.25 + (2(0.001))/0.025 + 0.001/4.000`
= 0.0016 + 0.08 + 0.00025
= 0.08185
From formula (iii),
% error in density = `(Deltarho)/rho xx 100`
= 0.08185 × 100
= 8.185%
Percentage error in density is 8.185%.
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