Combination Of Errors

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The density of a uniform sphere is found from its measured mass and radius; if the mass has a 1% error and the radius a 1.5% error, what is the percentage error in the calculated density?

RankGuru · Accepted Answer

Density is defined as mass divided by volume, and for a sphere the volume depends on the cube of the radius through V = (4/3)π r^3, so density ρ = m / [(4/3)π r^3]. Because the radius enters with a power of three in the denominator, its fractional error is multiplied by three when propagated, while the constant factor (4/3)π carries no uncertainty and drops out of the error analysis. The maximum percentage error therefore becomes (Δρ/ρ) = (Δm/m) + 3(Δr/r) = 1% + 3(1.5%) = 1% + 4.5% = 5.5%. The choice 2.5% wrongly treats the radius linearly, adding only 1.5% instead of 4.5%. The choice 4% omits the mass contribution or mishandles the factor of three. The choice 6.5% over-weights one of the terms. This follows directly from the NCERT rule that an exponent multiplies the corresponding fractional error. A sanity check supports the result: since volume scales as radius cubed, even a modest radius uncertainty dominates the total, which is exactly why the radius term contributes the larger 4.5%.

Combination Of Errors

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