If the error in measuring the radius of the sphere is 2 % and that in measuring its mass is 3 % , then the error in measuring the density of material of the sphere is

Question

If the error in measuring the radius of the sphere is 2 % and that in measuring its mass is 3 % , then the error in measuring the density of material of the sphere is, 5 %, 7 %, 9 %, 11 %

MARKS App · Accepted Answer

As density = m a s s v o l u m e ∴ ρ = M 4 3 π R 3 = 3 4 M π R 3 ∴ The percentage error in density is Taking "log" both sides we get ln ρ = ln 3 4 + ln M − ln π − 3 ln R differentiating, we get Δ ρ ρ = Δ M M − 3 Δ R R But as error is always added ⇒ Δ ρ ρ = Δ m m + 3 Δ R R ⇒ Δ ρ ρ × 100 % = Δ M M + 3 Δ R R × 100 % = 3 % + 3 2 % = 3 % + 6 % = 9 %

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