Combination Of Errors
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?
Select the correct option:
Solution
5.5%
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%.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
More combination of errors Practice Questions
In a pendulum experiment g is obtained from g = 4 π^2 L / T^2, with length measured to 0.5% accuracy...
In a pendulum experiment g is obtained from g = 4 π^2 L / T^2, with length measured to 0.5% accuracy...
A physical quantity is computed from the relation Z = A^2 B / C, where A, B and C carry percentage e...
A physical quantity is computed from the relation Z = A^2 B / C, where A, B and C carry percentage e...
The period of oscillation of a simple pendulum is T = 2π√(L/g). If the error in measurement of L is ...
The period of oscillation of a simple pendulum is T = 2π√(L/g). If the error in measurement of L is ...
About This Question
- Subject
- physics
- Chapter
- physics and measurement
- Topic
- combination of errors
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5.5%
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%.
This medium difficulty physics question is from the chapter physics and measurement, covering the topic of combination of errors. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse physics and measurement questions on RankGuru.