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Errors In Measurement

Easyphysics

A laboratory resistor is measured to have a value of 100 ohm with an absolute uncertainty of 2 ohm; what is the percentage error in this reading?

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About This Question

Subject
physics
Chapter
physics and measurement
Topic
errors in measurement
Difficulty
Easy
Year
2025
Tags
percentage errorabsolute errorrelative erroruncertaintyresistance

Solution

Correct Answer:

2%

Percentage error expresses the absolute uncertainty as a fraction of the measured value, scaled by one hundred, and it tells us the relative reliability of a measurement. The defining relation is percentage error = (absolute error / measured value) × 100. Substituting the absolute error of 2 ohm and the measured resistance of 100 ohm gives (2 / 100) × 100 = 2%. The choice 0.2% would result from forgetting to multiply the ratio by one hundred and instead reporting the bare fraction times ten. The choice 20% overstates the error tenfold, as if the uncertainty were 20 ohm rather than 2 ohm. The choice 0.02% is simply the dimensionless ratio without any percentage conversion and is far too small. This is the same definition of relative and percentage error introduced in the NCERT measurement chapter and used in every error-analysis problem. As a sanity check, an uncertainty that is one-fiftieth of the reading should indeed be a small but non-negligible two percent, which is consistent with a typical resistor tolerance band.

This easy difficulty physics question is from the chapter physics and measurement, covering the topic of errors in measurement. It appeared in the 2025 exam.

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