Combination Of Errors
In a pendulum experiment g is obtained from g = 4 π^2 L / T^2, with length measured to 0.5% accuracy and the time for fifty oscillations recorded over a total of 100 s using a stopwatch of least count 0.2 s; what is the maximum percentage error in g?
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Solution
0.9%
Determining the error in g requires propagating the uncertainties of both the length and the period through g = 4π^2 L / T^2, where the numerical factor 4π^2 is exact and contributes no error. The fractional error rule gives (Δg/g) = (ΔL/L) + 2(ΔT/T), the factor of two arising because the period appears squared. The length error is given directly as 0.5%. For the period, the timing uncertainty comes from the stopwatch least count of 0.2 s acting on the total measured interval of 100 s, so ΔT/T = 0.2/100 = 0.002, or 0.2%, since dividing total time by the number of oscillations does not change the fractional timing error. Substituting gives (Δg/g) = 0.5% + 2(0.2%) = 0.5% + 0.4% = 0.9%. The choice 0.7% omits the doubling of the period error. The choice 1.1% over-weights a term, and 1.4% double-counts the length contribution. This careful handling of the squared period is exactly what JEE Advanced timing problems demand. A check confirms it: timing many oscillations reduces the fractional error, keeping the total below one percent.
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About This Question
- Subject
- physics
- Chapter
- physics and measurement
- Topic
- combination of errors
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.9%
Determining the error in g requires propagating the uncertainties of both the length and the period through g = 4π^2 L / T^2, where the numerical factor 4π^2 is exact and contributes no error. The fractional error rule gives (Δg/g) = (ΔL/L) + 2(ΔT/T), the factor of two arising because the period appears squared. The length error is given directly as 0.5%. For the period, the timing uncertainty comes from the stopwatch least count of 0.2 s acting on the total measured interval of 100 s, so ΔT/T = 0.2/100 = 0.002, or 0.2%, since dividing total time by the number of oscillations does not change the fractional timing error. Substituting gives (Δg/g) = 0.5% + 2(0.2%) = 0.5% + 0.4% = 0.9%. The choice 0.7% omits the doubling of the period error. The choice 1.1% over-weights a term, and 1.4% double-counts the length contribution. This careful handling of the squared period is exactly what JEE Advanced timing problems demand. A check confirms it: timing many oscillations reduces the fractional error, keeping the total below one percent.
This hard difficulty physics question is from the chapter physics and measurement, covering the topic of combination of errors. It appeared in the 2025 exam.
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