Combination Of Errors

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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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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.

Combination Of Errors

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