Source: Extra Practice

A student measures the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s,2.56 s,2.42 s,2.71 s2.63\text{ s}, 2.56\text{ s}, 2.42\text{ s}, 2.71\text{ s} and 2.80 s2.80\text{ s}. If the student calculates the mean period, what is the relative error?

Options

Option A is correct

0.04

Option B

0.11

Option C

0.01

Option D

0.20

Explanation

Mean T=(2.63+2.56+2.42+2.71+2.80)/5=13.12/5=2.624≈2.62 sT = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5 = 13.12 / 5 = 2.624 \approx 2.62\text{ s}. Absolute errors: ∣2.62−2.63∣=0.01,∣2.62−2.56∣=0.06,∣2.62−2.42∣=0.20,∣2.62−2.71∣=0.09,∣2.62−2.80∣=0.18|2.62-2.63|=0.01, |2.62-2.56|=0.06, |2.62-2.42|=0.20, |2.62-2.71|=0.09, |2.62-2.80|=0.18. Mean absolute error ΔT=(0.01+0.06+0.20+0.09+0.18)/5=0.54/5=0.108≈0.11 s\Delta T = (0.01+0.06+0.20+0.09+0.18)/5 = 0.54/5 = 0.108 \approx 0.11\text{ s}. Relative error = ΔT/T=0.11/2.62≈0.0419\Delta T / T = 0.11 / 2.62 \approx 0.0419. Rounding to one or two significant figures common in error analysis, 0.040.04 is the most appropriate option.