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The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s,2.56s,2.42s,2.71sand2.80srespectively. The average absolute error is

a
0.1 s
b
0.11 s
c
0.01 s
d
1.0 s

detailed solution

Correct option is B

Average  value = 2.63+2.56+2.42+2.71+2.805                                 = 2.62 sNow  |ΔT1| = 2.63 − 2.62 = 0.01           |ΔT2| = 2.62 − 2.56 = 0.06           |ΔT3| = 2.62 − 2.42 = 0.20          |ΔT4| = 2.71 − 2.62 = 0.09          |ΔT5| = 2.80 − 2.62 = 0.18     Mean absolute errorΔT = |ΔT1|+|ΔT2| + |ΔT3|+|ΔT4|+|ΔT5|5  = 0.545 = 0.108 = 0.11s

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