Q.

The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s, 2.56s, 2.42s,2.71s and 2.80 s respectively. The average absolute error is

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a

0.1 s

b

0.11 s

c

0.01 s

d

1.0 s

answer is B.

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Detailed Solution

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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