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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