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Question

In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63s, 2.56s, 2.42s,2.71s, and 2.80s. The mean absolute error is:

A
0.11s
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B
0.12s
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C
0.13s
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D
0.14s
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Solution

The correct option is A 0.11s
The mean period of oscillation of the pendulum is
Tmean=ni=1Tin;Tmean= (2.63+2.56+2.42+2.71+2.80)5s
13.125s=2.624s=2.62s
(Rounded off to two decimal places)
The absolute errors in the measurement are
ΔT1=2.62s2.63s=0.01s;ΔT2=2.62s2.56s=0.06s
ΔT3=2.62s2.42s=0.20s;ΔT4=2.71s0.09s
ΔT5=2.62s2.80s=0.18s
Mean absolute error is
Tmean=ni=1|ΔTi|n
Tmean=0.01+0.06+0.20+0.09+0.185s
=0.545s=0.11s

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