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 A0.11s 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=2.62s−2.63s=0.01s;ΔT2=2.62s−2.56s=0.06s ΔT3=2.62s−2.42s=0.20s;ΔT4=−2.71s−0.09s ΔT5=2.62s−2.80s=−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