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

Options

A. 0.1 s
B. 0.11 s
C. 0.01 s
D. 1.0 s

Answer

B

Explanation

Average value
`(2.63 + 2.56 + 2.42 +2.71 + 2.80)/5`
`|DeltaT_(1)| = 2.63 - 2.62 = |0.01|`
`|DeltaT_(2)| = 2.56 - 2.62 = |0.06|`
`|DeltaT_(3)| = 2.42 - 2.62 = |0.20|`
`|DeltaT_(4)| = 2.71 - 2.62| = |0.92|`
`|DeltaT_(5)| = 2.80 -2.62= |0.18|`
Mean absoute error
`DeltaT= (|DeltaT_(1)|+|DeltaT_(2)|+|DeltaT_(3)|+|DeltaT_(4)|+|DeltaT_(5)|)/5`
` = (0.54)/5 = 0.108 = 0.11` sec