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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.A. `0.1 s`B. `0.11 s`C. `0.01 s`D. `1.0 s` |
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Answer» Correct Answer - b Average value `= (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/(5)` `= 2.62sec` Now `|Delta T_(1)| = 2.63 - 2.62 = 0.01` `|Delta T_(2)| = 2.62 - 2.56 = 0.06` `|Delta T_(3)| = 2.62 - 2.42 = 0.20` `|Delta T_(4)| = 2.71 - 2.62 = 0.09` `|Delta T_(5)| = 2.80 - 2.62 = 0.18` Mean absoltude `Delta T = (|Delta T_(1)| + |Delta T_(2)| + |Delta T_(3)| + |Delta T_(4)| + |Delta T_(5)|)/(5)` `= (0.54)/(5) = 0.108 = 0.11sec` |
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