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 $2.80 s$ respectively. The average absolute error is:

0.1 s

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0.11 s

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0.01 s

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1.0 s

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Solution

The correct option is C $0.11 s$
$T = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5}$
$\frac{13.12}{5} = 2.624 = 2.62$
$2.62 - 2.63 = - 0.01$
$2.62 - 2.56 = + 0.06$
$2.62 - 2.42 = + 0.20$
$2.62 - 2.71 = - 0.09$
$2.62 - 2.80 = - 0.18$
Average absolute error
$\frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5}$
$0.54 5 = 0.108 = 0.11 s$

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Similar questions

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

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2.56 s

2.42 s, 2.71 s

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. The mean absolute error is:

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

The correct option is C 0.11sT=2.63+2.56+2.42+2.71+2.80513.125=2.624=2.622.62−2.63=−0.012.62−2.56=+0.062.62−2.42=+0.202.62−2.71=−0.092.62−2.80=−0.18Average absolute error 0.01+0.06+0.20+0.09+0.1850.545=0.108=0.11s

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 $2.80 s$ respectively. The average absolute error is: