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In the measurment of the period of a simple pendulum the readings turn out to be (1) 2.63 s (2) 2.56 s (3) 2.42 s (4) 2.71 s (5) 2.80 s Calculate % error in the measurment -->

Answers (1)

@Sushant Hajare 

The mean period of oscillation will be 

T=2.63+2.56+2.42+2.71+2.8/5

=2.62 sec

The errors in the measurement wil be 

2.63-2.62=.01sec

2.56-2.62=-.06sec

2.42-2.62=-.2sec

2.71-2.62=.09sec

2.80-2.62=.18 sec

so mean of absolute error will be .11sec

Hence Period of oscillation will be (2.62±0.11)

Therefore T=2.6±.1s

relative error or percentage error is given by

=0.1/2.6*100

4%

Posted by

Safeer PP

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