Physics And Measurement Question 178
Question: 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
Options:
A) 0.1 s
B) 0.11 s
C) 0.01 s
D) 1.0 s
Show Answer
Answer:
Correct Answer: B
Solution:
Average value $ =\frac{2.63+2.56+2.42+2.71+2.80}{5} $
$ =2.62\sec $ 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 absolute error $ \Delta T=\frac{|\Delta T _{1}|+|\Delta T _{2}|+|\Delta T _{3}|+|\Delta T _{4}|+|\Delta T _{5}|}{5} $
$ =\frac{0.54}{5}=0.108=0.11sec $
