Physics And Measurement Question 189
Question: A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be
Options:
A) $ 92\pm 2s $
B) $ 92\pm 5.0s $
C) $ 92\pm 1.8s $
D) $ 92\pm 3s $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Measured time period of 100 oscillations are 90 sec, 91 sec, 95 sec and 92 sec.
Mean value of time= $ t _{m}=\frac{90+91+95+92}{4}=92\sec $ Absolute error in measurement $ | \Delta t _{1} |=| t _{m}-t _{1} |=2\sec $
$ | \Delta t _{2} |=| t _{m}-t _{2} |=1\sec $
$ | \Delta t _{3} |=| t _{m}-t _{3} |=3\sec $
$ | \Delta t _{4} |=| t _{m}-t _{4} |=0\sec $
Mean absolute error $ \Delta t _{mean}=\frac{2+1+3+0}{4}=1.5\sec $
But the least count of the measuring clock is 1 sec, so it cannot measure up to 0.5 second, so we have to round it off.
So mean error will be 2 second. Hence men time $ (92\pm 2sec) $ .
